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11,226 |
Typically, in the functor of points approach, one constructs the category of algebraic spaces by first constructing the category of locally representable sheaves for the global Zariski topology (Schemes) on $CRing^{op}$. That is, taking the full subcategory of $Psh(CRing^{op})$ which consists of objects $S$ such that $S$ is a sheaf in the global Zariski topology and $S$ has a cover by representables in the induced topology on $Psh(CRing^{op})$. This is the category of schemes. Then, one takes this category and equips it with the etale topology and repeats the construction of locally representable sheaves on this site (Sch with the etale topology) to get the category of algebraic spaces. Can we "skip" the category of schemes entirely by putting a different topology on $CRing^{op}$? My intuition is that since every scheme can be covered by affines, and every algebraic space can be covered by schemes, we can cut out the middle-man and just define algebraic spaces as locally representable sheaves for the global etale topology on $CRing^{op}$. If this ends up being the case, is there any sort of interesting further generalization before stacks, perhaps taking locally representable sheaves in a flat Zariski-friendly topology like fppf or fpqc? Some motivation: In algebraic geometry, all of our data comes from commutative rings in a functorial way (intentionally vague). All of the grothendieck topologies with nice notions of descent used in Algebraic geometry can be expressed in terms of commutative rings, e.g., the algebraic and geometric forms of Zariski's Main theorem are equivalent, we can describe etale morphisms in terms of etale ring maps, et cetera. What I'm trying to see is whether or not we can really express all of algebraic geometry as "left-handed commutative algebra + sheaves (including higher sheaves like stacks)". The functor of points approach for schemes validates this intuition in the simplest case, but does it actually generalize further? The main question is italicized, but feel free to tell me if I've incorrectly characterized something in the motivation or the background.
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Yes. The category of algebraic spaces is the smallest subcategory of the category of sheaves of sets on Aff, the opposite of the category of rings, under the etale topology which (1) contains Aff, (2) is closed under formation of quotients by etale equivalence relations, and (3) is closed under disjoint unions (indexed by arbitrary sets). An abstract context for such things is written down in "Algebraization of complex analytic varieties and derived categories" by Toen and Vaquie, which is available on the archive. Toen also has notes from a "master course" on stacks on his web page with more information. It might be worth pointing out that their construction of this category also goes by a two-step procedure, although in their case it's a single construction performed iteratively (and which stabilizes after two steps). This is unlike the approach using scheme theory in the literal sense, as locally ringed topological spaces, where the two steps are completely different. After the first step in T-V, you get algebraic spaces with affine diagonal. Also worth pointing out is that their approach is completely sheaf theoretic. The only input you need is a category of local models, a Grothendieck topology, and a class of equivalence relations. You then get algebraic spaces from the triple (Aff, etale, etale). But the general machine (which incidentally I believe is not in its final form) has nothing to do with commutative rings. I think it would be interesting to plug opposites of other algebraic categories into it.
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11,296 |
A set-theoretical reductionist holds that sets are the only abstract objects, and that (e.g.) numbers are identical to sets. (Which sets? A reductionist is a relativist if she is (e.g.) indifferent among von Neumann, Zermelo, etc. ordinals, an absolutist if she makes an argument for a priviledged reduction, such as identifying cardinal numbers with equivalence classes under equipotence). Contrasting views: classical platonism, which holds that (e.g.) numbers exist independently of sets; and nominalism, which holds that there are no abstract particulars. I'm interested in the relationship between "structuralism" as it is understood by philosophers of science and mathematics and the structuralist methodology in mathematics for which Bourbaki is well known. A small point that I'm hung up on is the place of set theory in Bourbaki structuralism. I'm weighing two readings. (1) conventionalism: Bourbaki used set theory as a convenient "foundation", a setting in which models of structures may be freely constructed, but "structure" as understood in later chapters is not essentially dependent on the formal theory of structure developed in Theory of Sets, (2) reductionism: sets provide a ground floor ontology for mathematics; mathematicians study structures in the realm of sets. In favor of conventionalism: (a) Leo Corry's arguments in “Nicolas Bourbaki and the Concept of Mathematical Structure” that the formal structures of Theory of Sets are to be distinguished from and play only a marginal role in the subsequent investigation of mathematical structure, (b) ordered pairs: definitions reducing pairs to sets like Kuratowski's bring "baggage" (i.e., extra structure) and Bourbaki used primitive ordered pairs in the first edition of Theory of Sets, showing no excess concern for complete reduction, (c) statements of Dieudonne to the Romanian Institute indicating chs. 1 and 2 are mostly to satisfy bothersome philosophers (like me I suppose) before getting on to topics of greater interest, (d) the discussion of axiomatics and structure in "The Architecture of Mathematics", placing no special emphasis on sets, (e) this interpretation serves my selfish philosophical agenda. In favor of reductionism: (a) linear ordering of texts suggests perceived logical dependence on Theory of Sets, (b) reductionism makes sense of unity of mathematics, (c) 1970 edition includes Kuratowski pairs, (d) makes sense of controversies over category theory, (e) makes sense of some outsider criticisms (e.g., Mac Lane in "Mathematical Models" that Bourbaki was dogmatic and stifling), (f) I fear that in leaning towards conventionalism I'm self-deceiving to serve my selfish philosophical agenda. Apologies: not sure this is MO appropriate, any answers may be anachronistic, probably no univocality of opinion among Bourbaki members, my views are based on popular expositions, interviews, and secondary literature and not close study of the primary texts. Discussion related to this question has recently occurred at n-category cafe , occasioned by Manin's recent claim that Bourbaki provided "pragmatic foundations". The conventionalist interpretation, I think, helps make sense of Manin's claim and would show some criticisms levelled toward Bourbakism to misapprehend their intention (if not their impact). I have Borel's "Twenty-Five Years With Bourbaki" which discusses Grothendieck and the controversy over the direction following the first six books. Corry makes the claim that the Theory of Sets approach had limitations in dealing with category theory. I would especially appreciate references or answers that help me better understand these issues in particular, which are accessible to a philosopher with some grad coursework in mathematics and with only a self taught rudimentary understanding of categories.
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First, most mathematicians don't really care whether all sets are "pure" -- i.e., only contain sets as elements -- or not. The theoretical justification for this is that, assuming the Axiom of Choice, every set can be put in bijection with a pure set -- namely a von Neumann ordinal. I would describe Bourbaki's approach as "structuralist", meaning that all structure is based on sets (I wouldn't take this as a philosophical position; it's the most familiar and possibly the simplest way to set things up), but it is never fruitful to inquire as to what kind of objects the sets contain. I view this as perhaps the key point of "abstract" mathematics in the sense that the term has been used for past century or so. E.g. an abstract group is a set with a binary law: part of what "abstract" means is that it won't help you to ask whether the elements of the group are numbers, or sets, or people, or what. I say this without having ever read Bourbaki's volumes on Set Theory, and I claim that this somehow strengthens my position! Namely, Bourbaki is relentlessly linear in its exposition, across thousands of pages: if you want to read about the completion of a local ring (in Commutative Algebra), you had better know about Cauchy filters on a uniform space (in General Topology). In places I feel that Bourbaki overemphasizes logical dependencies and therefore makes strange expository choices: e.g. they don't want to talk about metric spaces until they have "rigorously defined" the real numbers, and they don't want to do that until they have the theory of completion of a uniform space. This is unduly fastidious: certainly by 1900 people knew any number of ways to rigorously construct the real numbers that did not require 300 pages of preliminaries. However, I have never in my reading of Bourbaki (I've flipped through about five of their books) been stymied by a reference back to some previous set-theoretic construction. I also learned only late in the day that the "structures" they speak of actually get a formal definition somewhere in the early volumes: again, I didn't know this because whatever "structure-preserving maps" they were talking about were always clear from the context. Some have argued that Bourbaki's true inclinations were closer to a proto-categorical take on things. (One must remember that Bourbaki began in the 1930's, before category theory existed, and their treatment of mathematics is consciously "conservative": it's not their intention to introduce you to the latest fads.) In particular, apparently among the many unfinished books of Bourbaki lying on the shelf somewhere in Paris is one on Category Theory, written mostly by Grothendieck. The lack of explicit mention of the simplest categorical concepts is one of the things which makes their work look dated to modern eyes.
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11,301 |
The Mumford conjecture states that for each integer $n$, we have: the map $\mathbb{Q}[x_1,x_2,\dots] \to H^\ast(M_g ; \mathbb{Q})$ sending $x_i$ to the kappa class $\kappa_i$, is an isomorphism in degrees less than $n$, for sufficiently large $g$. Here $M_g$ denotes the moduli of genus $g$ curves, and the degree of $x_i$ is the degree of the kappa class $\kappa_i$. This conjecture was proved by Madsen-Weiss a few years ago. What are the heuristic or moral reasons for the conjecture? (EDIT: I am particularly interested in algebraic geometric reasons, if there are any. Though algebraic topologial reasons are very welcome too.) What lead Mumford to formulating the conjecture in the first place? I know very little about the Madsen-Weiss proof, but I know that it mainly uses algebraic topology methods. Are there any approaches to the conjecture which are more algebraic-geometric? Is there any analogous theorem or conjecture regarding the (topological) $K$-theory of $M_g$? Or the Chow ring of $M_g$? etc.
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All current proofs of Mumford's conjecture in fact prove a far stronger result, the "Strong Mumford conjecture", first formulated by Ib Madsen. This says the following (where by "moduli space" in the following we mean a homotopy type classifying concordance classes of surface bundles with perhaps some extra structure): there is a stable moduli space
$$\mathcal{M}_\infty := \mathrm{colim} \,\, \mathcal{M}_{g, 1}$$
where $\mathcal{M}_{g, 1}$ denotes the moduli space of genus $g$ surfaces with a single boundary component, and the colimit is formed by gluing on a torus with a single boundary component using the "pair of pants" product. There is also a space, usually called $\Omega^\infty MTSO(2)$, which classifies cobordism classes of "formal surface bundles": that is, codimension -2 submersions with an orientation of the (stable) vertical tangent bundle. As a surface bundle is a formal surface bundle, there is a map
$$\alpha: \mathcal{M}_\infty \to \Omega^\infty MTSO(2).$$
The strong Mumford conjecture says that this is an integral homology equivalence. For the record, there are currently four distinct known proofs, due to: The stable moduli space of Riemann surfaces: Mumford's conjecture , Madsen and Weiss, The homotopy type of the cobordism category , Galatius, Madsen, Tillmann and Weiss, Monoids of moduli spaces of manifolds , Galatius and myself, Madsen-Weiss for geometrically minded topologists , Eliashberg, Galatius and Mishachev. For part 1) of your question, from this point of view (I do not know what Mumford had in mind): the map $\alpha$ can be thought of as comparable to the map which compares holonomic sections to formal sections in the statement of Gromov's $h$-principle for a sheaf. The idea is then that given a "formal surface bundle" one may begin improving it to be more and more like a bundle, but in this process one cannot really control the genus of the fibres one ends up with. This is why it gives the infinite genus moduli space. This is more or less the approach to proving the conjecture that Eliashberg, Galatius and Mishachev take. The other approaches are less direct and use more algebraic topological machinery. So, for part 3) of your question: the map $\alpha$ is also an equivalence in any other (co)homology theory (by the Atiyah-Hirzebruch spectral sequence, for example). Thus the topological K-theory of $\mathcal{M}_\infty$ is "known" in the sense that it is the K-theory of the infinite loop space of a well-understood spectrum, which fits into various simple cofibration sequences and so on. On the other hand, it is "not known" in the sense that I don't think anyone knows what $K^0(\mathcal{M}_\infty)$ is as a group, though I once tried to compute it without success. On the other hand, even knowing this group, it does not necessarily tell you anything about what you are really interested in, $K^0(\mathcal{M}_g)$, because stability for ordinary homology does not imply stability in non-connective (co)homology theories.
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11,306 |
From the example $D_4$, $Q$, we see that the character table of a group doesn't determine the group up to isomorphism. On the other hand, Tannaka duality says that a group $G$ is determined by its representation ring $R(G)$. What is the additional information contained in $R(G)$ as opposed to the character table?
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If I'm not mistaken, the extra information is not contained in the representation ring, you have to look at the category of representations. In particular, you want to look at the representation category equipped with its forgetful functor to vector spaces. Then the group can be recovered as the automorphisms of this functor. Here's a blogpost I wrote which may be helpful: http://concretenonsense.wordpress.com/2009/05/16/tannaka%E2%80%93krein-duality/
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11,366 |
I ran into the following problem. I have some lengthy paper in which I develop some theory to attack some problem. While I was working on this paper, I got some nice result which might interest a bigger audience and can stand by itself (that is, it is interesting out of context of the big work, and has a fairly short proof). The problem: should I write one long paper or two papers, one long and the other short? If I write just one paper, the short result will strengthen the paper, and it will be, perhaps, more whole. On the other hand, for someone that is only interesting in the short one, it will be better if it appears in a separated paper.
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According to Gian-Carlo Rota , one of the secrets of mathematical success is to publish the same result many times.
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11,450 |
I wonder why the Artin conjecture is so important. The only reason I could figure out is that one could use the holomorphy of Artin L-series and Weil's converse theorem to show modularity of two-dimensional Galois representations. Are there other reasons?
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From a modern view-point, its importance stems from the relationship with modularity/automorphy. Namely, a stronger conjecture, due to Langlands, should be true:
if $\rho:G_K \to GL_n({\mathbb C})$ is a continuous irreducible representation, for some number field $K$, then there is a cuspidal automorphic representation $\pi$ of $GL_n({\mathbb A}_n)$ such that $\rho$ and $\pi$ have the same $L$-function; in short,
$\rho$ is determined by $\pi$. This is a non-abelian class field theory. In the case $K = {\mathbb Q}$ and $n = 2$,
it says that $2$-dimensional irred. cont. reps. of $G_{\mathbb Q}$ are classified by either
weight one holomorphic cuspidal eigenforms (if the rep. is odd) or Maass cuspidal eigenforms with $\lambda = 1/4$ (if the rep. is even). The odd case is now a fully proved theorem
(of Deligne--Serre to go from the modular forms to the Galois reps., and of Khare, Wintenberger, and Kisin to go the other way), while the even case is still open. (If the Maass form is dihedral, one can construct a corresponding dihedral Galois rep. --- this is due to Maass himself --- but otherwise that direction is open; if the image of $\rho$ is solvable, then one can construct the corresponding Maass form --- this is due to Langlands and Tunnell.) As you note, in the two-dimensional case over $\mathbb Q$, Weil's converse theorem shows that Langlands' conjecture is equivalent to the Artin conjecture. (In fact, proving this over any number field $K$ was one of the goals of the famous book of Jacquet--Langlands).
In general, the two are also very closely linked, so that no modern number theorist thinks about one without the other. In general, one working principle of modern number theory is that the only way to establish holomorphy of $L$-functions arising from Galois representations is by simultaneously proving automorphy.
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11,467 |
Pretty much the question in the title. If a grad student gets bad reviews as a TA, how much does that hurt them later? How much do good reviews help? What if the situation is more complex? (For instance, bad reviews when TAing, but good reviews when actually teaching/lecturing a summer course). Edit: I asked this question with the situation of a student hoping for a career at research universities in mind, however, I am also interested in other cases. Edit: In your answer, please mention what your background is: have you served on hiring committees? Are you reporting just what you've heard? Were you successful/unsuccessful in a job search and were told that your teaching evals did/did not make a difference?
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This is more of a comment than an answer, but I think it bears saying, based on some of the comments at the top. Anyone who is now a math grad student was not a typical member of an undergrad math class. While clearly the correlation between evaluations and good teaching is not perfect, it is not negligible either. In my experience (which includes many years of recieving evaluations, at several different institutions, many years of reading teaching letters, and other related considerations of this question), if someone gets very low evaluations, there is typically a good reason for it. While it may not be easy to get top evaluations, reasonable teaching almost always avoids terrible ones. And when one is recruiting post-docs, or more senior candidates, one wants to have some assurance that they won't be a teaching disaster. If they are a brilliant teacher, all the better,
but one is more concerned about avoiding disasters, since these lead to trouble and difficulty for everyone. If you are applying to research positions, if your teaching is average or better, this should be fine. If your evaluations are consistently and significantly below average, this is probably a sign that you need to put some effort into improving your teaching. Students normally react well to clear explanation, sincere treatment, some humility on the part of the lecturer, and a little effort to be organized. Negative evaluations often correlate with the violation of one or more of these principles. Very good teaching does much more, but just doing a reasonable job should be enough to keep your evaluations reasonable.
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11,480 |
Let $V=L$ denote the axiom of constructibility. Are there any interesting examples of set theoretic statements which are independent of $ZFC + V=L$? And how do we construct such independence proofs? The (apparent) difficulty is as follows: Let $\phi$ be independent of $ZFC + V=L$. We want models of $ZFC + V=L + \phi$ and $ZFC + V=L + \neg\phi$. An inner model doesn't work for either one of these since the only inner model of $ZFC + V = L$ is $L$ and whatever $ZFC$ can prove to hold in $L$ is a consequence of $ZFC + V=L$. Forcing models are of no use either, since all of them satisfy $V \neq L$.
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There are numerous examples of such statements. Let me organize some of them into several categories. First, there is the hierarchy of large cardinal axioms that are relatively consistent with V=L. See the list of large cardinals . All of the following statements are provably independent of ZFC+V=L, assuming the consistency of the relevant large cardinal axiom. There is an inaccessible cardinal. There is a Mahlo cardinal. There is a weakly compact cardinal. There is an indescribable cardinal. and so on, for all the large cardinals that happen to be relatively consistent with V=L. These are all independent of ZFC+V=L, assuming the large cardinal is consistent with ZFC, because if we have such a large cardinal in V, then in each of these cases (and many more), the large cardinal retains its large cardinal property in L, so we get consistency with V=L. Conversely, it is consistent with V=L that there are no large cardinals, since we might chop the universe off at the least inaccessible cardinal. Second, even for those large cardinal properties that are not consistent with V=L, we can still make the consistency statement, which is an arithmetic statement having the same truth value in V as in L. Con(ZFC) Con(ZFC+'there is an inaccessible cardinal') Con(ZFC+'there is a Mahlo cardinal') Con(ZFC+'there is a measurable cardinal') Con(ZFC+'there is a supercompact cardinal'). and so on, for any large cardinal property. Con(ZFC+large cardinal property). These are all independent of ZFC+V=L, assuming the large cardinal is consistent with ZFC, since on the one hand, if W is a model of ZFC+Con(ZFC+phi), then L W is a model of ZFC+V=L+Con(ZFC+phi), as Con(ZFC+phi) is an arithmetic statement. And on the other hand, by the Incompleteness theorem, there must be models of ZFC+¬Con(ZFC+phi), and the L of such a model will have ZFC+V=L+¬Con(ZFC+phi). Third, there is an interesting trick related to the theorem of Mathias that Dorais mentioned in his answer. For any statement phi, the assertion that there is a countable well-founded model of ZFC+phi is a Sigma 1 2 statement, and hence absolute between V and L. And the existence of a countable well-founded model of a theory is equivalent by the Lowenheim-Skolem theorem to the existence of a well-founded model of the theory. Thus, the truth of each of the following statements is the same in V as in L. There is a well-founded set model of ZFC. This is equivalent to the assertion: there is an ordinal α such that L α is a model of ZFC. There is a well-founded set model of ZFC with ¬CH. (This is also equivalent to the previous statement.) There is a well-founded set model of ZFC with Martin's Axiom. and so on. For all the statements known to be forceable, you can ask for a well-founded set model of the theory. There is a well-founded set model of ZFC with an inaccessible cardinal. There is a well-founded set model of ZFC with a measurable cardinal. There is a well-founded set model of ZFC with a supercompact cardinal. and the same for any large cardinal notion. These are all independent of ZFC+V=L, since they are independent of ZFC, and their truth is the same in V as in L. I find it quite remarkable that there can be a model of V=L that has a transitive model of ZFC+'there is a supercompact cardinal'. The basic lesson is that the L of a model with enormous large cardinals has very different properties and kinds of objects in it than a model of V=L arising elsewhere. And I believe that this gets to the heart of your question. Since all these statements are studied very much in set theory, and are very interesting, and are independent of ZFC+V=L, I find them to be positive instances of what was requested. However , how does this relate to Shelah's view in Dorais's excellent answer? He seems there to dismiss the entire class of consistency strength statements as combinatorics in disguise. What does he mean exactly? Since we set theorists are very interested in these statements, I don't think that he means to dismiss them as silly tricks with the Incompleteness theorem. Perhaps he means something like: to the extent that we believe that a large cardinal property LC is consistent, then we don't really want to consider the theory ZFC+V=L, but rather, the theory ZFC+V=L+Con(LC). That is, we aren't so interested in models having the wrong arithmetic theory, so we insist that Con(LC) if we are comitted to that. And none of the examples I have given exhibit independence from that corresponding theory.
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11,488 |
Is there a characterization of the class of varieties which can be described as an intersection of quadrics, apart from the taulogical one? Lots of varieties arise in this way (my favorite examples are the Grassmanianns and Schubert varieties and some toric varieties) and I wonder how far can one go.
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In fact the answer is in some sense tautological: every projective variety can be realized as a scheme-theoretic intersection of quadrics! See e.g. D. Mumford, "Varieties defined by quadratic equations", Questions on algebraic varieties, C.I.M.E. Varenna, 1969 , Cremonese (1970) pp. 29–100, for quantitative refinements of this question.
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11,502 |
[Une traduction française suit la version anglaise.] The question is only about elliptic curves $E$ over $\mathbb{Q}$ and concerns only the aspect (order of vanishing of $L(E,s)$ at $s=1$)$\ =\ $(rank of $E(\mathbb{Q})$). Let $r$ be the LHS and $d$ the RHS, so that (a special case of ) the Birch and Swinnerton-Dyer Conjecture is BSD?. $r=d$. By the end of the last millenium, we knew Theorem (1977--2000). If $\ r=0,1$, then $d=r$ ( and $\ \operatorname{Sha}(E)$ is finite ). Some years ago, I heard that there was some progress in proving $(r>0)\Longrightarrow (d>0)$ under the assumption of the finiteness of $\operatorname{Sha}(E)$. What is the current status of the Statement. Suppose that $\operatorname{Sha}(E)$ is finite. If $r>1$, then $d>0$ ? L'état actuel de la conjecture de Birch et Swinnerton-Dyer On s'interesse uniquement aux courbes abéliennes $A$ sur $\mathbf{Q}$ et à
l'aspect (ordre d'annulation de $L(A,s)$ en $s=1$)$\ =\ $(rang de $A(\mathbf{Q})$). Désignons par $r$ le membre de gauche et par $d$ le membre de droite, de sorte
que la conjecture de Birch et Swinnerton-Dyer prédit (en particulier) BSD? $r=d$. Vers la fin du millénaire dernier, on avait démontré le Théorème (1977--2000). Si $r=0,1$, alors $d=r$ ( et $\ \operatorname{Cha}(A)$ est fini ). Il y a quelques années, j'avais entendu dire qu'on a fait des progrès
concernant l'implication $(r>0)\Longrightarrow(d>0)$ sous l'hypothèse de la
finitude de Cha$(A)$. Quel est l'état actuel de l' Énoncé. Supposons que $\operatorname{Cha}(A)$ est fini . Si $r>1$, alors $d>0$ ?
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The parity conjecture is known, i.e. it is known that if the order of vanishing of the $L$-function is even/odd, then the corank of $p$-Selmer is of the same parity (and I think this is known for every $p$ at this point; Nekovar handled the good ordinary or multiplicative case,
and B.D. Kim the good supersingular case. T. and V. Dokchitser then gave a new proof that handled the general case). This would imply that if Sha(E) is finite (even after passing to the $p$-part of Sha for some prime $p$) and the $L$-function has odd order vanishing, then $E({\mathbb Q})$ has positive rank. There has also been recent work on establishing cases of positive even order vanishing, and trying to prove that the Selmer group has rank at least two. This has been investigated by both Bellaiche--Chenevier and Skinner--Urban. I don't know precise statements though, and I'm not sure if either pair of authors can handle elliptic curves. (In both cases, the arguments involve deforming along an eigenvariety, and there are problems at low weights. So they may only have results for modular forms of weight $k > 2$.) Incidentally, although it wasn't part of your question, I don't think anything new is known about finiteness of Sha, beyond what you recalled in your question.
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11,503 |
My question concerns the (completely deterministic) card game known as War , played by seven-year-olds everywhere, such as my son Horatio, and sometimes also by others, such as their fathers. The question is: Is the expected length of the game infinite? The Rules. (from http://en.wikipedia.org/wiki/War_(card_game) ) The deck is divided evenly among the two players, giving each a face-down stack. In unison, each player reveals the top card on his stack (a "battle"), and the player with the higher card takes both the cards played and moves them to the bottom of his stack. If the two cards played are of equal value, each player lays down three face-down cards and a fourth card face-up (a "war"), and the higher-valued card wins all of the cards on the table, which are then added to the bottom of the player's stack. In the case of another tie, the war process is repeated until there is no tie. A player wins by collecting all the cards. If a player runs out of cards while dealing the face-down cards of a war, he may play the last card in his deck as his face-up card and still have a chance to stay in the game. Let us assume that the cards are returned to the deck in a well-defined manner. For example, in the order that the cards are played, with the previous round's winner's cards going first (and a first player selected for the opening battle). On the Wikipedia page, they tabulate the results of 1 million simulated random games, reporting an average length game of 248 battles. But this does not actually answer the question, because it could be that there is a devious initial arrangement of the cards leading to a periodic game lasting forever. Since there are only finitely many shuffles, this devious shuffle will contribute infinitely to the Expected Value . Thus, the question really amounts to: Question. Is there a devious shuffle in War, which leads to an infinitely long game? Of course, the game described above is merely a special case of the more general game that might be called Universal War , played with N players using a deck of cards representing elements of a finite partial pre-order. Any strictly dominating card wins the trick; otherwise, there is war amongst the players whose cards were not strictly dominated. Does any instance of Universal War have infinite expected length?
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Yes, a game of War can continue endlessly. In particular, if the following hands are dealt and player 1's cards are always added before player 2's cards to the bottom of the winner's stack, then the resulting game of War will never end: Player 1:
10S JS KD 6C 6D 2S 7S AC 3S 8D 5C 5D 8H AD KH 2D 4S 7C 3H 3D 10C 4D KC 4H 6H 7D Player 2:
9H 4C QC 9S 10D QH 5H QS 10H 8C AH 8S JH QD JD 2C KS 9D 3C 5S 6S 7H 9C AS JC 2H
|
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|
11,517 |
In the course of doing mathematics, I make extensive use of computer-based calculations. There's one CAS that I use mostly, even though I occasionally come across out-and-out wrong answers. After googling around a bit, I am unable to find a list of such bugs. Having such a list would help us remain skeptical and help our students become skeptical. So here's the question: What are some mathematical bugs in computer algebra systems? Please include a specific version of the software that has the bug. Please note that I'm not asking for bad design decisions, and I'm not asking for a discussion of the relative merits of different CAS's.
|
I don't know any interesting bugs in symbolic algebra packages but I know a true, enlightening and entertaining story about something that looked like a bug but wasn't. $\def\sinc{\operatorname{sinc}}$ Define $\sinc x = (\sin x)/x$ . Someone found the following result in an algebra package: $\int_0^\infty dx \sinc x = \pi/2$ They then found the following results: $\int_0^\infty dx \sinc x \; \sinc (x/3)= \pi/2$ $\int_0^\infty dx \sinc x \; \sinc (x/3) \; \sinc (x/5)= \pi/2$ and so on up to $\int_0^\infty dx \sinc x \; \sinc (x/3) \; \sinc (x/5) \; \cdots \; \sinc (x/13)= \pi/2$ So of course when they got: $\int_0^\infty dx \sinc x \; \sinc (x/3) \sinc (x/5) \; \cdots \; \sinc (x/15)$$=
\frac{467807924713440738696537864469}{935615849440640907310521750000}\pi$ they knew they had to report the bug. The poor vendor struggled for a long time trying to fix it but eventually came to the stunning realisation that this result is correct. These are now known as Borwein Integrals . A video on this topic, titled "Researchers thought this was a bug," is on the 3Blue1Brown YouTube channel here .
|
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|
11,540 |
What are the most attractive Turing undecidable problems in mathematics? There are thousands of examples, so please post here only the most attractive, best examples. Some examples already appear on the Wikipedia page. Standard community wiki rules. One example per post please. I will accept the answer I find to be the most attractive, according to the following criteria: Examples must be undecidable in the sense of Turing computability. (Please not that this is not the same as the sense of logical independence; think of word problem , not Continuum Hypothesis.) The best examples will arise from natural mathematical questions. The best examples will be easy to describe, and understandable by most or all mathematicians. (Challenge) The very best examples, if any, will in addition have intermediate Turing degree, strictly below the halting problem. That is, they will be undecidable, but not because the halting problem reduces to them. Edit: This question is a version of a previous question by Qiaochu Yuan , inquiring which problems in mathematics are able to simulate Turing machines, with the example of the MRDP theorem on diophantine equations, as well as the simulation of Turing machines via PDEs. He has now graciously merged his question here.
|
The mortality problem for $3\times 3$ matrices: given a finite set $F$ of $3\times 3$ integer matrices, decide whether the zero matrix is a product of members of $F$ (with repetitions allowed). This was proved unsolvable by Michael Paterson,
Studies in Applied Mathematics 49 (1970), 105--107, doi: 10.1002/sapm1970491105 . The corresponding problem for $2\times 2$ matrices is apparently still open. Edit (11 September 2016): The problem for $2\times 2$ matrices is apparently still open, despite the solution of a seemingly similar problem mentioned in Igor Potapov's answer below.
|
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|
11,631 |
Obviously, graph invariants are wonderful things, but the usual ones (the Tutte polynomial, the spectrum, whatever) can't always distinguish between nonisomorphic graphs. Actually, I think that even a combination of the two I listed will fail to distinguish between two random trees of the same size with high probability. Is there a known set of graph invariants that does always distinguish between non-isomorphic graphs? To rule out trivial examples, I'll require that the problem of comparing two such invariants is in P (or at the very least, not obviously equivalent to graph isomorphism) -- so, for instance, "the adjacency matrix" is not a good answer. (Computing the invariants is allowed to be hard, though.) If this is (as I sort of suspect) in fact open, does anyone have any insight on why it should be hard? Such a set of invariants wouldn't require or violate any widely-believed complexity-theoretic conjectures, and actually there are complexity-theoretic reasons to think that something like it exists (specifically, under derandomization, graph isomorphism is in co-NP). It seems like it shouldn't be all that hard... Edit: Thorny's comment raises a good point. Yes, there is trivially a complete graph invariant, which is defined by associating a unique integer (or polynomial, or labeled graph...) to every isomorphism class of graphs. Since there are a countable number of finite graphs, we can do this, and we have our invariant. This is logically correct but not very satisfying; it works for distinguishing between finite groups, say, or between finite hypergraphs or whatever. So it doesn't actually tell us anything at all about graph theory. I'm not sure if I can rigorously define the notion of a "satisfying graph invariant," but here's a start: it has to be natural , in the sense that the computation/definition doesn't rely on arbitrarily choosing an element of a finite set. This disqualifies Thorny's solution, and I think it disqualifies Mariano's, although I could be wrong.
|
A complete graph invariant is computationally equivalent to a canonical labeling of a graph. A canonical labeling is by definition an enumeration of the vertices of every finite graph, with the property that if two graphs are isomorphic as unlabeled graphs, then they are still isomorphic as labeled graphs. If you have a black box that gives you a canonical labeling, then obviously that is a complete graph invariant. On the other hand, if you have a complete graph invariant for unlabeled graphs, then you also have one for partially labeled graphs. So given a black box that computes a complete graph invariant, you can assign the label 1 to the vertex that minimizes the invariant, then assign a label 2 to a second vertex than again minimizes the invariant, and so on. There are algorithms to decide graph isomorphism for certain types of graphs, or for all graphs but with varying performance, and there are algorithms for canonical labeling, again with varying performance. It is understood that graph isomorphism reduces to canonical labeling, but not necessarily vice versa. The distinction between the two problems is discussed in this classic paper by Babai and Luks. One natural canonical labeling of a graph is the one that is lexicographically first. I think I saw, although I don't remember where, a result that computing this canonical labeling for one of the reasonable lex orderings on labeled graphs is NP-hard. But there could well be a canonical labeling computable in P that doesn't look anything like first lex. As Douglas says, nauty is a graph computation package that includes a canonical labeling function. It is often very fast, but not always . Nauty uses a fancy contagious coloring algorithm. For a long time people thought that contagious coloring algorithms might in principle settle the canonical labeling and graph isomorphism problems, but eventually counterexamples were found in another classic paper by Cai, Furer, and Immerman. It was not clear at first whether this negative result would apply to nauty, but it seems that it does.
|
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|
11,669 |
Hi, Currently, I'm taking matrix theory, and our textbook is Strang's Linear Algebra. Besides matrix theory, which all engineers must take, there exists linear algebra I and II for math majors. What is the difference,if any, between matrix theory and linear algebra? Thanks!
|
Let me elaborate a little on what Steve Huntsman is talking about. A matrix is just a list of numbers, and you're allowed to add and multiply matrices by combining those numbers in a certain way. When you talk about matrices, you're allowed to talk about things like the entry in the 3rd row and 4th column, and so forth. In this setting, matrices are useful for representing things like transition probabilities in a Markov chain, where each entry indicates the probability of transitioning from one state to another. You can do lots of interesting numerical things with matrices, and these interesting numerical things are very important because matrices show up a lot in engineering and the sciences. In linear algebra, however, you instead talk about linear transformations, which are not (I cannot emphasize this enough) a list of numbers, although sometimes it is convenient to use a particular matrix to write down a linear transformation. The difference between a linear transformation and a matrix is not easy to grasp the first time you see it, and most people would be fine with conflating the two points of view. However, when you're given a linear transformation, you're not allowed to ask for things like the entry in its 3rd row and 4th column because questions like these depend on a choice of basis. Instead, you're only allowed to ask for things that don't depend on the basis, such as the rank, the trace, the determinant, or the set of eigenvalues. This point of view may seem unnecessarily restrictive, but it is fundamental to a deeper understanding of pure mathematics.
|
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|
11,674 |
I am not by any means an expert in category theory. Anyway whenever I have studied a concept in category theory I have always had the feeling that most of the subtleties introduced are artificial. For a few examples: -one does not usually consider isomorphic, but rather equivalent categories -universal objects are unique only up to a canonical isomorphism -the category of categories is really a 2-category, so some natural constructions do not yield functors into categories, but only pseudofunctors -cleavages of fibered categories do not always split .... My question is: can skeleta be used to simplify all this stuff? It looks like building everything using skeleta from the beginning would remove a lot of indeterminacies in these constructions. On the other hand it may be the case that this subtleties are really intrinsic, and so using skeleta, which are not canonically determined, would only move the difficulties around.
|
The first, and perhaps most important, point is that hardly any categories that occur in nature are skeletal . The axiom of choice implies that every category is equivalent to a skeletal one, but such a skeleton is usually artifical and non-canonical. Thus, even if using skeletal categories simplified category theory, it would not mean that the subtleties were artifical, but rather that the naturally occurring subtleties could be removed by an artificial construction (the skeleton). In fact, however, skeletons don't actually simplify much of anything in category theory. It is true, for instance, that any functor between skeletal categories which is part of an equivalence of categories is actually an isomorphism of categories. However, this isn't really useful because, as mentioned above, most interesting categories are not skeletal. So in practice, one would either still have to deal either with equivalences of categories, or be constantly replacing categories by equivalent skeletal ones, which is even more tedious (and you'd still need the notion of "equivalence" in order to know what it means to replace a category by an "equivalent" skeletal one). In all the other examples you mention, skeletal categories don't even simplify things that much. In general, not every pseudofunctor between 2-categories is equivalent to a strict functor, and skeletality won't help you here. Even if the hom-categories of your 2-categories are skeletal, there can still be pseudofunctors that aren't equivalent to strict ones, because the data of a pseudofunctor includes coherence isomorphisms that may not be identities. Similarly for cloven and split fibrations. A similar question was raised in the query box here : important data can be encoded in coherence isomorphisms even when they are automorphisms. The argument in CWM mentioned by Leonid is another good example of the uselessness of skeletons. Here's one final one that's bitten me in the past. You mention that universal objects are unique only up to (unique specified) isomorphism. So one might think that in a skeletal category, universal objects would be unique on the nose. This is actually false , because a universal object is not just an object, but an object together with data exhibiting its universal property, and a single object can have a given universal property in more than one way. For instance, a product of objects A and B is an object P together with projections P→A and P→B satisfying a universal property. If Q is another object with projections Q→A and Q→B and the same property, then from the universal properties we obtain a unique specified isomorphism P≅Q. Now if the category is skeletal, then we must have P=Q, but that doesn't mean the isomorphism P≅Q is the identity . In fact, if P is a product of A and B with the projections P→A and P→B, then composing these two projections with any automorphism of P produces another product of A and B, which happens to have the same vertex object P but has different projections. So assuming that your category is skeletal doesn't actually make anything any more unique.
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|
11,747 |
The question is about characterising the sets $S(K)$ of primes which split completely in a given galoisian extension $K|\mathbb{Q}$. Do recent results such as Serre's modularity conjecture (as proved by Khare-Wintenberger), or certain cases of the Fontaine-Mazur conjecture (as proved by Kisin), have anything to say about such subsets, beyond what Class Field Theory has to say ? I'll now introduce some terminology and recall some background. Let $\mathbb{P}$ be the set of prime numbers. For every galoisian extension $K|\mathbb{Q}$, we have the subset $S(K)\subset\mathbb{P}$ consisting of those primes which split (completely) in $K$. The question is about characterising such subsets; we call them galoisian subsets. If $T\subset\mathbb{P}$ is galoisian, there is a unique galoisian extension $K|\mathbb{Q}$ such that $T=S(K)$, cf. Neukirch (13.10). We say that $T$ is abelian if $K|\mathbb{Q}$ is abelian. As discussed here recently, a subset $T\subset\mathbb{P}$ is abelian if and only if it is defined by congruences. For example, the set of primes $\equiv1\pmod{l}$ is the same as $S(\mathbb{Q}(\zeta_l))$. "Being defined by congruences" can be made precise, and counts as a characterisation of abelian subsets of $\mathbb{P}$. Neukirch says that Langlands' Philosophy provides a characterisation of all galoisian subsets of $\mathbb{P}$. Can this remark now be illustrated by some striking example ? Addendum (28/02/2010) Berger 's recent Bourbaki exposé 1017 arXiv:1002.4111 says that cases of the Fontaine-Mazur conjecture have been proved by Matthew Emerton as well. I didn't know this at the time of asking the question, and the unique answerer did not let on that he'd had something to do with Fontaine-Mazur...
|
I think it is easiest to illustrate the role of the Langlands program (i.e. non-abelian class field theory) in answering this question by giving an example. E.g. consider the Hilbert class field $K$ of $F := {\mathbb Q}(\sqrt{-23})$ ; this is a degree 3 abelian extension of $F$ , and an $S_3$ extension of $\mathbb Q$ . (It is the splitting field of the polynomial $x^3 - x - 1$ .) The 2-dimensional representation of $S_3$ thus gives a representation $\rho:Gal(K/{\mathbb Q}) \hookrightarrow GL_2({\mathbb Q}).$ A prime $p$ splits in $K$ if and only if $Frob_p$ is the trivial conjugacy class
in $Gal(K{\mathbb Q})$ , if and only if $\rho(Frob_p)$ is the identity matrix, if and only
if trace $\rho(Frob_p) = 2$ . (EDIT: While $Frob_p$ is a 2-cycle, resp. 3-cycle, if and only if $\rho(Frob_p)$ has trace 0, resp. -1.) Now we have the following reciprocity law for $\rho$ : there is a modular form $f(q)$ , in fact
a Hecke eigenform, of weight 1 and level 23, whose $p$ th Hecke eigenvalue gives
the trace of $\rho(Frob_p)$ . (This is due to Hecke; the reason that Hecke could handle
this case is that $\rho$ embeds $Gal(K/{\mathbb Q})$ as a dihedral
subgroup of $GL_2$ , and so $\rho$ is in fact induced from an abelian character of the
index two subgroup $Gal(K/F)$ .) In this particular case, we have the following explicit formula: $$f(q) = q \prod_{n=1}^{\infty}(1-q^n)(1-q^{23 n}).$$ If we expand out this product as $f(q) = \sum_{n = 1}^{\infty}a_n q^n,$ then we find that $trace \rho(Frob_p) = a_p$ (for $p \neq 23$ ),
and in particular, $p$ splits completely in $K$ if and only if $a_p = 2$ .
(For example, you can check this way that the smallest split prime is $p = 59$ ;
this is related to the fact that $59 = 6^2 + 23 \cdot 1^2$ .).
(EDIT: While $Frob_p$ has order $2$ , resp. 3, if and only if $a_p =0$ , resp. $-1$ .) So we obtain a description of the set of primes that split in $K$ in terms of
the modular form $f(q)$ , or more precisely its Hecke eigenvalues (or what amounts
to the same thing, its $q$ -expansion). The Langlands program asserts that an analogous statement is true for any
Galois extension of number fields $E/F$ when one is given a continuous
representation $Gal(E/F) \hookrightarrow GL\_n(\mathbb C).$ This is known
when $n = 2$ and either the image of $\rho$ is solvable (Langlands--Tunnell) or $F = \mathbb Q$ and $\rho(\text{complex conjugation})$ is non-scalar (Khare--Wintenberger--Kisin).
In most other contexts it remains open.
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|
11,753 |
We are interested in tiling a rectangle with copies of a single tile (rotations and reflections are allowed). This is very easy to do, by cutting the rectangle into smaller rectangles. What happens when we ask that the pieces be non-rectangular? For an even number of pieces, this is easy again (cut it into rectangles, and then cut every rectangle in two through its diagonal. Other tilings are also easy to find). The interesting (and difficult) case is tiling with an odd number of non-rectangular pieces. Some questions: Can you give examples of such tilings? What is the smallest (odd) number of pieces for which it is possible? Is it possible for every number of pieces? ( e.g. , with five) There are two main versions of the problem: the polyomino case (when the tiles are made of unit squares), and the general case (when the tiles can have any shape). The answers to the above questions might be different in each case. It seems that it is impossible to do with three pieces (I have some kind of proof), and the smallest number of pieces I could get is $15$ , as shown above: (source) This problem is very useful for spending time when attending some boring talk, etc.
|
Golomb's book Polyominoes has a section on this. Call the smallest odd number of copies of a polyomino that can tile a rectangle its "odd-order". Then Golomb says there are polyominoes of odd order 1, 11, and 15+6t for all $t \ge 0$. The polyomino of odd order 11 is due to Klarner [1], and is illustrated here by Michael Reid . Reid has lots of pictures of tilings of rectangles with polyominoes . In particular the 15+6t family can be seen: here are polyominoes with odd-order 15 , odd-order 21 , odd-order 27 , and so on. Reid has shown [3] that other odd orders exist, including 35, 49, and 221, but I don't know if there's a general pattern. Finally, Stewart and Wormstein [2] proved that polyominoes of order 3 do not exist. (Stewart's book Another Fine Math You've got Me Into suggests that Wormstein is a fictional character.) [1] David A. Klarner, Packing a rectangle with congruent N-ominoes , J. Combin. Theory 7 (1969) 107-115,
[2] Stewart, Ian N. and Wormstein, Albert. Polyominoes of order 3 do not exist. J. Combin. Theory Series A 61 (1992) 130-136. [3] Michael Reid. Tiling Rectangles and Half Strips with Congruent Polyominoes.
J. Combin. Theory Series A 80 (1997) 106-123.
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|
11,774 |
For the definitions of the equivalence relations on algebraic cycles see http://en.wikipedia.org/wiki/Adequate_equivalence_relation . I want to know how far away from each other the equivalence relations on algebraic cycles are and what the intuition is for them. My impression is that rational equivalence gives much bigger Chow groups than algebraic equivalence, and that algebraic equivalence, homological equivalence and numerical equivalence are quite tight together. Take for example an elliptic curve. We have $CH^1(E) = \mathbb{Z} \times E(K)$, algebraic equivalence (take $C = E$) $\mathbb{Z}$ = numerical equivalence.
|
I will focus on complex projective varieties. Codimension one The situation in codimension one is considerably simpler than in higher codimensions.
Codimension one rational equivalence classes are parametrized by $Pic(X)= H^1(X,\mathcal O_X^{\ast})$ while algebraic equivalence classes are parametrized by the Neron-Severi group of $X$, which can be defined as the image of the Chern class map from $Pic(X)$ to $H^2(X,\mathbb Z)$. It follows that in codimension one the group of rational equivalence classes is a countable union of abelian varieties; the groups of algebraic equivalence classes and homological equivalence classes coincide, and are equal to $NS(X)$ a subgroup of $H^2(X,\mathbb Z)$; the group of numerical equivalence classes is the quotient of $NS(X)$ by its torsion subgroup. Higher codimension The higher codimension case, as pointed out by Tony Pantev, is considerably more complicate and algebraic and homological equivalence no longer coincide. Concerning rational equivalence, Mumford proved that the Chow group of zero cycles of surfaces admitting non-zero holomorphic $2$-forms are infinite dimensional , contradicting a conjecture by Severi. The paper is Mumford, D. Rational equivalence of $0$-cycles on surfaces. J. Math. Kyoto Univ. 9 1968. Warning The definitions of rational and algebraic equivalence at wikipedia are not correct.
I will commment below on the algebraic equivalence. There one can find the following definition. $Z ∼_{alg} Z'$ if there exists a curve $C$ and a
cycle $V$ on $X × C$ flat over C, such
that $$V \cap \left( X \times\lbrace c\rbrace \right) = Z \quad \text{ and }
\quad V \cap \left( X \times\lbrace c\rbrace \right) = Z' $$
for two points $c$ and $d$ on the
curve. This is not correct. The correct definition is $Z ∼_{alg} Z'$ if there exists a curve $C$ and a
cycle $V$ on $X × C$ flat over C, such
that $$V \cap \left( X \times\lbrace c\rbrace \right) - V \cap \left( X \times\lbrace d\rbrace \right) = Z - Z' $$
for two points $c$ and $d$ on the
curve. To construct an example of two algebraically equivalent divisors which do not satisfy the wikipedia definition let $X$ be a projective variety with $H^1(X,\mathcal O_X) \neq 0$ and
take a non-trivial line-bundle $\mathcal L$ over $X$ with zero Chern class.
If $Y = \mathbb P ( \mathcal O_X \oplus \mathcal L)$ then $Y$ contains two copies $X_0$ and $X_{\infty}$ of $X$ ( one for each factor of $\mathcal O_X \oplus \mathcal L$ ) which are algebraically equivalent but can't be deformed because their normal bundles are $\mathcal L$ and $\mathcal L^{\ast}$. This does not contradict the second definition because for sufficiently ample divisors $H$ it is clear $X_0 + H$ can be deformed into $X_{\infty} + H$.
|
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11,777 |
Given two non-isogenous elliptic curves $E_1$ and $E_2$ over $\mathbb{C}$.
Set $A:=E_1 \times E_2$. Given a nontrivial sheaf of quaternion algebras $D$ over $A$, what is the dimension of the vector space $H^1(A,D)$? If one thinks of $D$ as an element in the Brauer group $Br(A)$, then it is $2$-torsion, hence belongs to $Br(A)[2]$. Since the curves are non-isogenous there is an isomorphism $Pic(E_1)[2] \otimes Pic(E_2)[2] \to Br(A)[2]$. So there should be a connection between such quaternions and $2$-torsion line bundles on the curves, but i cannot find an explicit description for this isomorphism. If there is one, i thought one could use the Künneth formula to compute $H^1(A,D)$ in terms of the cohomology of the line bundles on the curves. For now i could only work out the bound $d=dim(H^1(A,D)) \geq 2$: using Hirzebruch-Riemann-Roch and simplifying terms one gets $d=c_2(D)+2$. After a result of M.Lieblich one has $c_2(D)\geq 0$. Does anyone see/have an explicit description of the isomorphism mentioned above? Is the idea using Künneth a promising approach to this problem at all? Or does anyone have another approach? Are there some calculations regarding this in the literature (i couldn't find one)? Another question in this context is: what is the image of such an algebra under the map $Br(A) \rightarrow Br(\mathbb{C}(A))$. This should be nontrivial $\mathbb{C}(A)$-quaternions, since the map "looking at the genric point $\eta$" is injective, i.e. $D_{\eta}$ is generated by elements $i,j$ with $i^2=a, j^2=b and ij=-ji$. But what are a resp. b? I think they should have something to do with functions h such that 2*Y=div(h), where Y defines one of the line bundles. Is this true?
|
I will focus on complex projective varieties. Codimension one The situation in codimension one is considerably simpler than in higher codimensions.
Codimension one rational equivalence classes are parametrized by $Pic(X)= H^1(X,\mathcal O_X^{\ast})$ while algebraic equivalence classes are parametrized by the Neron-Severi group of $X$, which can be defined as the image of the Chern class map from $Pic(X)$ to $H^2(X,\mathbb Z)$. It follows that in codimension one the group of rational equivalence classes is a countable union of abelian varieties; the groups of algebraic equivalence classes and homological equivalence classes coincide, and are equal to $NS(X)$ a subgroup of $H^2(X,\mathbb Z)$; the group of numerical equivalence classes is the quotient of $NS(X)$ by its torsion subgroup. Higher codimension The higher codimension case, as pointed out by Tony Pantev, is considerably more complicate and algebraic and homological equivalence no longer coincide. Concerning rational equivalence, Mumford proved that the Chow group of zero cycles of surfaces admitting non-zero holomorphic $2$-forms are infinite dimensional , contradicting a conjecture by Severi. The paper is Mumford, D. Rational equivalence of $0$-cycles on surfaces. J. Math. Kyoto Univ. 9 1968. Warning The definitions of rational and algebraic equivalence at wikipedia are not correct.
I will commment below on the algebraic equivalence. There one can find the following definition. $Z ∼_{alg} Z'$ if there exists a curve $C$ and a
cycle $V$ on $X × C$ flat over C, such
that $$V \cap \left( X \times\lbrace c\rbrace \right) = Z \quad \text{ and }
\quad V \cap \left( X \times\lbrace c\rbrace \right) = Z' $$
for two points $c$ and $d$ on the
curve. This is not correct. The correct definition is $Z ∼_{alg} Z'$ if there exists a curve $C$ and a
cycle $V$ on $X × C$ flat over C, such
that $$V \cap \left( X \times\lbrace c\rbrace \right) - V \cap \left( X \times\lbrace d\rbrace \right) = Z - Z' $$
for two points $c$ and $d$ on the
curve. To construct an example of two algebraically equivalent divisors which do not satisfy the wikipedia definition let $X$ be a projective variety with $H^1(X,\mathcal O_X) \neq 0$ and
take a non-trivial line-bundle $\mathcal L$ over $X$ with zero Chern class.
If $Y = \mathbb P ( \mathcal O_X \oplus \mathcal L)$ then $Y$ contains two copies $X_0$ and $X_{\infty}$ of $X$ ( one for each factor of $\mathcal O_X \oplus \mathcal L$ ) which are algebraically equivalent but can't be deformed because their normal bundles are $\mathcal L$ and $\mathcal L^{\ast}$. This does not contradict the second definition because for sufficiently ample divisors $H$ it is clear $X_0 + H$ can be deformed into $X_{\infty} + H$.
|
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|
11,784 |
Are there some fun applications of the theory of representations of finite groups? I would like to have some examples that could be explained to a student who knows what is a finite group but does not know much about what is a repersentation (say knows the definition). The standard application that is usually mentioned is Burnside's theorem http://en.wikipedia.org/wiki/Burnside_theorem . The application may be of any kind, not necessarely in math. But math applications are of course very wellcome too!!! It will be very helpfull also if you desribe a bit this application.
|
An example from Kirillov's book on representation theory: write numbers 1,2,3,4,5,6 on the faces of a cube, and keep replacing (simultaneously) each number by the average of its neighbours. Describe (approximately) the numbers on the faces after many iterations. Another example I like to use in the beginning of a group reps course: write down the multiplication table in a finite group, and think of it as of a square matrix whose entries are formal variables corresponding to elements of the group. Then the determinant of this matrix is a polynomial in these variables. Describe its decomposition into irreducibles. This question, which Frobenius was asked by Dedekind, lead him to invention of group characters. A function in two variables can be uniquely decomposed as a sum of a symmetric and antisymmetric (skew-symmetric) function. What happens for three and more variables - what types of symmetries do exist there?
|
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|
11,885 |
I'm looking for a nontrivial, but not super difficult question concerning Fibonacci numbers. It should be at a level suitable for an undergraduate course. Here is a (not so good) example of the sort of thing I am looking for. a) Prove that every positive integer can be represented in binary over the basis of Fibonacci numbers. That is, show that for all $n$, there exist bits $x_1,\ldots,x_k$ such that $n = \sum_{i=1}^kx_iF_i$. b) Give an algorithm to increment such numbers in constant amortized time. Any ideas for better ones?
|
An accessible (and interesting) thing to look at with Fibonacci numbers is their periodicity modulo various integers, especially primes and prime powers. One example of an accessible result is that if $k(p)$ is the period of the Fibonacci numbers modulo a prime $p$, then $k(p)\mid p^2-1$. You can get sharper results by examining whether or not 5 is a quadratic residue mod $p$ (think of the importance of $\frac{1\pm\sqrt{5}}{2}$ to the Fibonacci numbers). You can prove things about this periodicity directly, or reduce the 2x2 matrix which Gowers mentions modulo $p$ and get the same thing, depending on what you'd like to emphasize to your students. Some good resources for this subject are http://en.wikipedia.org/wiki/Pisano_period http://euclid.math.temple.edu/~renault/fibonacci/fib.html Another neat thing about the Fibonacci numbers is their appearance as sums of "diagonals" in Pascal's Triangle, as in this picture: However, this fact is provable simply by induction, so maybe this is too easy for what you have in mind.
|
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|
11,978 |
I was very surprised when I first encountered the Mertens conjecture . Define $$ M(n) = \sum_{k=1}^n \mu(k) $$ The Mertens conjecture was that $|M(n)| < \sqrt{n}$ for $n>1$, in contrast to the Riemann Hypothesis, which is equivalent to $M(n) = O(n^{\frac12 + \epsilon})$ . The reason I found this conjecture surprising is that it fails heuristically if you assume the Mobius function is randomly $\pm1$ or $0$. The analogue fails with probability $1$ for a random $-1,0,1$ sequence where the nonzero terms have positive density. The law of the iterated logarithm suggests that counterexamples are large but occur with probability 1. So, it doesn't seem surprising that it's false, and that the first counterexamples are uncomfortably large. There are many heuristics you can use to conjecture that the digits of $\pi$, the distribution of primes, zeros of $\zeta$ etc. seem random. I believe random matrix theory in physics started when people asked whether the properties of particular high-dimensional matrices were special or just what you would expect of random matrices. Sometimes the right random model isn't obvious, and it's not clear to me when to say that an heuristic is reasonable. On the other hand, if you conjecture that all naturally arising transcendentals have simple continued fractions which appear random, then you would be wrong, since $e = [2;1,2,1,1,4,1,1,6,...,1,1,2n,...]$, and a few numbers algebraically related to $e$ have similar simple continued fraction expansions. What other plausible conjectures or proven results can be framed as heuristically false according to a reasonable probability model?
|
Just run across this question, and am surprised that the first example
that came to mind was not mentioned: Fermat's "Last Theorem" is heuristically true for $n > 3$,
but heuristically false for $n=3$ which is one of the easier
cases to prove. if $0 < x \leq y < z \in (M/2,M]$ then $|x^n + y^n - z^n| < M^n$.
There are about $cM^3$ candidates $(x,y,z)$ in this range
for some $c>0$ (as it happens $c=7/48$), producing values of
$\Delta := x^n+y^n-z^n$ spread out on
the interval $(-M^n,M^n)$ according to some fixed distribution
$w_n(r) dr$ on $(-1,1)$ scaled by a factor $M^n$ (i.e.,
for any $r_1,r_2$ with $-1 \leq r_1 \leq r_2 \leq 1$
the fraction of $\Delta$ values in $(r_1 M^n, r_2 M^n)$
approaches $\int_{r_1}^{r_2} w_n(r) dr$ as $M \rightarrow \infty$). This suggests that any given value of $\Delta$, such as $0$,
will arise about $c w_n(0) M^{3-n}$ times. Taking $M=2^k=2,4,8,16,\ldots$
and summing over positive integers $k$ yields a rapidly divergent sum
for $n<3$, a barely divergent one for $n=3$, and a rapidly convergent
sum for $n>3$. Specifically, we expect the number of solutions of $x^n+y^n=z^n$
with $z \leq M$ to grow as $M^{3-n}$ for $n<3$ (which is true and easy),
to grow as $\log M$ for $n=3$ (which is false), and to be finite for $n>3$
(which is true for relatively prime $x,y,z$ and very hard to prove [Faltings]). More generally, this kind of analysis suggests that for $m \geq 3$
the equation $x_1^n + x_2^n + \cdots + x_{m-1}^n = x_m^n$
should have lots of solutions for $n<m$,
infinitely but only logarithmically many for $n=m$,
and finitely many for $n>m$. In particular, Euler's conjecture
that there are no solutions for $m=n$ is heuristically false for all $m$.
So far it is known to be false only for $m=4$ and $m=5$. Generalization in a different direction suggests that any cubic
plane curve $C: P(x,y,z)=0$ should have infinitely many rational points.
This is known to be true for some $C$ and false for others;
and when true the number of points of height up to $M$ grows as
$\log^{r/2} M$ for some integer $r>0$ (the rank of the elliptic curve),
which may equal $2$ as the heuristic predicts but doesn't have to.
The rank is predicted by the celebrated conjecture of Birch and
Swinnerton-Dyer, which in effect refines the heuristic by accounting
for the distribution of values of $P(x,y,z)$ not just
"at the archimedean place" (how big is it?) but also "at finite places"
(is $P$ a multiple of $p^e$?). The same refinement is available for equations in more variables,
such as Euler's generalization of the Fermat equation;
but this does not change the conclusion (except for equations such as
$x_1^4 + 3 x_2^4 + 9 x_3^4 = 27 x_4^4$,
which have no solutions at all for congruence reasons),
though in the borderline case $m=n$ the expected power of $\log M$ might rise. Warning : there are subtler obstructions that may prevent a surface from
having rational points even when the heuristic leads us to expect
plentiful solutions and there are no congruence conditions that
contradict this guess. An example is the Cassels-Guy cubic
$5x^3 + 9y^3 + 10z^3 + 12w^3 = 0$, with no nonzero rational solutions
$(x,y,z,w)$: Cassels, J.W.S, and Guy, M.J.T.:
On the Hasse principle for cubic surfaces, Mathematika 13 (1966), 111--120.
|
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|
11,981 |
I have a large set of data points which have 18 dimensions to them. I know that these data points must follow a strict polynomial formula with no possible variance. Given that I have enough data (I assure you I have), how do I go about building a regression model to find the general formula of the lowest polynomials for the equation which gives these data points?
|
Just run across this question, and am surprised that the first example
that came to mind was not mentioned: Fermat's "Last Theorem" is heuristically true for $n > 3$,
but heuristically false for $n=3$ which is one of the easier
cases to prove. if $0 < x \leq y < z \in (M/2,M]$ then $|x^n + y^n - z^n| < M^n$.
There are about $cM^3$ candidates $(x,y,z)$ in this range
for some $c>0$ (as it happens $c=7/48$), producing values of
$\Delta := x^n+y^n-z^n$ spread out on
the interval $(-M^n,M^n)$ according to some fixed distribution
$w_n(r) dr$ on $(-1,1)$ scaled by a factor $M^n$ (i.e.,
for any $r_1,r_2$ with $-1 \leq r_1 \leq r_2 \leq 1$
the fraction of $\Delta$ values in $(r_1 M^n, r_2 M^n)$
approaches $\int_{r_1}^{r_2} w_n(r) dr$ as $M \rightarrow \infty$). This suggests that any given value of $\Delta$, such as $0$,
will arise about $c w_n(0) M^{3-n}$ times. Taking $M=2^k=2,4,8,16,\ldots$
and summing over positive integers $k$ yields a rapidly divergent sum
for $n<3$, a barely divergent one for $n=3$, and a rapidly convergent
sum for $n>3$. Specifically, we expect the number of solutions of $x^n+y^n=z^n$
with $z \leq M$ to grow as $M^{3-n}$ for $n<3$ (which is true and easy),
to grow as $\log M$ for $n=3$ (which is false), and to be finite for $n>3$
(which is true for relatively prime $x,y,z$ and very hard to prove [Faltings]). More generally, this kind of analysis suggests that for $m \geq 3$
the equation $x_1^n + x_2^n + \cdots + x_{m-1}^n = x_m^n$
should have lots of solutions for $n<m$,
infinitely but only logarithmically many for $n=m$,
and finitely many for $n>m$. In particular, Euler's conjecture
that there are no solutions for $m=n$ is heuristically false for all $m$.
So far it is known to be false only for $m=4$ and $m=5$. Generalization in a different direction suggests that any cubic
plane curve $C: P(x,y,z)=0$ should have infinitely many rational points.
This is known to be true for some $C$ and false for others;
and when true the number of points of height up to $M$ grows as
$\log^{r/2} M$ for some integer $r>0$ (the rank of the elliptic curve),
which may equal $2$ as the heuristic predicts but doesn't have to.
The rank is predicted by the celebrated conjecture of Birch and
Swinnerton-Dyer, which in effect refines the heuristic by accounting
for the distribution of values of $P(x,y,z)$ not just
"at the archimedean place" (how big is it?) but also "at finite places"
(is $P$ a multiple of $p^e$?). The same refinement is available for equations in more variables,
such as Euler's generalization of the Fermat equation;
but this does not change the conclusion (except for equations such as
$x_1^4 + 3 x_2^4 + 9 x_3^4 = 27 x_4^4$,
which have no solutions at all for congruence reasons),
though in the borderline case $m=n$ the expected power of $\log M$ might rise. Warning : there are subtler obstructions that may prevent a surface from
having rational points even when the heuristic leads us to expect
plentiful solutions and there are no congruence conditions that
contradict this guess. An example is the Cassels-Guy cubic
$5x^3 + 9y^3 + 10z^3 + 12w^3 = 0$, with no nonzero rational solutions
$(x,y,z,w)$: Cassels, J.W.S, and Guy, M.J.T.:
On the Hasse principle for cubic surfaces, Mathematika 13 (1966), 111--120.
|
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|
12,009 |
One the proofs that I've never felt very happy with is the classification of finitely generated abelian groups (which says an abelian group is basically uniquely the sum of cyclic groups of orders $a_i$ where $a_i|a_{i+1}$ and a free abelian group). The proof that I know, and am not entirely happy with goes as follows: your group is finitely presented, so take a surjective map from a free abelian group. The kernel is itself finitely generated (this takes a little argument in and of itself; note that adding a new generator to a subgroup of free abelian group either increases dimension after tensoring with $\mathbb{Q}$ or descreases the size of the torsion of the quotient), so our group is the cokernel of a map between finite rank free groups. Now, (and here's the part I dislike) look at the matrix for this map, and remember that it has a Smith normal form . Thus, our group is the quotient of a free group by a diagonal matrix where the non-zero entries are $a_i$ as above. I really do not think I should have to algorithmically reduce to Smith normal form or anything like that, but know of no proof that doesn't do that. By the way, if you're tempted to say "classification of finitely generated modules for PIDs!" make sure you know a proof of that that doesn't use Smith normal form first.
|
I reject the premise of the question. :-) It is true, as Terry suggests, that there is a nice dynamical proof of the classification of finite abelian groups. If $A$ is finite, then for every prime $p$ has a stable kernel $A_p$ and a stable image $A_p^\perp$ in $A$, by definition the limits of the kernel and image of $p^n$ as $n \to \infty$. You can show that this yields a direct sum decomposition of $A$, and you can use linear algebra to classify the dynamics of the action of $p$ on $A_p$. A similar argument appears in Matthew Emerton's proof. As Terry says, this proof is nice because it works for finitely generated torsion modules over any PID. In particular, it establishes Jordan canonical form for finite-dimensional modules over $k[x]$, where $k$ is an algebraically closed field. My objection is that finite abelian groups look easier than finitely generated abelian groups in this question. The slickest proof of the classification that I know is one that assimilates the ideas of Smith normal form. Ben's question is not entirely fair to Smith normal form, because you do not need finitely many relations. That is, Smith normal form exists for matrices with finitely many columns, not just for finite matrices. This is one of the tricks in the proof that I give next. Theorem. If $A$ is an abelian group with $n$ generators, then it is a direct sum of at most $n$ cyclic groups. Proof. By induction on $n$. If $A$ has a presentation with $n$ generators and no relations, then $A$ is free and we are done. Otherwise, define the height of any $n$-generator presentation of $A$ to be the least norm $|x|$ of any non-zero coefficient $x$ that appears in some relation. Choose a presentation with least height, and let $a \in A$ be the generator such that $R = xa + \ldots = 0$ is the pivotal relation. (Pun intended. :-) ) The coefficient $y$ of $a$ in any other relation must be a multiple of $x$, because otherwise if we set $y = qx+r$, we can make a relation with coefficient $r$. By the same argument, we can assume that $a$ does not appear in any other relation. The coefficient $z$ of another generator $b$ in the relation $R$ must also be a multiple of $x$, because otherwise if we set $z = qx+r$ and replace $a$ with $a' = a+qb$, the coefficient $r$ would appear in $R$. By the same argument, we can assume that the relation $R$ consists only of the equation $xa = 0$, and without ruining the previous property that $a$ does not appear in other relations. Thus $A \cong \mathbb{Z}/x \oplus A'$, and $A'$ has $n-1$ generators. □ Compare the complexity of this argument to the other arguments supplied so far. Minimizing the norm $|x|$ is a powerful step. With just a little more work, you can show that $x$ divides every coefficient in the presentation, and not just every coefficient in the same row and column. Thus, each modulus $x_k$ that you produce divides the next modulus $x_{k+1}$. Another way to describe the argument is that Smith normal form is a matrix version of the Euclidean algorithm. If you're happy with the usual Euclidean algorithm, then you should be happy with its matrix form; it's only a bit more complicated. The proof immediately works for any Euclidean domain; in particular, it also implies the Jordan canonical form theorem. And it only needs minor changes to apply to general PIDs.
|
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|
12,019 |
Let's say you have a planar tiling. Quite often these tilings are introduced as geometrical objects with metrics; each tile having coordinates assigned to its vertices. The tiling has an associated graph: the nodes of the graph are the vertices of the tiles, etc. What kind of additional value is in general provided by insisting the tile vertices to have coordinates? Does not discussing the properties of the tiling graph topology give enough information? I understand that the answer depends on the type of the tiling in question. If one takes graphs as the starting point, then what would be some natural ways to define infinite planar graph periodicity so that there would exist periodic planar tilings corresponding to a given graph? Can Penrose tiling be defined by its graph topology and can some of its general properties, like aperiodicity, be described purely by its tiling graph without relying on the angles and edge length of the tiles? (If You find this post fuzzy or ignorant, that would be because the poster is not a
professional mathematician.) Pontus
|
I reject the premise of the question. :-) It is true, as Terry suggests, that there is a nice dynamical proof of the classification of finite abelian groups. If $A$ is finite, then for every prime $p$ has a stable kernel $A_p$ and a stable image $A_p^\perp$ in $A$, by definition the limits of the kernel and image of $p^n$ as $n \to \infty$. You can show that this yields a direct sum decomposition of $A$, and you can use linear algebra to classify the dynamics of the action of $p$ on $A_p$. A similar argument appears in Matthew Emerton's proof. As Terry says, this proof is nice because it works for finitely generated torsion modules over any PID. In particular, it establishes Jordan canonical form for finite-dimensional modules over $k[x]$, where $k$ is an algebraically closed field. My objection is that finite abelian groups look easier than finitely generated abelian groups in this question. The slickest proof of the classification that I know is one that assimilates the ideas of Smith normal form. Ben's question is not entirely fair to Smith normal form, because you do not need finitely many relations. That is, Smith normal form exists for matrices with finitely many columns, not just for finite matrices. This is one of the tricks in the proof that I give next. Theorem. If $A$ is an abelian group with $n$ generators, then it is a direct sum of at most $n$ cyclic groups. Proof. By induction on $n$. If $A$ has a presentation with $n$ generators and no relations, then $A$ is free and we are done. Otherwise, define the height of any $n$-generator presentation of $A$ to be the least norm $|x|$ of any non-zero coefficient $x$ that appears in some relation. Choose a presentation with least height, and let $a \in A$ be the generator such that $R = xa + \ldots = 0$ is the pivotal relation. (Pun intended. :-) ) The coefficient $y$ of $a$ in any other relation must be a multiple of $x$, because otherwise if we set $y = qx+r$, we can make a relation with coefficient $r$. By the same argument, we can assume that $a$ does not appear in any other relation. The coefficient $z$ of another generator $b$ in the relation $R$ must also be a multiple of $x$, because otherwise if we set $z = qx+r$ and replace $a$ with $a' = a+qb$, the coefficient $r$ would appear in $R$. By the same argument, we can assume that the relation $R$ consists only of the equation $xa = 0$, and without ruining the previous property that $a$ does not appear in other relations. Thus $A \cong \mathbb{Z}/x \oplus A'$, and $A'$ has $n-1$ generators. □ Compare the complexity of this argument to the other arguments supplied so far. Minimizing the norm $|x|$ is a powerful step. With just a little more work, you can show that $x$ divides every coefficient in the presentation, and not just every coefficient in the same row and column. Thus, each modulus $x_k$ that you produce divides the next modulus $x_{k+1}$. Another way to describe the argument is that Smith normal form is a matrix version of the Euclidean algorithm. If you're happy with the usual Euclidean algorithm, then you should be happy with its matrix form; it's only a bit more complicated. The proof immediately works for any Euclidean domain; in particular, it also implies the Jordan canonical form theorem. And it only needs minor changes to apply to general PIDs.
|
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|
12,045 |
The obvious ones are 0 and $e^{-x^2}$ (with annoying factors), and someone I know suggested hyperbolic secant. What other fixed points (or even eigenfunctions) of the Fourier transform are there?
|
The following is discussed in a little more detail on pages 337-339 of Frank Jones's book "Lebesgue Integration on Euclidean Space" (and many other places as well). Normalize the Fourier transform so that it is a unitary operator $T$ on $L^2(\mathbb{R})$. One can then check that $T^4=1$. The eigenvalues are thus $1$, $i$, $-1$, and $-i$. For $a$ one of these eigenvalues, denote by $M_a$ the corresponding eigenspace. It turns out then that $L^2(\mathbb{R})$ is the direct sum of these $4$ eigenspaces! In fact, this is easy linear algebra. Consider $f \in L^2(\mathbb{R})$. We want to find $f_a \in M_a$ for each of the eigenvalues such that $f = f_1 + f_{-1} + f_{i} + f_{-i}$. Using the fact that $T^4 = 1$, we obtain the following 4 equations in 4 unknowns: $f = f_1 + f_{-1} + f_{i} + f_{-i}$ $T(f) = f_1 - f_{-1} +i f_{i} -i f_{-i}$ $T^2(f) = f_1 + f_{-1} - f_{i} - f_{-i}$ $T^3(f) = f_1 - f_{-1} -i f_{i} +i f_{-i}$ Solving these four equations yields the corresponding projection operators. As an example, for $f \in L^2(\mathbb{R})$, we get that $\frac{1}{4}(f + T(f) + T^2(f) + T^3(f))$ is a fixed point for $T$.
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|
12,081 |
(asked by Nathaniel Hellerstein on the Q&A board at JMM) Is there a "half-exponential" function $h(x)$ such that $h(h(x))=e^x$? Is it unique? Is it analytic? Related question: Is there an invertible smooth function $E$ such that $E(x+1)=e^{E(x)}$? Is it unique? If so, then we can take $h(x)=E(E^{-1}(x)+1/2)$.
|
There is no entire (holomorphic everywhere) function $f(z)$ with $f(f(z)) = e^z$. To see this, one can use Picard's little theorem, and do case analysis. For example, $f(z)$ cannot take all values, for then so does $f(f(z))$, while $e^z$ omits zero. The case that $f(z)$ has an omitted value can also be excluded, this is not difficult. EDIT: Perhaps I should be more explicit. If $f(z)$ omits a value, that has to be zero, for $e^z$ takes all other values. Thus there is an entire function $h(z)$ such that $f(z) = e^{h(z)}$ (the complex plane is simply connected). Now $
e^{h(e^{h(z)})} = e^z
$ and so $h(e^{h(z)}) = z + 2{\pi}ik$ for some fixed integer $k$. Since the right hand side takes all values, so does the left hand side. So $h(z)$ takes the two values $0$ and $2{\pi}i$, say $h(a) = 0$ and $h(b) = 2{\pi}i$. Now $
a + 2{\pi}ik = h(e^{h(a)}) = h(e^0) = h(e^{2{\pi}i}) = h(e^{h(b)}) = b + 2{\pi}ik
$ and so $a = b$. Contradiction! Alternatively, use the theorem of Polya that if $f(z)$ and $g(z)$ are entire functions, then
$f(g(z))$ is of infinite order unless (i) $f(z)$ is of finite order and $g(z)$ is a polynomial, or (ii) $f(z)$ has order zero and $g(z)$ is of finite order. The exponential function has order $1$. EDIT: Here $f(f(z)) = e^z$ yields the absurd conclusion that $e^z$ is a polynomial in case (i). While in case (ii) we find that $f(z)$ is an entire function of order zero with
$f(f(z)) = e^z$. But an entire function of order zero that is not a polynomial takes every value infinitely often (by the Hadamard factorization theorem), so we are led to the absurd conclusion that $e^z$ takes the value zero infinitely often.
|
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|
12,085 |
I would like to ask about examples where experimentation by computers has led to major mathematical advances. A new look Now as the question is five years old and there are certainly more examples of mathematical advances via computer experimentation of various kinds, I propose to consider contributing new answers to the question. Motivation I am aware about a few such cases and I think it will be useful to gather such examples together. I am partially motivated by the recent polymath5 which, at this stage, have become an interesting experimental mathematics project. So I am especially interested in examples of successful "mathematical data mining"; and cases which are close in spirit to the experimental nature of polymath5. My experience is that it can be, at times, very difficult to draw useful insights from computer data. Summary of answers according to categories (Added Oct. 12, 2015) To make the question a useful resource (and to allow additional answers), here is a quick summery of the answers according to categories. (Links are to the answers, and occasionally to an external link from the answer itself.) 1) Mathematical conjectures or large body of work arrived at by examining experimental data - Classic The Prime Number Theorem ; Birch and Swinnerton-Dyer conjectures ; Shimura-Taniyama-Weil conjecture ; Zagier's conjectures on polylogarithms ; Mandelbrot set ; Gosper Glider Gun ( answer ), Lorenz attractor ; Chebyshev's bias ( answer ) ; the Riemann hypothesis ; the discovery of the Feigenbaum constant ; (related) Feigenbaum-Coullet-Tresser universality and Milnor's Hairiness conjecture ; Solving numerically the so-called Fermi--Pasta--Ulam chain and then of its continuous limit, the Korteweg--de Vries equation 2) Mathematical conjectures or large body of work arrived at by examining experimental data - Current " Maeda conjecture "; the work of Candès and Tao on compressed sensing ; Certain Hankel determinants ; Weari-Phelan structure; the connection of multiple zeta values to renormalized Feynman integrals ; Thistlethwaite's discovery of links with trivial Jones polynomial; The Monstrous Moonshine; McKay's account on experimentation leading to mysterious "numerology" regarding the monster. ( link to the answer ); Haiman conjectures on the quotient ring by diagonal invariants 3) Computer-assisted proofs of mathematical theorems Kepler's conjecture ; a new way to tile the plane with a pentagon: advances regarding bounded gaps between primes following Zhang's proof; Cartwright and Steger's work on fake projective planes ; the Seifert-Weber dodecahedral space is not Haken ; the four color theorem, the proof of the nonexistence of a projective plane of order 10; Knuth's work on a class of projective planes; The search for Mersenne primes; Rich Schwartz 's work ; The computations done by the 'Atlas of Lie groups' software of Adams, Vogan, du Cloux and many other; Cohn-Kumar proof for the densest lattice pacing in 24-dim; Kelvin's conjecture ; (NEW) $R(5,5) \le 48$ and $R(4,5)=25$ ; 4) Computer programs that interactively or automatically lead to mathematical conjectures. Graffiti 5) Various computer programs which allow proving automatically theorems or generating automatically proofs in a specialized field. Wilf-Zeilberger formalism and software; FLAGTOOLS 6) Computer programs (both general purpose and special purpose) for verification of mathematical proofs. The verification of a proof of Kepler's conjecture. 7) Large databases and other tools Sloane's online encyclopedia for integers sequences; the inverse symbolic calculator . 8) Resources: Journal of experimental mathematics; Herb Wilf 's article: Mathematics, an experimental science in the Princeton Companion to Mathematics, genetic programming applications a fairly comprehensive website experimentalmath.info ; discovery and experimentation in number theory; Doron Zeilberger's classes called "experimental mathematics":math.rutgers.edu/~zeilberg/teaching.html; V.I. Arnol'd's two books on the subject of experimental mathematics in Russian, Experimental mathematics , Fazis, Moscow, 2005, and Experimental observation of mathematical facts , MCCME, Moscow, 2006 Answers with general look on experimental mathematics: Computer experiments allow new avenues for productive strengthening of a problem (A category of experimental mathematics). Bounty: There were many excellent answers so let's give the bounty to Gauss... Related question: Where have you used computer programming in your career as an (applied/pure) mathematician? , What could be some potentially useful mathematical databases? Results that are easy to prove with a computer, but hard to prove by hand ; What advantage humans have over computers in mathematics? Conceptual insights and inspirations from experimental and computational mathematics
|
Computer experiments (in the early 1960s!) led Birch and Swinnerton-Dyer to the formulation of their conjecture, which stimulated the development of much of arithmetic geometry.
|
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|
12,092 |
It is well known that Beck's theorem for Comonad is equivalent to Grothendieck flat descent theory on scheme. There are several version of derived noncommutative geometry. I wonder whether someone developed the triangulated version of Beck's theorem. And What does it mean,if exists?
|
There isn't a descent theory for derived categories per se - one can't glue objects in the derived category of a cover together to define an object in the base. (Trying to apply the usual Barr-Beck to the underlying plain category doesn't help.) But I think the right answer to your question is to use an enriched version of triangulated categories (differential graded or $A_\infty$ or stable $\infty$-categories), for which there is a beautiful Barr-Beck and descent theory, due to Jacob Lurie. (This is discussed at length in the n-lab I believe, and came up recently on the n-category cafe (where I wrote basically the same comment here ..)
This is proved in DAG II: Noncommutative algebra . In the comonadic form it goes like this. Given an adjunction between $\infty$-categories (let's call the functors pullback and pushforward, to mimic descent), if we have pullback is conservative (it respects isomorphisms), and pullback respects certain limits (namely totalizations of cosimplicial objects,
which are split after pullback) then the $\infty$-category downstairs is equivalent to comodules over the comonad
(pullback of pushforward). (There's an opposite monadic form as well)
This can be verified in the usual settings where you expect descent to hold.
In other words if you think of derived categories as being refined to $\infty$-categories (which have the derived category as their homotopy category), then everything you might want to hold does. So while derived categories don't form a sheaf (stack), their refinements do:
you can recover a complex (up to quasiisomorphism) from a collection of complexes on a cover, identification on overlaps, coherences on double overlaps, coherences of coherences on triple overlaps etc.
More formally: define a sheaf as a presheaf $F$ which has the property that
for an open cover $U\to X$, defining a Cech simplicial object $U_\bullet=\{U\times_X U\times_X U\cdots\times_X U\}$, then $F(X)$ is the totalization of the cosimplicial object $F(U_\bullet)$. Then enhanced derived categories form sheaves (in appropriate topologies) as you would expect. This is of course essential to having a good theory of noncommutative algebraic geometry!
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|
12,100 |
In many formulation of Class Field theory, the Weil group is favored as compared to the Absolute Galois group. May I asked why it is so? I know that Weil group can be generalized better to Langlands program but is there a more natural answer? Also we know that the abelian Weil Group is the isomorphic image of the reciprocity map of the multiplicative group (in the local case) and of the idele-class group (in the global case). Is there any sense in which the "right" direction of the arrow is the inverse of the reciprocity map? Please feel free to edit the question into a form that you think might be better.
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The Weil group appears for several reasons. Firstly: if $K$ is a non-archimedean local field with residue field $k$,
the local reciprocity law induces an embedding
$K^{\times} \hookrightarrow G_K^{ab}.$ The image consists of all
elements in $G_k$ whose image is an integral power of Frobenius.
This is the abelianized Weil group; it just appears naturally. Secondly: suppose that $K$ is a global field of positive characteristic,
i.e. the function field of a curve over a finite field $k$. Then the
global reciprocity map identifies the idele class group of $K$ with
a subgroup of $G_K^{ab}$ consisting of elements which act on $k$
by integral powers of Frobenius. So again, it is the abelianized Weil group
that appears. Thirdly: suppose that $E$ is an elliptic curve over a quadratic imaginary
field $K$
with complex multipliction by $\mathcal O$, the ring of integers in $K$. (Thus I am implicity fixing $K$ to
have class number one, but this is not so important for what I am going to say next.)
If $\ell$ is a prime, then the $\ell$-adic Tate module is then free of rank one over
$\mathcal O_{\ell}$ (the $\ell$-adic completion of $\mathcal O$), and the $G_K$-action
on this Tate module induces a character $\psi_{\ell}:G_K^{ab} \rightarrow \mathcal O_{\ell}^{\times}$. There is a sense in which the various $\psi_{\ell}$ are indepenent of $\ell$,
but what is that sense? Well, suppose that $\wp$ is a prime of $K$, not dividing $\ell$ and at which
$E$ has good reduction. Then the value of $\psi_{\ell}$ on $Frob_{\wp}$ is indepenent
of $\ell$, in the sense that its value is an element of $\mathcal O$, and this value
is independent of $\ell$.
More generally, provided that $\wp$ is prime to $\ell$,
the restriction of $\psi_{\ell}$ to the local Weil group at $\wp$ is independent of $\ell$
(in the sense that the value at a lift of Frobenius will be an algebraic integer that
is independent of $\ell$, and its restriction to inertia at $\wp$ will be a finite image
representation, hence defined over algebraic integers, which again is then independent
of $\ell$). Note that independence of $\ell$ doesn't make sense for $\psi_{\ell}$
on the full local Galois group at $\wp$, since on this whole group it will
certainly take values that are not algebraic, but rather just some $\ell$-adic
integers, which can't be compared with one another as $\ell$ changes. Now there is also a sense in which the $\psi_\ell$, as global Galois characters, are independent of $\ell$. Indeed, we can glue together the
various local Weil group representations to get a representation $\psi$ of the global Weil
group $W_K$. Since it is abelian, this will just be an idele class character $\psi$,
or what is also called a Hecke character or Grossencharacter. It will take values
in complex numbers. (At the finite places it even takes algebraic number values, but
when we organize things properly at the infinite places, we are forced to think of it
as complex valued.) Note that $\psi$ won't factor through the connected component group, i.e. it won't be
a character of $G_K^{ab}$. It is not a Galois character, but a Weil group character.
It stores in one object the information contained in a whole collection of
$\ell$-adic Galois characters, and gives a precise sense to the idea that these various $\ell$-adic characters are independent of $\ell$. This is an important general role of Weil groups. Fourthly: The Hecke character $\psi$ above will be an algebraic Hecke character, i.e.
at the infinite places, it will involve raising to integral powers. But we can also
raise real numbers to an arbitrary complex power $s$, and so there are Hecke characters
that do not come from the preceding construction (or ones like it); in other words,
there are non-algebraic, or non-motivic, Hecke characters. But they are abelian characters
of the global Weil group, and they have a meaning; the variable $s$ to which we can raise
real numbers is the same variable $s$ as appears in $\zeta$- or $L$-functions. In summary: Because Weil groups are "less completed", or "less profinite", than Galois
groups, they play an important role in describing how a system of $\ell$-adic representations can be independent of $\ell$. Also, they allow one to describe
phenomena which are automorphic, but not motivic (i.e. which correspond to non-integral
values of the $L$-function variable $s$). (They don't describe all automorphic phenomena,
though --- one would need the entire Langlands group for that.)
|
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|
12,169 |
A (discrete) group is amenable if it admits a finitely additive probability measure (on all its subsets), invariant under left translation. It is a basic fact that every abelian group is amenable. But the proof I know is surprisingly convoluted. I'd like to know if there's a more direct proof. The proof I know runs as follows. Every finite group is amenable (in a unique way). This is trivial. $\mathbb{Z}$ is amenable. This is not trivial as far as I know; the proof I know involves choosing a non-principal ultrafilter on $\mathbb{N}$. This means that $\mathbb{Z}$ is amenable in many different ways, i.e. there are many measures on it, but apparently you can't write down any measure 'explicitly' (without using the Axiom of Choice). The direct product of two amenable groups is amenable. This isn't exactly trivial, but the measure on the product is at least constructed canonically from the two given measures. Every finitely generated abelian group is amenable. This follows from 1--3 and the classification theorem. The class of amenable groups is closed under direct limits (=colimits over a directed poset). This is like step 2: it seems that there's no canonical way of constructing a measure on the direct limit, given measures on each of the groups that you start with; and the proof involves choosing a non-principal ultrafilter on the poset. Every abelian group is amenable. This follows from 4 and 5, since every abelian group is the direct limit of its finitely generated subgroups. Is there a more direct proof? Is there even a one-step proof? Update Yemon Choi suggests an immediate simplification: replace 1 and 4 by 1'. Every quotient of an amenable group is amenable. This is simple: just push the measure forward. 4'. Every f.g. abelian group is amenable, by 1', 2 and 3. This avoids using the classification theorem for f.g. abelian groups. Tom Church mentions the possibility of skipping steps 1--3 and going straight to 4. If I understand correctly, this doesn't use the classification theorem either. The argument is similar to the one for $\mathbb{Z}$: one still has to choose an ultrafilter on $\mathbb{N}$. (One also constructs a Følner sequence on the group, a part of the argument which I didn't mention previously but was there all along). Yemon, Tom and Mariano Suárez-Alvarez all suggest using one or other alternative formulations of amenability. I'm definitely interested in answers like that, but it also reminds me of the old joke: Tourist: Excuse me, how do I get to Edinburgh Castle from here? Local: I wouldn't start from here if I were you. In other words, if a proof of the amenability of abelian groups uses a different definition of amenability than the one I gave, then I want to take the proof of equivalence into account when assessing the simplicity of the overall proof. Jim Borger points out that if, as seems to be the case, even the proof that $\mathbb{Z}$ is amenable makes essential use of the Axiom of Choice, then life is bound to be hard. I take his point. However, one simplification to the 6-step proof that I'd like to see is a merging of steps 2 and 5. These are the two really substantial steps, but they're intriguingly similar. None of the answers so far seem to make this economy. That is, every proof suggested seems to involve two separate Følner-type arguments.
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Here is a simpler argument, combining 1--6 into one step. Let $G$ be a countable abelian group generated by $x_1,x_2,\ldots$. Then a Følner sequence is given by taking $S_n$ to be the pyramid consisting of elements which can be written as $a_1x_2+a_2x_2+\cdots+a_nx_n$ with $\lvert a_1\rvert\leq n,\lvert a_2\rvert\leq n-1,\ldots,\lvert a_n\rvert\leq 1$. The invariant probability measure is then defined by $\mu(A)=\underset{\omega}{\lim}\lvert A\cap S_n\rvert / \lvert S_n\rvert$ as usual. A more natural way to phrase this argument is: The countable group $\mathbb{Z}^\infty$ is amenable. All countable abelian groups are amenable, because amenability descends to quotients. But I would like to emphasize that there is really only one step here, because the proof for $\mathbb{Z}^\infty$ automatically applies to any countable abelian group. This two-step approach is easier to remember, though. (The ideas here are the same as in my other answer, but I think this formulation is much cleaner.) 2016 Edit: Here is an argument to see that $S_n$ is a Følner sequence. It is quite pleasant to think about precisely where commutativity comes into play. Fix $g\in G$ and any finite subset $S\subset G$. We first analyze the size of the symmetric difference $gS\bigtriangleup S$. Consider the equivalence relation on $S$ generated by the relation $x\sim y$ if $y=x+g$ (which is itself neither symmetric, reflexive, or transitive). We will call an equivalence class under this relation a "$g$-string". Every $g$-string consists of elements $x_1,\ldots,x_k\in S$ with $x_{j+1}=x_j+g$. The first key observation is that $\lvert gS\bigtriangleup S\rvert$ is at most twice the number of $g$-strings . Indeed, if $z\in S$ belongs to $gS\bigtriangleup S$, then $z$ must be the "leftmost endpoint" of a $g$-string; if $z\notin S$ belongs to $gS\bigtriangleup S$, then $z-g$ must be the "rightmost endpoint" of a $g$-string; and each $g$-string has at most 2 such endpoints (it could have 1 if the endpoints coincide, or 0 if $g$ has finite order). Our goal is to prove for all $g\in G$ that $\frac{\lvert gS_n \bigtriangleup S_n\rvert}{\lvert S_n\rvert}\to 0$ as $n\to \infty$. Since $\lvert abS\bigtriangleup S\rvert\leq\lvert abS\bigtriangleup bS\rvert+\lvert bS\bigtriangleup S\rvert= \lvert aS\bigtriangleup S\rvert+\lvert bS\bigtriangleup S\rvert$, it suffices to prove this for all $g_i$ in a generating set. By the observation above, to prove that $\frac{\lvert g_iS_n \bigtriangleup S_n\rvert}{\lvert S_n\rvert}\to 0$, it suffices to prove that $\frac{\text{# of $g_i$-strings in $S_n$}}{\lvert S_n\rvert}\to 0$. Equivalently, we must prove that the reciprocal $\frac{\lvert S_n\rvert}{\#\text{ of $g_i$-strings in $S_n$}}$ diverges, or in other words that the average size of a $g_i$-string in $S_n$ diverges . We now use the specific form of our sets $S_n=\{a_1g_1+\cdots+a_ng_n\,|\, \lvert a_i\rvert\leq n-i\}$. For any $i$ and any $n$, set $k=n-i$ (so that $\lvert a_i\rvert\leq k$ in $S_n$). The second key observation is that every $g_i$-string in $S_n$ has cardinality at least $2k+1$ unless $g_i$ has finite order. Indeed given $x\in S_n$, write it as $x=a_1g_1+\cdots+a_ig_i+\cdots+a_ng_n$; then the elements $a_1g_1+\cdots+bg_i+\cdots+a_ng_n\in S_n$ for $b=-k,\ldots,-1,0,1,\ldots,k$ belong to a single $g_i$-string containing $x$. If $g_i$ does not have finite order, these $2k+1$ elements must be distinct . This shows that the minimum size of a $g_i$-string in $S_n$ is $2n-2i+1$, so for fixed $g_i$ the average size diverges as $n\to \infty$. When $g_i$ has finite order $N$ this argument does not work (a $g_i$-string has maximum size $N$, so the average size cannot diverge). However once $N<2k+1$, the subset containing the $2k+1$ elements above is closed under multiplication by $g_i$. In other words, once $n\geq i+N/2$ the set $S_n$ is $g_i$-invariant, so $\lvert g_iS_n\bigtriangleup S_n\rvert=0$. I'm grateful to David Ullrich for pointing out that this claim is not obvious, since the quotient of a Følner sequence need not be a Følner sequence (Yves Cornulier gives an example here ).
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|
12,200 |
How does one prove that on $S^n$ (with the standard connection) any geodesic between two fixed points is part of a great circle? For the special case of $S^2$ I tried an naive approach of just writing down the geodesic equations (by writing the Euler-Lagrange equations of the length function) and solving them to gain some insights but even if the equations are solvable I can't see how to show that they are great circles. (the solutions are some pretty complicated functions which don't give me much insight) I checked the article on Great Circles on Wolfram Mathworld for a coordinate geometry approach to it but that article looked quite cryptic to me! One knows that on compact semi-simple lie groups any one-parameter subgroup generates a geodesic and $S^n$ is the quotient of 2 compact semi-simple lie groups $SO(n+1)/SO(n)$. Is this line of thought useful for this question? =================================================================================
After some of the responses came let me put in "a" way of seeing the above for $S^2$ (wonder if it is correct). If $\theta$ and $\phi$ are the standard coordinates on $S^2$ then the equations for the curve are $$\ddot{\theta} = \dot{\phi}^2 sin(\theta)cos(\theta)$$
$$\dot{\phi}sin^2{\theta} = k$$ where $k$ is some constant set by the initial data of the curve. Now given the initial point I can choose my coordinate system such that the the initial data looks like $\dot{\phi}=0$, $\theta = \text{some constant}$, $\dot{\theta}=\text{some constant}$, $\phi = \text{some constant}$. Then the differential equations tell me that the $k=0$ and the only way it can happen for times is by having , $$\dot{\phi} = 0$$ Which clearly gives me a longitude in this coordinate system.
Hence the geodesic equation gives as a solution a great circle. Surely not an elegant proof like Bar's reference. But I hope this is correct. {As a friend of mine pointed out that this set of coordinates is motivated by the fact that the way the "energy" of the curve is being parametrized the z-component of the angular momentum is conserved which is in fact my second Euler-Lagrange equations}
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Although Jose has made the essentially the same point, I just want to elaborate (this is really just a comment, but I always run out of room in the comment box). What nobody else has mentioned explicitly is that you should have trouble solving the Euler-Lagrange equation for the length functional. The Euler-Lagrange equation is a second order ODE, but a highly degenerate one. And you know this before you even start. Why? Well, suppose you have a solution. Then if you reparameterize that curve using any arbitrary parameterization (i.e., any monotone function of the original parameter), the newly parameterized curve is still a solution to the Euler-Lagrange equation. That means that the ODE has an infinite dimensional space of solutions and it is nothing like any ODE we learned about in our ODE courses or textbooks. A trick is needed to get around this, namely to use the so-called energy functional $E[\gamma] = \int_0^1 |\gamma'(t)|^2 dt$ (which is not invariant under reparameterization of the curve) instead of the length functional (which is). The Holder inequality shows that a minimum of the energy functional is necessarily a minimum of the length functional that is parameterized by a constant times arclength, i.e. a constant speed geodesic. The Euler-Lagrange equation for the energy functional is a nice nondegenerate 2nd order ODE that can be handled by standard ODE techniques and theorems. As for the standard sphere, there are many different ways to solve for the geodesics. To review the ways already suggested in other answers: 1) I recommend that you first do it without the machinery of Riemannian geometry and using only the Euclidean structure of $R^{n+1}$. Using the discussion above, you should be able to show that a curve on the unit sphere is a constant speed geodesic if and only if its acceleration vector is always normal to the sphere. You should then be able to work out the solutions to this ODE. The suggestion that you assume one point is the north pole and the other lies in a co-ordinate plane is a good one and makes the ODE easy to solve. 2) The other way is to do it all intrinsically. Here, I recommend using stereographic co-ordinates and assuming one point is the origin in those co-ordinates. Again, everything becomes very easy in that situation. 3) And the third way is to view the sphere as a homogeneous space and use formulas for that situation. I don't remember the details myself, but I learned them from the book by Cheeger and Ebin. I recommend that you work through all 3 different ways, as well as any other way you can find. As others have noted, the calculations for geodesics on hyperbolic space are identical, except that you are working with a "unit sphere" in Minkowski instead of Euclidean space. There is even a notion of stereographic projection (but onto what?). This is also fun to work out carefully. Finally, I do want to note that after you work this all out and have it all in your head, it's a really beautiful picture and story. And if you find the right angle, it's all very simple, so you can work out the details yourself and not rely on reading a book line-by-line or having someone else show you all the details. Try to get the essential ideas and necessary tricks (like using the energy functional) from books, lectures, or teachers, but try to work everything else out from scratch (i.e., minimal reliance on theorems you can't prove yourself).
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|
12,236 |
Let $X$ be a regular scheme (all local rings are regular). Let $Y,Z$ be two closed subschemes defined by ideals sheaves $\mathcal I,\mathcal J$. Serre gave a beautiful formula to count the intersection multiplicity of $Y,Z$ at a generic point $x$ of $Y\cap Z$ as: $$\sum_{i\geq 0} (-1)^i\text{length}_{\mathcal O_{X,x}} \text{Tor}_i^{\mathcal O_{X,x}}(\mathcal O_{X,x}/\mathcal I_x, \mathcal O_{X,x}/\mathcal J_x)$$ It takes quite a bit of work to show that this is the right definition (even that the sum terminates is a non-trivial theorem of homological algebra): it is non-negative, vanishes if the dimensions don't add up correctly, positivity etc. In fact, some cases are still open as far as I know. See here for some reference. I have heard one of the great things about Lurie's thesis is setting a framework for derived algebraic geometry. In fact, in the introduction he used Serre formula as a motivation (it is pretty clear from the formula that a "derived" setting seems natural). However, I could not find much about it aside from the intro, and Serre formula was an old flame of mine in grad school. So my (somewhat vague): Question : Does any of the desired properties of Serre formula follow naturally from Lurie's work? If so (since things are rarely totally free in math), where did we actually pay the price (in terms of technical work to establish the foundations)? EDIT: Clark's answer below greatly clarifies and gives more historical context to my question, highly recommend!)
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There are a number of comments to make about Serre's intersection formula and its relation to derived algebraic geometry. First, we should be a little more cautious about attribution. The idea of using "derived rings" to give an intrinsic version of the Serre intersection formula is not recent. The idea goes back at least to thoughts of Deligne, Kontsevich, Drinfeld, and Beilinson in the 1980s (and possibly earlier). These ideas have been made precise in a number of ways, in particular in work of Kapranov & Ciocan-Fontaine, and Toën & Vezzosi. EDIT : As Ben-Zvi reminded me below, one should also mention Behrend and Behrend-Fantechi on DG schemes and virtual fundamental classes. Of course Lurie's work has been the most comprehensive and powerful in its treatment of the foundations of DAG, but it's important to understand that his work arose in the context of these fascinating ideas. Now, just to provide a little context, let me try to recall how Serre's formula arises from DAG considerations. Let's start by using the notation above, but let's assume for simplicity that $X$, $Y$, and $Z$ are all local schemes. (Some of the technicalities of DAG arise in making sheaf theory work with some sort of "derived rings," so our discussion will be easier if we ignore that for now.) So we write $X=\mathrm{Spec}(A)$, $Y=\mathrm{Spec}(B)$, and $Z=\mathrm{Spec}(C)$ for local rings $A$, $B$, and $C$. Now if our aim is to intersect $Y$ and $Z$ in $X$, we know how to do that algebro-geometrically. We form the fiber product $Y\times_XZ=\mathrm{Spec}(B\otimes_AC)$. The tensor product that appears here is really the thing we're going to alter. To do that, we're going to regard $B$ and $C$ as (discrete) simplicial (commutative) $A$-algebras , and we're going to form the derived tensor product. This produces a new simplicial commutative ring $B\otimes^{\mathbf{L}}_AC$ whose homotopy groups are exactly the groups $\mathrm{Tor}^A_i(B,C)$. The intersection multiplicity is simply the length of $B\otimes^{\mathbf{L}}_AC$ as a simplicial $A$-module. As Ben Webster says, the real joy of DAG is in thinking of the geometry of our new derived ring $B\otimes^{\mathbf{L}}_AC$ as a single unit instead of thinking only of its disembodied homotopy groups. The question you're asking seems to be: does thinking geometrically about this gadget help us to prove Serre's multiplicity conjectures in a more conceptual manner? The short answer is: I don't know. I do not think a new proof of any of these has been announced using DAG (and it's definitely not in any of Lurie's papers), and in any case I do not think DAG has the potential to make the conjectures "easy." But let me see if I can make a case for the following idea: revisiting Serre's original method of reduction to the diagonal in the context of DAG. Recall that, if $k$ is a field, if $A$ is a $k$-algebra, and if $M$ and $N$ are $A$-modules, then $$M\otimes_AN=A\otimes_{A\otimes_kA}(M\otimes_kN).$$
Hence to understand $\mathrm{Tor}^A_{\ast}(M,N)$, it suffices to understand $\mathrm{Tor}^{A\otimes_kA}_{\ast}(A,-)$. This allowed Serre to reduce to the case of the diagonal in $\mathrm{Spec}(A\otimes_kA)$. The key point here is that everything is flat over $k$, so Serre could only use this to prove the multiplicity conjectures for $A$ essentially of finite type over a field. Observe that the same equality holds if we work in the derived setting: if $M$ and $N$ are simplicial $A$-modules, and $A$ is an $R$-algebra, then the derived tensor product of $M$ and $N$ over $A$ can be computed as
$$A\otimes^{\mathbf{L}}_{A\otimes^{\mathbf{L}}_RA}(M\otimes^{\mathbf{L}}_RN).$$
The gadget on the right (or, strictly speaking, its homotopy) has a name familiar to toplogists; it's the Hochschild homology $\mathrm{HH}^R(A,M\otimes^{\mathbf{L}}_RN)$. The hope is that we've chosen $R$ cleverly enough that $B\otimes^{\mathbf{L}}_RC$ is "less complicated" than $B\otimes^{\mathbf{L}}_AC$. (More precisely, we want the $\mathrm{Tor}$-amplitude of $M$ and $N$ to decrease when we think of them as $R$-modules. There's a particular way of building $R$, but let me skip over this point.) Has our situation improved? Perhaps only a little: we've turned our problem of looking at the derived intersection $Y\times^h_XZ$ into the study of the derived intersection of the diagonal inside $X\times^h_RX$ with some simpler derived subscheme $Y\times^h_RZ$ thereof. But now we can try to iterate this, working inductively. I don't know whether this can be made to work, of course.
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|
12,284 |
In dimensions 1 and 2 there is only one, respectively 2, compact Kaehler manifolds with zero first Chern class, up to diffeomorphism. However, it is an open problem whether or not the number of topological types of such manifolds of dimension 3 (Calabi-Yau threefolds) is bounded. I would like to ask what is known in this direction. In particular, is it known that the Euler characteristic or the total Betti number of Calabi-Yau threefolds can't be arbitrarily large? Are there any mathematical (or physical?) reasons to expect either answer? As a side question: I remember having heard several times something like "Calabi-Yau 3-folds parametrize (some kind of) vacua in string theory" but was never able to make precise sense of this. So any comments on this point or references accessible to mathematicians would be very welcome.
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This is a very good question, and I would really love to know the answer since its current state seems to be quite obscure. Below is just a collection of remarks, surely not the full answer by any means. I would like to argue that for the moment there is no any deep mathematical reason to think that the Euler number of CY 3-folds is bounded. I don't believe either that there is any physical intuition on this matter. But there is some empirical information, and I will describe it now, starting by speaking about "how many topological types of CY manifolds we know for the moment". As far as I understand for today the construction of Calabi-Yau 3-folds, that brought by far the largest amount of examples is the construction of Batyrev. He starts with a reflexive polytope in dimension 4, takes the corresponding toric 4-fold, takes a generic anti-canonical section and obtains this way a Calabi-Yau orbifold. There is always a crepant resolution. So you get a smooth Calabi-Yau. Reflexive polytopes in dimension 4 are classified the number is 473,800,776. I guess, this number let Miles Reid to say in his article "Updates on 3-folds" in 2002 http://arxiv.org/PS_cache/math/pdf/0206/0206157v3.pdf , page 519 "This gives some 500,000,000 families of CY 3-folds, so much more impressive than a mere infinity (see the website http://tph16.tuwien.ac.at/~kreuzer/CY/ ). There are certainly many more; I believe there are infinitely many families, but the contrary opinion is widespread" A problem with the number 500,000,000 in this phrase is that it seems more related to the number of CY orbifolds, rather than to the number of CY manifolds obtained by resolving them. Namely, the singularities that appear in these CY orbifolds can be quite involved and they have a lot of resolutions (I guess at least thousands sometimes), so the meaning of 500,000,000 is not very clear here. This summer I asked Maximillian Kreuzer (one of the persons who actually got this number 473,800,776 of polytopes), a question similar to what you ask here. And he said that he can guarantee that there exist at least 30108 topological types of CY 3-folds. Why? Because for all these examples you can calculate Hodge numbers $h^{1,1}$ and $h^{2,1}$, and you get 30108 different values. Much less that 473,800,776. As for more refined topological invariants (like multiplication in cohomology) according to him, this was not really studied, so unfortunately 30108 seems to be the maximal number guarantied for today. But I would really love to know that I am making a mistake here, and there is some other information. Now, it seems to me that the reason, that some people say, that the Euler characteristics of CY 3-folds could be bounded is purely empirical. Namely, the search for CY 3-folds is going for 20 years already. Since then a lot of new families were found. We know that mirror symmetry started with this symmetric table of numbers "($h^{1,1}, h^{2,1})$", and the curious fact is that, according to Maximillian, what happened to this table in 20 years -- it has not got any wider in 20 years, it just got denser. The famous picture can be found on page 9 of the following notes of Dominic Joyce http://people.maths.ox.ac.uk/~joyce/SympGeom2009/SGlect13+14.pdf . So, this means that we do find new families of CY manifolds, all the time. But the values of their Hodge numbers for some reason stay in the same region. Of course this could easily mean that we are just lacking a good construction. Final remark is that in the first version of this question it was proposed to consider complex analytic manifolds with $c_1=0$. If we don't impose condition of been Kahler, then already in complex dimension 2 there is infinite number of topological types, given by Kodaira surfaces, they are elliptic bundles over elliptic curve. In complex dimension 3 Tian have shown that for every $n>1$ there is a holomorphic structure on the connected sum of n copies of $S^3\times S^3$, with a non-vanishing holomorphic form. Surely these manifolds are non-Kahler. So if you want to speak about any finiteness, you need to discuss say, Kahler 3-folds with non-vanishing holomorphic volume form, but not all complex analytic ones.
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12,303 |
Why is one interested in the mod p reduction of modular curves and Shimura varieties? From an article I learned that this can be used to prove the Eichler-Shimura relation which in turn proves the Hasse-Weil conjecture for modular curves. Are there similar applications for Shimura varieties?
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The Eichler-Shimura relation doesn't just prove the Hasse-Weil conjecture for modular curves. It e.g. attaches Galois representations to modular forms of weight 2. More delicate arguments (using etale cohomology with non-constant coefficients, machinery that wasn't available to Shimura) attaches Galois representations to higher weight modular forms. These ideas have had many applications (e.g. eventually they proved FLT). In summary: computing the mod p reduction of modular curves isn't just for Hasse-Weil. Doing the same for Shimura varieties is technically much harder because one runs into problems both geometric and automorphic. But the upshot, in some sense, is the same: if one can resolve these issues (which one can for, say, many unitary Shimura varieties nowadays, but by no means all Shimura varieties) then one can hope to attach Galois representations to automorphic forms on other reductive groups, and also to compute the L-function of the Shimura variety in terms of automorphic forms. Why would one want to do these things? Let me start with attaching Galois reps to auto forms. These sorts of ideas are what have recently been used to prove the Sato-Tate conjecture. Enough was known about the L-functions attached to automorphic forms on unitary groups to resolve the analytic issues, and so the main issue was to check that the symmetric powers of the Galois representations attached to an elliptic curve were all showing up in the cohomology of Shimura varieties. Analysing the reduction mod p of these varieties was just one of the many things that needed doing in order to show this (although it was by no means the hardest step: the main technical issues were I guess in the "proving R=T theorems", similar to the final step in the FLT proof being an R=T theorem; the L-function ideas came earlier). But to answer your original question, yes: if you're in the situation where you understand the cohomology of the Shimura variety well enough, then analysing the reduction of the variety will tell you non-trivial facts about the L-function of the Shimura variety. Note however that the link isn't completely formal. Mod p reduction of the varieties only gives you an "Eichler-Shimura relation", and hence a polynomial which will annihiliate the Frobenius element acting on the etale cohomology. To understand the L-function you need to know the full characteristic polynomial of this Frobenius element. For GL_2 one is lucky in that the E-S poly is the char poly, simply because there's not enough room for it to be any other way. This sort of argument breaks down in higher dimensions. As far as I know these questions are still very open for most Shimura varieties. So in summary, for general Shimura varieties, you can still hope for an Eichler-Shimura relation, but you might not actually be able to compute the L-function in terms of automorphic forms as a consequence.
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12,342 |
From time to time, when I write proofs, I'll begin with a claim and then prove the contradiction. However, when I look over the proof afterwards, it appears that my proof was essentially a proof of the contrapositive, and the initial claim was not actually important in the proof. Can all claims proven by reductio ad absurdum be reworded into proofs of the contrapositive? If not, can you give some examples of proofs that don't reduce? If not all reductio proofs can be reduced, is there any logical reason why not? Is reductio stronger or weaker than the contrapositive? Edit: Just another minor question (of course this is optional and will not affect me choosing an answer): If they are equivalent, then why would you bother using reductio? And another bonus question (Like the above, does not influence how I choose the answer to accept.)
Are the two techniques intuitionistically equivalent?
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Although the other answers correctly explain the basic logical equivalence of the two proof methods, I believe an important point has been missed: With good reason , we mathematicians prefer a direct proof of an implication over a proof by contradiction, when such a proof is available. (all else being equal) What is the reason? The reason is the fecundity of the proof, meaning our ability to use the proof to make further mathematical conclusions. When we prove an implication (p implies q) directly, we assume p, and then make some intermediary conclusions r 1 , r 2 , before finally deducing q. Thus, our proof not only establishes that p implies q, but also, that p implies r 1 and r 2 and so on. Our proof has provided us with additional knowledge about the context of p, about what else must hold in any mathematical world where p holds. So we come to a fuller understanding of what is going on in the p worlds. Similarly, when we prove the contrapositive (¬q implies ¬p) directly, we assume ¬q, make intermediary conclusions r 1 , r 2 , and then finally conclude ¬p. Thus, we have also established not only that ¬q implies ¬p, but also, that it implies r 1 and r 2 and so on. Thus, the proof tells us about what else must be true in worlds where q fails. Equivalently, since these additional implications can be stated as (¬r 1 implies q), we learn about many different hypotheses that all imply q. These kind of conclusions can increase the value of the proof, since we learn not only that (p implies q), but also we learn an entire context about what it is like in a mathematial situation where p holds (or where q fails, or about diverse situations leading to q). With reductio, in contrast, a proof of (p implies q) by contradiction seems to carry little of this extra value. We assume p and ¬q, and argue r 1 , r 2 , and so on, before arriving at a contradiction. The statements r 1 and r 2 are all deduced under the contradictory hypothesis that p and ¬q, which ultimately does not hold in any mathematical situation. The proof has provided extra knowledge about a nonexistent, contradictory land. (Useless!) So these intermediary statements do not seem to provide us with any greater knowledge about the p worlds or the q worlds, beyond the brute statement that (p implies q) alone. I believe that this is the reason that sometimes, when a mathematician completes a proof by contradiction, things can still seem unsettled beyond the brute implication, with less context and knowledge about what is going on than would be the case with a direct proof. Edit: For an example of a proof where we are led to false expectations in a proof by contradiction, consider Euclid's proof that there are infinitely many primes. In a common proof by contradiction, one assumes that p 1 , ..., p n are all the primes. It follows that since none of them divide the product-plus-one p 1 ...p n +1, that this product-plus-one is also prime. This contradicts that the list was exhaustive. Now, many beginners falsely expect after this argument that whenever p 1 , ..., p n are prime, then the product-plus-one is also prime. But of course, this isn't true, and this would be a misplaced instance of attempting to extract greater information from the proof, misplaced because this is a proof by contradiction, and that conclusion relied on the assumption that p 1 , ..., p n were all the primes. If one organizes the proof, however, as a direct argument showing that whenever p 1 , ..., p n are prime, then there is yet another prime not on the list, then one is led to the true conclusion, that p 1 ...p n +1 has merely a prime divisor not on the original list.
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12,347 |
Let $X$ be a smooth projective algebraic variety over $\mathbb{C}$. Then I think that someone (Serre?) showed that the Cohomological Etale Brauer Group agrees with the torsion part of the Analytic Brauer Group $H^{2}(X,\mathcal{O}^{\times})$. This latter group is calculated in the classical (metric) topology on the associated complex manifold with the sheaf of nowhere vanishing holomorphic functions. However there can easily be non-torsion elements in $H^{2}(X,\mathcal{O}^{\times})$: for instance consider the image in $H^{3}(X,\mathbb{Z}) \cap (H^{(2,1)}(X) \oplus H^{(1,2)}(X))$. Could there be a topology more refined than etale but defined algebraically which can see these non-torsion classes? Notice that one can also ask the question for any $H^{i}(X,\mathcal{O}^{\times})$. For $i=0,1$ the Zariski and etale work fine. Why do things break down for $i>1$?
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Although the other answers correctly explain the basic logical equivalence of the two proof methods, I believe an important point has been missed: With good reason , we mathematicians prefer a direct proof of an implication over a proof by contradiction, when such a proof is available. (all else being equal) What is the reason? The reason is the fecundity of the proof, meaning our ability to use the proof to make further mathematical conclusions. When we prove an implication (p implies q) directly, we assume p, and then make some intermediary conclusions r 1 , r 2 , before finally deducing q. Thus, our proof not only establishes that p implies q, but also, that p implies r 1 and r 2 and so on. Our proof has provided us with additional knowledge about the context of p, about what else must hold in any mathematical world where p holds. So we come to a fuller understanding of what is going on in the p worlds. Similarly, when we prove the contrapositive (¬q implies ¬p) directly, we assume ¬q, make intermediary conclusions r 1 , r 2 , and then finally conclude ¬p. Thus, we have also established not only that ¬q implies ¬p, but also, that it implies r 1 and r 2 and so on. Thus, the proof tells us about what else must be true in worlds where q fails. Equivalently, since these additional implications can be stated as (¬r 1 implies q), we learn about many different hypotheses that all imply q. These kind of conclusions can increase the value of the proof, since we learn not only that (p implies q), but also we learn an entire context about what it is like in a mathematial situation where p holds (or where q fails, or about diverse situations leading to q). With reductio, in contrast, a proof of (p implies q) by contradiction seems to carry little of this extra value. We assume p and ¬q, and argue r 1 , r 2 , and so on, before arriving at a contradiction. The statements r 1 and r 2 are all deduced under the contradictory hypothesis that p and ¬q, which ultimately does not hold in any mathematical situation. The proof has provided extra knowledge about a nonexistent, contradictory land. (Useless!) So these intermediary statements do not seem to provide us with any greater knowledge about the p worlds or the q worlds, beyond the brute statement that (p implies q) alone. I believe that this is the reason that sometimes, when a mathematician completes a proof by contradiction, things can still seem unsettled beyond the brute implication, with less context and knowledge about what is going on than would be the case with a direct proof. Edit: For an example of a proof where we are led to false expectations in a proof by contradiction, consider Euclid's proof that there are infinitely many primes. In a common proof by contradiction, one assumes that p 1 , ..., p n are all the primes. It follows that since none of them divide the product-plus-one p 1 ...p n +1, that this product-plus-one is also prime. This contradicts that the list was exhaustive. Now, many beginners falsely expect after this argument that whenever p 1 , ..., p n are prime, then the product-plus-one is also prime. But of course, this isn't true, and this would be a misplaced instance of attempting to extract greater information from the proof, misplaced because this is a proof by contradiction, and that conclusion relied on the assumption that p 1 , ..., p n were all the primes. If one organizes the proof, however, as a direct argument showing that whenever p 1 , ..., p n are prime, then there is yet another prime not on the list, then one is led to the true conclusion, that p 1 ...p n +1 has merely a prime divisor not on the original list.
|
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|
12,352 |
In Samuel James Patterson's article titled Gauss Sums in The Shaping of Arithmetic after C. F. Gauss’s Disquisitiones Arithmeticae , Patterson says "Hecke [proved] a beautiful theorem on the different of k, namely that the class of the absolute different in the ideal class group is a square. This theorem - an analogue of the fact that the Euler characteristic of a Riemann surface is even - is the crowning moment (coronidis loco) in both Hecke's book and Andre Weil's Basic Number Theory." About the same matter, J.V. Armitage says (in his review of the 1981 translation of Hecke's book): "That beautiful theorem deservedly occupies the 'coronidis loco' in Weil's Basic number theory and was the starting point for the work on parity problems in algebraic number theory and algebraic geometry, which has borne such rich fruit in the past fifteen years." What is a reference for learning about the parity
problems that Armitage alludes to? It can be impossible to verbalize the reasons for aesthetic preferences, but Why might Weil, Patterson and Armitage
have been so favorably impressed by
the theorem that the ideal class of
the different of a number field is a
square in the ideal class group? Weil makes no comment on why he chose to end Basic Number Theory with the above theorem. It should be borne in mind that Weil's book covers the class number formula and all of class field theory, so that the standard against which the above theorem is being measured in the above quotes is high!
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It is not hard to see that if $L/K$ is an extension of number fields, then the
discriminant of $L/K$, which is an ideal of $K$, is a square in the ideal class group of $K$.
Hecke's theorem lifts this fact to the different. (Recall that the discriminant is the norm of the different.) If you recall that the inverse different $\mathcal D_{L/K}^{-1}$ is equal to $Hom_{\mathcal O_K}(\mathcal O_L,\mathcal O_K),$ you see that the inverse different is the relative dualizing sheaf of $\mathcal O_L$ over $\mathcal O_K$; it is analogous to the canonical bundle of a curve (which is the dualizing sheaf of the curve over the ground field). Saying that
$\mathcal D_{L/K}$, or equivalently $\mathcal D_{L/K}^{-1}$, is a square is the same as saying that there is a rank 1 projective $\mathcal O_L$-module $\mathcal E$ such that
$\mathcal E^{\otimes 2} \cong \mathcal D_{L/K}^{-1}$, i.e. it says that one can take a square
root of the dualizing sheaf. In the case of curves, this is the existence of theta characteristics. Thus, apart from anything else (and as indicated in the quotation given in the question),
Hecke's theorem significantly strengthens the analogy between rings of integers in number fields and algebraic curves. If you want to think more arithmetically, it is a kind of reciprocity law. It expresses in some way a condition on the ramification of an arbitrary extension of number fields: however the ramification occurs, overall it must be such that the different ramified primes balance out in some way in order to have $\prod_{\wp} \wp^{e_{\wp}}$ be trivial in the class group
mod $2$ (where $\wp^{e_{\wp}}$ is the local different at a prime $\wp$). (And to go back to the analogy: this is supposed to be in analogy with the fact that if $\omega$ is any meromorphic differential on a curve, then the sum of the orders of all the divisors and poles of $\omega$ is even.) Note that Hecke proved his theorem as an application of quadratic reciprocity in an arbitrary number field.
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12,366 |
In the following suppose L/K is a finite Galois extension of number fields, (maybe it works for other cases also, I don't know) By the Chebotorev density theorem when Gal(L/K) is cyclic, there are infinitely many primes in K that stay inert during this extension (cf Janus p136, Algerbaic Number Fields.) When L/K is non cyclic, an exercise from Neukirch (somewhere in Chap I) says there are at most finitely many primes that stay inert. I want to say that there are none. The reason is by a cycle description from Janus, p101, Prop 2.8, In short, that proposition says when $\delta:=Frob(\frac{L/K}{\beta})$, $\beta|p$ is a prime in L, consider $\delta$ act on the cosets of H in G, H=Gal(L/E), $K\subset E\subset L$, then every cycle of length i corresponds to a prime factor in E with residue degree i. In particular, for inert guys we want there is only one cycle in the action. When we take H to be trivial, E=L is Galois over K, and the cosets are just the elements of G themselves.
So we want that there exist an element (the Frobenius element above p) act transitively on G, thus G is cyclic. I wonder if this is true, then more people should have been aware of it. If it is not, is there a counter example?
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If $L/K$ is a finite, Galois extension of number fields such that $\text{Gal}(L/K)$ is not cyclic, then no prime of K remains inert L. Indeed, one always has an isomorphism $D_p/I_p\cong \text{Gal}(L_p/K_p)$ of the Decomposition group modulo the Inertia group with the Galois group of the corresponding residue field extension. The latter group is the Galois group of a finite extension of finite fields, hence is cyclic. If the prime p were to remain inert in L, then by definition the Inertia group would be trivial and the Decomposition group would be all of $\text{Gal}(L/K)$ . But this would imply that $L/K$ was a cyclic extension - a contradiction. [ Edit ] I can't help but mention a cute application of this. Let $n$ be any positive integer for which $(\mathbb{Z}/n\mathbb{Z})^*$ is not cyclic. Then the cycotomic polynomial $\Phi_n(x)$ is reducible modulo $p$ for every rational prime $p$ . Indeed, suppose that $p$ is a rational prime for which $\Phi_n(x)$ is irreducible modulo $p$ . Then by the Dedekind-Kummer theorem, $p$ is inert in the cyclotomic field $\mathbb{Q}(\zeta_n)$ . Then the Galois group of the residue class field extension, which is cyclic, is isomorphic to the Decomposition group, which in this case is $\text{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})\cong(\mathbb{Z}/n\mathbb{Z})^*$ . But the latter group is not cyclic - contradiction. Thus $\Phi_n(x)$ is reducible modulo $p$ for all rational primes $p$ .
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12,394 |
There have been a couple questions recently regarding metric spaces, which got me thinking a bit about representation theorems for finite metric spaces. Suppose $X$ is a set equipped with a metric $d$. I had initially assumed there must be an $n$ such that $X$ embeds isometrically into $\mathbb{R}^n$, but the following example shows that this doesn't quite work: Take for $X$ the vertex set of any graph, and let $d(x,y)$ be the length (in steps) of the shortest path connecting $x$ to $y$. Then for a minimal path connecting $x$ to $y$, the path must map to a straight line in $\mathbb{R}^n$. This is because $\mathbb{R}^n$ has the property that equality in the triangle inequality implies colinearity. So take a graph such that $x$ and $y$ have $d(x,y) \geq 2$ and two minimal paths between them; a plain old square will do the trick. The two minimal paths must each get mapped to the same line in $\mathbb{R}^n$, so the map cannot be an isometry (nor even an embedding, for that matter). This gives us one obstruction to representability: a finite metric space cannot be representable unless it satisfies the property $d(x,y) = d(x,z) + d(z,y) \wedge d(x,y) = d(x,z') + d(z',y) \wedge d(x,z) = d(x,z') \implies z = z'$. In the graph case this means "unique shortest paths"; I'm not clear if there is a snappy characterization like that in the general case. Question #1: Is this the only obstruction to representability? In a slightly different direction, you could get around the problem above by trying to represent the finite metric spaces on some surface instead of $\mathbb{R}^n$. This at least works in the graph case by replacing the points with little discs and the edges with very fat ribbons of length 1. Then compactifying the whole thing should give a surface into which the graph embeds isometrically. This suggests the answer to Question #2a: Does every finite metric space have a representation on a surface? is yes, as long as the answer to Question #2b: Does every finite metric space have a global scaling which embeds $\epsilon$-isometrically into a graph? is also yes. The $\epsilon$ is to take care of finite metric spaces with irrational distances. Of course, there is also the important Question #0: Is there some standard place I should have looked for all this?
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Since the paper referred to by Hagen Knaf is published by Springer, it may not be available to one and all. The (publicly viewable) MathSciNet reference is: MR355836 . It's a very short paper (7 pages) and the main theorem is: Theorem A metric space can be embedded in Euclidean n-space if and only if the metric space is flat and of dimension less than or equal to n. Clearly, the two terms "flat" and "dimension" need expanding. To define these, Morgan considers simplices in the metric space; that is, an n-simplex is simply an ordered (n+1)-tuple of elements of the metric space. Given such an n-tuple, say $(x_0, \dots, x_n)$, Morgan defines $D(x_0,\dots,x_n)$ to be the determinant of the matrix whose $(i,j)$th entry is $$
\frac{1}{2}\left(d(x_0,x_i)^2 + d(x_0,x_j)^2 - d(x_i,x_j)^2\right)
$$ A metric space is flat if this is positive for all simplices. If it is flat, its dimension (if this exists) is the largest n for which there is an n-simplex with this quantity positive. The argument (on a skim read) is quite cunning. Rather than go for a one-shot embedding, Morgan defines a map into $\mathbb{R}^n$ of the appropriate dimension and then uses that map to define an inner product on $\mathbb{R}^n$ with respect to which the map is an embedding. Standard linear algebra then completes the argument. Now finite dimension, in this sense, is clearly very strong. It basically says that the metric is controlled by n points. The flatness condition says that those n points embed properly (and presumably rules out the examples in the question and the first answer). But then, that's probably to be expected since embeddability into Euclidean space is similarly a strong condition.
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12,401 |
Has fuzzy logic been commercially applied in finance fields and has it been successful ?
I have got knowledge that it has been applied in Algorithmic trading and operational risk, but I want to know what are the other fields where they have been applied successfully on a commercial basis
|
Since the paper referred to by Hagen Knaf is published by Springer, it may not be available to one and all. The (publicly viewable) MathSciNet reference is: MR355836 . It's a very short paper (7 pages) and the main theorem is: Theorem A metric space can be embedded in Euclidean n-space if and only if the metric space is flat and of dimension less than or equal to n. Clearly, the two terms "flat" and "dimension" need expanding. To define these, Morgan considers simplices in the metric space; that is, an n-simplex is simply an ordered (n+1)-tuple of elements of the metric space. Given such an n-tuple, say $(x_0, \dots, x_n)$, Morgan defines $D(x_0,\dots,x_n)$ to be the determinant of the matrix whose $(i,j)$th entry is $$
\frac{1}{2}\left(d(x_0,x_i)^2 + d(x_0,x_j)^2 - d(x_i,x_j)^2\right)
$$ A metric space is flat if this is positive for all simplices. If it is flat, its dimension (if this exists) is the largest n for which there is an n-simplex with this quantity positive. The argument (on a skim read) is quite cunning. Rather than go for a one-shot embedding, Morgan defines a map into $\mathbb{R}^n$ of the appropriate dimension and then uses that map to define an inner product on $\mathbb{R}^n$ with respect to which the map is an embedding. Standard linear algebra then completes the argument. Now finite dimension, in this sense, is clearly very strong. It basically says that the metric is controlled by n points. The flatness condition says that those n points embed properly (and presumably rules out the examples in the question and the first answer). But then, that's probably to be expected since embeddability into Euclidean space is similarly a strong condition.
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12,426 |
Background Assuming ZFC is consistent, then by downward Löwenheim–Skolem, there is a countable model (M,$\in$) of ZFC. Since the universe M is countable, we may as well think of it as actually being the set of natural numbers, so $\in$ will be some binary relation on the natural numbers. Can such a relation ever be computable? Partial results One can show that the class of binary relations $R$ on the natural numbers such that $(\mathbb{N},R) \models ZFC$ forms a $\Pi_0^1$ class, and will be nonempty so long as ZFC is consistent. This already gives us some interesting results. For example, by the low basis theorem, there is a low $R$ such that $(\mathbb{N},R) \models ZFC$. But I have been unable to determine whether such a function can be made computable; the best I can do is show that if such a function is computable, then there is no effective way of finding, given a finite set D of natural numbers, the element n such that D={m : mRn}.
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The Tennenbaum phenomenon is amazing, and that is totally correct, but let me give a direct proof using the idea of computable inseparability . Theorem . There is no computable model of ZFC. Proof: Suppose to the contrary that M is a computable model of ZFC. That is, we assume that the underlying set of M is ω and the membership relation E of M is computable. First, we may overcome the issue you mention at the end of your question, and we can computably get access to what M thinks of as the
n th natural number, for any natural number n. To see this, observe first that there is a particular natural number z, which M believes
is the natural number 0, another natural number N, which M believes to the set of all natural numbers, and another natural number s, which M
believes is the successor function on the natural numbers. By decoding what it means to evaluate a function in set theory using ordered pairs, We
may now successively compute the function i(0)=z and i(n+1) = the unique number that M believes is the successor function s of i(n). Thus,
externally, we now have computable access to what M believes is the n th natural number. Let me denote i(n) simply by n . (We
could computably rearrange things, if desired, so that these were, say, the odd numbers). Let A, B be any computably inseparable sets. That is, A and B are disjoint computably
enumerable sets having no computable separation. (For example, A is the set of TM programs halting with output 0 on input 0, and B is the set of
programs halting with output 1 on input 0.) Since A and B are each computably enumerable, there are programs p 0 and p 1 that
enumerate them (in our universe). These programs are finite, and M agrees that p 0 and p 1 are TM programs that
enumerate a set of what it thinks are natural numbers. There is some particular natural number c that M thinks is the set of natural numbers
enumerated by p 0 before they are enumerated by p 1 . Let A + = { n | n E c }, which is the set
of natural numbers n that M thinks are enumerated into M's version of A before they are enumerated into M's version of B. This is a computable
set, since E is computable. Also, every member of A is in A + , since any number actually enumerated into A will be seen by M to have
been so. Finally, for the same reason, no member of B is in A + , because M can see that they are enumerated into B by a (standard)
stage, when they have not been enumerated into A. Thus, A + is a computable separation of A and B, a contradiction. QED Essentially this argument also establishes the version of Tennenbaum's theorem mentioned by Anonymous, that there is no computable nonstandard
model of PA. But actually, Tennenbaum proved a stronger result, showing that neither plus nor times individually is computable in a nonstandard model of PA. And this takes a somewhat more subtle argument. Edit (4/11/2017). The question was just bumped by another edit, and so I thought it made sense to update my answer here by mentioning a generalization of the result. Namely, in recent joint work, M. T. Godziszewski and J. D. Hamkins, Computable quotient presentations of models of arithmetic and set theory , click through for pdf at the arxiv. we prove that not only is there no computable model of ZFC, but also there is no computable quotient presentation of a model of ZFC, and indeed there is no c.e. quotient presentation of such a model, by an equivalence relation of any complexity. That is, there is no c.e. relation $\hat\in$ and equivalence relation $E$, of any complexity, which is a congruence with respect to $\hat\in$, such that the quotient $\langle\mathbb{N},\hat\in\rangle/E$ is a model of ZFC.
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|
12,436 |
I know there was a question about good algebraic geometry books on here before, but it doesn't seem to address my specific concerns. **
Question
** Are there any well-motivated introductions to scheme theory? My idea of what "well-motivated" means are specific enough that I think it warrants a detailed example. **
Example of what I mean by well motivated
** The only algebraic geometry books I have seen which cover schemes seem to leave out essential motivation for definitions. As a test case, look at Hartshorne's definition of a separated morphism: Let $f:X \rightarrow Y$ be a morphism of schemes. The diagonal morphism is the unique morphism $\Delta: X \rightarrow X \times_Y X$ whose composition with both projection maps $\rho_1,\rho_2: X \times_Y X \rightarrow X$ is the identity map of $X$. We say that the morphism $f$ is separated if the diagonal morphism is a closed immersion. Hartshorne refers vaguely to the fact that this corresponds to some sort of "Hausdorff" condition for schemes, and then gives one example where this seems to meet up with our intuition. There is (at least for me) little motivation for why anyone would have made this definition in the first place. In this case, and I would suspect many other cases in algebraic geometry, I think the definition actually came about from taking a topological or geometric idea, translating the statement into one which only depends on morphisms (a more category theoretic statement), and then using this new definition for schemes. For example translating the definition of a separated morphism into one for topological spaces, it is easy to see why someone would have made the original definition. Use the same definition, but say topological spaces instead of schemes, and say "image is closed" instead of closed immersion, i.e. Let $f:X \rightarrow Y$ be a morphism of topological spaces. The diagonal morphism is the unique morphism $\Delta: X \rightarrow X \times_Y X$ whose composition with both projection maps $\rho_1,\rho_2: X \times_Y X \rightarrow X$ is the identity map of $X$. We say that the morphism $f$ is separated if the image of the diagonal morphism is closed. After unpacking this definition a little bit, we see that a morphism $f$ of topological spaces is separated iff any two distinct points which are identified by $f$ can be separated by disjoint open sets in $X$. A space $X$ is Hausdorff iff the unique morphism $X \rightarrow 1$ is separated. So here, the topological definition of separated morphism seems like the most natural way to give a morphism a "Hausdorff" kind of property, and translating it with only very minor tweaking gives us the "right notion" for schemes. Is there any book which does this kind of thing for the rest of scheme theory? Are people just expected to make these kinds of analogies on their own, or glean them from their professors? I am not entirely sure what kind of posts should be community wiki - is this one of them?
|
I would say that the book you're looking for is probably "The Geometry of Schemes" by Eisenbud and Harris. It is very concrete and geometric, and motivates things well (though I don't think it does so in quite the detail of proving that a topological space is Hausdorff iff $X\to 1$ is separated, but I believe it does discuss separatedness and why it is good and why it captures the intuition of Hausdorff space)
|
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12,462 |
how does limsup and liminf for a sequence of sets, apply to probability theory. any real world examples would be much appreciated
|
For a sequence of subsets $A_n$ of a set $X$, the $\limsup A_n$ $= \cap_{N=1}^\infty ( \cup_{n\ge N} A_n )$ and $\liminf A_n$ $= \cup_{N=1}^\infty (\cap_{n \ge N} A_n)$. If $ x \in \limsup A_n$ then $x$ is in all of the $\cup_{n\ge N} A_n$, which means no matter how large you pick $N$ you will find an $A_n$ with $n>N$ of which $x$ is a member. Thus members of $\limsup A_n$ are those elements of $X$ that are members of infinitely many of the $A_n$'s. If $A_n$ are thought of as events (in the sense of probability) $\limsup A_n$ will be another event. It corresponds exactly to the occurance of infinitely many of the $A_n$'s. This is why $\limsup A_n$ is sometimes written $x \in A_n$ infinitely often. Similarly, if $x\in \liminf A_n$ then $x$ is in one of $\cap_{n\ge N} A_n$, which means $x \in A_n$ for all $n > N$. Thus, for $x$ to be in the $\liminf$, it must be in all of the $A_n$, with finitely many exceptions. This is how the phrase "ultimately all of them" comes up. Both of these operations, similar to their counterparts in metric spaces, concern the tail of the sequence $\{A_n\}$. I.e., neither changes if an initial portion of the sequence is truncated. As a previous response pointed out, often the sets $A_n$ are defined to track the deviation of a sequence of random variables from a candidate limit by setting $A_n = \{x: |Y_n(x) -Y(x)| \ge \epsilon\}$. The members of $\limsup A_n$ then represents those sequences that every now and then deviate $\epsilon$ away from $Y(x)$, which is solely determined by the tail of the sequence $Y_n$. Here is a conceptual game that can be partially understood using these concepts: We have a deck of cards, on the face of each card an integer is printed; thus the cards are $\{1,2,3...\}.$ At the nth round of this game, the first $n^2$ cards are taken, they are shuffled. You pick one of them. If your pick is 1, you win that round. Let $A_n$ denote the event that you win the nth round. The complement $A_n^c$ of $A_n$ will represent that you lose the $n^{th}$ round. The event $\limsup A_n$ represents those scenarios in which you win infinitely many rounds. The complement of this event is $\liminf A_n^c$, and this represents those scenarios in which you ultimately lose all of the rounds. By the Borel Cantelli Lemma $P(\limsup A_n)$ $=0$ or equivalently $P(\liminf A_n^c)=1$. Thus, a player of this game will deterministically experience that there comes a time, after which he never wins.
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|
12,638 |
Do you find it a good idea to take lecture notes (even detailed lecture notes) in mathematical lectures? Related question: Digital Pen for Math: Your Experiences?
|
I usually bring a pad of paper with me to talks, but don't take notes. I do write down things I want to revisit later, and sometimes it turns into full blown note-taking. Taking notes reduces my ability to concentrate on the lecture, so if it's a really difficult lecture or if the material is totally new to me, taking notes is a sure way to not get much out of the lecture (unless I commit a large chunk of time to reviewing the notes afterwards). If I know I'm going to review the material carefully (e.g. if I'm taking a class and I'm putting a lot of effort into it), then I live-TeX notes, which has a number of advantages: It's much faster to strike a key than it is to write a character by hand, so I TeX much better notes than I can take by hand. I find that I can even write down lots of things the speaker says but doesn't write. I can throw the notes up on my web page, which makes other people in the same class happy. I can grep (or otherwise search) my notes. If I have the time, I can edit the notes and end up with an awesome reference (e.g. I put a lot of effort into editing my Lie groups and Stacks notes). More often, I review my notes but not edit them very much, in which case they're still pretty nice to have later on. If you're considering live-TeXing, try it! It's a lot easier than you might think. Also, look at this page of advice I wrote up when I found myself repeating it to many people. Edit: By the way, I post all my source , and you're welcome to look at it if you want to see how I set up my headers or manage my files. If I have an svn repository for the notes, you can get them with the command svn checkout svn://sheafy.net/courses/halg_sp2008 (to get my homological algebra notes from Peter Teichner's course for example). If there's no svn repository, you can download the tgz (tar+gzip) file and uncompress it.
|
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|
12,657 |
This is an elementary question, but a little subtle so I hope it is suitable for MO. Let $T$ be an $n \times n$ square matrix over $\mathbb{C}$. The characteristic polynomial $T - \lambda I$ splits into linear factors like $T - \lambda_iI$, and we have the Jordan canonical form: $$ J = \begin{bmatrix} J_1 \\\ & J_2 \\\ & & \ddots \\\ & & & J_n \end{bmatrix}$$ where each block $J_i$ corresponds to the eigenvalue $\lambda_i$ and is of the form $$ J_i = \begin{bmatrix} \lambda_i & 1 \\\ & \lambda_i & \ddots \\\ & & \ddots & 1 \\\ & & & \lambda_i \end{bmatrix}$$ and each $J_i$ has the property that $J_i - \lambda_i I$ is nilpotent, and in fact has kernel strictly smaller than $(J_i - \lambda_i I)^2$, which shows that none of these Jordan blocks fix any proper subspace of the subspace which they fix. Thus, Jordan canonical form gives the closest possible to a diagonal matrix. The elements in the superdiagonals of the Jordan blocks are the obstruction to diagonalization. So far, so good. What I want to prove is the assertion that "Almost all square matrices over $\mathbb{C}$ is diagonalizable". The measure on the space of matrices is obvious, since it can be identified with $\mathbb{C}^{n^2}$. How to prove, perhaps using the above Jordan canonical form explanation, that almost all matrices are like this? I am able to reason out the algebra part as above, but is finding difficulty in the analytic part. All I am able to manage is the following. The characteristic equation is of the form $$(x - \lambda_1)(x - \lambda_2) \cdots (x - \lambda_n)$$ and in the space generated by the $\lambda_i$'s, the measure of the set in which it can happen that $\lambda_i = \lambda_j$ when $i \neq j$, is $0$: this set is a union of hyperplanes, each of measure $0$. But here I have cheated, I used only the characteristic equation instead of using the full matrix. How do I prove it rigorously?
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The discriminant of the characteristic polynomial of a matrix depends polynomially on the coefficients of the matrix, and its vanishing detects precisely the existence of multiple eigenvalues. Therefore the set where the discriminant does not vanish is contained in the set of diagonalizable matrices. Now the set where a non-zero polynomial vanishes is very, very thin (in many senses: it does not contain open sets, it has zero Lebesgue measure, etc) so in consequence the set of diagonalizable matrices is correspondingly thick.
|
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|
12,684 |
I am a recent PhD specializing in algebraic geometry. But I also want to do research in some other areas of math (e.g. quantum computation, compressed sensing, and PDE's). What would be a good way of learning these so that they can become a research interest? I have very little background in physics. My plan/goal is to begin research in one of these areas by next February. I do not want to be limited in the areas in which I do my research (e.g. only doing research in algebraic geometry).
|
It may not be relevant to your situation, but a bit of advice I was once given, which I think is a good one, is that moving fields is an excellent idea but it is even better to make the path continuous. Given the interconnectedness of mathematics, this advice is easier to follow than it at first seems.
|
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12,688 |
I always had trouble remembering this. Is it true that a curve over a non-algebraically-closed field is normal implies that it's non-singular? How about a 1 dimensional scheme? How about dimension 2? I think I heard once that surfaces over a non-algebraically closed field is normal does imply that it's non-singular. Is it true for2 dimensional schemes? What is the reason that these theorems are true for small dimensions, but fail for higher dimensions?
|
For curves over a field $k$, normal implies regular. (The point is that a normal Noetherian local ring
of dimension one is automatically regular, i.e. a DVR.) If $k$ is not
perfect, then it might not be smooth over $k$. The reason is that in this case it is possible to have a regular local $k$-algebra of dimension one whose base-change to $\overline{k}$ is no longer regular. (On the other hand,
smoothness over $k$ is preserved by base-change, since it is a determinental condition
on Jacobians.) Here is a (somewhat cheap) example: let $l$ be a non-separable extension of $k$ (which
exists since $k$ is not perfect), and let $X = \text{Spec} l[t],$ though of as a $k$-scheme.
This will be regular, but not smooth over $k$. In dimension 2, even over an algebraically closed field, normal does not imply regular
(and so in particular, does not imply smooth). Normal is equivalent to having the singular locus be of codimension 2 or higher (so for a surface, just a bunch of points) (this is what Serre calls R_1) together with the condition that if a rational function on some open subset has no poles in codimension one, it is in fact regular on that open set (this is Serre's condition S_2). For a surface in ${\mathbb P}^3$, which is necessarily cut out by a single equation, the
condition $S_2$ is automatic (this is true of any local complete intersection in a smooth variety), so normal is equivalent to the singular locus being 0-dimensional. For surfaces in higher dimensional projective space, $R_1$ and $S_2$ are independent
conditions; either can be satisfied without the other. And certainly both together (i.e. normality) are still weaker than smoothness. From Serre's criterion (normal is equivalent to $R_1$ and $S_2$) you can see that normality
just involves conditions in codimension one or two. Thus for curves it says a lot,
for surfaces it says something, but it diverges further from smoothness the higher the dimension of the variety is. Edit: As Hailong pointed out in a comment (now removed), I shouldn't say that S_2 is a condition only in dimension 2;
one must check it all points. Never the less, at some sufficiently vague level, the spirit
of the preceding remark is true: $R_1$ and $S_2$ capture less and less information about the local structure of the variety, the higher the dimension of the variety.
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|
12,732 |
There are a lot of theorems in basic homological algebra, such as the five lemma or the snake lemma, that seem like they'd be more easily proven by computer than by hand. This led me to consider the following question: is the theory of categories decidable? More specifically, I was wondering whether or not statements about abelian categories can be determined true or false in finite time. Also, if they can be determined to be false, is it possible to explicitly describe a counterexample? If it is known to be decidable, is anything known about the complexity? (Other decidable theories often have multiply-exponential time complexities.) If it is known to be undecidable, say by embedding the halting problem, then can I change my assumptions a bit and make it decidable? (For example, maybe I shouldn't be looking at abelian categories after all.) Thanks in advance. Edit : It appears a clarification is needed. My goal was to consider the minimal theory that could state things like the five lemma, but not necessarily prove them. For example, I want to say: If in an abelian category, you have a bunch of maps $0\to A \to B \to C\to 0$ and $0\to A' \to B' \to C'\to 0$ which make up two short exact sequences and some more maps $a:A\to A'$, $b:B\to B'$, $c:C\to C'$ which commute with the previous maps, and $a$ and $c$ are isomorphisms, then $b$ is an isomorphism, too. Sentences of this form would be inputs to a program, which decides if this statement is in fact true in ZFC (or your other favorite axiomatization of category theory). The point here is that I am restricting the sentences one can input into the program, but keeping ZFC or whatnot as my framework. I hoped (perhaps naively) that if I restricted the class of sentences, it might be decidable whether or not these statements were true. For example, I imagined that every such theorem is either proven by diagram chasing, or it is possible to find a concrete example of maps among, say, R-modules that contradict the result.
|
Thanks for clarifying your question. The formulation that
you and Dorais give seems perfectly reasonable. You have a
first order language for category theory, where you can
quantify over objects and morphisms, you can compose
morphisms appropriately and you can express that a given
object is the initial or terminal object of a given
morphism. In this language, one can describe various finite
diagrams, express whether or not they are commutative, and
so on. In particular, one can express that composition is
associative, etc. and describe what it means to be a
category in this way. The question now becomes: is this theory decidable? In
other words, is there a computable procedure to determine,
given an assertion in this language, whether it holds in
all categories? The answer is No . One way to see this is to show even more: one cannot even
decide whether a given statement is true is true in all
categories having only one object. The reason is that group
theory is not a decidable theory. There is no computable
procedure to determine whether a given statement in the
first order language of group theory is true in all groups.
But the one-point categories naturally include all the
groups (and we can define in a single statement in the
category-theoretic language exactly what it takes for the
collection of morphisms on that object to be a group).
Thus, if we could decide category theory, then we could
decide the translations of the group theory questions into
category theory, and we would be able to decide group
theory, which we can't. Contradiction. The fundamental obstacle to decidability here, as I
mentioned in my previous answer (see edit history), it the
ability to encode arithmetic. The notion of a strongly
undecidable
structure is key for proving various theories are undecidable. A
strongly undecidable theory is a finitely axiomatizable
theory, such that any theory consistent with it is
undecidable. Robinson proved that there is a strongly
undecidable theory of arithmetic, known as Robinson's Q. A strongly undecidable structure is a structure modeling a
strongly undecidable theory. These structures are amazing,
for any theory true in a strongly undecidable structure is
undecidable. For example, the standard model of arithmetic,
which satisfies Q, is strongly undecidable. If A is
strongly undecidable and interpreted in B, then it follows
that B is also strongly undecidable. Thus, we can prove
that graph theory is undecidable, that ring theory is
undecidable and that group theory is undecidable, merely by
finding a graph, a ring or a group in which the natural
numbers is interpreted. Tarski found a strongly undecidable
group, namely, the group G of permutations of the integers
Z. It is strongly undecidable because the natural numbers
can be interpreted in this group. Basically, the number n
is represented by translation-by-n. One can identify the
collection of translations, as exactly those that commute
with s = translation-by-1. Then, one can define addition as
composition (i.e. addition of exponents) and the divides
relation is definable by: i divides j iff anything that
commutes with s i also commutes with
s j . And so on. I claim similarly that there is a strongly undecidable
category. This is almost immediate, since every group can
be viewed as the morphisms of a one-object category, and
the group is interpreted as the morphisms of this category.
Thus, the category interprets the strongly undecidable
group, and so the category is also strongly undecidable. In
particular, any theory true in the category is also
undecidable. So category theory itself is undecidable.
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12,804 |
A cardinal $\lambda$ is weakly inaccessible, iff a. it is regular (i.e. a set of cardinality $\lambda$ can't be represented as a union of sets of cardinality $<\lambda$ indexed by a set of cardinality $<\lambda$) and b. for all cardinals $\mu<\lambda$ we have $\mu^+<\lambda$ where $\mu^+$ is the successor of $\mu$. Strongly inaccessible cardinals are defined in the same way, with $\mu^+$ replaced by $2^\mu$. Usually one also adds the condition that $\lambda$ should be uncountable. As far as I understand, a "large cardinal" is a weakly inaccessible cardinal with some extra properties. In set theory one considers various "large cardinal axioms", which assert the existence of large cardinals of various kinds. Notice that these axioms are quite different from, say the Continuum Hypothesis. In particular, one can't deduce the consistency of ZFC + there exists at least one (uncountable) weakly inaccessible cardinal from the consistency of ZFC, see e.g. Kanamori, The Higher Infinite, p.19. I.e., assuming ZFC is consistent, these axioms can not be shown independent of ZFC. The "reasonable" large cardinal axioms seem to be ordered according to their consistency strength, as explained e.g. here http://en.wikipedia.org/wiki/Large_cardinal . This is not a theorem, just an observation. A list of axioms according to their consistency strength can be found e.g. on p. 472 of Kanamori's book mentioned above. (Noticeably, it starts with "0=1", which is a very strong axiom indeed.) Large cardinals appear to occur seldom in "everyday" mathematics. One such instance when they occur is when one tries to construct the foundations of category theory. One of the ways to do that (and the one that seems (to me) to be the most attractive) is to start with the set theory and to add Grothendieck's Universe axiom, which states that every set is an element of a Grothendieck universe. (As an aside remark, let me mention another application of large cardinal axioms: incredibly, the fastest known solution of the word problem in braid groups originated from research on large cardinal axioms; the proof is independent of the existence of large cardinals, although the first version of the proof did use them. See Dehornoy, From large cardinals to braids via distributive algebra, Journal of knot theory and ramifications, 4, 1, 33-79.) Translated into the language of cardinals, the Universe axiom says that for any cardinal there is a strictly larger strongly inaccessible cardinal. I have heard several times that this is pretty low on the above consistency strength list, but was never able to understand exactly how low. So I would like to ask: does the existence of a (single) large cardinal of some kind imply (or is equivalent to) the Universe axiom?
|
A Grothendieck universe is known in set theory as the set V κ for a (strongly) inaccessible cardinal κ. They are exactly the same thing. Thus, the existence of a Grothendieck universe is exactly equivalent to the existence of one inaccessible cardinal. These cardinals and the corresponding universes have been studied in set theory for over a century. The Grothendieck Universe axiom (AU) is the assertion that every set is an element of a universe in this sense. Thus, it is equivalent to the assertion that the inaccessible cardinals are unbounded in the cardinals. In other words, that there is a proper class of inaccessible cardinals. This is the axiom you sought, which is exactly equivalent to AU. In this sense, the axiom AU is a statement in set theory, having nothing necessarily to do with category theory. The large cardinal axioms are fruitfully measured in strength not just by direct implication, but also by their consistency strength. One large cardinal property LC 1 is stronger than another LC 2 in consistency strength if the consistency of ZFC with an LC 1 large cardinal implies the consistency of ZFC with an LC 2 large cardinal. Measured in this way, the AU axiom has a stronger consistency strength than the existence of any finite or accessible number of inaccessible cardinals, and so one might think it rather strong. But actually, it is much weaker than the existence of a single Mahlo cardinal, the traditional next-step-up in the large cardinal hierarchy. The reason is that if κ is Mahlo, then κ is a limit of inaccessible cardinals, and so V κ will satisfy ZFC plus the AU axiom. The difference between AU and Mahloness has to do with the thickness of the class of inaccessible cardinals. For example, strictly stronger than AU and weaker than a Mahlo cardinal is the assertion that the inaccessible cardinals form a stationary proper class, an assertion known as the Levy Scheme (which is provably equivconsistent with some other interesting axioms of set theory, such as the boldface Maximality Principle, which I have studied a lot). Even Mahlo cardinals are regarded as rather low in the large cardinal hierarchy, far below the weakly compact cardinals, Ramsey cardinals, measurable cardinals, strong cardinals and supercompact cardinals. In particular, if δ is any of these large cardinals, then δ is a limit of Mahlo cardinals, and certainly a limit of strongly inaccessible cardinals. So in particular, V δ will be a model of the AU axiom. Rather few of the large cardinal axioms imnply AU directly, since most of them remain true if one were to cut off the universe at a given inaccessible cardinal, a process that kills AU. Nevertheless, implicit beteween levels of the large caridnal hiearchy are the axioms of the same form as AU, which assert an unbounded class of the given cardinal. For example, one might want to have unboundedly many Mahlo cardinals, or unboundedly many measurable cardinals, and so on. And the consistency strength of these axioms is still below the consistency strength of a single supercompact cardinal. The hierarchy is extremely fine and intensely studied. For example, the assertion that there are unboundedly many strong cardinals is equiconsistent with the impossiblity to affect projective truth by forcing. The existence of a proper class of Woodin cardinals is particularly robust set-theoretically, and all of these axioms are far stronger than AU. There are natural weakenings of AU that still allow for almost all if not all of what category theorists do with these universes. Namely, with the universes, it would seem to suffice for almost all category-theoretic purposes, if a given universe U were merely a model of ZFC, rather than V κ for an inaccessible cardinal κ. The difference is that U is merely a model of the Power set axiom, rather than actually being closed under the true power sets (and similarly using Replacement in place of regularity). The weakening of AU I have in mind is the axiom that asserts that every set is an element of a transitive model of ZFC. This assertion is strictly weaker in consistency strength thatn even a single inaccessible cardinal. One can get much lower, if one weakens the concept of universe to just a fragment of ZFC. Then one could arrive at a version of AU that was actually provable in ZFC, but which could be used for most all of the applications in cateogory theory to my knowledge. In this sense, ZFC itself is a kind of large cardinal axiom relative to the weaker fragments of ZFC.
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12,810 |
Here is a double series I have been having trouble evaluating:
$$S=\sum_{k=0}^{m}\sum_{j=0}^{k+m-1}(-1)^{k}{m \choose k}\frac{[2(k+m)]!}{(k+m)!^{2}}\frac{(k-j+m)!^{2}}{(k-j+m)[2(k-j+m)]!}\frac{1}{2^{k+j+m+1}}\text{.}$$ I am confident that $S=0$ for any $m>0$. In fact, I have no doubt. I have done lots of algebraic manipulation, attempted to "convert" it to a hypergeometric series, check tables (Gradshteyn and Ryzhik), etc., but I have not been able to get it into a form from which I can prove zero equivalence. Here is another form of the sum (well, I hope at least) that might be easier to work with: $$S=\sum_{k=0}^{m}\sum_{j=0}^{k+m-1}(-1)^{k}\frac{m!}{k!(m-k)!}\frac{(k+m-1-j)!}{(k+m)!}\frac{(k+m-1/2)!}{(k+m-1/2-j)!}\frac{1}{2^{k-j+m+1}}\text{.}$$ I have read Concrete Mathematics and $A=B$, and looked at Gosper's and Zeilberger's work for some hints, but no cigar. Note: $0!=1$ and $n!=n(n-1)!$ for $n\in\mathbb{N}\cup\{0\}$. For $n\in\mathbb{R}^+$, $n!=n\Gamma(n)$ where $\Gamma\colon\: \mathbb{C}\to\mathbb{C}$ and, for $\Re z>0$ and $z\notin\mathbb{Z}^{-}$, $$\Gamma\colon\: z\mapsto \int_0^\infty t^{z-1}\mathrm{e}^{-t}\,\mathrm{d}t\text{.}$$
which can be analytically extended to $\mathbb{C}$ via the recurrence $\Gamma(z+1)=z\Gamma(z)$.
|
Olivier Gerard just told me about this wonderful website!
Regarding the question it can be done in one nano-second using the Maple
package http://www.math.rutgers.edu/~zeilberg/tokhniot/MultiZeilberger accopmaying my article with Moa Apagodu http://www.math.rutgers.edu/~zeilberg/mamarim/mamarimhtml/multiZ.html Here is the command: F:=(-1)^k*binomial(m,k)*(k+m-1-j)!/(k+m)!*simplify((k+m-1/2)!/(k+m-1/2-j)!)/2^(k-j): lprint(MulZeil(F,[j,k],m,M,{})[1]); and here is the output:
-1/4*(2*m+1)/(m+1)+M (Note that I had to divide the summand by 1/2^(m+1) if you don't you get FAIL,
the prgram does not like extraneous factors) Translated to humaneze we have that (my S(m) is hte original S(m) times 2^(m+1))
S(m+1)=(2m+1)/(m+1)S(m) Since S(1)=0 (check!) This is a completely rigorous proof. P.S. The proof can be gotten by finding the so-called multi-certificate lprint(MulZeil(F,[j,k],m,M,{})[2]); -Doron Zeilbeger
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|
12,814 |
I know this should be pretty simple, but right now the only way I can see how to prove it is to sit down and write out explicit formulae for the group law, and see that everything works out. What's the geometric or abstract-nonsense reason why the abelian group structure of elliptic curves behaves nicely under homomorphisms?
|
This follows by a rigidity property of certain morphisms. It is important to note that elliptic curves are complete, that is, proper and integral schemes. Then we have the following "Rigidity Lemma" (see Mumford's Abelian Varieties, the beginning of chapter II (page 43 of the old edition), for example): Let $X$ be a complete variety, $Y$ and $Z$ any varieties, and $f:X\times Y\rightarrow Z$ a morphism such that for some $y_0\in Y$, $f(X\times\{y_0\})$ is a single point $z_0$ of $Z$. Then there is a morphism $g:Y\rightarrow Z$ such that if $p_2:X\times Y\rightarrow Y$ is the projection, $f=g\circ p_2$. How does this help? Well, elliptic curves have a distinguished point (the origin), and a morphism of elliptic curves is a morphism of their underlying schemes taking one distinguished point to the other. We'd like to show that any such morphism is actually compatible with the group law. So consider a morphism of elliptic curves $f:E_1\rightarrow E_2$, and let $\Phi:E_1\times E_1\rightarrow E_2$ be defined by $\Phi(x,y)=f(x+y)-f(x)-f(y)$. Then $\Phi(E_1\times\{0\})=0$ so by the lemma I quoted, there is a morphism $g:E_1\rightarrow E_2$ such that $\Phi=g\circ p_2$. And since $\Phi(\{0\}\times E_1)=0$, as well, $g$ must be zero. But then $\Phi$ itself must be zero. None of this relied on $E_1$ and $E_2$ being elliptic curves; it works exactly the same way for abelian varieties.
|
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|
12,865 |
As everybody knows, the ZFC axioms may serve as a foundation for (almost)
all of contemporary mathematics, and it is also well-known that several results
are "indecidable" in ZFC, which means that they cannot be proved or disproved within
ZFC. It is therefore natural to look for "new axioms" to add to ZFC and make it a
stronger system. But by Godel's second incompleteness theorem, the consistency of
ZFC cannot be deduced from ZFC itself. Therefore, we may add the axiom "ZFC is consistent" and obtain
a new system $ ZFC_1 $ consisting of "ZFC+(ZFC is consistent)". We may
iterate this, and define $ ZFC_2 $ as "$ZFC_1$+($ZFC_1$ is consistent)", etc,
and we may even define $ZFC_{\omega}$, or $ZFC_{\alpha}$ for any
ordinal $\alpha$. This seems a little too easy, so my question to logicians is : is this construction
completely irrelevant to logic ans set theory questions ? If so why? Is it true that the results which are classically independent of ZFC are also
independent of $ZFC_1$, $ZFC_2$, $ZFC_{\omega}$ etc ?
|
Such constructions are interesting! However, they are often done with PA instead of ZFC (see note). For an interesting discussion, I recommend Torkel Franzén's book Inexhaustibility: a non-exhaustive treatment (Lecture Notes in Logic 16, ASL, 2004). You can also read this excellent blog article by Mike O'Connor. Note: The following is explained in Mike O'Connor's article, but I think I need to clarify why ZFC is not the ideal base theory to do this and why PA is a better candidate. The idea is that Con(T) is usually understood as an arithmetical statement. More precisely, given a recursive presentation of the theory T the statement Con(T) is arithmetical formalization of "there is no proof of a contradiction from T" which is encoded using Gödel numbers for proofs and formulas. (This is the messy part of Gödel's Theorem.) This is why PA, or more generally any recursively axiomatizable extension of PA, provides a more natural environment for the analysis of such statements. For example, instead of ZFC, you may as well use the purely arithmetical part of ZFC. There is also an even more fundamental problem with transfinite iterates. Given a recursive presentation of a theory T, the iterates T 0 = T, T 1 = T 0 + Con(T 0 ), T 2 = T 1 + Con(T 1 ), etc. Can be continued into the transfinite, but only to a limited extent. It is easy to give a recursive presentation of T ω or T ω+ω+3 but there are only countably many ordinals for which this works. Indeed, these iterates are better defined in terms of ordinal notations than in terms of proper ordinals. Ordinal notations can go pretty far up, but there are clear limitations. These difficulties and their implications are discussed in great detail in Franzén's book. As Mike O'Connor explains, it is natural to go further and extend these to subsystems of second-order arithmetic , but there and in set theory the appropriate principles to study are reflection principles and large cardinal axioms which have better semantic interpretations and take advantage of their richer surroundings.
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|
12,943 |
For any set X, let S X be the symmetric group on
X, the group of permutations of X. My question is: Can there be two nonempty sets X and Y with
different cardinalities, but for which S X is
isomorphic to S Y ? Certainly there are no finite examples, since the symmetric
group on n elements has n! many elements, so the finite
symmetric groups are distinguished by their size. But one cannot make such an easy argument in the infinite
case, since the size of S X is 2 |X| ,
and the exponential function in cardinal arithmetic is not
necessarily one-to-one. Nevertheless, in some set-theoretic contexts, we can still
make the easy argument. For example, if the Generalized
Continuum Hypothesis holds, then the answer to the question
is No, for the same reason as in the finite case, since the
infinite symmetric groups will be characterized by their
size. More generally, if κ < λ implies
2 κ < 2 λ for all
cardinals, (in another words, if the exponential function
is one-to-one, a weakening of the GCH), then again
S κ is not isomorphic to
S λ since they have different
cardinalities. Thus, a negative answer to the question is
consistent with ZFC. But it is known to be consistent with ZFC that
2 κ = 2 λ for some
cardinals κ < λ. In this case, we will
have two different cardinals κ < λ, whose
corresponding symmetric groups S κ and
S λ nevertheless have the same cardinality. But can we still
distinguish these groups as groups in some other (presumably more group-theoretic) manner? The smallest instance of this phenomenon occurs under Martin's Axiom plus ¬CH, which implies 2 ω =
2 ω 1 . But also, if one just forces ¬CH by adding Cohen reals over a model of GCH, then again 2 ω =
2 ω 1 . (I am primarily interested in what happens with AC. But if
there is a curious or weird counterexample involving
¬AC, that could also be interesting.)
|
According to the Schreier–Ulam–Baer theorem, the nontrivial normal subgroups of $S(X)$ are (i) the subgroup $S_\mathrm{fin}(X)$ of permutations of $X$ of finite support, (ii) the subgroup $A_\mathrm{fin}(X)$ of $S_\mathrm{fin}(X)$ of even permutations, and (iii) for each cardinal $\kappa$ the subgroups $S_{<\beta}(X)$ and $S_{\leq\beta}(X)$ of permutations which move strictly less than $\beta$ points and at most $\beta$ points, respectively. Since, as you said, a cardinal is determined by the order type of the set of cardinals below it, looking at the lattice of normal subgroups of $S(X)$, then, lets you guess the cardinal of $X$.
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|
12,973 |
It is easy to see that in ZFC, any non-empty set $S$ admits a group structure: for finite $S$ identify $S$ with a cyclic group, and for infinite $S$, the set of finite subsets of $S$ with the binary operation of symmetric difference forms a group, and in ZFC there is a bijection between $S$ and the set of finite subsets of $S$, so the group structure can be taken to $S$. However, the existence of this bijection needs the axiom of choice. So my question is Can it be shown in ZF that for any non-empty set $S$ there exists a binary operation $\ast$ on $S$ making $(S,\ast)$ into a group?
|
In ZF, the following are equivalent: (a) For every nonempty set there is a binary operation making it a group (b) Axiom of choice Non trivial direction [(a) $\to$ (b)]: The trick is Hartogs' construction which gives for every set $X$ an ordinal $\aleph(X)$ such that there is no injection from $\aleph(X)$ into $X$. Assume for simplicity that $X$ has no ordinals. Let $\circ$ be a group operation on $X \cup \aleph(X)$. Now for any $x \in X$ there must be an $\alpha \in \aleph(X)$ such that $x \circ \alpha \in \aleph(X)$ since otherwise we get an injection of $\aleph(X)$ into $X$. Using $\circ$, therefore, one may inject $X$ into $(\aleph(X))^{2}$ by sending $x \in X$ to the $<$-least pair $(\alpha, \beta)$ in $(\aleph(X))^{2}$ such that $x \circ \alpha = \beta$. Here, $<$ is the lexicographic well-ordering on the product $(\aleph(X))^{2}$. This induces a well-ordering on $X$.
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|
13,005 |
The key step in Kontsevich's proof of deformation quantization of Poisson manifolds is the so-called formality theorem where 'a formal complex' means that it admits a certain condition. I wonder why it is called 'formal'. I only found the definition of Sullivan in Wikipedia: 'formal manifold is one whose real homotopy type is a formal consequence of its real cohomology ring'. But still I am confused because most of articles I found contain the same sentence only and I cannot understand the meaning of 'formal consequence'. Does anyone know the history of this concept?
|
I would guess that the terminology goes back to the work of Sullivan and Quillen on rational homotopy theory. You should probably also look at the paper of Deligne-Griffiths-Morgan-Sullivan on the real homotopy theory of Kähler manifolds. Actually, I think that at least some familiarity with the DGMS paper is an important prerequisite for understanding many of Kontsevich's papers. I am not totally sure, but I believe that the definitions are as follows: A differential graded algebra $(A,d)$ is called formal if it is quasi-isomorphic (in general, if we work in the category of dg algebras and not, say, the category of A-infinity algebras, we need a "zig-zag" of quasi-isomorphisms) to $H^\ast(A,d)$ considered as a dg algebra with zero differential. A space X is called formal (over the rationals resp. the reals) if its cochain dg algebra $C^\ast(X)$ (with rational resp. real coefficients) with the standard differential is a formal dg algebra. One of the things I'm not sure about is whether in the definition we should require $H^\ast(A,d)$ to be commutative; but for spaces this is not an issue since $H^\ast(X)$ is always (graded-)commutative. The DGMS paper proves that if X is a compact Kähler manifold, then the de Rham dg algebra consisting of (real, $C^\infty$ ) differential forms on X with the standard de Rham differential is a formal dg algebra. The phrase "the real (resp. rational) homotopy type of X is a formal consequence of the real (resp. rational) cohomology ring of X", which appears in e.g. the DGMS paper, simply means that the real (resp. rational) homotopy theory of X is determined by (and is probably explicitly and algorithmically computable from?) the cohomology ring of X. In other words, if X and Y are formal (over the rationals resp. the reals) and have isomorphic (rational resp. real) cohomology rings, then their respective (rational resp. real) homotopy theories are the same (and are explicitly computable, if we know the cohomology ring(s)?). For example, the ranks of their homotopy groups will be equal. Actually I am not totally sure whether what I said in the last paragraph is true. I think it's true when X and Y are simply connected. I'm not sure about what happens more generally. In the context of rational homotopy theory, I think the term "formal" is fine, for the reasons I've explained above. Perhaps in the more general context of dg algebras, the use of the term "formal" makes less sense. However, I think that it is still reasonable, for the following reasons. Let me use the more "modern" language of A-infinity algebras. In general, it is not true that a dg algebra $(A,d)$ is quasi-isomorphic to $H^\ast(A,d)$ considered as a dg algebra with zero differential. However, it is a "standard" fact (Kontsevich-Soibelman call this the "homological perturbation lemma" (for example, it's buried somewhere in this paper ), and you can find it in the operads literature as the "transfer theorem") that you can put an A-infinity structure on $H^\ast(A,d)$ which makes $A$ and $H^\ast(A,d)$ quasi-isomorphic as A-infinity algebras. The A-infinity structure manifests itself as a series of $n$ -ary products satisfying various compatibilities. Intuitively at least, these $n$ -ary products should be thought of as being analogous to Massey products in topology. So $H^\ast(A,d)$ with this A-infinity structure does carry some "homotopy theoretic" information. In this language then, a dg algebra $(A,d)$ is formal if it is quasi-isomorphic, as an A-infinity algebra, to $H^\ast(A,d)$ with all higher products zero. In other words, all of the "Massey products" vanish*, and thus the only remaining "homotopy theoretic" information is that coming from the ordinary ring structure on $H^\ast(A,d)$ . *Don Stanley notes correctly that vanishing of Massey products is weaker than formality. However, I believe that triviality of the A-infinity structure is equivalent to formality. In the language of the DGMS paper, which does not use the A-infinity language, they say that formality is equivalent to the vanishing of Massey products "in a uniform way". I believe this uniform vanishing is the same as triviality of A-infinity structure. From the paper: ... a minimal model is a formal consequence of its cohomology ring if, and only if, all the higher order products vanish in a uniform way. and also [Choosing a quasi-isomorphism from a minimal dg algebra to its cohomology] is a way of saying that one may make uniform choices so that the forms representing all Massey products and higher order Massey products are exact . This is stronger than requiring each individual Massey product or higher order Massey product to vanish. The latter means that, given one such product, choices may be made to make the form representing it exact, and there may be no way to do this uniformly. (Sorry for the proliferation of parentheses, and sorry for my lack of certainty on all of this, I have not thought about this in a while. People should definitely correct me if I'm wrong on any of this.)
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|
13,008 |
What is the expected value of the determinant over the uniform distribution of all possible 1-0 NxN matrices? What does this expected value tend to as the matrix size N approaches infinity?
|
If $N \ge 2$, then the expected value is $0$ since interchanging two rows preserves the distribution but negates the determinant.
|
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|
13,072 |
This question is prompted by another one . I want to motivate the definition of a scheme for people who know about manifolds(smooth, or complex analytic). So I define a manifold in the following way. Defn: A smooth $n$-manifold is a pair $(X, \mathcal{O}_X)$, where $X$ is a topological space and $\mathcal{O}_X$ is a sheaf of rings on it, such that, every point $x \in X$ has a neighborhood $U_x$ which is homeomorphic to an open set $V_x$ in $\mathbb{R}^n$ and $\mathcal{O}_X$ restricted to $U_x$ is isomorphic to the sheaf of ring of smooth functions on $V_x$ and its open subsets. This agrees with the usual definition using charts and atlases, for all except the requirement that a manifold is a (separable) Hausdorff space. But indeed it seems that many things in differential topology can be proven without using the Hausdorff property. In a fleeting conversation in a brief encounter, a personal I shall refer to as O., informed me that even Stokes' theorem can be done this way. But I am unable to ask O. again about this, as he is not physically around. If the above is true, then this is a really good point to mention when introducing schemes to a person who knows about differentiable manifolds. So my main question: Is it true that the proof of Stokes' theorem for smooth manifolds can be proven without the Hausdorff condition on the manifold? If so, is it done in any well-known reference? Aside question: What are some crucial propositions/theorems in differential topology that use Hausdorff condition, except those involving imbedding in some $\mathbb{R}^m$ for high enough $m$, for which of course Hausdorffness is a necessary condition(together with separable property)? Tom Church answers below that partitions of unity does not work, for instance on the example of the line with the double point. However I believe that one can still make sense of integration of differential forms even with such pathologies, because by introducing a measure for instance, we can ignore sets of measure $0$.
|
The existence of flows in the direction of a vector field seems to require Hausdorff; indeed, consider the vector field $\frac{\partial}{\partial x}$ on the line-with-two-origins. We have no global existence of a flow for any positive t, even if we make our space compact (that is, considering the circle-with-one-point-doubled). If the nonexistence of the flow is not visibly clear, consider instead the real line with the interval [0,1] doubled. Also, partitions of unity do not exist; for example, in the line with two origins, take the open cover by "the line plus the first origin" and "the line plus the second origin". There is no partition of unity subordinate to this cover (the values at each origin would have to be 1). For me, a basic example of the beauty of this function-theoretic approach is the definition of a vector field as a derivation $D\colon C^\infty(M)\to C^\infty(M)$. The proof that such a derivation defines a vector field hinges upon the fact that $Df$ near a point p only depends on $f$ near the point p. To prove this fact you use the fineness of your sheaf $\mathcal{O}_X$, i.e. the existence of partitions of unity. (It is true though that the failure of fineness in the non-Hausdorff case is of a different sort and might not break this particular theorem.) I feel that the existence of partitions of unity, and the implications thereof, is one of the basic fundamentals of approaching smooth manifolds through their functions; more importantly, a good handle on how partitions of unity are used is important to understand the differences that arise when the same approach is extended to more rigid functions (holomorphic, algebraic, etc.). Now that the question has been edited to ask specifically about Stokes' theorem, let me say a bit more. Stokes' theorem will be false for non-Hausdorff manifolds, because you can (loosely speaking) quotient out by part of your manifold, and thus part of its homology, without killing all of it. For the simplest example, consider dimension 1, where Stokes' theorem is the fundamental theorem of calculus. Let $X$ be the forked line, the 1-dimensional (non-Hausdorff) manifold which is the real line with the half-ray $[0,\infty)$ doubled. For nonnegative $x$, denote the two copies of $x$ by $x^\bullet$ and $x_\bullet$, and consider the submanifold $M$ consisting of $[-1,0) \cup [0^\bullet,1^\bullet] \cup [0_\bullet,1_\bullet]$. The boundary of $M$ consists of the three points $[-1]$ (with negative orientation), $[1^\bullet]$ (with positive orientation), and $[1_\bullet]$ (with positive orientation); to see this, just note that every other point is a manifold point. Consider the real-valued function on $X$ given by "$f(x)=x$" (by which I mean $f(x^\bullet)=f(x_\bullet)=x$). Its differential is the 1-form which we would naturally call $dx$. Now consider $\int_M dx$; it seems clear that this integral is 3, but I don't actually need this. Stokes' theorem would say that $\int_M dx=\int_M df = \int_{\partial M}f=f(1^\bullet)+f(1_\bullet)-f(-1)=1+1-(-1)=3$. This is all fine so far, but now consider the function given by $g(x)=x+10$. Since $dg=dx$, we should have $\int_M dx=\int_M dg=\int_{\partial M}g=g(1^\bullet)+g(1_\bullet)-g(-1)=11+11-9=13$. Contradiction. It's possible to explain this by the nonexistence of flows (instead of $df$, consider the flux of the flow by $\nabla f$). But also note that Stokes' theorem, i.e. homology theory, is founded on a well-defined boundary operation. However, without the Hausdorff condition, open submanifolds do not have unique boundaries, as for example $[-1,0)$ inside $X$, and so we can't break up our manifolds into smaller pieces. We can pass to the Hausdorff-ization as Andrew suggests by identifying $0^\bullet$ with $0_\bullet$, but now we lose additivity. Recall that $M$ was the disjoint union of $A=[-1,0)$ and $B=[0^\bullet,1^\bullet] \cup [0_\bullet,1_\bullet]$. So in the quotient $\partial [A] = [0]-[-1]$ and $\partial [B] = [1^\bullet]-[0]+[1_\bullet]-[0]=[1^\bullet]+[1_\bullet]-2[0]$, which shows that $\partial [M]\neq \partial [A]+\partial [B]$. This is inconsistent with any sort of Stokes formalism. Finally, I'd like to point out that Stokes' theorem aside, even rather nice non-Hausdorff manifolds can be significantly more complicated than we might want to deal with. One nice example is the leaf-space of the foliation of the punctured plane by the level sets of the function $f(x,y)=xy$. The leaf-space looks like the union of the lines $y=x$ and $y=-x$, except that the intersection has been blown up to four points, each of which is dense in this subset. In general, any finite graph can be modeled as a non-Hausdorff 1-manifold by blowing up the vertices, and in higher dimensions the situation is even more confusing. So for any introductory explanation, I would strongly recommend requiring Hausdorff until the students have a lot more intuition about manifolds.
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13,074 |
Just a d*mb question on Lie algebras: Given a Dynkin diagram of a root system (or a Cartan Matrix), how do I know which combination of simple roots are roots? Eg. Consider the root system of G_2, let a be the short root and b be the long one, it is clear that a, b, b+a, b+2a, b+3a are positive roots. But it is not clear to me that 2b+3a is a root just from the Dynkin diagram.
|
Here's an answer in the simply-laced case. Its proof, and generalization to non-simply-laced, are left to the reader. 1) Start with a simple root, and think of it as a labeling of the Dynkin diagram with a 1 there and 0s elsewhere. 2) Look for a vertex whose label is < 1/2 the sum of the surrounding labels. Increment that label. You've found a root! 3) Go back to (2), unless there is no such vertex anymore. You've found the highest root! Take the union over all such games, and you get all the positive roots. Include their negatives, and you have all roots. If you start with a non-Dynkin-diagram, the game doesn't terminate. This is part of a way to classify the Dynkin diagrams. [EDIT: yes it can terminate. There's a variant, where you replace the vertex label by the sum of the surrounding, minus the original label. That game terminates only for Dynkin diagrams.] BTW at the highest root, most of the vertices have labels = 1/2 the surrounding sum. If you put in a new vertex, connected to those vertices with > 1/2 the sum, you get the affine Dynkin diagram.
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13,089 |
I think/hope this is okay for MO. I often find that textbooks provide very little in the way of motivation or context. As a simple example, consider group theory. Every textbook I have seen that talks about groups (including some very basic undergrad level books) presents them as abstract algebraic structures (while providing some examples, of course), then spends a few dozen pages proving theorems, and then maybe in some other section of the book covers some Galois Theory. This really irks me. Personally I find it very difficult to learn a topic with no motivation, partly just because it bores me to death. And of course it is historically backwards; groups arose as people tried to solve problems they were independently interested in. They didn't sit down and prove a pile of theorems about groups and then realize that groups had applications. It's also frustrating because I have to be completely passive; if I don't know what groups are for or why anyone cares about them, all I can do is sit and read as the book throws theorems at me. This is true not just with sweeping big picture issues, but with smaller things too. I remember really struggling to figure out why it was supposed to matter so much which subgroups were closed under conjugation before finally realizing that the real issue was which subgroups can be kernels of homomorphisms, and the other thing is just a handy way to characterize them. So why not define normal subgroups that way, or at least throw in a sentence explaining that that's what we're really after? But no one does. I've heard everyone from freshmen to Fields Medal recipients complain about this, so I know I'm not alone. And yet these kinds of textbooks seem to be the norm. So what I want to know is: Why do authors write books like this? And: How do others handle this situation? Do you just struggle through? Get a different book? Talk to people? (Talking to people isn't really an option for me until Fall...) Some people seem legitimately to be able to absorb mathematics quite well with no context at all. How?
|
By now the advice I give to students in math courses, whether they are math majors or not, is the following: a) The goal is to learn how to do mathematics, not to "know" it. b) Nobody ever learned much about doing something from either lectures or textbooks. The standard examples I always give are basketball and carpentry. Why is mathematics any different? c) Lectures and textbooks serve an extremely important purpose: They show you what you need to learn. From them you learn what you need to learn. d) Based on my own experience as both a student and a teacher, I have come to the conclusion that the best way to learn is through "guided struggle". You have to do the work yourself, but you need someone else there to either help you over obstacles you can't get around despite a lot of effort or provide you with some critical knowledge (usually the right perspective but sometimes a clever trick) you are missing. Without the prior effort by the student, the knowledge supplied by a teacher has much less impact. A substitute for a teacher like that is a working group of students who are all struggling through the same material. When I was a graduate student, we had a wonderful working seminar on Sunday mornings with bagels and cream cheese, where I learned a lot about differential geometry and Lie groups with my classmates. ADDED: So how do you learn from a book? I can't speak for others, but I have never been able to read a math book forwards. I always read backwards. I always try to find a conclusion (a cool definition or theorem) that I really want to understand. Then I start working backwards and try to read the minimum possible to understand the desired conclusion. Also, I guess I have attention deficit disorder, because I rarely read straight through an entire proof or definition. I try to read the minimum possible that's enough to give me the idea of what's going on and then I try to fill the details myself. I'd rather spend my time writing my own definition or proof and doing my own calculations than reading what someone else wrote. The honest and embarrassing truth is that I fall asleep when I read math papers and books. What often happens is that as I'm trying to read someone else's proof I ask myself, "Why are they doing this in such a complicated way? Why couldn't you just....?" I then stop reading and try to do it the easier way. Occasionally, I actually succeed. More often, I develop a greater appreciation for the obstacles and become better motivated to read more. WHAT'S THE POINT OF ALL THIS? I don't think the solution is changing how math books are written. I actually prefer them to be terse and to the point. I fully agree that students should know more about the background and motivation of what they are learning. It annoys me that math students learn about calculus without understanding its real purpose in life or that math graduate students learn symplectic geometry without knowing anything about Hamiltonian mechanics. But it's not clear to me that it is the job of a single textbook to provide all this context for a given subject. I do think that your average math book tries to cover too many different things. I think each math book should be relatively short and focus on one narrowly and clearly defined story. I believe if you do that, it will be easier to students to read more different math books.
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13,106 |
I've attempted going past basic number theory several times, and always got lost in its vastness. Do any of you, perhaps, know a good review that pieces together the many concepts involved (Hecke algebras, SL 2 (ℤ), Fuchsian groups, L-functions, Tate's thesis, Ray class groups, Langlands program, Fourier analysis on number fields, cohomological versions of CFT, Iwasawa theory, modular forms, ...)? Thanks.
|
The book you are looking for exists!! And indeed it contains ALL the buzzwords in your question! It is Manin/Panchishkin's "Introduction to Modern Number Theory". This is a survey book that starts with no prerequisites, contains very few proofs, but nicely explains the statements of central theorems and the notions occurring therein and gives motivations for the questions that are being pursued. You should take a look, at least it can help you decide what you want to study in more detail.
|
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13,130 |
The analytic continuation and functional equation for the Riemann zeta function were proved in Riemann's 1859 memoir "On the number of primes less than a given magnitude." What is the earliest reference for the analytic continuation and functional equation of Dirichlet L-functions? Who first proposed that they might satisfy a Riemann hypothesis? Dirichlet did none of these things; his paper dates from 1837, and as far as I know he only considered his L-functions as functions of a real variable.
|
Riemann was the first person who brought complex analysis into the game, but if you ask just about functional equations then he was not the first. In the 1840s, there were proofs of the functional equation for the $L$-function of the nontrivial character mod 4, relating values at $s$ and $1-s$ for real $s$ between 0 and 1, where the $L$-function is defined by its Dirichlet series. In particular, this happened before Riemann's work on the zeta-function. The proofs were due independently to Malmsten and Schlomilch. Eisenstein had a proof as well (unpublished) which was found in his copy of Gauss' Disquisitiones. It involves Poisson summation. Eisenstein's proof is dated 1849 and Weil suggested that this might have motivated Riemann in his work on the zeta-function. For more on Eisenstein's proof, see Weil's "On Eisenstein's Copy of the Disquisitiones" pp. 463--469 of "Alg. Number Theory in honor of K. Iwasawa" Academic Press, Boston, 1989.
|
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|
13,167 |
In Lecture Notes in Mathematics 1488, Lawvere writes the introduction to the Proceedings for a 1990 conference in Como. In this article, Lawvere, the inventor of Toposes and Algebraic Theories, discusses two ancient philosophical "categories": that of BEING and that of BECOMING. And he's serious. While some of the motivation for this article is to understand these two ancient even mystical topics, the actual content is almost purely mathematical. Lawvere makes definitions and claims in the manner of a serious mathematician engaged in deliberate but casual explanation of ideas. I want to understand this article, but it is difficult. The definitions seem to be written for someone with a bit more background or expertise on topos theory and its application. Q: I am writing to ask whether anyone here has read or understood this article (or parts of it). I'm interested in your thoughts on it. Have the ideas been written formally?
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The notion of a "category of Being" that Lawvere discusses there is the notion that more recently he has been calling a category of cohesion . I'll try to illuminate a bit what's going on . I'll restrict to the case that the category is a topos and say cohesive topos for short. This is a topos that satisfies a small collection of simple but powerful axioms that are supposed to ensure that its objects may consistently be thought of as geometric spaces built out of points that are equipped with "cohesive" structure (for instance topological structure, or smooth structure, etc.). So the idea is to axiomatize big toposes in which geometry may take place. Further details and references can be found here: http://nlab.mathforge.org/nlab/show/cohesive+topos . Let's walk through the article: One axiom on a cohesive topos $\mathcal{E}$ is that the global section geometric morphism $\Gamma : \mathcal{E} \to \mathcal{S}$ to the given base topos $\mathcal{S}$ has a further left adjoint $\Pi_0 := \Gamma_! : \mathcal{E} \to \mathcal{S}$ to its inverse image $\Gamma^{\ast}$, which I'll write $\mathrm{Disc} := \Gamma^{\ast}$, for reasons discussed below. This extra left adjoint has the interpretation that it sends any object $X$ to the set $\Pi_0(X)$ "of connected components". What Lawvere calls a connected object in the article (p. 4) is hence one that is sent by $\Pi_0$ to the terminal object. Another axiom is that $\Pi_0$ preserves finite products. This implies by the above that the collection of connected objects is closed under finite products. This appears on page 6. What he mentions there with reference to Hurewicz is that given a topos with such $\Pi_0$, it becomes canonically enriched over the base topos in a second way, a geometric way. I believe that this, like various other aspects of cohesive toposes, lives up to its full relevance as we make the evident step to cohesive $\infty$-toposes. More details on this are here http://nlab.mathforge.org/nlab/show/cohesive+(infinity,1)-topos (But notice that this, while inspired by Lawvere, is not due to him.) In this more encompassing context the extra left adjoint $\Pi_0$ becomes $\Pi_\infty$ which I just write $\Pi$: it sends, one can show, any object to its geometric fundamental $\infty$-groupoid, for a notion of geometric paths intrinsic to the $\infty$-topos. The fact that this preserves finite products then says that there is a notion of concordance of principal $\infty$-bundles in the $\infty$-topos. The next axiom on a cohesive topos says that there is also a further right adjoint $\mathrm{coDisc} := \Gamma^! : \mathcal{S} \to \mathcal{E}$ to the global section functor. This makes in total an adjoint quadruple $$
(\Pi_0 \dashv \mathrm{Disc} \dashv \Gamma \dashv \mathrm{coDisc}) :=
(\Gamma_! \dashv \Gamma^* \dashv \Gamma_* \dashv \Gamma^!) : \mathcal{E} \to \mathcal{S}
$$ and another axiom requires that both $\mathrm{Disc}$ as well as $\mathrm{coDisc}$ are full and faithful. This is what Lawvere is talking about from the bottom of p. 12 on. The downward functor that he mentions is $\Gamma : \mathcal{E} \to \mathcal{S}$. This has the interpretation of sending a cohesive space to its underlying set of points, as seen by the base topos $\mathcal{S}$. The left and right adjoint inclusions to this are $\mathrm{Disc}$ and $\mathrm{coDisc}$. These have the interpretation of sending a set of points to the corresponding space equipped with either discrete cohesion or codiscrete (indiscrete) cohesion . For instance in the case that cohesive structure is topological structure, this will be the discrete topology and the indiscrete topology, respectively, on a given set. Being full and faithful, $\mathrm{Disc}$ and $\mathrm{coDisc}$ hence make $\mathcal{S}$ a subcategory of $\mathcal{E}$ in two ways (p. 7), though only the image of $\mathrm{coDisc}$ will also be a subtopos, as he mentions on page 7. (This has, by the way, an important implication that Lawvere does not seem to mention: it implies that we are entitled to the corresponding quasi-topos of separated bipresheaves, induced by the second topology that is induced by the sub-topos. That, one can show, may be identified with the collection of concrete sheaves, hence concrete cohesive spaces (those whose cohesion is indeed supported on their points). In the case of the cohesive topos for differential geometry, the concrete objects in this sense are precisely the diffeological spaces . ) He calls the subtopos given by the image of $\mathrm{coDisc} : \mathcal{S} \to \mathcal{E}$ that of "pure Becoming" further down on p. 7, whereas the subcategory of discrete objects he calls that of "non Becoming". The way I understand this terminology (which may not be quite what he means) is this: whereas any old $\infty$-topos is a collection of spaces with structure , a cohesive $\infty$-topos comes with the extra adjoint $\Pi$, which I said has the interpretation of sending any space to its path $\infty$-groupoid. Therefore there is an intrinsic notion of geometric paths in any cohesive $\infty$-topos. This allows notably to define parallel transport along paths and higher paths, hence a kind of dynamics . In fact there is differential cohomology in every cohesive $\infty$-topos. Now, in a discrete object there are no non-trivial paths (formally because $\Pi \; \mathrm{Disc} \simeq \mathrm{Id}$ by the fact that $\mathrm{Disc}$ is full and faithful), so there is "no dynamics" in a discrete object hence "no becoming", if you wish. Conversely in a codiscrete object every sequence of points whatsoever counts as a path, hence the distinction between the space and its "dynamics" disappears and so we have "pure becoming", if you wish. Onwards. Notice next that every adjoint triple induces an adjoint pair of a comonad and a monad. In the present situation we get $$
(\mathrm{Disc} \;\Gamma \dashv \mathrm{coDisc}\; \Gamma) : \mathcal{E} \to \mathcal{E}
$$ This is what Lawvere calls the skeleton and the coskeleton on p. 7. In the $\infty$-topos context the left adjoint $\mathbf{\flat} := \mathrm{Disc} \; \Gamma$ has the interpretation of sending any object $A$ to the coefficient for cohomology of local systems with coefficients in $A$. The paragraph wrapping from page 7 to 8 comments on the possibility that the base topos $\mathcal{S}$ is not just that of sets, but something richer. An example of this that I am kind of fond of is that of super cohesion (in the sense of superalgebra and supergeometry): the topos of smooth super-geometry is cohesive over the base topos of bare super-sets. What follows on page 9 are thoughts of which I am not aware that Lawvere has later formalized them further. But then on the bottom of p. 9 he gets to the axiomatic identification of infinitesimal or formal spaces in the cohesive topos. In his most recent article on this what he says here on p. 9 is formalized as follows: he says an object $X \in \mathcal{E}$ is infinitesimal if the canonical morphism $\Gamma X \to \Pi_0 X$ is an isomorphism. To see what this means, suppose that $\Pi_0 X = *$, hence that $X$ is connected. Then the isomorphism condition means that $X$ has exactly one global point. But $X$ may be bigger: it may be a formal neighbourhood of that point, for instance it may be $\mathrm{Spec} \;k[x]/(x^2)$. A general $X$ for which $\Gamma X \to \Pi_0 X$ is an iso is hence a disjoint union of formal neighbourhoods of points. Again, the meaning of this becomes more pronounced in the context of cohesive $\infty$-toposes: there objects $X$ for which $\Gamma X \simeq * \simeq \Pi X$ have the interpretation of being formal $\infty$-groupoids , for instance formally exponentiated $L_\infty$-algebras. And so there is $\infty$-Lie theory canonically in every cohesive $\infty$-topos. I'll stop here. I have more discussion of all this at: http://nlab.mathforge.org/schreiber/show/differential+cohomology+in+a+cohesive+topos
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|
13,205 |
One often hears in popular explanations of the failure to find a "Grand Unified Theory" that "Gravity goes off to infinity, but cutting off the edges gives us wrong answers", and other similar mathematically vague statements. Clearly, this issue has some kind of mathematical explanation, but I'm not really qualified to read the corresponding physics work, so I'm wondering what actually fails mathematically.
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Other people have said that the problem is that GR isn't renormalizable. I want to explain what that means in measure-theoretic terms. What I say won't be 100% rigorous, but it should get the general story across. Quantum field theories are generally defined using a Feynman path integral measure. This measn that you compute correlation functions of observables by summing over all histories of your system, weighting each history by e^{-S}, where S is an functional on the space of fields, called the action. In a field theory, these histories are functions on some spacetime manifold. Just as you define the ordinary integral as a limit of Riemann sums, you define the Feynman path integral as a limit of "regularized" path integral measures. There are a lot of ways to do this; one of the most popular is the lattice regularization. We choose a finite set of points in spacetime, living on a lattice whose nearest neighbors are a distance a apart. Then we choose a "microscopic" action on our space of fields and discretize it, replacing derivatives with finite difference quotients, and approximate the path integral's weighted sum over histories with a sum over functions defined on the lattice. Let's grant for a moment that, for any fixed lattice spacing a, this procedure defines a quantum theory, meaning that you can use the moments of the measure to compute correlation functions for your observables. Let's also grant that the expectation values of observables satisfy classical equations of motion (so that when we approach the classical limits, and the probability distributions become concentrated at their expected value, we get deterministic evolution of these values). We usually want to take a continuum limit, making the lattice spacing smaller and smaller. If we do this, while keeping fixed the coupling constants in the microscopic action we used to define our measure, in most cases, we run into a problem: the coupling constants in the classical equations of motion depend on the lattice spacing a, and become infinite as a goes to zero. So that's probably not the limit we want. And indeed, in most circumstances, we know the classical physics, and are trying to find a quantum theory that reproduces it. So we'll try something else: make the coupling constants in the microscopic action depend on the lattice spacing, and hope that we can tune them in a way that keeps the classical physics fixed. Sometimes this works; sometimes it doesn't. It can happen that there isn't any way of tuning the coupling constants so as to reproduce a nice classical limit; when this happens, the theory is said to be non-renormalizable. This appears to be what happens in General Relativity, if you use the discretized Einstein-Hilbert action as your microscopic action.
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|
13,320 |
Since this forum is densely populated with algebraists, I think I'll ask it here. I'm teaching intermediate level algebra this semester and I'd like to entertain my students with some clever applications of group theory. So, I'm looking for problems satisfying the following 4 conditions 1) It should be stated in the language having nothing whatsoever to do with groups/rings/other algebraic notions. 2) It should have a slick easy to explain (but not necessarily easy to guess) solution using finite (preferrably non-abelian) groups. 3) It shouldn't have an obvious alternative elementary solution (non-obvious alternative elementary solutions are OK). 4) It should look "cute" to an average student (or, at least, to a person who is curious about mathematics but has no formal education). An example I know that, in my opinion, satisfies all 4 conditions is the problem of tiling a given region with given polyomino (with the solution that the boundary word should be the identity element for the tiling to be possible and various examples when it is not but the trivial area considerations and standard colorings do not show it immediately) I'm making it community wiki but, of course, you are more than welcome to submit more than one problem per post. Thanks in advance!
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In the TV show "Futurama", there's an episode named "The Prisoner of Benda" in which two of the characters swap bodies using a machine with a fundamental flaw: no two people can use it to swap bodies twice. This means they can't simply use the machine again to swap back. They spend the rest of the episode trying to return to their original bodies. Eventually, a scientist (who also happens to be a Harlem globetrotter) figures out how to help them using group theory. With a little reflection, one can see that this problem can be recast into one about the symmetric group.
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|
13,322 |
A very important theorem in linear algebra that is rarely taught is: A vector space has the same dimension as its dual if and only if it is finite dimensional. I have seen a total of one proof of this claim, in Jacobson's "Lectures in Abstract Algebra II: Linear Algebra". The proof is fairly difficult and requires some really messy arguments about cardinality using, if I remember correctly, infinite sequences to represent $\mathbb{N}\times\mathbb{N}$ matrices. Has anyone come up with a better argument in the 57 years since Jacobson's book was published, or is the noted proof still the only way to prove this fact? Edit: For reference, the proof is on pages 244-248 of Jacobson's Lectures in Abstract Algebra: II. Linear Algebra .
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Here is a simple proof I thought, tell me if anything is wrong. First claim. Let $k$ be a field, $V$ a vector space of dimension at least the cardinality of $k$ and infinite. Then $\operatorname{dim}V^{*} >\operatorname{dim}V$. Indeed let $E$ be a basis for $V$. Elements of V* correspond bijectively to functions from $E$ to $k$, while elements of $V$ correspond to such functions with finite support. So the cardinality of $V^{*}$ is $k^E$, while that of $V$ is, if I'm not wrong, equal to that of $E$ (in this first step I am assuming $\operatorname{card} k \le \operatorname{card} E$). Indeed $V$ is a union parametrized by $\mathbb{N}$ of sets of cardinality equal to $E$. In particular $\operatorname{card} V < \operatorname{card} V^{*}$, so the same inequality holds for the dimensions. Second claim. Let $h \subset k$ two fields. If the thesis holds for vector spaces on $h$, then it holds for vector spaces on $k$. Indeed let $V$ be a vector space over $k$, $E$ a basis. Functions with finite support from $E$ to $h$ form a vector space $W$ over $h$ such that $V$ is isomorphic to the extension of $W$, i.e. to $W\otimes_h k$. Every functional from $W$ to $h$ extends to a functional from $V$ to $k$, hence $$\operatorname{dim}_k V = \operatorname{dim}_h W < \operatorname{dim}_h W^* \leq \operatorname{dim}_k V^*.$$ Putting the two claims together and using the fact that every field contains a field at most denumerable yields the thesis.
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13,428 |
Kidding, kidding. But I do have a question about an $n$-line outline of a proof of the first case of FLT, with $n$ relatively small. Here's a result of Eichler (remark after Theorem 6.23 in Washington's Cyclotomic Fields): If $p$ is prime and the $p$-rank of the class group of $\mathbb{Q}(\zeta_p)$ satisfies $d_p<\sqrt{p}-2$, then the first case of FLT has no non-trivial solutions. Once you know Herbrand-Ribet and related stuff, the proof of this result is even rather elementary. The condition that $d_p<\sqrt{p}-2$ seems reminiscent of rank bounds used with Golod-Shafarevich to prove class field towers infinite. More specifically, a possibly slightly off (and definitely improvable) napkin calculation gives me that for $d_p>2+2\sqrt{(p-1)/2}$, the $p$-th cyclotomic field $\mathbb{Q}(\zeta_p)$ has an infinite $p$-class field tower. It's probably worth emphasizing at this point that by the recent calculation of Buhler and Harvey, the largest index of irregularity for primes less than 163 million is a paltry 7. So it seems natural to me to conjecture, or at least wonder about, a relationship between the unsolvability of the first case of FLT and the finiteness of the $p$-class field tower over $\mathbb{Q}(\zeta_p)$. Particularly compelling for me is the observation that regular primes (i.e., primes for which $d_p=0$) are precisely the primes for which this tower has length 0, and have obvious historical significance in the solution of this problem. In fact, the mechanics of the proof would probably/hopefully be to lift the arithmetic to the top of the (assumed finite) p-Hilbert class field tower, and then use that its class number is prime to $p$ to make arguments completely analogously to the regular prime case. I haven't seen this approach anywhere. Does anyone know if it's been tried and what the major obstacles are, or demonstrated why it's likely to fail? Or maybe it works, and I just don't know about it? Edit : Franz's answer indicates that even for a relatively simple Diophantine equation (and relatively simple class field tower), moving to the top of the tower introduces as many problems as it rectifies. This seems pretty compelling. But if anyone has any more information, I'd still like to know if anyone knows or can come up with an example of a Diophantine equation which does benefit from this approach.
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Suppose you have a diophantine problem whose solution is connected with the structure of the p-class group of a number field K. Then you have the following options: Use ideal arithmetic in the maximal order O K Replace O K by a suitable ring of S-integers with trivial p-class group Replace K by the Hilbert class field, which (perhaps) has trivial p-class group. Experience with descent on elliptic curves has shown me that ultimately, the equations you have to solve in methods 1 and 2 are the same; moreover, the approach using ideals is a lot less technical than using factorial domains in S-integers (the class group relations come back in through the larger rank of the group of S-units). I am certain that the route via the Hilbert class field is even more technical: again, the unit group in the class field will produce more difficulties than a trivial class group will eliminate. Edit. As an example illustrating my point in a very simple example, let me solve the diophantine equation $x^2 + 5y^2 = z^2$ in several different ways. I will always assume that $\gcd(x,y) = 1$. 1. Elementary Number Theory The basic idea is factoring: from $5y^2 = (z+x)(z-x)$. Since $d = \gcd(z-x,z+x) = \gcd(2z,z+x) \mid 2$ we have $d = 1$ or $d = 2$; moreover we clearly have $z-x > 0$.
This gives $z+x = da^2$, $z-x = 5db^2$ or $z+x = 5da^2$, $z-x = db^2$. Solving for $x$ and $z$ yields
$$ x = \pm \frac d2 (a^2 - 5b^2), \quad y = dab. $$ 2. Parametrization Set $X = \frac xz$ and $Y = \frac yz$; then $X^2 + 5Y^2 = 1$. Take the line $Y = t(X+1)$ through the obvious point $(-1,0)$; the second point of intersection is given by
$$ X = \frac{1-5t^2}{1+5t^2}, \quad Y = \frac{2t}{1+5t^2}. $$
Dehomogenizing using $t = \frac ba$ and $X = \frac xz$ etc. gives
the projective parametrization
$$ (x:y:z) = (a^2-5b^2:2ab:a^2+5b^2). $$
If $ab$ is odd, all coordinates are even, and we find
$$ x = \frac12(a^2 - 5b^2), \quad y = ab; $$
if $a$ or $b$ is even we get
$$ x = a^2 - 5b^2, \quad y = 2ab $$
as above. 3. Algebraic Number Theory Consider the factorization
$$ (x + y\sqrt{-5}\,)(x - y\sqrt{-5}\,) = z^2 $$
in the ring of integers of the number field $K = {\mathbb Q}(\sqrt{-5}\,)$.
The class number of $K$ is $2$, and the ideal class is generated by
the prime ideal ${\mathfrak p} = (2,1+\sqrt{-5}\,)$. The ideal $(x + y\sqrt{-5}, x - y\sqrt{-5}\,)$ is either $(1)$ or
${\mathfrak p}$; thus
$$ (x + y\sqrt{-5}\,) = {\mathfrak a}^2, \quad (x - y\sqrt{-5}\,) =
{\mathfrak b}^2 $$
in the first and
$$ (x + y\sqrt{-5}\,) = {\mathfrak p}{\mathfrak a}^2, \quad
(x - y\sqrt{-5}\,) = {\mathfrak p}{\mathfrak b}^2 $$
in the second case. The second case is impossible since the left hand side as well as
${\mathfrak a}^2$ are principal, but ${\mathfrak p}$ is not. We
could have seen this immediately since $x$ and $y$ cannot both be odd. In the first case, assume first that ${\mathfrak a} = (a + b\sqrt{-5}\,)$
is principal. Since the only units in ${\mathcal O}_K$ are $\pm 1$,
this gives $x + y \sqrt{-5} = \pm(a+b\sqrt{-5}\,)^2$ and hence
$$ x = \pm (a^2 - 5b^2), \quad y = \pm 2ab. $$
If ${\mathfrak a}$ is not principal, then
${\mathfrak p}{\mathfrak a} = (a+b\sqrt{-5}\,)$ is,
and from $({\mathfrak p}{\mathfrak a})^2 = 2(x+y\sqrt{-5}\,)$ we
similarly get
$$ x = \pm \frac12(a^2 - 5b^2), \quad y = \pm ab. $$ 4. S-Integers The ring $R = {\mathbb Z}[\sqrt{-5}\,]$ is not a UFD, but $S = R[\frac12]$ is;
in fact, $S$ is even norm-Euclidean for the usual norm in $S$
(the norm is the same as in $R$ except that powers of $2$ are dropped).
It is also easily seen that $S^\times = \langle -1, 2 \rangle$. From
$$ (x + y\sqrt{-5}\,)(x - y\sqrt{-5}\,) = z^2 $$
and the fact that the factors on the left hand side are
coprime we deduce that $x + y\sqrt{-5} = \varepsilon \alpha^2$ for some unit
$\varepsilon \in S^\times$ and some $\alpha \in S$. Subsuming squares into
$\alpha$ we may assume that $\varepsilon \in \{\pm 1, \pm 2\}$. Setting
$\alpha = \frac{a + b\sqrt{-5}}{2^t}$, where we may assume that $a$
and $b$ are not both even, we get
$$ x + y \sqrt{-5} = \varepsilon \frac{a^2 - 5b^2 + 2ab\sqrt{-5}}{2^{2t}}. $$
It is easily seen that we must have $t = 0$ and $\varepsilon = \pm 1$ or
$t = 1$ and $\varepsilon = \pm 2$; a simple calculation then yields the
same formulas as above. 5. Hilbert Class Fields The Hilbert class field of $K$ is given by $L = K(i)$. It is not
too difficult to show that $L$ has class number $1$ (actually it is
norm-Euclidean), and that its unit group is generated by $i = \sqrt{-1}$
and $\omega = \frac{1+\sqrt{5}}2$ (we only need to know that these
units and their product are not squares). From
$$ (x + y\sqrt{-5}\,)(x - y\sqrt{-5}\,) = z^2 $$
and the fact
that the factors on the left hand side are coprime in ${\mathcal O}_K$
we deduce that $x + y \sqrt{-5} = \varepsilon \alpha^2$. Subsuming
squares into $\alpha^2$ we may assume that
$\varepsilon \in \{1, i, \omega, i\omega \}$. Applying the nontrivial
automorphism of $L/K$ to $x + y \sqrt{-5} = \varepsilon \alpha^2$ we find
$\varepsilon \alpha^2 = \varepsilon' {\alpha'}^2$. Since the ideal
${\mathfrak a} = (\alpha)$ is fixed and since $L/K$ is unramified,
the ideal ${\mathfrak a}$ must be an ideal in ${\mathcal O}_K$.
Thus either ${\mathfrak a} = (a+b\sqrt{-5}\,)$ is principal in $K$,
or ${\mathfrak p} {\mathfrak a} = (a+b\sqrt{-5}\,)$ is; in the second
case we observe
that ${\mathfrak p} = (1+i)$ becomes principal in ${\mathcal O}_L$. Thus either
$$ x + y \sqrt{-5} = (a+b\sqrt{-5}\,)^2 \quad \text{or} \quad
x + y \sqrt{-5} = i \Big(\frac{a+b\sqrt{-5}}{1+i}\,\Big)^2, $$
giving us the same formulas as above. Avoiding ideal arithmetic in $K$ and only using the fact that
${\mathcal O}_L$ is a UFD seems to complicate the proof even more. Edit 2 For good measure . . . 6. Hilbert 90 Consider, as above, the equation $X^2 + 5Y^2 = 1$.
It shows that the element $X + Y \sqrt{-5}$ has norm $1$;
by Hilbert 90, we must have
$$ X + Y \sqrt{-5} = \frac{a+b\sqrt{-5}}{a-b\sqrt{-5}}
= \frac{a^2 - 5b^2 + 2ab\sqrt{-5}}{a^2 + 5b^2}. $$
Dehomogenizing via $X = \frac xz$ and $Y = \frac yz$ yields the same
projective parametrization as above, and we end up with the
familiar formulas. 7. Binary Quadratic Forms The equation $x^2 + 5y^2 = z^2$ tells us that the form $Q_0(X,Y) = X^2 + 5Y^2$
with fundamental discriminant $\Delta = -20$ represents a square;
this implies that $Q_0$ lies in the principal genus (which is trivial
since $Q_0$ is the principal form), and that the representations of
$z^2$ by $Q_0$ come from composing representations of $z$ by forms
$Q_1$ with $Q_1^2 \sim Q_0$ with themselves. There are only two forms with discriminant $\Delta$ whose square is
equivalent to $Q_0$: the principal form $Q_0$ itself and the form
$Q_1(X,Y) = 2X^2 + 2XY + 3Y^2$. Thus either
$$ z = Q_0(a,b) = a^2 + 5b^2 \quad \text{or} \quad
z = Q_1(a,b) = 2a^2 + 2ab + 3b^2. $$
The formulas for Gauss composition of forms immediately provide us with
expressions for $x$ and $y$ in terms of $a$ and $b$, but they can also
be checked easily by hand. In the first case, we get
$$ x^2 + 5y^2 = (a^2 + 5b^2)^2 = (a^2 - 5b^2)^2 + 5(2ab)^2, $$
and in the second case we can reduce the equations to this one
by observing that $2Q_1(a,b) = A^2 + 5b^2$ with $A = 2a+b$, which gives
$$ x^2 + 5y^2 = \frac14\Big(A^2 + 5b^2\Big)^2
= \Big(\frac{A^2 - 5b^2}2\Big)^2 + 5(Ab)^2. $$
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|
13,526 |
This afternoon I was speaking with some graduate students in the department and we came to the following quandary; Is there a geometric interpretation of the trace of a matrix? This question should make fair sense because trace is coordinate independent. A few other comments. We were hoping for something like: "determinant is the volume of the parallelepiped spanned by column vectors." This is nice because it captures the geometry simply, and it holds for any old set of vectors over $\mathbb{R}^n$ . The divergence application of trace is somewhat interesting, but again, not really what we are looking for. Also, after looking at the wiki entry, I don't get it. This then requires a matrix function, and I still don't really see the relationship. One last thing that we came up with; the trace of a matrix is the same as the sum of the eigenvalues. Since eigenvalues can be seen as the eccentricity of ellipse, trace may correspond geometrically to this. But we could not make sense of this.
|
If your matrix is geometrically projection (algebraically $A^2=A$) then the trace is the dimension of the space that is being projected onto. This is quite important in representation theory.
|
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|
13,555 |
I want to know if there's any book that categorizes problems by subjects of Functional Analysis. I'm studying Functional Analysis now a days and I really need to solve some problems in order to assure myself that I've really understood the concepts and definitions. For example: problems related to the Hahn-Banach theorem or Banach Spaces or Hilbert Spaces or related subjects.
|
Another classical book is Theorems and problems in functional analysis by Kirillov and Gvishiani.
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13,619 |
How could a free (i.e. free content) alternative for MathSciNet and Zentralblatt be created? Comments Some mathematicians have stopped writing reviews for MathSciNet because they feel their output should be freely available. (The Pricing for MathSciNet is not high, but it is not the point.) Two related questions: Are there any good websites... and Errata database? ;
see also r-forum , nLab and wikademic . What can be done (based on answers below) One thing that can be really useful and doable is to create (and maintain) a database of articles (and maybe abstracts), where you can find all the articles that were referring to a given one. Once it is done we can add lists of errors --- it will add something new and valuable for the project (but this will take a while). The above two things might be already enough for practical purpose. It will be even better if it will attract enough reviewers to the project.
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Everyone I know in the AMS would like to make MR/MathSci free, but the problem is that it costs millions of dollars to produce and maintain (it requires a large staff in Ann Arbor and elsewhere, including many mathematicians), and no one has managed to find any other way to pay for it*. This is certainly something the mathematicians in the AMS are aware of and have thought about. The AMS attempts to make it as widely available as possible given the constraint that it has to be paid for. As far as I know, the revenue from MR/MathSci only pays to support it, not any of the AMS's other activities [not so; see below.] Posters don't seem to realize the huge effort that goes into maintaining a project like this (for example, every article has to be assigned to a reviewer, and every review has to edited). Certainly, I don't believe a free alternative would be able to come anywhere near the quality MathSci maintains, so my answer is no, a free alternative to MR/MathSci is not possible. Perhaps free supplements to MathSci could be useful, but anything that drew potential reviewers away from MR/MathSci would harm, not help, what is an extremely valuable resource. *Of course, the intelligent thing would be for the funding agencies in the wealthy countries (US,EU,Japan,...) to pay for MR/MathSci directly, so that it could be distributed freely, but getting them to do this seems to be hopeless. Added: In response to Anton Petrunin's comment, here are some numbers. The AMS employs 15 mathematical editors (i.e., mathematicians) and a total staff of over 70 at Mathematical Reviews (in Ann Arbor). The total direct cost of producing MR/MathSci in 2008 was 6,569,000USD. However, contrary to what I thought, the AMS does use the revenue from MR/MathSci and its other publications to support a large part of its other activities (Member and professional services, general and administrative expenses). In 2008, about 24% of total publication revenue was used in this way. See 2008 report and ad .
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|
13,638 |
I consider a game to be mathematical if there is interesting mathematics (to a mathematician) involved in the game's structure, optimal strategies, practical strategies, analysis of the game results/performance. Which popular games are particularly mathematical by this definition? Motivation : I got into backgammon a bit over 10 years ago after overhearing Rob Kirby say to another mathematician at MSRI that he thought backgammon was a game worth studying. Since then, I have written over 100 articles on the mathematics of backgammon as a columnist for a backgammon magazine. My target audience is backgammon players, not mathematicians, so much of the material I cover is not mathematically interesting to a mathematician. However, I have been able to include topics such as martingale decomposition, deconvolution, divergent series, first passage times, stable distributions, stochastic differential equations, the reflection principle in combinatorics, asymptotic behavior of recurrences, $\chi^2$ statistical analysis, variance reduction in Monte Carlo simulations, etc. I have also made a few videos for a poker instruction site, and I am collaborating on a book on practical applications of mathematics to poker aimed at poker players. I would like to know which other games can be used similarly as a way to popularize mathematics, and which games I am likely to appreciate more as a mathematician than the general population will. Other examples: go bridge Set. Non-example : I do not believe chess is mathematical, despite the popular conception that chess and mathematics are related. Game theory says almost nothing about chess. The rules seem mathematically arbitrary. Most of the analysis in chess is mathematically meaningless, since positions are won, drawn, or tied (some minor complications can occur with the 50 move rule), and yet chess players distinguish strong moves from even stronger moves, and usually can't determine the true value of a position. To me, the most mathematical aspect of chess is that the linear evaluation of piece strength is highly correlated which side can win in the end game. Second, there is a logarithmic rating system in which all chess players say they are underrated by 150 points. (Not all games have good rating systems.) However, these are not enough for me to consider chess to be mathematical. I can't imagine writing many columns on the mathematics of chess aimed at chess players. Non-example : I would exclude Nim. Nim has a highly mathematical structure and optimal strategy, but I do not consider it a popular game since I don't know people who actually play Nim for fun. To clarify, I want the games as played to be mathematical. It does not count if there are mathematical puzzles you can describe on the same board. Does being a mathematician help you to learn the game faster, to play the game better, or to analyze the game more accurately? (As opposed to a smart philosopher or engineer...) If mathematics helps significantly in a game people actually play, particularly if interesting mathematics is involved in a surprising way, then it qualifies to be in this collection. If my criteria seem horribly arbitrary as some have commented, so be it, but this seems in line with questions like Real world applications of math, by arxive subject area? or Cocktail party math . I'm asking for applications of real mathematics to games people actually play. If someone is unconvinced that mathematics says anything they care about, and you find out he plays go, then you can describe something he might appreciate if you understand combinatorial game theory and how this arises naturally in go endgames.
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Set is a card game that is very mathematical. Set is played with a deck with $81$ cards. Each card corresponds to a point in affine $4$ -space over $\mathbb Z/3$ , with $3$ possible colors, shadings, shapes, and counts. The players must identify Sets, sets of $3$ cards corresponding to collinear points. Sets are also triples of cards which add up to the $0$ -vector . The three cards pictured form a Set. A natural question which arises during play is how many cards you can deal out without producing a Set. There can be $9$ cards in a codimension- $1$ subspace which do not contain a Set, corresponding to a nondegenerate conic in affine $3$ -space such as $z=x^2+y^2$ . There can be at most $20$ cards not containing a Set , corresponding to a nondegenerate conic in the projective $3$ -space containing $10$ points. Lest anyone think it is "just a game", this has led to quite a bit of research , including applications to computational complexity of matrix multiplication.
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13,647 |
This is a very basic question of course, and exposes my serious ignorance of analytic number theory, but what I am looking for is a good intuitive explanation rather than a formal proof (though a sufficiently short formal proof could count as an intuitive explanation). So, for instance, a proof that estimated a contour integral and thereby showed that the number of zeros inside the contour was greater than zero would not count as a reason. A brief glance at Wikipedia suggests that the Hadamard product formula could give a proof: if there are no non-trivial zeros then you get a suspiciously nice formula for ζ(s) itself. But that would feel to me like formal magic. A better bet would probably be Riemann's explicit formula, but that seems to require one to know something about the distribution of primes. Perhaps a combination of the explicit formula and the functional equation would do the trick, but that again leaves me feeling as though something magic has happened. Perhaps magic is needed. A very closely related question is this. Does the existence of non-trivial zeros on the critical strip imply anything about the distribution of prime numbers? I know that it implies that the partial sums of the Möbius and Liouville functions cannot grow too slowly, and it's really this that I want to understand.
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Here's a variation on the arguments using the explicit formula. If there were no zeros in the critical strip then the explicit formula would say $$\psi_0(x) = x - \frac{\zeta'(0)}{\zeta(0)} - \frac12 \log(1-x^{-2}),$$ where $\psi_0(x)$ is the same as the usual $\psi(x)$ except when $x$ takes integer values $\psi_0(x)$ is the average of the left and right limits of $\psi(x)$. The right hand side is continuous for $x >0$ while the left hand side is not. The argument generalizes to show that there cannot be finitely many zeros in the critical strip, or even an infinitude that do not grow "too quickly." Precisely, if $$\sum_{\rho} \frac{x^{\rho}}{\rho}$$ converged fast enough to form a continuous function of $x$ then the same argument would carry over.
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|
13,682 |
I'm planning a course for the general public with the general theme of "Mathematical ideas that have changed history" and I would welcome people's opinions on this topic. What do you think have been the most influential mathematical ideas in terms of what has influenced science/history or changed the way humans think, and why? I won't expect my audience to have any mathematical background other than high-school. My thoughts so far are: non-Euclidean geometry, Cantor's ideas on uncountability, undecidability, chaos theory and fractals, the invention of new number systems (i.e. negative numbers, zero, irrational, imaginary numbers), calculus, graphs and networks, probability theory, Bayesian statistics. My apologies if this has already been discussed in another post.
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Turing's work on computability, extending those of Goedel and the other early logicians, paved the way for the development of modern computers. Before Turing and Goedel, the concept of computability was murky. It was Turing who realized that there could be a universal computer---a computer whose hardware does not have to be separately modified for every change in application. Although we all take this for granted now, as we install various programs on our laptop computers, the mathematical idea of it was and is profound. Turing's early work introduced the formal concept of subroutines in computation, computational languages, and so on, which laid the groundwork for the later development of computers as we know them.
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13,817 |
This is a followup to a previous question What is the right definition of the Picard group of a commutative ring? where I was worried about the distinction between invertible modules and rank one projective modules over an arbitrary commutative ring. I was worrying too much, because of the following theorem [Bourbaki, Commutative Algebra, Section II.5.2, Theorem 1]: Let $R$ be a commutative ring and $M$ a finitely generated $R$-module. The following are equivalent: (i) $M$ is projective. (ii) $M$ is finitely presented and locally free in the weaker sense: $\forall \mathfrak{p} \in \operatorname{Spec}(R), \ M_{\mathfrak{p}} \cong R_{\mathfrak{p}}^{r(\mathfrak{p})}$. (iii) $M$ is locally free in the weaker sense and its rank function $\mathfrak{p} \mapsto r(\mathfrak{p})$ is locally constant on $\operatorname{Spec}(R)$. (iv) $M$ is locally free in the stronger sense: there exist $f_1,\ldots,f_n \in R$, generating the unit ideal, such that for each $i$, $M_{f_i}$ is a free $R_{f_i}$-module. (v) For every maximal ideal $\mathfrak{m}$ of $R$, there exists $f \in R \setminus \mathfrak{m}$ such that $M_f$ is a free $R_f$-module. This answers my previous question, because the rank function of an invertible module is identically one. In order to really feel like I understand what's going on here, I would like to see an example of a finitely generated locally free [in the weaker sense of (ii) above] module which is not projective. Thus $R$ must be non-Noetherian. The wikipedia article on projective modules contains some nice information, in particular sketching an example of such a module over a Boolean ring. For a Boolean ring though the localization at every prime ideal is simply $\mathbb{Z}/2\mathbb{Z}$, so it is not too surprising that there are more locally free modules than projectives. I would like to see an example with $R$ an integral domain, if possible. It would be especially nice if you can give a reference to one of the standard texts on commutative algebra which contains such an example or at least a citation of such an example.
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It is impossible to produce an example of a finitely generated flat $R$-module that is not projective when $R$ is an integral domain. See: Cartier, "Questions de rationalité des diviseurs en géométrie algébrique," here , Appendice, Lemme 5, p. 249. Also see Bourbaki Algèbre Chapitre X (Algèbre Homologique, "AH") X.169 Exercise Sect. 1, No. 13. I also sketch an alternate proof that there are no such examples for $R$ an integral domain below. Observe that, for finitely generated $R$-modules $M$, being locally free in the weaker sense is equivalent to being flat [Bourbaki, AC II.3.4 Pr. 15, combined with AH X.169 Exercise Sect. 1, No. 14(c).]. ($R$ doesn't have to be noetherian for this, though many books seem to assume it.) There's a concrete way to interpret projectivity for finitely generated flat modules. We begin by translating Bourbaki's criterion into the language of invariant factors. For any finitely generated flat $R$-module $M$ and any nonnegative integer $n$, the $n$-th invariant factor $I_n(M)$ is the annihilator of the $n$-th exterior power of $M$. Lemma. (Bourbaki's criterion) A finitely generated flat $R$-module $M$ is projective if and only if, for any nonnegative integer $n$, the set $V(I_n(M))$ is open in $\mathrm{Spec}(R)$. This openness translates to finite generation. Proposition. If $M$ is a finitely generated flat $R$-module, then $M$ is projective iff its invariant factors are finitely generated. Corollary. The following conditions are equivalent for a ring $R$: (1) Every flat cyclic $R$-module is projective. (2) Every finitely generated flat $R$-module is projective. Corollary. Over an integral domain $R$, every finitely generated flat $R$-module is projective. Corollary. A flat ideal $I$ of $R$ is projective iff its annihilator is finitely generated. Example. Let me try to give an example of a principal ideal of a ring $R$ that is locally free in the weak sense but not projective. Of course my point is not the nature of this counterexample itself, but rather the way in which one uses the criteria above to produce it. Let $S:=\bigoplus_{n=1}^{\infty}\mathbf{F}_2$, and let $R=\mathbf{Z}[S]$. (The elements of $R$ are thus expressions $\ell+s$, where $\ell\in\mathbf{Z}$ and $s=(s_1,s_2,\dots)$ of elements of $\mathbf{F}_2$ that eventually stabilize at $0$.) Consider the ideal $I=(2+0)$. I first claim that for any prime ideal $\mathfrak{p}\in\mathrm{Spec}(R)$, the $R_{\mathfrak{p}}$-module $I_{\mathfrak{p}}$ is free of rank $0$ or $1$. There are three cases: (1) If $x\notin\mathfrak{p}$, then $I_{\mathfrak{p}}=R_{\mathfrak{p}}$. (2) If $x\in\mathfrak{p}$ and $\mathfrak{p}$ does not contain $S$, then $I_{\mathfrak{p}}=0$. (3) Finally, if both $x\in\mathfrak{p}$ and $S\subset\mathfrak{p}$, then $I_{\mathfrak{p}}$ is a principal ideal of $R_{\mathfrak{p}}$ with trivial annihilator. It remains to show that $I$ is not projective as an $R$-module. But its annihilator is $S$, which is not finitely generated over $R$. [This answer was reorganized on the recommendation of Pete Clark.]
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13,843 |
We have a natural number $n>1$. We want to determine whether there exist
natural numbers $a, k>1$ such that $n = a^k$. Please suggest a polynomial-time algorithm.
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This can be done in "essentially linear time." Check out Daniel Bernstein's website: http://cr.yp.to/arith.html Especially note his papers labeled [powers] and [powers2].
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13,851 |
I have a slight interest in both the inverse Galois problem and in the Monster group. I learned some time ago that all of the sporadic simple groups, with the exception of the Mathieu group $M_{23}$, have been proven to be Galois groups over $\mathbb{Q}$. In particular, the Monster group has been proven to be a Galois group over $\mathbb{Q}$. What techniques are used to prove such an assertion? Is proving that $M_{23}$ is also Galois over $\mathbb{Q}$ within reach? I assume that the same techniques do not apply, for it is a much more manageable group than the Monster.
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the monster is a nice example of how the so-called rigidity method for the inverse Galois problem works. There is a lot of beautiful mathematics behind this, I will sketch the different steps. A general remark: it is known that every profinite group, i.e. every group which could be a Galois group of some field extension, is indeed the Galois group of some Galois extension. This is still quite elementary (a result of Leptin, also proved by Waterhouse). To make the inverse Galois question more interesting, we should therefore consider a fixed base field $K$ . We can't hope that any profinite group will still be a Galois group over $K$ - every Galois group over $K$ is a quotient of the absolute Galois group of $K$ , and therefore the cardinality of a Galois group over $K$ is bounded from above, whereas it is easy to see that there are profinite groups which are "strictly bigger" in cardinality. So a very reasonable question is indeed to ask whether every finite group is a Galois group over some fixed base field $K$ . The most natural case is to ask the question for $K = \mathbb{Q}$ , but also other base fields can be considered - for example, for $K = \mathbb{C}(t)$ the inverse Galois conjecture is true. In fact, the inverse Galois problems for different base fields $K$ are sometimes closely linked; the method which I will sketch below is a perfect illustration for this, since we will have the consider four different base fields: $\mathbb{C}(t)$ , $\overline{\mathbb{Q}}(t)$ , $\mathbb{Q}(t)$ , and of course $\mathbb{Q}$ . (1) Start with the fact that each finite group $G$ can be realized as a Galois group over $\mathbb{C}(t)$ . This follows from the theory of coverings of Riemann-surfaces; if $G$ can be generated by $n - 1$ elements, then we can realize $G$ as a quotient of the fundamental group of the punctured Riemann sphere $\pi_1^{\text{top}}(\mathbb{P}^1(\mathbb{C}) \setminus \{P_1,P_2,\,\cdots,P_n\})$ , where we choose the points $P_1,P_2,\,\cdots,P_n$ to be rational. (2) We use the theory of the étale fundamental group to get an isomorphism $\pi_1(\mathbb{P}^1(\mathbb{C}) \setminus \{P_1,P_2,\,\cdots,P_n\}) = \pi_1(\mathbb{P}^1(\overline{\mathbb{Q}}) \setminus \{P_1,P_2,\,\cdots,P_n\})$ , which allows us to realize $G$ as a Galois group over $\overline{\mathbb{Q}}(t)$ . [ $\pi_1$ is the étale fundamental group - here the profinite completion of the topological version.] [Of course, this is already advanced material; see the Wikipedia article for background. For a proper introduction to the theory, there is SGA 1 by Grothendieck; and the recent book "Galois groups and fundamental groups" by Tamas Szamuely is a very gentle introduction (and does all this in detail).] (3) There exists an exact sequence $1 \to \pi_1(\mathbb{P}^1(\overline{\mathbb{Q}}) \setminus \{P_1,P_2,\,\cdots,P_n\}) \to \pi_1(\mathbb{P}^1(\mathbb{Q}) \setminus \{P_1,P_2,\,\cdots,P_n\}) \to \text{Gal}(\overline{\mathbb{Q}}|\mathbb{Q}) \to 1$ (this is a very fundamental result; see again the books I mentioned) and basically we now want to extend a surjective homomorphism from $\pi_1(\mathbb{P}^1(\overline{\mathbb{Q}}) \setminus \{P_1,P_2,\,\cdots,P_n\})$ to $G$ to a surjective homomorphism from $\pi_1(\mathbb{P}^1(\mathbb{Q}) \setminus \{P_1,P_2,\,\cdots,P_n\})$ to $G$ . Of course this depends heavily on the structure of the group $G$ . This works for many finite simple groups; the construction is quite general, but for particular groups there is always some technical work to do to show that the method applies. In particular it works for finite groups with a trivial centre, and a rigid system of rational conjugacy classes - these are quite technical conditions, of course, and I will just state the definitions. An $n$ -tuple of conjugacy classes $C_1,C_2,\,\cdots,C_n$ of $G$ is rigid if there exists $(g_1,g_2,\,\cdots,g_n) \in G^n$ such that the $g_i$ generate $G$ , $g_1g_2\cdots g_n = 1$ and $g_i \in C_i$ , and if moreover $G$ acts transitively on the set of all such $n$ -tuples $(g_1,g_2,\,\cdots,g_n)$ . A conjugacy class $C$ of $G$ is rational if $g \in C$ implies $g^m \in C$ for all $m$ coprime to the order of $G$ . I won't explain why precisely these conditions give you what you want, since it is really technical. Szamuely explains this very clearly. The conditions can be generalized, but that doesn't make it more readable... [References: section 4.8 in Szamuely's book I mentioned above, and also Serre's wonderful book "Topics in Galois theory", which should maybe be called "Topics in inverse Galois theory" :)] (4) The previous step allows us to descend from $\overline{\mathbb{Q}}(t)$ to $\mathbb{Q}(t)$ , i.e. to realize $G$ as a Galois group of a regular extension - another technical notion which I won't explain, but it is not unimportant - of $\mathbb{Q}(t)$ . To descend from $\mathbb{Q}(t)$ to $\mathbb{Q}$ , there is Hilbert's irreducibility theorem , or some slight generalization (I don't remember exactly). According to Thompson, the Monster has a rigid system of three rational conjugacy classes of orders 2, 3 and 29. So the method will apply; of course, I guess that it will be very hard to construct these conjugacy classes, and it is clear that the classification of finite simple groups has played a very big role in these developments. (But I am not a group theorist, so anyone who knows how this works is welcome to give additional information about this construction :)) So I hope this gives you an idea; I wrote this up in a hurry, so suggestions to make this clearer or more coherent (or of course corrections of details which I got wrong) are always welcome.
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13,896 |
Dick Lipton has a blog post that motivated this question. He recalled the Stark-Heegner Theorem : There are only a finite
number of imaginary quadratic fields
that have unique factorization. They
are $\sqrt{d}$ for $d \in \{-1,-2,-3,-7,-11,-19,-43,-67,-163 \}$ . From Wikipedia (link in the theorem statement above): It was essentially proven by Kurt Heegner in 1952, but Heegner's proof had some minor gaps and the theorem was not accepted until Harold Stark gave a complete proof in 1967, which Stark showed was actually equivalent to Heegner's. Heegner "died before anyone really understood what he had done". I am also reminded of Grassmann 's inability to get his work recognized. What are some other examples of important correct work being rejected by the community? NB. There was a complementary question here before.
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Tarski ran into some trouble when he tried to publish his result that the Axiom of Choice is equivalent to the statement that an infinite set $X$ has the same cardinality as $X \times X$. From Mycielski : "He tried to publish his theorem in the Comptes Rendus but Frechet and Lebesgue refused to present it. Frechet wrote that an implication between two well known propositions is not a new result. Lebesgue wrote that an implication between two false propositions is of no interest. And Tarski said that after this misadventure he never tried to publish in the Comptes Rendus."
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13,942 |
Cherry Kearton, Bayer-Fluckiger and others have results that say the monoid of isotopy classes of smooth oriented embeddings of $S^n$ in $S^{n+2}$ is not a free commutative monoid provided $n \geq 3$. The monoid structure I'm referring to is the connect sum of knots. Bayer-Fluckiger has a result in particular that says you can satisfy these equations $$a+b=a+c, \ \ \ \ b \neq c$$
where $a,b,c$ are isotopy classes of knots and $+$ is connect sum. When $n=1$ it's an old result of Horst Schubert's that the monoid of knots is free commutative on countably-infinite many generators. What I'm wondering is, does anyone have an idea of how difficult it might be to compute the structure of the group completion of the monoid of knots, say, for $n \geq 3$? That's not really my question for the forum, though. It's this: Do people have good examples where it's "easy" to compute the group-completion of a commutative monoid, but for which the monoid itself is still rather mysterious? Meaning, one where rather minimal amounts of information are required to compute the group completion? Presumably there are examples where it's painfully difficult to say anything about the group completion? For example, can it be hard to say if there's torsion in the group completion?
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Do people have good examples where it's "easy" to compute the group-completion of a commutative monoid, but for which the monoid itself is still rather mysterious? This happens all the time in K-theory $K^0(X)$, both algebraic and topological. Perhaps it is even the reason that K-theory is a useful tool. For a striking algebraic example, take $X = \mathbb{A}^n_k$ where $k$ is a field. Then $K^0(X)$ is the group completion of the commutative monoid $M$ of isomorphism classes of finitely generated projective modules over $R = k[x_1, \ldots, x_n]$. In 1955 Serre asked whether every such module was free, i.e., whether $M = \mathbb{N}$. This question became known as Serre's conjecture . Serre proved in 1957 that every finitely generated projective $R$-module is stably free, i.e., $K^0(X) = \mathbb{Z}$. However, it was not until 1976 that Quillen and Suslin independently proved Serre's original conjecture. So between 1957 and 1976, $M$ was an example of a commutative monoid whose group completion was known but which itself was not known. This is only a historical example, because $M = \mathbb{N}$ turns out to be very simple; however, it illustrates the difficulty of the question in general. A topological example where the commutative monoid is not so simple is given by $KO^0(S^n)$. Let us take $n$ congruent to 3, 5, 6, or 7 modulo 8, so that $KO^0(S^n) = \mathbb{Z}$ by Bott periodicity (the generator being given by the trivial one-dimensional real vector bundle). Let $T$ be the tangent bundle to $S^n$. In $KO^0(S^n)$, of course, the class of $T$ is equal to its dimension $n$. But if we let $M$ be the commutative monoid of isomorphism classes of finite-dimensional real vector bundles on $S^n$ (so that $KO^0(S^n)$ is the group completion of $M$) then the class of $T$ is not equal to the class of the trivial $n$-dimensional vector bundle unless $S^n$ is parallelizable, which only happens when $n$ is equal to (0 or 1 or) 3 or 7. So for all other values of $n$, $M$ is not simply $\mathbb{N}$; there are extra vector bundles which get killed by the group completion process. Understanding these monoids $M$ for all $n$ amounts to understanding the homotopy groups of all the groups $O(m)$, which I expect is not much easier than understanding unstable homotopy groups of spheres. Finally, Pete's example of the monoid of cardinalities of at most countable sets and its absorbing element also makes an appearance in K-theory; here it is called the Eilenberg swindle and it explains why we restrict ourselves to finitely-generated projective modules.
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14,054 |
Does there exist anywhere a comprehensive list of small genus modular curves $X_G$, for G a subgroup of GL(2,Z/(n))$? (say genus <= 2), together with equations? I'm particularly interested in genus one cases, and moreso in split/non-split cartan, with or without normalizers. Ken Mcmurdy has a list here for $X_0(N)$, and Burcu Baran writes down equations for all $X_{ns}^+(p)$ of genus <=2 in this preprint.
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No, there does not exist a comprehensive list of equations: the known equations are spread out over several papers, and some people (e.g. Noam Elkies, John Voight; and even me) know equations which have not been published anywhere. When I have more time, I will give bibliographic data for some of the papers which give lists of some of these equations. Some names of the relevant authors: Ogg, Elkies, Gonzalez, Reichert. In my opinion, it would be a very worthy service to the number theory community to create an electronic source for information on modular curves (including Shimura curves) of low genus, including genus formulas, gonality, automorphism groups, explicit defining equations...In my absolutely expert opinion (that is, I make and use such computations in my own work, but am not an especially good computational number theorist: i.e., even I can do these calculations, so I know they're not so hard), this is a doable and even rather modest project compared to some related things that are already out there, e.g. William Stein's modular forms databases and John Voight's quaternion algebra packages. It is possible that it is a little too easy for our own good, i.e., there is the sense that you should just do it yourself. But I think that by current standards of what should be communal mathematical knowledge, this is a big waste of a lot of people's time. E.g., by coincidence I just spoke to one of my students, J. Stankewicz, who has spent some time implementing software to enumerate all full Atkin-Lehner quotients of semistable Shimura curves (over Q) with bounded genus. I assigned him this little project on the grounds that it would be nice to have such information, and I think he's learned something from it, but the truth is that there are people who probably already have code to do exactly this and I sort of regret that he's spent so much time reinventing this particular wheel. (Yes, he reads MO, and yes, this is sort of an apology on my behalf.) Maybe this is a good topic for the coming SAGE days at MSRI? Addendum : Some references: Kurihara, Akira
On some examples of equations defining Shimura curves and the Mumford uniformization.
J. Fac. Sci. Univ. Tokyo Sect. IA Math. 25 (1979), no. 3, 277--300. $ \ $ Reichert, Markus A. Explicit determination of nontrivial torsion structures of elliptic curves over quadratic number fields. Math. Comp. 46 (1986), no. 174, 637--658. http://alpha.math.uga.edu/~pete/Reichert86.pdf $ \ $ Gonzàlez Rovira, Josep Equations of hyperelliptic modular curves. Ann. Inst. Fourier (Grenoble) 41 (1991), no. 4, 779--795. http://alpha.math.uga.edu/~pete/Gonzalez.pdf $ \ $ Noam Elkies, equations for some hyperelliptic modular curves, early 1990's. [So far as I know, these have never been made publicly available, but if you want to know an equation of a modular curve, try emailing Noam Elkies!] $ \ $ Elkies, Noam D. Shimura curve computations. Algorithmic number theory (Portland, OR, 1998), 1--47, Lecture Notes in Comput. Sci., 1423, Springer, Berlin, 1998. http://arxiv.org/abs/math/0005160 $ \ $ An algorithm which was used to find explicit defining equations for $X_1(N)$ , $N$ prime, can be found in Pete L. Clark, Patrick K. Corn and the UGA VIGRE Number Theory Group, Computation On Elliptic Curves With Complex Multiplication , preprint. http://alpha.math.uga.edu/~pete/TorsCompv6.pdf This is just a first pass. I probably have encountered something like 10 more papers on this subject, and I wasn't familiar with some of the papers that others have mentioned.
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14,058 |
I'm interested in the structures of categories like $Rep(GL_n), Rep(SL_n)$, etc. of algebraic representations of an algebraic group. I understand that there should be some relation between these and the categories of representations of the corresponding Lie algebras.
However, it's not as intuitive to me what's going on here as with the case of, say, a Lie group, perhaps because the notion of a "tangent vector" is somewhat different. So, how does one switch between the categories $Rep(G)$ and $Rep(\mathfrak{g})$ for $G$ an algebraic group and $\mathfrak{g}$ its Lie algebra---are there functors in each direction? Can this be used to prove that $Rep(G)$ is semisimple when $G$ is reductive?
In another direction, can the structure of $Rep(\mathfrak{g})$ as known from the representation theory of, say, semisimple Lie algebras give the structure of $Rep(G)$?
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If $G$ is semisimple simply connected in characteristic zero, the differential at $1$ gives an equivalence of (tensor) categories $Rep(G)\to Rep({\mathfrak g})$. If $G$ is not semisimple, this is not the case, but this functor is always fully faithful (i.e. an equivalence onto a full subcategory) if $G$ is connected. The essential image of this functor can be described explicitly. Namely, consider the Levi decomposition ${\mathfrak g}={\mathfrak l}\ltimes {\mathfrak u}$, where ${\mathfrak l}$ is reductive and ${\mathfrak u}=Lie(U)$, where $U$ is the unipotent radical of $G$. Then the image is those finite dimensional representations of ${\mathfrak g}$ for which ${\mathfrak u}$ acts nilpotently, and the weights for ${\mathfrak l}$ are integral (i.e. descend to chartacters of the maximal torus).
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14,076 |
Let $k$ be a field. I am interested in sufficient criteria for $f \in k[x,y]$ to be irreducible. An example is Theorem A of this paper (Brindza and Pintér, On the irreducibility of some polynomials in two variables , Acta Arith. 1997). Does anyone know of similar results in the same vein? How about criteria over fields other than the complex numbers?
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A trick which works surprisingly often in my experience: If the Newton polytope of $f$ can not be written as a Minkowski sum of two smaller polytopes, then $f$ is irreducible. I think of this as a generalization of Eisenstein's criterion. It is surprisingly easy to test whether a lattice polygon in $\mathbb{R}^2$ can be written as a Minkowski sum of smaller lattice polygon. Let $P$ be a lattice polytope. Travel around $\partial P$ and write down the vectors pointing from each lattice point to the next lattice point; call this sequence of vectors $v(P)$ . For example, if our polynomial is $a y^2 + b y + c xy + d + e x + f x^2 + g x^3$ , with $adg \neq 0$ , then the lattice points on the boundary are $(0,2)$ , $(0,1)$ , $(0,0)$ , $(1,0)$ , $(2,0)$ , $(3,0)$ so $v(P) =(\ (0,-1),\ (0,-1),\ (1,0),\ (1,0),\ (1,0),\ (-3,2)\ )$ . (Note that $(1,1)$ is not on the boundary of the triangle.) It turns out that $v(A + B)$ is simply the sequences $v(A)$ and $v(B)$ , interleaved by sorting their slopes . So, if $P$ can be written as the Minkowski sum $A+B$ , we must be able to partition $v(P)$ into two disjoint sub-sequences, each of which sums to zero. In the above example, this can't be done, so any polynomial of the form $a y^2 + b y + c xy + d + e x + f x^2 + g x^3$ , with $adg \neq 0$ is irreducible. As an example of a polynomial which could factor, look at $a y^2 + by + c xy + dx + e x^2$ . So the boundary is $(0,2)$ , $(0,1)$ , $(1,0)$ , $(2,0)$ , $(1,1)$ with $v(P) = (\ (0,-1),\ (1,-1),\ (1,0),\ (-1,1),\ (-1, 1)\ )$ . This is the interleaving of $(\ (0,-1),\ (1,0),\ (-1, 1)\ )$ and $(\ (1,-1),\ (-1,1)\ )$ , so a polynomial with this Newton polytope could factor.
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14,177 |
We know that any smooth projective curve can be embedded (closed immersion) in $\mathbb{P}^3$. By definition a projective scheme over $k$ admits an embedding into some $\mathbb{P}^n$. Can we create an upper bound for the $n$ required (perhaps by strengthening the hypotheses) necessary to create an embedding of a smooth projective dimension $k$ scheme into $\mathbb{P}^n$ much like Whitney's theorem tells us we can embed an $n$ dimensional manifold in $\mathbb{R}^{2n}$?
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Over an algebraically closed field, any projective smooth variety of dimension $n$ can be embedded in $\mathbb P^{2n+1}$. This is elementary and can be found in Shafarevich's Basic Algebraic Geometry, Chapter II, §5.4 .
Of course specific varieties might be embedded in projective spaces of lower dimension. For an abelian variety however we have a very satisfying complete description of the situation: For $n=1$, we can embed any abelian (=elliptic) curve in $\mathbb P^{2}$. For $n=2$, some abelian surfaces but not all of them can be embedded in $\mathbb P^{4}$.
The others can only be embedded in $\mathbb P^{5}$. This is due to Horrocks-Mumford. For $n\geq 3$, no abelian variety of dimension $n$ can be embedded in $\mathbb P^{2n}$. They can only be embedded in $\mathbb P^{2n+1}$. (This theorem was proved by Van de Ven.) Summary and references (added later) $\quad$ Over an algebraically closed field every projective smooth variety of dimension $n$ can be embedded in $\mathbb P^{2n+1}$. The embedding dimension $2n+1$ is sharp in the sense that for every $n$ there is a projective smooth variety of dimension $n$ not embeddable in $\mathbb P^{2n}$. [For $n=1$ the sharpness is due to the fact that smooth curves do not embed in $\mathbb P^{2}$ unless their genus is of the form $(d-1)(d-2)/2$. For $n\geq 2$ the sharpness is due to the discussion of abelian varieties above] G. Horrocks and D. Mumford. A rank 2 vector bundle on P4 with
15,000 symmetries. Topology 12 (1973), 63-81 A. Van de Ven. On the embedding of abelian varieties in projective spaces. Ann. Mat. Pura Appl. (4), 103:127–129, 1975.
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|
14,219 |
I'm looking for a "conceptual" explanation to the question in the title. The standard proofs that I've seen go as follows: use the Schubert cell decomposition to get a basis for cohomology and show that the special Schubert classes satisfy Pieri's formula. Then use the fact that basic homogeneous symmetric functions $h_n$ are algebraically independent generators of the ring of symmetric functions, to get a surjective homomorphism from the ring of symmetric functions to the cohomology ring. Pieri's rule can be shown with some calculations, but is there any reason a priori to believe that tensor product multiplicities for the general linear group should have anything at all to do with the Grassmannian? Maybe a more specific question: is it possible to prove that these coefficients are the same without calculating them beforehand? One motivation for asking is that the cohomology ring of the type B and type C Grassmannians ${\bf OGr}(n,2n+1)$ and ${\bf IGr}(n,2n)$ are described by (modified) Schur P- and Q-functions which seem to have nothing(?) to do with the representation theory of the orthogonal and symplectic groups ${\bf SO}(2n+1)$ and ${\bf Sp}(2n)$. So as far as I can tell the answer isn't just because general linear groups and Grassmannians are "type A" objects.
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There are several rings-with-bases to get straight here. I'll explain that, then describe three serious connections (not just Ehresmann's Lesieur's proof as recounted in the OP). The wrong one is $Rep(GL_d)$ , whose basis is indexed by decreasing sequences in ${\mathbb Z}^d$ . That has a subring $Rep(M_d)$ , representations of the Lie monoid of all $d\times d$ matrices, whose basis is indexed by decreasing sequences in ${\mathbb N}^d$ , or partitions with at most $d$ rows. That is a quotient of $Rep({\bf Vec})$ , the Grothendieck ring of algebraic endofunctors of ${\bf Vec}$ , whose basis (coming from Schur functors) is indexed by all partitions. Obviously any such functor will restrict to a rep of $M_d$ (not just $GL_d$ ); what's amazing is that the irreps either restrict to $0$ (if they have too many rows) or again to irreps! Harry Tamvakis' proof is to define a natural ring homomorphism $Rep({\bf Vec}) \to H^*(Gr(d,\infty))$ , applying a functor to the tautological vector bundle, then doing a Chern-Weil trick to obtain a cohomology class. (It's not just the Euler class of the resulting huge vector bundle.)
The Chern-Weil theorem is essentially the statement that Harry's map takes alternating powers to special Schubert classes. So then it must do the right thing, but to know that he essentially repeats the Ehresmann proof. Kostant studied $H^* (G/P)$ in general, in "Lie algebra cohomology and
generalized Schubert cells", by passing to the compact
picture $H^* (K/L)$ , then to de Rham cohomology, then taking $K$ -invariant
forms, which means $L$ -invariant forms on the tangent space $Lie(K)/Lie(L)$ .
Then he complexifies that space to $Lie(G)/Lie(L_C)$ , and identifies that
with $n_+ \oplus n_-$ , where $n_+$ is the nilpotent radical of $Lie(P)$ .
Therefore forms on that space is $Alt^* (n_+) \otimes Alt^* (n_-)$ . Now, there are two things left to do to relate this space to $H^* (G/P)$ . One is to take cohomology of this complex (which is hard, but he describes the differential), and the other
is to take $L$ -invariants as I said. Luckily those commute. Kostant degenerates the differential so as to make sense on each factor separately (at the cost of not quite getting $H^* (G/P)$ ). Theorem: (1) Once you take cohomology, $Alt^* (n_+)$ is a multiplicity-free $L$ -representation. So when you tensor it with its dual and take $L$ -invariants, you get a canonical basis by Schur's lemma. (2) This basis is the degeneration of the Schubert basis. Theorem: (1) If $P$ is (co?)minuscule, the differential is zero, so you can skip the take-cohomology step. That is, $Alt^* (n_+)$ is already a multiplicity-free $L$ -rep. The Schur's lemma basis has structure constants coming from representation theory. (2) In the Grassmannian case, the degeneration doesn't actually affect the answer, so the product of Schubert classes does indeed come from representation theory. I believe the degenerate product on $H^*(G/P)$ is exactly the one described by [Belkale-Kumar]. It's fun to see what's going on in the Grassmannian case -- $L = U(d) \times U(n-d)$ , $n_+ = M_{d,n-d}$ , and $Alt^* (n_+)$ contains each partition (or rather, the $U(d)$ -irrep corresponding) fitting inside that rectangle tensor its transpose (or rather, the $U(n-d)$ -irrep). I think this is going to be the closest to what you want, for other groups' Grassmannians. (No, 3. Silly site software!)
Belkale has the best (least decategorified) proof I've seen (which appeared separately as _Invariant theory of $GL(n)$ and intersection theory of Grassmannians _ IMRN 2004 Issue 69 (2004) 3709–3721, https://doi.org/10.1155/S107379280414155X ). He takes three Schubert cycles meeting transversely, and for each point of intersection, constructs an actual invariant vector inside the corresponding triple product of representations. The set of such vectors is then a basis.
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14,235 |
By Heisenberg group I mean the group with presentation $H$ generated by $x$ and $y$ such that $x$ and $y$ commute with $xyx^{-1}y^{-1}$. Is there an infinite chain of subgroups
$H > H_1 > H_2 > \dots$ such that the index $[H_i: H_{i+1}]< n$ for some $n\ ?$ Thanks
|
There are several rings-with-bases to get straight here. I'll explain that, then describe three serious connections (not just Ehresmann's Lesieur's proof as recounted in the OP). The wrong one is $Rep(GL_d)$ , whose basis is indexed by decreasing sequences in ${\mathbb Z}^d$ . That has a subring $Rep(M_d)$ , representations of the Lie monoid of all $d\times d$ matrices, whose basis is indexed by decreasing sequences in ${\mathbb N}^d$ , or partitions with at most $d$ rows. That is a quotient of $Rep({\bf Vec})$ , the Grothendieck ring of algebraic endofunctors of ${\bf Vec}$ , whose basis (coming from Schur functors) is indexed by all partitions. Obviously any such functor will restrict to a rep of $M_d$ (not just $GL_d$ ); what's amazing is that the irreps either restrict to $0$ (if they have too many rows) or again to irreps! Harry Tamvakis' proof is to define a natural ring homomorphism $Rep({\bf Vec}) \to H^*(Gr(d,\infty))$ , applying a functor to the tautological vector bundle, then doing a Chern-Weil trick to obtain a cohomology class. (It's not just the Euler class of the resulting huge vector bundle.)
The Chern-Weil theorem is essentially the statement that Harry's map takes alternating powers to special Schubert classes. So then it must do the right thing, but to know that he essentially repeats the Ehresmann proof. Kostant studied $H^* (G/P)$ in general, in "Lie algebra cohomology and
generalized Schubert cells", by passing to the compact
picture $H^* (K/L)$ , then to de Rham cohomology, then taking $K$ -invariant
forms, which means $L$ -invariant forms on the tangent space $Lie(K)/Lie(L)$ .
Then he complexifies that space to $Lie(G)/Lie(L_C)$ , and identifies that
with $n_+ \oplus n_-$ , where $n_+$ is the nilpotent radical of $Lie(P)$ .
Therefore forms on that space is $Alt^* (n_+) \otimes Alt^* (n_-)$ . Now, there are two things left to do to relate this space to $H^* (G/P)$ . One is to take cohomology of this complex (which is hard, but he describes the differential), and the other
is to take $L$ -invariants as I said. Luckily those commute. Kostant degenerates the differential so as to make sense on each factor separately (at the cost of not quite getting $H^* (G/P)$ ). Theorem: (1) Once you take cohomology, $Alt^* (n_+)$ is a multiplicity-free $L$ -representation. So when you tensor it with its dual and take $L$ -invariants, you get a canonical basis by Schur's lemma. (2) This basis is the degeneration of the Schubert basis. Theorem: (1) If $P$ is (co?)minuscule, the differential is zero, so you can skip the take-cohomology step. That is, $Alt^* (n_+)$ is already a multiplicity-free $L$ -rep. The Schur's lemma basis has structure constants coming from representation theory. (2) In the Grassmannian case, the degeneration doesn't actually affect the answer, so the product of Schubert classes does indeed come from representation theory. I believe the degenerate product on $H^*(G/P)$ is exactly the one described by [Belkale-Kumar]. It's fun to see what's going on in the Grassmannian case -- $L = U(d) \times U(n-d)$ , $n_+ = M_{d,n-d}$ , and $Alt^* (n_+)$ contains each partition (or rather, the $U(d)$ -irrep corresponding) fitting inside that rectangle tensor its transpose (or rather, the $U(n-d)$ -irrep). I think this is going to be the closest to what you want, for other groups' Grassmannians. (No, 3. Silly site software!)
Belkale has the best (least decategorified) proof I've seen (which appeared separately as _Invariant theory of $GL(n)$ and intersection theory of Grassmannians _ IMRN 2004 Issue 69 (2004) 3709–3721, https://doi.org/10.1155/S107379280414155X ). He takes three Schubert cycles meeting transversely, and for each point of intersection, constructs an actual invariant vector inside the corresponding triple product of representations. The set of such vectors is then a basis.
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14,278 |
I remember reading Weil's "Basic Number Theory" and giving up after a while. Now I find myself thinking of it (thanks to some comments by Ben Linowitz). Right from the very beginning, Weil uses the fact that when you have a locally compact topolgocal group $G$ and a locally compact subgroup $H$ , in addition to the Haar measures on $G$ and $H$ , there exists a "Haar measure" on the coset space $G/H$ , with some properties. For instance, the upper half plane $\mathbb H$ is the quotient $\operatorname{SL}_2(\mathbb R)/{\operatorname{SO}_2(\mathbb R)}$ and the usual measure there which gives rise to the usual hyperbolic metric, is arising in this way. I originally assumed this theorem and went ahead(but not much) with that book. I want to have a reference for the above theorem. A reference which is not written by Weil. I find him very hard to penetrate. This should exclude Bourbaki's "Integration", as I supppose it would be heavily influenced by him, and thus is a horrible book (note to Harry: this is personal opinion; spare me the brickbats). I had originally seen the construction of Haar measure on H. Royden's "Real Analysis", in which he is not considering any quotients.
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The book I always look at for such things is Nachbin, The Haar Integral, which is short, and has a whole chapter on Integration on Locally Compact Homogeneous Spaces. And a plus: he gives you a choice of reading the proof of the existence and uniqueness of the Haar integral according to Weil or according to Henri Cartan.
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|
14,311 |
I was reading this question on why algebraic geometry looks at prime ideals instead of only maximal ideals, and I understand Anton's answer, but I'm a little confused as to how this fits with Hilbert's Nullstellensatz - affine algebraic sets are in bijection with radical ideals, not prime ideals, and it seems like we'd want the extra information we'd get by looking at "RadSpec(R)" (my own imagined notation). Also, the preimage of a radical ideal is radical, so there isn't the same objection as to maximal ideals - "RadSpec" would also be a contravariant functor. So, why not radical ideals instead of only prime ideals, and what kinds of things could we say about RadSpec(R) even if they aren't very interesting?
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The short answer is that every radical ideal is the intersection of the prime ideals containing it, so that the pullback map on Specs determines the pullback map on your wouldbe RadSpecs. Note that a similar reason underlies the success of looking only at maximal ideals in classical algebraic geometry: a finitely generated algebra over a field is a Hilbert-Jacobson ring: every radical ideal is the intersection of the maximal ideals containing it. I won't rule out the possibility that RadSpec could be useful for something, though...
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14,314 |
The fact that a commutative ring has a natural topological space associated with it is still a really interesting coincidence. The entire subject of Algebraic geometry is based on this simple fact. Question: Are there other categories of algebraic objects that have interesting natural topologies that carry algebraic data like the Zariski topology on a ring (spectrum)? If they exist, what are they and how are they used?
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Yes, there are plenty of such things. [In the following, "compact" implies "locally compact" implies "Hausdorff".] 1) To a Boolean algebra, one associates its Stone space, a compact totally disconnected space. (Via the correspondence between Boolean algebras and Boolean rings, this is a special case of the Zariski topology -- but with a distinctive flavor -- that predates it.) 2) To a non-unital Boolean ring one associates its Stone space, a locally compact totally
disconnected space. 3) To a commutative C*-algebra with unit, one associates its Gelfand spectrum, a compact space. 4) To a commutative C*-algebra without unit, one associates its Gelfand spectrum, a locally compact space. 6) To a commutative Banach ring [or a scheme over a non-Archimedean field, or...] one
associates its Berkovich spectrum (the bounded multiplicative seminorms). 7) To a commutative ring R, one associates its real spectrum (prime ideals, plus orderings on the residue domain.) 8) To a field extension K/k, one associates its Zariski Riemann surface (equivalence classes of valuations on K which are trivial on k). This is by no means a complete list... Addendum : I hadn't addressed the second part of your question, i.e., explaining what these things are used for. Briefly, the analogy to the Zariski spectrum of a commutative ring is tight enough to give the correct impression of the usefulness of these other spectra/spaces: they are topological spaces associated (cofunctorially) to the algebraic (or algebraic-geometric, topological algebraic, etc.) objects in question. They carry enough information to be useful in the study of the algebraic objects themselves (sometimes, e.g. in the case of Stone and Gelfand spaces, they give complete information, i.e., an anti-equivalence of categories, but not always). In some further cases, one can get the anti-equivalence by adding further structure in a very familiar way: one can attach structure sheaves to these guys and thus get a class of "model spaces" for a certain species of locally ringed spaces -- e.g., Berkovich spectra glue together to give Berkovich analytic spaces.
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14,338 |
I'm community wikiing this, since although I don't want it to be a discussion thread, I don't think that there is really a right answer to this. From what I've seen, model theorists and logicians are mostly opposed to GCH, while on the other end of the spectrum, some functional analysis depends on GCH, so it is much better tolerated among functional analysts. In fact, I considered myself very much +GCH for a while, but Joel and Francois noted some interesting stuff about forcing axioms, (the more powerful ones directly contradict CH). What is the general opinion on GCH in the mathematical community (replace GCH with CH where necessary)? Does it happen to be that CH/GCH doesn't often come up in algebra? Please don't post just post "I agree with +-CH". I'd like your assessment of the mathematical community's opinion. Maybe your experiences with mathematicians you know, etc. Even your own experiences or opinion can work. I am just not interested in having 30 or 40 one line answers. Essentially, I'm not looking for a poll. Edit: GCH=Generalized Continuum Hypothesis
CH= Continuum Hypothesis CH says that $\aleph_1=\mathfrak{c}$. That is, the successor cardinal of $\aleph_0$ is the continuum. The generalized form (GCH) says that for any infinite cardinal $\kappa$, we have $\kappa^+=2^\kappa$, that is, there are no cardinals strictly between $\kappa$ and $2^\kappa$. Edit 2 (Harry): Changed the wording about FA. If it still isn't true, and you can improve it, feel free to edit the post yourself and change it.
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There is definitely a not-CH tendency among set theorists with a strong Platonist bent, and my impression is that this is the most common view. Many of these set theorists believe that the large cardinal hierarchy and the accompanying uniformization consequences are pointing us towards the final, true set theory, and that the various forcing axioms, such as PFA, MM etc. are a part of it. Another large group of set theorists working in the area of inner model theory have GCH in all the most important models that they study, and regard GCH as one of the attractive regularity features of those inner models. There is a far smaller group of set theorists (among whom I count myself) with a multiverse perspective, who take the view that set theory is really about studying all the possible universes that we might live in, and studying their inter-relations. For this group, the CH question is largely settled by the fact that we understand in a very deep way how to move fom the CH universes to the not-CH universes and vice versa, by the method of forcing. They are each dense in a sense in the collection of all set-theoretic universes.
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14,341 |
A recent answer motivated me to post about this. I've always had a vague, unpleasant feeling that somehow lattice theory has been completely robbed of the important place it deserves in mathematics - lattices seem to show up everywhere, the author or teacher says "observe that these ____ form a complete lattice" or something similar, and then moves on, never to speak of what that might imply. But, not currently knowing anything about them, I can't be sure. What would be a good place to learn about lattice theory, especially its implications for "naturally occurring" lattices (subgroups, ideals, etc.)?
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A good, user-friendly, modern, introductory textbook is Davey and Priestley's Introduction to Lattices and Order . Incidentally, Gian-Carlo Rota used to say much the same thing as you, Zev: that lattice theory had been robbed of its rightful place in mathematics.
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14,356 |
Bourbaki used a very very strange notation for the epsilon-calculus consisting of $\tau$s and $\blacksquare$. In fact, that box should not be filled in, but for some reason, I can't produce a \Box. Anyway, I was wondering if someone could explain to me what the linkages back to the tau mean, what the boxes mean. Whenever I read that book, I replace the $\tau$s with Hilbert's $\varepsilon$s. I mean, they went to an awful lot of trouble to use this notation, so it must mean something nontrivial, right? You can see it on the first page of the google books link I've posted. I'm not sure if it's supposed to be intentionally vague, but they never introduce any metamathematical rules to deal with linkages except for the criteria of substitution, which pretty much cannot interact with $\tau$ terms. Also, of course, since it's a book written to be completely a pain in neck to read, they use a hilbert calculus, and even worse, without primitive equality for determining whether or not two assemblies are equivalent.
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Let me address the part of the question about "what the linkages back to the tau mean, what the boxes mean." The usual notation for using Hilbert's epsilon symbol is that one writes $(\varepsilon x)\phi(x)$ to mean "some (unspecified) $x$ satisfying $\phi$ (if one exists, and an arbitrary object otherwise)." If, like Bourbaki, one wants to avoid quantifiers in the official notation and use $\varepsilon$ instead (specifically, expressing $(\exists x)\phi(x)$ as $\phi((\varepsilon x)\phi(x))$), then any non-trivial formula will contain lots of $\varepsilon$'s, applied to lots of variables, all nested together in a complicated mess. To slightly reduce the complication, let me suppose that bound variables have been renamed so that each occurrence of $\varepsilon$ uses a different variable. Bourbaki's notation (even more complicated, in my opinion) is what you would get if you do the following for each occurrence of $\varepsilon$ in the formula. (1) Replace this $\varepsilon$ with $\tau$. (2) Erase the variable that comes right after the $\varepsilon$. (3) Replace all subsequent occurrences of that variable with a box. (4) Link each of those boxes to the $\tau$ you wrote in (1). So $(\varepsilon x)\phi(x)$ becomes $\tau\phi(\square)$ with a link from the $\tau$ to the boxes (as many boxes as there were $x$'s in $\phi(x)$). One might wonder why Bourbaki does all this. As far as I know, the point of the boxes and links is that there are no bound variables in the official notation; they've all been replaced by boxes. So Bourbaki doesn't need to define things like free and bound occurrences of a variable. Where I (and just about everybody else) would say that a variable occurs free in a formula, Bourbaki can simply say the variable occurs in the formula. I suspect that Bourbaki chose to use Hilbert's $\varepsilon$ operator as a clever way of getting the axiom of choice and the logical quantifiers all at once. And I have no idea why they changed $\varepsilon$ to $\tau$ (although, while typing this answer, I noticed that I'd much rather type tau than varepsilon).
|
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14,371 |
I am now supposed to organize a tiny lecture course on algebraic geometry for undergraduate students who have an interest in this subject. I wonder whether there are some basic algebraic geometry texts considering the level of undergraduate students who have not learnt commutative algebra or homological algebra; they just know linear algebra and basic abstract algebra. I am looking for some textbooks which provide a lot of examples (more computations using linear algebra and calculus). Actually, I am also looking for some textbooks based on very basic mathematics but which talk a little bit about a modern view point. Thanks in advance!
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An invitation to algebraic geometry by Karen Smith is excellent; it is very intuitive, and does everything over the complex numbers. For absolute newcomers, this is probably the best introduction. Algebraic curves by William Fulton is a classic, quite easily readable for beginners, and free available online in pdf! (He recently published the third edition on his site.) The books by Reid, Miranda and Hulek are also good. Reid does many explicit examples.
|
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|
14,404 |
Has somebody translated J.-P. Serre's "Faisceaux algébriques cohérents" into English? At least part of it? In a fit of enthusiasm, I started translating it and started TeXing. But after section 8, I got tired and stopped. However if somebody else already took the trouble, I would be most grateful. I do not know a word of French(except maybe faisceau), and forgot whatever I learned in the process of translation very quickly. This is made community wiki, as I do not want to get into rep issues. Please feel free to close this if you think this qn is inappropriate for MO(I have added my own vote for closing, in case this helps). I would be happy to receive answers in comments.
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Together with some help from my friend, I translated FAC into English. I didn't have so much time to proofread it, so probably there are some mistakes. It can be found here: FAC , Source .
|
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|
14,456 |
Lagrange proved that every positive integer is a sum of 4 squares. Are there general results like this for rings of integers of number fields? Is this class field theory? Explicitly, suppose a number field is formally real. Denote its ring of integers by $Z$. Is it true that for every algebraic integer $x$ in $Z$ either $x$ or $-x$ is a sum of squares?
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To address the particularities of this question for number fields, the basic theorem is attributed to Hilbert, Landau and Siegel. First of all, any nonzero sum of squares in a number field has to be totally positive (that is, it is positive in all real embeddings). Hilbert (1902) conjectured that in any number field, a totally positive element is a sum of 4 squares in the number field. This was proved by Landau (1919) for quadratic fields and by Siegel (1921) for all number fields. This sounds superficially like a direct extension of Lagrange's theorem, but there is a catch: it is about field elements, not algebraic integers as sums of squares of algebraic integers. A totally positive algebraic integer in a number field $K$ need not be a sum of 4 squares of algebraic integers in $K$. The Hilbert-Landau-Siegel theorem only says it is a sum of 4 squares of algebraic numbers in $K$. For instance, in $\mathbf{Q}(i)$ all elements are totally positive in a vacuous sense (no real embeddings), so every element is a sum of four squares. As an example,
$$
i = \left(\frac{1+i}{2}\right)^2 + \left(\frac{1+i}{2}\right)^2.
$$
This shows $i$ is a sum of two squares in $\mathbf{Q}(i)$.
It is impossible to write $i$ as a finite sum of squares in ${\mathbf Z}[i]$ since
$$
(a+bi)^2 = a^2 - b^2 + 2abi
$$
has even imaginary part when $a$ and $b$ are in $\mathbf{Z}$. Thus any finite sum of squares in $\mathbf{Z}[i]$ has even imaginary part, so such a sum can't equal $i$.
Therefore it is false that every totally positive algebraic integer in a number field is a sum of 4 squares (or even any number of squares) of algebraic integers . Here are some further examples: In $\mathbf{Q}(\sqrt{2})$, $5 + 3\sqrt{2}$ is totally positive since
$5+3\sqrt{2}$ and $5-3\sqrt{2}$ are both positive. So it must be a sum of at most four squares in this field by Hilbert's theorem, and with a little fiddling around you find
$$
5 + 3\sqrt{2} = (1+\sqrt{2})^2 + \left(1 + \frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2.
$$
It is impossible to write $5 + 3\sqrt{2}$ as a sum of squares in the ring of integers $\mathbf{Z}[\sqrt{2}]$ because of the parity obstruction we saw for $i$ as a sum of squares in $\mathbf{Z}[i]$: the coefficient of $\sqrt{2}$ in $5 + 3\sqrt{2}$ is odd. In $\mathbf{Q}(\sqrt{2})$, $\sqrt{2}$ is not totally positive (it becomes negative when we replace $\sqrt{2}$ with $-\sqrt{2}$), so it can't be a sum of squares in this field. But in the larger field $\mathbf{Q}(\sqrt{2},i)$, everything is totally positive in a vacuous sense so everything is a sum of at most four squares in this field by the Hilbert-Landau-Siegel theorem. And looking at $\sqrt{2}$ in $\mathbf{Q}(\sqrt{2},i)$, we find
$$
\sqrt{2} = \left(1 + \frac{1}{\sqrt{2}}\right)^2 + i^2 + \left(\frac{i}{\sqrt{2}}\right)^2.
$$ Hilbert made his conjecture on totally positive numbers being sums of four squares as a theorem, in his Foundations of Geometry . It is Theorem 42. He says the proof is quite hard, and no proof is included. A copy of the book (in English) is available at the time I write this as http://math.berkeley.edu/~wodzicki/160/Hilbert.pdf . See page 83 of the file (= page 78 of the book). Siegel's work on this theorem/conjecture was done just before the Hasse-Minkowski theorem was established in all number fields (by Hasse), and the former can be regarded as a special instance of the latter. Indeed, for nonzero $\alpha$ in a number field $K$, consider the quadratic form $$Q(x_1,x_2,x_3,x_4,x_5) = x_1^2+x_2^2+x_3^2+x_4^2-\alpha{x}_5^2.$$ To say $\alpha$ is a sum of four squares in $K$ is equivalent to saying $Q$ has a nontrivial zero over $K$. (In one direction, if $\alpha$ is a sum of four squares over $K$ then $Q$ has a nontrivial zero over $K$ where $x_5 = 1$. In the other direction, if $Q$ has a nontrivial zero over $K$ where $x_5 \not= 0$ then we can scale and make $x_5 = 1$, thus exhibiting $\alpha$ as a sum of four squares in $K$. If $Q$ has a nontrivial zero over $K$ where $x_5 = 0$ then the sum of four squares quadratic form represents 0 nontrivially over $K$ and thus it is universal over $K$, so it represents $\alpha$ over $K$.) By Hasse-Minkowski, $Q$ represents 0 nontrivially over $K$ if and only if it represents 0 nontrivially over every completion of $K$. Since any nondegenerate quadratic form in five or more variables over a local field or the complex numbers represents 0 nontrivially, $Q$ represents 0 nontrivially over $K$ if and only it represents 0 nontrivially in every completion of $K$ that is isomorphic to ${\mathbf R}$. The real completions of $K$ arise precisely from embeddings $K \rightarrow {\mathbf R}$. For $t \in {\mathbf R}^\times$, the equation
$x_1^2+x_2^2+x_3^2+x_4^2-t{x}_5^2 =0$ has a nontrivial real solution if and only if $t > 0$, so $Q$ has a nontrivial representation of 0 in every real completion of $K$ if and only if $\alpha$ is positive in every embedding of $K$ into ${\mathbf R}$, which is what it means for $\alpha$ to be totally positive. (Strictly speaking, to be totally positive in a field means being positive in every ordering on the field. The orderings on a number field all arise from embeddings of the number field into $\mathbf R$, so being totally positive in a number field is the same as being positive in every real completion.) Siegel's paper is "Darstellung total positiver Zahlen durch Quadrate, Math. Zeit. 11 (1921), 246--275, and can be found online at http://gdz.sub.uni-goettingen.de/en/dms/loader/img/?PPN=PPN266833020_0011&DMDID=DMDLOG_0022 .
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14,514 |
Let G be a nontrivial finite group. Does there exist an irreducible representation of G of dimension greater than or equal to the cardinality of G? [Edited for clarity. -- PLC]
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EDIT: Part 4 added. EDIT2: Second proof of Part 4 added. 1. The answer is no (as long as we are working over a field - of any characteristic, algebraically closed or not). If $k$ is a field and $G$ is a finite group, then the dimension of any irreducible representation $V$ of $G$ over $k$ is $\leq \left|G\right|$. This is actually obvious: Take any nonzero vector $v\in V$; then, $k\left[G\right]v$ is a nontrivial subrepresentation of $V$ of dimension $\leq\dim\left(k\left[G\right]\right)=\left|G\right|$. Since our representation $V$ was irreducible, this subrepresentation must be $V$, and hence $\dim V\leq\left|G\right|$. 2. Okay, we can do a little bit better: Any irreducible representation $V$ of $G$ has dimension $\leq\left|G\right|-1$, unless $G$ is the trivial group. Same proof applies, with one additional step: If $\dim V=\left|G\right|$, then the map $k\left[G\right]\to V,\ g\mapsto gv$ must be bijective (in fact, it is surjective,
since $k\left[G\right]v=V$, and it therefore must be bijective since $\dim\left(k\left[G\right]\right)=\left|G\right|=\dim V$), so it is an isomorphism of representations (since it is $G$-equivariant), and thus $V\cong k\left[G\right]$. But $k\left[G\right]$ is not an irreducible representation, unless $G$ is the trivial group (in fact, it always contains the $1$-dimensional trivial representation). 3. Note that if the base field $k$ is algebraically closed and of characteristic $0$, then we can do much better: In this case, an irreducible representation of $G$ always has dimension $<\sqrt{\left|G\right|}$ (in fact, in this case, the sum of the squares of the dimensions of all irreducible representations is $\left|G\right|$, and one of these representations is the trivial $1$-dimensional one). However, if the base field is not necessarily algebraically closed and of arbitrary characteristic, then the bound $\dim V\leq \left|G\right|-1$ can be sharp (take cyclic groups). 4. There is a way to improve 2. so that it comes a bit closer to 3. : Theorem 1. If $V_1$, $V_2$, ..., $V_m$ are $m$ pairwise nonisomorphic irreducible representations of a finite-dimensional algebra $A$ over a field $k$ (not necessarily algebraically closed, not necessarily of characteristic $0$), then $\dim V_1+\dim V_2+...+\dim V_m\leq\dim A$. (Of course, if $A$ is the group algebra of some finite group $G$, then $\dim A=\left|G\right|$, and we get 2. as a consequence.) First proof of Theorem 1. At first, for every $i\in\left\lbrace 1,2,...,m\right\rbrace$, the (left) representation $V_i^{\ast}$ of the algebra $A^{\mathrm{op}}$ (this representation is defined by $a\cdot f=\left(v\mapsto f\left(av\right)\right)$ for any $f\in V_i^{\ast}$ and $a\in A$) is irreducible (since $V_i$ is irreducible) and therefore isomorphic to a quotient of the regular (left) representation $A^{\mathrm{op}}$ (since we can choose some nonzero $u\in V_i^{\ast}$, and then the map $A^{\mathrm{op}}\to V_i^{\ast}$ given by $a\mapsto au$ must be surjective, because its image is a nonzero subrepresentation of $V_i^{\ast}$ and therefore equal to $V_i^{\ast}$ due to the irreducibility of $V_i^{\ast}$). Hence, by duality, $V_i$ is isomorphic to a subrepresentation of the (left) representation $A^{\mathrm{op}\ast}=A^{\ast}$ of $A$. Hence, from now on, let's assume that $V_i$ actually is a subrepresentation of $A^{\ast}$ for every $i\in\left\lbrace 1,2,...,m\right\rbrace$. Now, let us prove that the vector subspaces $V_1$, $V_2$, ..., $V_m$ of $A^{\ast}$ are linearly disjoint, i. e., that the sum $V_1+V_2+...+V_m$ is actually a direct sum. We will prove this by induction over $m$, so let's assume that the sum $V_1+V_2+...+V_{m-1}$ is already a direct sum. It remains to prove that $V_m\cap \left(V_1+V_2+...+V_{m-1}\right)=0$. In fact, assume the contrary. Then, $V_m\cap \left(V_1+V_2+...+V_{m-1}\right)=V_m$ (since $V_m\cap \left(V_1+V_2+...+V_{m-1}\right)$ is a nonzero subrepresentation of $V_m$, and $V_m$ is irreducible). Thus, $V_m\subseteq V_1+V_2+...+V_{m-1}$. Consequently, $V_m$ is isomorphic to a subrepresentation of the direct sum $V_1\oplus V_2\oplus ...\oplus V_{m-1}$ (because the sum $V_1+V_2+...+V_{m-1}$ is a direct sum, according to our induction assumption). Now, according to Theorem 2.2 and Remark 2.3 of Etingof's "Introduction to representation theory" , any subrepresentation of the direct sum $V_1\oplus V_2\oplus ...\oplus V_{m-1}$ must be a direct sum of the form $r_1V_1\oplus r_2V_2\oplus ...\oplus r_{m-1}V_{m-1}$ for some nonnegative integers $r_1$, $r_2$, ..., $r_{m-1}$. Hence, every irreducible subrepresentation of the direct sum $V_1\oplus V_2\oplus ...\oplus V_{m-1}$ must be one of the representations $V_1$, $V_2$, ..., $V_{m-1}$. Since we know that $V_m$ is isomorphic to a subrepresentation of the direct sum $V_1\oplus V_2\oplus ...\oplus V_{m-1}$, we conclude that $V_m$ is isomorphic to one of the representations $V_1$, $V_2$, ..., $V_{m-1}$. This contradicts the non-isomorphy of the representations $V_1$, $V_2$, ..., $V_m$. Thus, we have proven that the sum $V_1+V_2+...+V_m$ is actually a direct sum. Consequently, $\dim V_1+\dim V_2+...+\dim V_m=\dim\left(V_1+V_2+...+V_m\right)\leq \dim A^{\ast}=\dim A$, and Theorem 1 is proven. Second proof of Theorem 1. I just learnt the following simpler proof of Theorem 1 from §1 Lemma 1 in Crawley-Boevey's "Lectures on representation theory and invariant theory" : Let $0=A_0\subseteq A_1\subseteq A_2\subseteq ...\subseteq A_k=A$ be a composition series of the regular representation $A$ of $A$. Then, by the definition of a composition series, for every $i\in \left\lbrace 1,2,...,k\right\rbrace$, the representation $A_i/A_{i-1}$ of $A$ is irreducible. Let $T$ be an irreducible representation of $A$. We are going to prove that there exists some $I\in \left\lbrace 1,2,...,k\right\rbrace$ such that $T\cong A_I/A_{I-1}$ (as representations of $A$). In fact, let $I$ be the smallest element $i\in \left\lbrace 1,2,...,k\right\rbrace$ satisfying $A_iT\neq 0$ (such elements $i$ exist, because $A_kT=AT=T\neq 0$). Then, $A_IT\neq 0$, but $A_{I-1}T=0$. Now, choose some vector $t\in T$ such that $A_It\neq 0$ (such a vector $t$ exists, because $A_IT\neq 0$), and consider the map $f:A_I\to T$ defined by $f\left(a\right)=at$ for every $a\in A_I$. Then, this map $f$ is a homomorphism of representations of $A$. Since it maps the subrepresentation $A_{I-1}$ to $0$ (because $f\left(A_{I-1}\right)=A_{I-1}t\subseteq A_{I-1}T=0$), it gives rise to a map $g:A_I/A_{I-1}\to T$, which, of course, must also be a homomorphism of representations of $A$. Since $A_I/A_{I-1}$ and $T$ are irreducible representations of $A$, it follows from Schur's lemma that any homomorphism of representations from $A_I/A_{I-1}$ to $T$ is either an isomorphism or identically zero. Hence, $g$ is either an isomorphism or identically zero. But $g$ is not identically zero (since $g\left(A_I/A_{I-1}\right)=f\left(A_I\right)=A_It\neq 0$), so that $g$ must be an isomorphism, i. e., we have $T\cong A_I/A_{I-1}$. So we have just proven that (1) For every irreducible representation $T$ of $A$, there exists some $I\in \left\lbrace 1,2,...,k\right\rbrace$ such that $T\cong A_I/A_{I-1}$ (as representations of $A$). Denote this $I$ by $I_T$ in order to make it clear that it depends on $T$. So we have $T\cong A_{I_T}/A_{I_T-1}$ for each irreducible representation $T$ of $A$. Applying this to $T=V_i$ for every $i\in\left\lbrace 1,2,...,m\right\rbrace$, we see that $V_i\cong A_{I_{V_i}}/A_{I_{V_i}-1}$ for every $i\in\left\lbrace 1,2,...,m\right\rbrace$. Hence, the elements $I_{V_1}$, $I_{V_2}$, ..., $I_{V_m}$ of the set $\left\lbrace 1,2,...,k\right\rbrace$ are pairwise distinct (because $I_{V_i}=I_{V_j}$ would yield $V_i\cong A_{I_{V_i}}/A_{I_{V_i}-1}=A_{I_{V_j}}/A_{I_{V_j}-1}\cong V_j$, but the representations $V_1$, $V_2$, ..., $V_m$ are pairwise nonisomorphic), and thus $\sum\limits_{i=1}^{m}\dim\left(A_{I_{V_i}}/A_{I_{V_i}-1}\right)=\sum\limits_{\substack{j\in\left\lbrace 1,2,...,k\right\rbrace ;\ \\ \text{there exists }\\ i\in\left\lbrace 1,2,...,m\right\rbrace \\ \text{ such that }j=I_{V_i}}}\dim\left(A_j/A_{j-1}\right)$
$\leq \sum\limits_{j\in\left\lbrace 1,2,...,k\right\rbrace}\dim\left(A_j/A_{j-1}\right)$ (since $\dim\left(A_j/A_{j-1}\right)\geq 0$ for every $j$, so that adding more summands cannot decrease the sum)
$=\sum\limits_{j=1}^{k}\dim\left(A_j/A_{j-1}\right)=\sum\limits_{j=1}^{k}\left(\dim A_j-\dim A_{j-1}\right)$. Since $\dim\left(A_{I_{V_i}}/A_{I_{V_i}-1}\right)=\dim V_i$ for each $i$ (due to $A_{I_{V_i}}/A_{I_{V_i}-1}\cong V_i$) and $\sum\limits_{j=1}^{k}\left(\dim A_j-\dim A_{j-1}\right)=\dim A$ (in fact, the sum $\sum\limits_{j=1}^{k}\left(\dim A_j-\dim A_{j-1}\right)$ is a telescopic sum and simplifies to $\dim A_k-\dim A_0=\dim A-\dim 0=\dim A-0=\dim A$), this inequality becomes $\sum\limits_{i=1}^{m}\dim V_i\leq\dim A$. This proves Theorem 1.
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14,518 |
This is related to Anweshi's question about theories of noncommutative geometry . Let's start out by saying that I live, mostly, in a commutative universe. The only noncommutative rings I have much truck with are either supercommutative, almost commutative (filtered, with commutative associated graded), group algebras or matrix algebras, none of which really show many of the true difficulties of noncommutative things. So, here's my (somewhat pithy) question: what's noncommutative geometry good for? To be a bit more precise, I have a vague sense that $C^*$ stuff is supposed to work well in quantum mechanics, but I'm somewhat more interested in more algebraic noncommutative geometry. What sorts of problems does it solve that we can't solve without leaving the commutative world? Why should, say, a complex algebraic geometer learn some noncommutative geometry?
|
Charles, a couple of reasons why a complex algebraic geometer (certainly someone who is
interested in moduli spaces of vector bundles, as your profile tells me) might
at least keep
an open verdict on the stuff NC-algebraic geometers (NCAGers from now on) are trying to do. in recent years ,a lot of progress has been made towards understanding moduli spaces of
semi-stable representations of 'formally smooth' algebras (think 'smooth in the NC-world).
in particular when it comes to their etale local structure and their rationality.
for example, there is this book , by someone. this may not seem terribly relevant to you until you realize that some of the more
interesting moduli spaces in algebraic geometry are among those studied. for example, the
moduli space of semi-stable rank n bundles of degree 0 over a curve of genus g is the moduli
space of representations of a certain dimension vector over a specific formally smooth algebra,
as Aidan Schofield showed. he also applied this to rationality results about these spaces. likewise, the moduli space of semi-stable rank n vectorbundles on the projective plane with Chern
classes c1=0 and c2=n is birational to that of semi-simple n-dimensional representations of the free algebra
in two variables. the corresponding rationality problem has been studied by NCAG-ers (aka 'ringtheorists'
at the time) since the early 70ties (work by S.A. Amitsur, Claudio Procesi and Ed Formanek). by their results, we NCAGers,
knew that the method of 'proof' by Maruyama of their stable rationality in the mid 80ties, couldn't possibly work. it's rather ironic that the best rationality results on these moduli spaces (of bundles over the
projective plane) are not due to AGers but to NCAGers : Procesi for n=2, Formanek for n=3 and 4 and
Bessenrodt and some guy for n=5 and 7. together with a result by Aidan Schofield these results show
that this moduli space is stably rational for all divisors n of 420. further, what a crepant resolution of a quotient singularity is to you, is to NCAGers the moduli space of certain
representations of a nice noncommutative algebra over the singularity. likewise, when you AGers mumble 'Deligne-Mumford stack', we NCAGers say 'ah! a noncommutative algebra'.
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|
14,529 |
The Riemann–Stieltjes integral $\int_a^b f(x)\,dg(x)$ is a generalization of the Riemann integral. It is e.g. heavily used as a starting point for stochastic integration. The approximating Riemann–Stieltjes sums are analogous to the Riemann sums $\sum_{i=0}^{n-1} f(c_i)(g(x_{i+1})-g(x_i))$ where $c_i$ is in the $i$ -th subinterval $[x_i,x_{i+1}]$ . The Riemann-sums can be very intuitively visualized by rectangles that approximate the area under the curve.
See e.g. Wikipedia:Riemann sum .
Unfortunately, I cannot find respective intuitive visualizations of the Riemann–Stieltjes sums. My question: Could anyone provide me with some literature, pictures, links, or especially tools (e.g. Mathematica or even Excel) with which I could play around to get a similar intuition for this more general kind of integral?
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It's fairly easy to visualize the Riemann–Stieltjes integral $\int_a^b f(t)\,dg(t)$ [I changed the name of the integration variable for convenience below] if $f\ge0$ and $g$ is nondecreasing. Just draw the graph of the curve $(x,y)=(g(t),f(t))$. The integral is just the area below the curve. (Whenever $g$ makes a jump at some $t_0$, fill in the gap by setting $y=f(t_0)$ there.) The Riemann–Stieltjes sums are now easy to visualize as sums of areas of rectangles (details left as an exercise).
|
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|
14,568 |
The Poisson summation says, roughly, that summing a smooth $L^1$-function of a real variable at integral points is the same as summing its Fourier transform at integral points(after suitable normalization). Here is the wikipedia link. For many years I have wondered why this formula is true. I have seen more than one proof, I saw the overall outline, and I am sure I could understand each step if I go through them carefully. But still it wouldn't tell me anything about why on earth such a thing should be true. But this formula is exceedingly important in analytic number theory. For instance, in the book of Iwaniec and Kowalski, it is praised to high heavens. So I wonder what is the rationale of why such a result should be true.
|
It is a special case of the trace formula. Both sides are the trace of the same operator.
|
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14,574 |
There are certain things in mathematics that have caused me a pleasant surprise -- when some part of mathematics is brought to bear in a fundamental way on another, where the connection between the two is unexpected. The first example that comes to my mind is the proof by Furstenberg and Katznelson of Szemeredi's theorem on the existence of arbitrarily long arithmetic progressions in a set of integers which has positive upper Banach density, but using ergodic theory. Of course in the years since then, this idea has now become enshrined and may no longer be viewed as surprising, but it certainly was when it was first devised. Another unexpected connection was when Kolmogorov used Shannon's notion of probabilistic entropy as an important invariant in dynamical systems. So, what other surprising connections are there out there?
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As well known as the connection is, I am constantly amazed by the power of analytical geometry (developed by Descartes and Fermat) to make connections between geometrical ideas and algebraic ideas. It seems remarkable to me that so much geometrical information (as for example in the case of the conic sections) can be represented so succinctly (via quadratic equations in two variables). The geometry suggests things to think about in algebra and the algebra suggests things to think about in geometry. It is just amazing!!
|
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14,587 |
For all those who are unlikely to have answers to my questions, I provide some Background: In some sense, pure motives are generalisations of smooth projective varieties. Every Weil cohomology theory factors through the embedding of smooth projective varieties into the category of pure Chow motives . Pure effective motives In the definition of pure motives (say over a field k), the last step is to take the category of pure effective motives and formally invert the Lefschetz motive L. The category of pure effective motives is the pseudo-abelian envelope of a category of correspondence classes, which has as objects smooth projective varieties over k and as morphisms X → Y cycle classes in X×Y of dimension dim X (think of it as a generalisation of morphisms, where morphisms are included as their graphs), where an adequate equivalence relation is imposed, to have a well-defined composition (therefore the word "classes"). When the adequate relation is rational equivalence, the resulting category is called the category of pure effective Chow motives. In each step of the construction, the monoidal structure of one step defines a monoidal structure on the next step. For more background see Ilya's question about the yoga of motives . Definition of the Lefschetz motive The Lefschetz motive L is defined as follows: For each point p in P¹ (1-dimensional projective space over k), there is the embedding morphism Spec k → P¹, which can be composed with the structural morphism P¹ → Spec k to yield an endomorphism of P¹. This is an idempotent, since the other composition Spec k → Spec k is the identity. The category of effective pure motives is pseudo-abelian, so every idempotent has a kernel and thus [P¹] = [Spec k] + [something] =: 1+L, where the summand [something] is now named Lefschetz motive L. Properties The definition of L doesn't depend on the choice of the point p. From nLab and Kahn's leçons I learned that the inversion of the Lefschetz motive is what makes the resulting monoidal category a rigid monoidal category - while the category of pure effective motives is not necessarily rigid. In the category of pure motives, the inverse $L^{-1}$ is called T, the Tate motive. Questions: These questions are somehow related to each other: Why is this motive L called "Lefschetz"? Why is its inverse $L^{-1}$ called "Tate"? Why is it exactly this construction that "rigidifies" the category? Would another construction work, too - or is this somehow universal? How should I think of L geometrically? I have almost no background in number theory, so even if you have good answers, it may remain totally unclear to me, why the name "Tate" intervenes. I expect however, that the name "Lefschetz" has something to do with the Lefschetz trace formula. I guess that the procedure of inverting L is the only one which makes the category rigid, in some universal way, but I have no idea, why. In addition, I guess there is no "geometric" picture of L. If I made any mistakes in the background section, feel free to edit. As I'm currently taking a first course on motives, I may now have asked something completely stupid. If this is the case, please point me politely to some document which will then enlighten me or, at least, let me ascend to a higher level of confusion. UPDATE: Thanks so far for the answers, questions 1-4 are now clear to me. It remains, if the "rigidification" could be accomplished by another construction - maybe some universal way to turn a monoidal category into a rigid one? Then one could later identify the Lefschetz motive as some kind of a generator of the kernel of the rigidification functor. The geometric intuition, to think of L as a curve and of $L^{\otimes d}$ as a d-dimensional manifold, remains fuzzy, but I have the hope that this becomes clear when I have worked a little bit on the classical Lefschetz/Poincare theorems and the proof of Weil conjectures for Betti cohomology (is this hope justified?).
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The motive $L$ is called Lefschetz because it is the cycle class of the point in ${\mathbb P}^1$, and so underlies (in a certain sense) the Lefschetz theorems about the cohomology of
projective varieties. To understand this better, you may want to read about how the hard Lefschetz theorem for varieties over finite fields follows from the Riemann hypothesis, as well as a discussion of Grothendieck's standard conjectures and how they relate to the Weil
conjectures. The motvie $L^{-1}$, when converted into an $\ell$-adic Galois representation, is precisely the $\ell$-adic Tate module of the roots of unity. Tensoring by this Galois representation is traditionally called Tate twisting, and so the motive underlying this Galois representation is called the Tate motive. One needs to have $L^{-1}$ at hand in order for the category to admit duals. If one were working with just usual singular cohomology, this wouldn't be necessary; Poincare duality would pair $H^i$ with $H^{\text{top}-i}$ into $H^{\text{top}}$, which would be
isomorphic with ${\mathbb Q}$ via the fundamental class. But motivically, if $X$ (smooth, connected, projective) has dimension $d$, so that the top dimension is $2d$, then $H^{\text{top}} = L^{\otimes d}$, and so $H^i$ and $H^{2d-i}$ pair into
$L^{\otimes d}$. To get a pairing into $\mathbb Q$ (the trivial 1-dim'l motive) we need
to be able to tensor by powers of $L^{-1}$. Traditionally tensoring by the $n$th tensor power of $L^{-1}$ is denoted $(n)$; so we find e.g. that
$H^i$ pairs with $H^{2d -i}(d)$ into ${\mathbb Q}$, and we have our duality. You can see from the fact that cup product pairs cohomology into powers of $L$ that inverting $L$ is precisely what is needed in order to obtain duals. Finally, one should think of $L$ as the fundamental class of a curve,
think of $L^{\otimes d}$ as the fundamental class of a smooth projective $d$-dimensional
manifold, and also become comfortable with Poincare duality and the Lefschetz theorems;
these are the basic ideas which will help give solid geometric sense to motivic constructions.
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{
"source": [
"https://mathoverflow.net/questions/14587",
"https://mathoverflow.net",
"https://mathoverflow.net/users/956/"
]
}
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