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17,946 |
In 1986 G.X. Viennot published "Heaps of pieces, I : Basic definitions and combinatorial lemmas" where he developed the theory of heaps of pieces, from the abstract: a geometric interpretation of Cartier-Foata's commutation monoid. This theory unifies and simplifies many other works in Combinatorics : bijective proofs in matrix algebra (MacMahon Master theorem, inversion matrix formula, Jacobi identity, Cayley-Hamilton theorem), combinatorial theory for general (formal) orthogonal polynomials, reciprocal of Rogers-Ramanujan identities, graph theory (matching and chromatic polynomials). In the references the subsequent articles "Heaps of pieces, 4 and 5" are listed as "in preparation" where the applications of the theory to solving the directed animal problem and statistical physics are supposed to be developed. I know that these parts of the theory have appeared in literature but I am sort of puzzled as to why the series of papers was not continued (searching for Heaps of pieces II or III or IV doesn't give results). Is there any survey of the full theory somewhere else? Also, since I didn't feel like asking this in a separate question, is there any paper that proves classical theorems of dimers (Kasteleyn's theorem, Aztec diamond etc.) using Viennot's theory?
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Ok, this I know. Viennot basically invents lots of great stuff but rarely publishes his work. About "heaps of pieces" - this is a pretty little theory with very few original consequences. It is really equivalent to Cartier-Foata partially commutative monoid (available here ). See Christian Krattenthaler's article for the connection and details. See there also some references to other recent papers. Now, for many of Viennot's unpublished results, see his "Orthogonal polynomials..." book, which was unavailable for years, but is now on his web page . See there also several of his video lectures (mostly in French), where he outlined some interesting bijections based on the heaps (some related to various lattice animals were new to me, even if he may have come up with them some years ago - take a look). The reciprocal of R-R identities is an elegant single observation which he published separately: MR0989236 . It really does not reprove the R-R identities, just gives a new combinatorial interpretation for one side using heaps of dimers (various related results were obtained by Andrews-Baxter a bit earlier). About some recent applications outside of enumerative combinatorics. Philippe Marchal describes here that heaps of pieces easily imply David Wilson's theorem on Loop-erased random walks giving random spanning trees. Ellenberg and Tymoczko give a beautiful application to the diameter bound of certain Cayley graphs. Finally, in my paper with Matjaz Konvalinka, we use a heaps-of-pieces style bijection to give the "book proof" of the (non- and q-commutative) MacMahon Master theorem . On your followup question regarding Kasteleyn's theorem and the Aztec diamond theorem - no, these are results of different kind, heaps don't really apply, at least as far as I know.
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17,953 |
Let $\mathcal C$ be a category. Recall that a morphism $f : X \to Y$ is epi if $$\circ f: \hom(Y,Z) \to \hom(X,Z)$$ is injective for each object $Z \in \mathcal C$. ($f$ is mono if $f\circ : \hom(Z,X) \to \hom(Z,Y)$ is injective.) Let $\mathcal C,\mathcal D$ be categories. Then $\hom(\mathcal C,\mathcal D)$, the collectional of all functors $\mathcal C \to \mathcal D$, is naturally a category, where the morphisms are natural transformations : if $F,G: \mathcal C \to \mathcal D$ are functors, a natural transformation $\alpha: F \Rightarrow G$ assigns a morphism $\alpha(x) : F(x) \to G(x)$ in $\mathcal D$ for each object $x \in \mathcal C$, and if $f: x \to y$ is a morphism in $\mathcal C$, then $\alpha(y) \circ F(f) = G(f) \circ \alpha(x)$ as morphisms in $\mathcal D$. Given a natural transformation, can I check whether it is epi (or mono) by checking pointwise? I.e.: is a natural transformation $\alpha$ epi (mono) iff $\alpha(x)$ is epi (mono) for each $x$? If not, is there an implication in one direction between whether a natural transformation is epi and whether it is pointwise-epi? A more general question, one that I never really learned, is what types of properties of a functor are "pointwise" in that they hold for the functor if they hold for the functor evaluated at each object. E.g.: is the (co)product of functors the pointwise (co)product?
|
Theo, the answer is basically "yes". It's a qualified "yes", but only very lightly qualified. Precisely: if a natural transformation between functors $\mathcal{C} \to \mathcal{D}$ is pointwise epi then it's epi. The converse doesn't always hold, but it does if $\mathcal{D}$ has pushouts. Dually, pointwise mono implies mono, and conversely if $\mathcal{D}$ has pullbacks. The context for this --- and an answer to your more general question --- is the slogan (Co)limits are computed pointwise. You have, let's say, two functors $F, G: \mathcal{C} \to \mathcal{D}$, and you want to compute their product in the functor category $\mathcal{D}^\mathcal{C}$. Assuming that $\mathcal{D}$ has products, the product of $F$ and $G$ is computed in the simplest possible way, the 'pointwise' way: the value of the product $F \times G$ at an object $A \in \mathcal{C}$ is simply the product $F(A) \times G(A)$ in $\mathcal{D}$. The same goes for any other shape of limit or colimit. For a statement of this, see for instance 5.1.5--5.1.8 of these notes . (It's probably in Categories for the Working Mathematician too.) See also sheet 9, question 1 at the page linked to. For the connection between monos and pullbacks (or epis and pushouts), see 4.1.31. You do have to impose this condition that $\mathcal{D}$ has all (co)limits of the appropriate shape (pushouts in the case of your original question). Kelly came up with some example of an epi in $\mathcal{D}^\mathcal{C}$ that isn't pointwise epi; necessarily, his $\mathcal{D}$ doesn't have all pushouts.
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17,960 |
Hi all! Google published recently questions that are asked to candidates on interviews. One of them caused very very hot debates in our company and we're unsure where the truth is. The question is: In a country in which people only want
boys every family continues to have
children until they have a boy. If
they have a girl, they have another
child. If they have a boy, they stop.
What is the proportion of boys to
girls in the country? Despite that the official answer is 50/50 I feel that something wrong with it. Starting to solve the problem for myself I got that part of girls can be calculated with following series: $$\sum_{n=1}^{\infty}\frac{1}{2^n}\left (1-\frac{1}{n+1}\right )$$ This leads to an answer: there will be ~61% of girls. The official solution is: This one caused quite the debate, but
we figured it out following these
steps: Imagine you have 10 couples who have 10 babies. 5 will be girls. 5
will be boys. (Total babies made: 10,
with 5 boys and 5 girls) The 5 couples who had girls will have 5 babies. Half (2.5) will be
girls. Half (2.5) will be boys. Add
2.5 boys to the 5 already born and 2.5 girls to the 5 already born. (Total
babies made: 15, with 7.5 boys and 7.5
girls.) The 2.5 couples that had girls will have 2.5 babies. Half (1.25) will
be boys and half (1.25) will be girls.
Add 1.25 boys to the 7.5 boys already
born and 1.25 girls to the 7.5 already
born. (Total babies: 17.5 with 8.75
boys and 8.75 girls). And so on, maintianing a 50/50 population. Where the truth is?
|
The proportion of girls in one family is a biased estimator of the proportion of girls in a population consisting of many families because you are underweighting the families with a large number of children. If there were just 1 family, then your formula would be wrong, but the average of the percentage of girls you would observe would be $$\sum_{n=0}^\infty \frac{1}{2^{n+1}} \bigg(\frac{n}{n+1}\bigg) = 1-\log2 = 30.69\%.$$ Half of the time, you would observe $0\%$ girls. If you have multiple families, the average of the observed percentage of girls in the population will increase. For 2 families, the average percentage of girls would be $$\sum_{n=0}^\infty \frac{n+1}{2^{n+2}} \bigg(\frac{n}{n+2}\bigg) = \log 4 - 1 = 38.63\%.$$ More generally, the average percentage for $k$ families is $$\sum_{n=0}^\infty \frac{n+k-1 \choose k-1}{2^{n+k}} \bigg(\frac{n}{n+k}\bigg) = \frac{k}{2}\bigg(\psi(\frac{k+2}2)-\psi(\frac{k+1}2)\bigg)$$ where $\psi$ is the digamma function which satisfies $$ \psi(m) = -\gamma + \sum_{i=1}^{m-1} \frac1i = -\gamma + H_{m-1}$$
$$ \psi(m+\frac12) = -\gamma -2\log 2 + \sum_{i=1}^m \frac{2}{2i-1}.$$ With a little work, one can verify that this goes to $1/2$ as $k\to \infty$. So, for a large population such as a country, the official answer of $1/2$ is approximately correct, although the explanation is misleading. In particular, for $10$ couples, the expected percentage of girls is $10 \log 2 - 1627/252 = 47.51\%$ contrary to what the official answer suggests. With $k$ families, the expected proportion is about $1/2 - 1/4k$. It is not enough to argue that the expected number of boys equals the expected number of girls, since we want $E[G/(G+B)] \ne E[G]/E[G+B].$ Expectation is linear, but not multiplicative for dependent variables, and $G$ and $G+B$ are not independent even though $G$ and $B$ are.
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18,002 |
Suppose $G_i$ are finite groups for $i=1,2$ and G is the direct product of $G_i$. If V is a finite dimensional irreducible representation of $G$, then it is well known that $V$ is a tensor product of $V_i$,$i=1,2$ and each $V_i$ is an irreducible representation of $G_i$. The question I have is when $V$ is given, is there a canonical way to construct $V_i$ from $V$?
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I agree with David Speyer's answer, and furthermore there is no canonical way to construct $V_i$ from $V$. This is a subtle and oft-overlooked point in representation theory, in my opinion. Many texts prove that an irrep of $G_1 \times G_2$ is isomorphic to a tensor product of an irrep of $G_1$ with an irrep of $G_2$. The typical slick proof relies on character theory -- kind of a cheat, in my view, since it only says something about isomorphism classes. Here's a categorical explanation of the theorem: Let $G_1$ and $G_2$ be finite groups, and let $\pi$ be an irrep of $G_1 \times G_2$ on a complex vector space $V$. Then, for every pair $(\rho_1, W_1)$, $(\rho_2, W_2)$ of representations of $G_1, G_2$, one gets a complex vector space:
$$H_\pi(\rho_1, \rho_2) := Hom_G(\rho_1 \boxtimes \rho_2, \pi).$$
In fact, this extends to a contravariant functor:
$$H_\pi: Rep_{G_1} \times Rep_{G_2} \rightarrow Vec.$$
Here we use categories of finite-dimensional complex representations and vector spaces. This is also functorial in $\pi$, yielding a functor:
$$H: Rep_G \rightarrow [ Rep_{G_1} \times Rep_{G_2}, Vec ],$$
where the right side of the arrow denotes the category of functors (for categories enriched in $Vec$). What this demonstrates is that the canonical thing is to take representations of $G$ to objects of an appropriate functor category related to $Rep_{G_1}$ and $Rep_{G_2}$. By Yoneda's lemma (for categories enriched in $Vec$), there is an embedding of categories:
$$Rep_{G_1} \times Rep_{G_2} \hookrightarrow [ Rep_{G_1} \times Rep_{G_2}, Vec ].$$ It turns out -- and this is where some finiteness is important, and a proof necessarily uses some counting, character theory, or the like -- that for any irrep $\pi$ of $Rep_G$, the functor $H_\pi \in [Rep_{G_1} \times Rep_{G_2}, Vec]$ is representable. It is not uniquely representable, but it is uniquely representable up to natural isomorphism. Practically, what this means is that given an irrep $\pi$ of $G$, there exists an isomorphism $\iota: \pi \rightarrow \pi_1 \boxtimes \pi_2$ for some irreps $\pi_1, \pi_2$ of $G_1, G_2$, respectively. The pair $(\pi_1, \pi_2)$ is not unique, but the triple $(\iota, \pi_1, \pi_2)$ is unique up to unique isomorphism. This is usually good enough.
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18,034 |
If you have two manifolds $M^m$ and $N^n$, how does one / can one decompose the diffeomorphisms $\text{Diff}(M\times N)$ in terms of $\text{Diff}(M)$ and $\text{Diff}(N)$? Is there anything we can say about the structure of this group? I have looked in some of my textbooks, but I haven't found any actual discussion of the manifolds $\text{Diff}(M)$ of a manifold $M$ other than to say they are "poorly understood." Can anyone point me to a source that discusses the manifold $\text{Diff}(M)$? My background is in physics, and understanding the structure of these kinds of groups is important for some of the the things we do, but I haven't seen any discussion of this in my differential geometry textbooks.
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The homotopy-type of the group of diffeomorphisms of a manifold are fairly well understood in dimensions $1$ , $2$ and $3$ . For a sketch of what's known see Hatcher's "Linearization in three-dimensional topology," in: Proc. Int. Congress of. Math., Helsinki, Vol. I (1978), pp. 463-468. Similarly, the finite subgroups of $Diff(M)$ are well understood in dimensions $3$ and lower. Hatcher's paper is a good reference for that as well, when combined with a few semi-recent theorems. If you're interested in general subgroups of $Diff(M)$ , there's still a fair bit of discussion going on just for subgroups of $Diff(S^1)$ , as it contains a pretty rich collection of subgroups. In high dimensions there's not much known. For example, nobody knows if $Diff(S^4)$ has any more than two path-components. See for example this little blurb. Some of the rational homotopy groups of $Diff(S^n)$ are known for $n$ large enough. I wrote a survey on what's known about the spaces $Diff(S^n)$ , and spaces of smooth embeddings of one sphere in another $Emb(S^j,S^n)$ a few years ago: A family of embedding spaces (Budney, 2006) . Getting back to your earlier question, groups of diffeomorphisms of connect-sums can be pretty compicated objects. In dimension $2$ it's already interesting. For example, $Diff(S^1 \times S^1)$ has the homotopy-type of $S^1 \times S^1 \times GL_2(\mathbb Z)$ . Diff of a connect-sum of $g$ copies of $S^1 \times S^1$ has the homotopy-type of a discrete group provided $g>1$ , this is called the mapping class group of a surface of genus $g$ . It's a pretty complicated and heavily-studied object. In the genus $g=2$ case this group is fairly similar to the braid group on $6$ strands. In dimension $3$ , it's an old theorem of Hatcher's that $Diff(S^1 \times S^2)$ doesn't have the homotopy-type of a finite-dimensional CW-complex, as it has the homotopy-type of $O_2 \times O_3 \times \Omega SO_3$ . I've been spending a lot of time recently, studying the homotopy-type of $Diff(M)$ when $M$ is the complement of a knot in $S^3$ , and knot complements in general. The paper of mine I linked to goes into some detail on this. From the perspective of differential geometry, the homotopy-type of $Diff(S^n)$ is rather interesting as it's closely related to the homotopy-type of the space of "round Riemann metrics" on $S^n$ . This is a classic construction, is outlined in my paper but it goes like this: $Diff(S^n)$ has the homotopy type of a product $O_{n+1} \times Diff(D^n)$ where the diffeomorphisms of $D^n$ are required to be the identity on the boundary -- this is a local linearization argument. $Diff(D^n)$ has the homotopy-type of the space of round metrics on $S^n$ . The idea is that any two round metrics are related by a diffeomorphism of $S^n$ . So $Diff(S^n)$ acts transitively on the space of round metrics (with a fixed volume, say), and the stabilizer of a round metric is $O_{n+1}$ basically by the definition of a round metrics. Kind of silly but fundamental. edit: I should add, there are some nice theorems about $\pi_0 Diff(S^1 \times D^n)$ for $n$ at least 5, and similarly $\pi_0 Diff( (S^1)^n )$ , due to Hatcher and Wagoner. They derive their results in some sense indirectly, by getting a strong understanding of the pseudo-isotopy diffeomorphisms of $S^1 \times D^n$ . One way to think about their work, is that the isotopy-classes of diffeomorphisms of $S^1 \times D^n$ (fixing the boundary pointwise) is governed by three groups: (1) $\pi_0 Diff(D^n)$ , (2) $\pi_0 Diff(D^{n+1})$ and (3) $\pi_0 Emb(D^n, S^1 \times D^n) / Diff(D^n)$ . Think of this last group as the isotopy-classes of embedded n-discs in $S^1 \times D^n$ that agree with a standard linear embedding on the boundary. i.e. these are submanifolds without parametrization. It turns out this is a group with a stacking construction. Hatcher and Wagoner show that $\pi_0 Diff(S^1 \times D^n)$ is the direct sum of these three groups, with this group of embedded discs being an infinitely-generated $2$ -torsion group. Recently David Gabai and I were able to give a weak analogue to this Hatcher-Wagoner theorem but in dimension $4$ , i.e. for $\pi_0 Diff(S^1 \times D^3)$ . While we have not managed any new results about $\pi_0 Diff(D^4)$ , we can show $\pi_0 Emb(D^3, S^1 \times D^3)$ is infinitely-generated, even rationally. From a certain perspective our embeddings of $D^3$ in $S^1 \times D^3$ are fairly similar to the Hatcher-Wagoner embeddings, but we rely on slightly different geometry than they do. Roughly speaking, our embeddings stem more from the ability for $S^2 \sqcup S^1$ being able to link in $S^4$ , while Hatcher and Wagoner's construction has more to do with $S^i \sqcup S^j$ for $i+j=n$ being able to Hopf link in $S^n$ when $i,j \geq 3$ . In a vague sense that is the key difference between our diffeomorphisms being rationally independent, while theirs are $2$ -torsion.
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18,041 |
Peter May said famously that algebraic topology is a subject poorly served by its textbooks. Sadly, I have to agree. Although we have a freightcar full of excellent first-year algebraic topology texts - both geometric ones like Allen Hatcher's and algebraic-focused ones like the one by Rotman and more recently, the beautiful text by tom Dieck (which I'll be reviewing for MAA Online in 2 weeks, watch out for that!) - there are almost no texts which bring the reader even close to the frontiers of the subject. GEOMETRIC topology has quite a few books that present its modern essentials to graduate student readers - the books by Thurston, Kirby and Vassiliev come to mind - but the vast majority of algebraic topology texts are mired in material that was old when Ronald Reagan was President of The United States. This is partly due to the youth of the subject, but I think it's more due to the sheer vastness of the subject now. Writing a cutting edge algebraic topology textbook - TEXTBOOK, not MONOGRAPH - is a little like trying to write one on algebra or analysis. The fields are so gigantic and growing, the task seems insurmountable. There are only 2 "standard" advanced textbooks in algebraic topology and both of them are over 30 years old now: Robert Switzer's Algebraic Topology: Homology And Homotopy and George Whitehead's Elements of Homotopy Theory . Homotopy theory in particular has undergone a complete transformation and explosive expansion since Whitehead wrote his book. (That being said, the fact this classic is out of print is a crime.) There is a recent beautiful textbook that's a very good addition to the literature, Davis and Kirk's Lectures in Algebraic Topology - but most of the material in that book is pre-1980 and focuses on the geometric aspects of the subject. We need a book that surveys the subject as it currently stands and prepares advanced students for the research literature and specialized monographs as well as makes the subject accessible to the nonexpert mathematician who wants to learn the state of the art but not drown in it. The man most qualified to write that text is the man to uttered the words I began this post with. His beautiful concise course is a classic for good reason; we so rarely have an expert give us his "take" on a field. It's too difficult for a first course, even for the best students, but it's "must have" supplementary reading. I wish Dr. May - perhaps when he retires - will find the time to write a truly comprehensive text on the subject he has had such a profound effect on. Anyone have any news on this front of future advanced texts in topology? I'll close this box and throw it open to the floor by sharing what may be the first such textbook available as a massive set of online notes. I just discovered it tonight; it's by Garth Warner of The University Of Washington and available free for download at his website. I don't know if it's the answer, but it sure looks like a huge step in the right direction. Enjoy. And please comment here. http://www.math.washington.edu/~warner/TTHT_Warner.pdf
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Good lord, Charles, was the reposting of this an invitation for another advertisement from me?
``More concise algebraic topology. Localization, completion, and model categories'', by Kate Ponto
and myself, is available for purchase and will be formally and officially published next month. I have a copy in my hand, and the final version is 514 pages (including Bibliography and Index).
Still 65 dollars (and don't fall for pirate editions on the web). It is not perfect, of course. (I know of one careless mistake every reader will catch and one subtle mistake almost no reader will catch). I offer $10 to any reader finding a mistake I don't know about, even misprints. The book is intended to help fill the gap (and another, more calculational, follow up to Concise is planned). The first half covers localization and completion and is more technical than I hoped simply because so much detail was needed to fill out the theory as it was left in the great sources from the early 1970's (Bousfield-Kan, Sullivan, Hilton-Mislin-Roitberg, etc), especially about fracture theorems. The second half is an introduction to model category theory, and it has a number of idiosyncratic features, such as emphasis on the trichotomy of Quillen, Hurewicz, and mixed model structures on spaces and chain complexes. The order is deliberate: novices should see a worked example of serious homotopical algebra before starting on categorical homotopy theory. There is a bonus track on Hopf algebras for algebraic topologists and a brief primer
on spectral sequences. There are example applications sprinkled around, although more might have been desirable. The book is quite long enough as it is. Merry Christmas all.
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18,084 |
Wikipedia defines the Jaccard distance between sets A and B as $$J_\delta(A,B)=1-\frac{|A\cap B|}{|A\cup B|}.$$ There's also a book claiming that this is a metric. However, I couldn't find any explanation of why $J_\delta$ obeys the triangle inequality. The naive approach of writing the inequality with seven variables (e.g., $x_{001}$ thru $x_{111}$ , where $x_{101}$ is the number of elements in $(A\cap C) \backslash B$ ) and trying to reduce it seems hopeless for pen and paper. In fact it also seems hopeless for Mathematica, which is trying to find a counterexample for 20 minutes and is still running. (It's supposed to say if there isn't any.) Is there a simple argument showing that this is a distance? Somehow, it feels like the problem shouldn't be difficult and I'm missing something.
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The trick is to use a transform called the Steinhaus Transform. Given a metric $(X, d)$ and a fixed point $a \in X$ , you can define a new distance $D'$ as $$D'(x,y) = \frac{2D(x,y)}{D(x,a) + D(y,a) + D(x,y)}.$$ It's known that this transformation produces a metric from a metric. Now if you take as the base metric $D$ the symmetric difference between two sets and empty set as $a$ , what you end up with is the Jaccard distance (which actually is known by many other names as well). For more information and references, check out Ken Clarkson's survey Nearest-neighbor searching and metric space dimensions (Section 2.3).
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18,089 |
I am aware that assigning the type of Type to be Type (rather than stratifying to a hierarchy of types) leads to an inconsistency. But does this inconsistency allow the construction of a well-typed term with no normal form, or does it actually allow a proof of False? Are these two questions equivalent?
|
Girard's paradox constructs a non-normalizing proof of False. You could read Hurken's "A simplification of Girard's paradox", or maybe Kevin Watkin's formalization in Twelf . In general, these questions are not equivalent, though they often coincide. A "reasonable" type theory will by inspection have no normal proofs of False, and so then normalization implies consistency. The inverse (non-normalization => proof of False) is much less obvious, and it is certainly possible to construct reasonable paraconsistent type theories, where non-termination is confined under a monad and does not result in a proof of False.
|
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18,094 |
If $p_n$ is the $n$'th prime, let $A_n(x) = x^n + p_1x^{n-1}+\cdots + p_{n-1}x+p_n$. Is $A_n$ then irreducible in $\mathbb{Z}[x]$ for any natural number $n$?
I checked the first couple of hundred cases using Maple, and unless I made an error in the code those were all irreducible.
I have thought about this for a long time now, and asked many others, with no answer yet.
|
I will prove that $A_n$ is irreducible for all $n$, but most of the credit goes to Qiaochu. We have
$$(x-1)A_n = b_{n+1} x^{n+1} + b_n x^n + \cdots + b_1 x - p_n$$
for some positive integers $b_{n+1},\ldots,b_1$ summing to $p_n$. If $|x| \le 1$, then
$$|b_{n+1} x^{n+1} + b_n x^n + \cdots + b_1 x| \le b_{n+1}+\cdots+b_1 = p_n$$
with equality if and only $x=1$, so the only zero of $(x-1)A_n$ inside or on the unit circle is $x=1$. Moreover, $A_n(1)>0$, so $x=1$ is not a zero of $A_n$, so every zero of $A_n$ has absolute value greater than $1$. If $A_n$ factors as $B C$, then $B(0) C(0) = A_n(0) = p_n$, so either $B(0)$ or $C(0)$ is $\pm 1$. Suppose that it is $B(0)$ that is $\pm 1$. On the other hand, $\pm B(0)$ is the product of the zeros of $B$, which are complex numbers of absolute value greater than $1$, so it must be an empty product, i.e., $\deg B=0$. Thus the factorization is trivial. Hence $A_n$ is irreducible.
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18,100 |
I am interested in theorems with unexpected conclusions. I don't mean
an unintuitive result (like the existence of a space-filling curve), but
rather a result whose conclusion seems disconnected from the
hypotheses. My favorite is the following. Let $f(n)$ be the number of
ways to write the nonnegative integer $n$ as a sum of powers of 2, if
no power of 2 can be used more than twice. For instance, $f(6)=3$
since we can write 6 as $4+2=4+1+1=2+2+1+1$. We have
$(f(0),f(1),\dots) = $ $(1,1,2,1,3,2,3,1,4,3,5,2,5,3,4,\dots)$. The
conclusion is that the numbers $f(n)/f(n+1)$ run through all the
reduced positive rational numbers exactly once each. See A002487 in
the On-Line Encyclopedia of Integer Sequences for more
information. What are other nice examples of "unexpected conclusions"?
|
My favorite example of this phenomenon is Goodstein's Theorem . Take any positive number $a_2$ , such as the number $73$ , and write it in complete base $2$ , which means write it as a sum of powers of $2$ , but write the exponents also in this way. $$a_2 = 73 = 64 + 8 + 1 = 2^{2^2+2} + 2^{2+1} + 1.$$ Now, obtain $a_3$ by replacing all $2$ 's with $3$ 's, and subtracting $1$ . So in this case, $$a_3 = 3^{3^3+3} + 3^{3+1} + 1 - 1 = 3^{3^3+3} + 3^{3+1}.$$ Similarly, write this in complete base $3$ , replace $3$ 's with $4$ 's, and substract one, to get $$a_4= 4^{4^4+4} + 4^{4+1} - 1 = 4^{4^4+4} + 3\cdot 4^4 + 3\cdot 4^3 + 3\cdot 4^2 + 3\cdot 4 + 3.$$ And so on. The surprising conclusion is that: Goodstein's Theorem. For any initial positive integer $a_2$ , there is $n>2$ for which $a_n=0$ . That is, although it seems that the sequence is always growing larger, eventually it hits zero. Our initial impression that this process should proceed to ever larger numbers is simply not correct. Indeed, a stronger version of the theorem, with essentially the same proof, allows one to replace the current base with any desired larger base, rather than merely increasing it by $1$ at each step. To prove Goodstein's theorem, one can use the transfinite ordinals to measure the complexity of the numbers that arise. For example, our initial number $a_2=73$ is associated with the ordinal $\omega^{\omega^\omega+\omega}+\omega^{\omega+1}+1$ . One then proves that these associated ordinals strictly descend at each step, because of the subtracting one part. Thus, it must hit zero, and the only way this happens is if the number itself is zero. One can see that we had to split up the complexity of the number in moving from $a_3$ to $a_4$ , and this causes the associated ordinal to strictly descend, even though in this case the actual number happened to get larger. Eventually, the proof goes, the complexity drops low enough that the base exceeds the number, and from this point on, one is just subtracting one endlessly. This conclusion is very surprising. But this theorem actually packs a one-two punch! Because not only is the theorem itself surprising, but then there is the following surprise follow-up theorem: Theorem. Goodstein's theorem is not provable in the usual Peano Axioms PA of arithmetic. That is, the statement of Goodstein's theorem is independent of PA. It was a statement about finite numbers that is provable in ZFC, but not in PA.
|
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|
18,313 |
I am currently doing a self study on algebraic geometry but my ultimate goal is to study more on elliptic curves. Which are the most recommended textbooks I can use to study? I need something not so technical for a junior graduate student but at the same time I would wish to get a book with authority on elliptic curves. Thanks
|
Silverman and Tate to start, then Silverman , and finally Silverman again . These are basically canonical references for the subject.
|
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|
18,336 |
Background Model categories are an axiomization of the machinery underlying the study of topological spaces up to homotopy equivalence. They consist of a category $C$ , together with three distinguished classes of morphism: Weak Equivalences, Fibrations, and Cofibrations. There are then a series of axioms this structure must satisfy, to guarantee that the classes behave analogously to the topological maps they are named after. The axioms can be found here . (As far as I know...) The main practical advantage of this machinery is that it gives a rather concrete realization of the localization category $C/\sim$ where the Weak Equivalences have been inverted, which generalizes the homotopy category of topological spaces. The main conceptual advantage is that it is a first step towards formalizing the concept of "a category enriched over topological spaces". A discussion of examples and intuition can be found at this question . The Question The examples found in the answers to Ilya's question , as well as in the introductory papers I have read, all have a model category structure that could be expected. They are all examples along the lines of topological spaces, derived categories, or simplicial objects, which are all conceptually rooted in homotopy theory and so their model structures aren't really surprising. I am hoping for an example or two which would elicit disbelief from someone who just learned the axioms for a model category. Along the lines of someone who just learned what a category being briefly skeptical that any poset defines a category, or that ' $n$ -cobordisms' defines a category.
|
Here is an example that surprised me at some time in the past.
Bisson and Tsemo introduce a nontrivial model structure on the topos of directed graphs.
Here a directed graph is simply a $4$-tuple $(V,E,s,t)$ where an arc $e \in E$ starts at $s(e) \in V$ and ends at $t(e) \in V$. Fibrations are maps that induce a surjection on the set of outgoing arcs of each vertex, cofibrations are embeddings obtained by attaching a bunch of trees, and weak equivalences are maps that induce a bijection on the sets of cycles. In this model structure fibrant objects are graphs without sinks and
cofibrant objects are graphs with exactly one incoming arc for every vertex.
Cofibrant replacement replaces a graph by the disjoint union of its cycles
with the obvious morphism into the original graph. We have a chain of inclusions of categories $A\to B \to C\to D$, where $D$ is the topos of directed graphs, $C$ is the full subcategory of $D$ consisting of all graphs with exactly
one incoming arc for each vertex, $B$ is the full subcategory of $C$ consisting of all graphs with exactly one outgoing arc for each vertex, and $A$ is the full subcategory of $B$ consisting of all graphs such that $s=t$. Each functor is a part of a Quillen adjunction and total left and right derived
functors compute nontrivial information about graphs under consideration. Two finite graphs are homotopy equivalent iff they are isospectral iff their zeta-functions coincide.
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18,352 |
As a non-native English speaker (and writer) I always had the problem of understanding the distinction between a 'Theorem' and a 'Proposition'. When writing papers, I tend to name only the main result(s) 'Theorem', any auxiliary result leading to this Theorem a 'Lemma' (and, sometimes, small observations that are necessary to prove a Lemma are labeled as 'Claim'). I avoid using the term 'Proposition'. However, sometimes a paper consists of a number important results (which by all means earn to be called 'Theorem') that are combined to obtain a certain main result. Hence, another term such as 'Proposition' might come in handy, yet I don't know whether it suits either the main or the intermediate results. So, my question is: When to use 'Theorem' and when to use 'Proposition' in a paper?
|
The way I do it is this: main results are theorems, smaller results are called propositions.
A Lemma is a technical intermediate step which has no standing as an independent result.
Lemmas are only used to chop big proofs into handy pieces.
|
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|
18,421 |
In an unrelated thread Sam Nead intrigued me by mentioning a formalized proof of the Jordan curve theorem. I then found that there are at least two, made on two different systems. This is quite an achievement, but is it of any use for a mathematician like me? (No this is not what I am asking, the actual question is at the end.) I'd like to trust the theorems I use. To this end, I can read a proof myself (the preferred way, but sometimes hard to do) or believe experts (a slippery road). If I knew little about topology but occasionally needed the Jordan theorem, a machine-verified proof could give me a better option (and even if I am willing to trust experts, I could ensure that there are no hidden assumptions obvious to experts but unknown to me). But how to make sure that a machine verified the proof correctly? The verifying program is too complex to be trusted. A solution is of course that this smart program generates a long, unreadable proof that can be verified by a dumb program (so dumb that an amateur programmer could write or check it). I mean a program that performs only primitive syntax operations like "plug assertions 15 and 28 into scheme 9". This "dumb" part should be independent of the "smart" part. Given such a system, I could check axioms, definitions and the statement of the theorem, feed the dumb program (whose operation I can comprehend) with these formulations and the long proof, and see if it succeeds. That would convince me that the proof is indeed verified. However I found no traces of this "dumb" part of the system. And I understand that designing one may be hard. Because the language used by the system should be both human-friendly (so that a human can verify that the definitions are correct) and computer-friendly (so that a dumb program can parse it). And the definitions should be chosen carefully - I don't want to dig through a particular construction of the reals from rationals to make sure that this is indeed the reals that I know. Sorry for this philosophy, here is the question at last. Is there such a "dumb" system around? If yes, do formalization projects use it? If not, do they recognize the need and put the effort into developing it? Or do they have other means to make their systems trustable? UPDATE: Thank you all for interesting answers. Let me clarify that the main focus is interoperability with a human mathematician (who is not necessarily an expert in logic). It seems that this is close to interoperability between systems - if formal languages accepted by core checkers are indeed simple, then it should be easy to automatically translate between them. For example, suppose that one wants to stay within symbolic logic based on simple substitutions and axioms from some logic book. It seems easy to write down these logical axioms plus ZF axioms, basic properties (axioms) of the reals and the plane, some definitions from topology, and finally the statement of the Jordan curve theorem. If the syntax is reasonable, it should be easy to write a program verifying that another stream of bytes represents a deduction of the stated theorem from the listed axioms. Can systems like Mizar, Coq, etc, generate input for such a program? Can they produce proofs verifiable by cores of other systems?
|
Is there such a "dumb" system around? If yes, do formalization projects use it? If not, do they recognize the need and put the effort into developing it? Or do they have other means to make their systems trustable? This is called the "de Bruijn criterion" for a proof assistant -- just as you say, we want a simple proof checker, which should be independent of the other machinery. The theorem provers which most directly embody this methodology are those in the LCF tradition, such as Isabelle and HOL/Light. They actually work by generating proof objects via whatever program you care to write, and sending that to a small core to check. Systems based on dependent type theory (such as Coq) tend to have more complex logical cores (due to the much greater flexibility of the underlying logic), but even here a core typechecker can fit in a couple of thousand lines of code, which can easily be ( and have been ) understood and reimplemented.
|
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|
18,423 |
I've got a group $G$ that I'm trying to prove is free. I already know that $G$ is torsion-free. Moreover, I can "almost" prove what I want : I can find a finite index subgroup $G'$ of $G$ that is definitely free. This leads me to the following question. Can anyone give me an example of a torsion-free group $G$ that is not free but contains a free subgroup of finite index? I've tried pretty hard to find groups like this, but i can't seem to avoid introducing torsion. Thanks!
|
It's a theorem of Stallings and Swan that a group of cohomological dimension one is free. By a theorem of Serre, torsion-free groups and their finite index subgroups have the same cohomological dimension. So, a torsion-free group is free if and only if one of its finite index subgroups are free. (Here are the references. For Stallings-Swan, see John R. Stallings, "On torsion-free groups with infinitely many ends", Annals of Mathematics 88 (1968), 312–334. and Richard G. Swan, "Groups of cohomological dimension one", Journal of Algebra 12 (1969), 585–610. Serre's theorem is in Brown's book "Cohomology of Groups.")
|
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|
18,440 |
This summer my wife and one of my friends (who are both programmers and undergraduate math majors, but have not learned any algebraic geometry) want to learn some algebraic geometry from me, and I want to learn how to program from them, so we were planning on working on some computational algebraic geometry together. While there are several books which we could work through, I thought it might be more fun and productive if we had the goal of developing a usable new algorithm, or at least implementing an algorithm which no one has implemented before. I do not have any ideas, but I thought that some mathoverflowers might have had an idea for an algorithm they would like to see implemented but have never had the time to work through the details. Keep in mind that my wife and friend will have to learn any mathematics past a first course in topology and abstract algebra as we go. So does anyone have any ideas for an algorithm they would like to use which is within the reach of my "team" to implement within a summer? We are planning on working on this stuff between 2 and 3 hours a day for about 3 months.
|
Just a thought, but maybe you should have a look at sage . It's a big open source project that is currently under very active development. If you're interested in contributing, I would suggest that you post to the sage-devel Google group with this same question. Some thoughts for things to do would be to improve the support for relative extensions of number fields and for function fields.
|
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|
18,496 |
Let $R$ be a Noetherian domain, and let $\mathfrak{p}$ be a prime ideal; consider the completion $\hat R_{\mathfrak{p}}$ of $R$ at $\mathfrak{p}$ (the inverse limit of the system of quotients $R/\mathfrak{p}^n$). If $R$ is a PID, it is easy to see that $\hat R_{\mathfrak{p}}$ is a domain. Someone asked in sci.math if $\hat R_{\mathfrak{p}}$ would always be a domain. I thought it would, but looking at Eisenbud's "Commutative Algebra", I found a reference to a theorem of Larfeldt and Lech that says that if $A$ is any finite-dimensional algebra over a field $k$, then there is a Noetherian local integral domain $R$ with maximal ideal $\mathfrak{m}$ such that $\hat{R_{\mathfrak{M}}}\cong A[[x_1,\ldots,x_n]]$ for some $n$; and so this completion will not be a domain if $A$ is not a domain. I would like to know an example directly, if possible. Does someone know an easy example of a noetherian domain $R$ and a prime ideal $\mathfrak{p}$ such that $\hat R_{\mathfrak{p}}$ is not a domain? Thanks in advance.
|
Let $R=\mathbb{C}[x,y]/(y^2-x^2(x-1))$. This is the nodal cubic in the plane. Look at the prime $\mathfrak{p}=(x,y)$, corresponding to the nodal point. The completion here is isomorphic to $\mathbb{C}[[x,y]]/(xy)$.
|
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|
18,539 |
As I understand it, Lions and DiPerna demonstrated existence and uniqueness for the Boltzmann equation . Moreover, this paper claims that Appropriately scaled families of
DiPerna–Lions renormalized solutions
of the Boltzmann equation are shown to
have fluctuations whose limit points
(in the weak $L^1$ topology) are governed
by a Leray solution of the limiting
Navier–Stokes equations. Probably there is a lot of other work along these lines. But I am not well-versed enough in these areas to go through the literature easily, and so I hope someone can give a very high-level answer to my question: Why does renormalizing the Boltzmann
equation not (yet?) give existence and
uniqueness for Navier-Stokes?
|
Okay, after figuring out which paper you were trying to link to in the third link, I decided that it is better to just give an answer rather then a bunch of comments. So... there are several issues at large in your question. I hope I can address at least some of them. The "big picture" problem you are implicitly getting at is the Hilbert problem of hydrodynamical limit of the Boltzmann equations: that intuitively the ensemble behaviour at the large, as model by a fluid as a vector field on a continuum, should be derivable from the individual behaviour of particles, as described by kinetic theory. Very loosely tied to this is the problem of global existence and regularity of Navier-Stokes. If your goal is to solve the Navier-Stokes problem using the hydrodynamic limit, then you need to show that (a) there are globally unique classical solutions to the the Boltzmann equations and (b) that they converge in a suitably regular norm, in some rescaling limit, to a solution of Navier-Stokes. Neither step is anywhere close to being done. As far as I know, there are no large data, globally unique, classical solutions to the Boltzmann equation. Period. If we drop some of the conditions, then yes: for small data (perturbation of Maxwellian), the recent work of Gressman and Strain (0912.0888) and Ukai et al (0912.1426) solve the problem for long-range interactions (so not all collision kernels are available). If you drop the criterion of global, there are quite a bit of old literature on local solutions, and if you drop the criterion of unique and classical, you have the DiPerna-Lions solutions (which also imposes an angular-cutoff condition that is not completely physical). The work of Golse and Saint-Raymond that you linked to establishes the following: that the weak solution of DiPerna-Lions weakly converges to the well-known weak solutions of Leray for the Navier-Stokes problem. While this, in some sense, solve the problem of Hilbert, it is rather hopeless for a scheme trying to show global properties of Navier-Stokes: the class of Leray solutions are non-unique. As I see it, to go down this route, you'd need to (i) prove an analogue of DiPerna-Lions, or to get around it completely differently, and arrive at global classical and unique solutions for Boltzmann. This is a difficult problem, but I was told that a lot of very good people are working on it. Then you'd need (ii) also to prove an analogue of Golse-Saint-Raymond in a stronger topology, or you can use Golse-Saint-Raymond to first obtain a weak-limit that is a Leray solution, and then show somehow that regularity is preserved under this limiting process. This second step is also rather formidable. I hope this somewhat answers your question.
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|
18,588 |
Then one can construct a model for the inverse limit by taking all the compatible sequences.
This is a subspace of a product of compact spaces. This product is compact by Tychonoff. If all the spaces are Hausdorff, then this is even a closed subspace. However, if the spaces are not Hausdorff, it needn't be a closed subspace. If you take a two point space with the trivial topology as $X_n$ and constant structure maps, you will get as the inverse limit the space of all constant sequences, which is not a closed subspace of the infinite product, as the infinite product also has the trivial topology. But the space is again compact. So I am wondering, whether there is a generalization of the proof of Tychonoff's theorem, that applies directly to inverse limits.
|
What does this example do ... All spaces are on set $\{1,2,\dots\}$. Space $X_n$ has topology that makes $\{1,2,\dots,n\}$ discrete and $\{n+1,\dots\}$ indiscrete. Of course $X_n$ is compact non-Hausdorff. Map $X_{n+1} \to X_n$ by the "identity". Inverse limit is ... ???
|
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|
18,593 |
With which notation do you feel uncomfortable?
|
There is a famous anecdote about Barry Mazur coming up with the worst notation possible at a seminar talk in order to annoy Serge Lang. Mazur defined $\Xi$ to be a complex number and considered the quotient of the conjugate of $\Xi$ and $\Xi$: $$\frac{\overline{\Xi}}{\Xi}.$$ This looks even better on a blackboard since $\Xi$ is drawn as three horizonal lines.
|
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|
18,636 |
This question is inspired from here , where it was asked what possible determinants an $n \times n$ matrix with entries in {0,1} can have over $\mathbb{R}$. My question is: how many such matrices have non-zero determinant? If we instead view the matrix as over $\mathbb{F}_2$ instead of $\mathbb{R}$, then the answer is $(2^n-1)(2^n-2)(2^n-2^2) \dots (2^n-2^{n-1}).$ This formula generalizes to all finite fields $\mathbb{F}_q$, which leads us to the more general question of how many $n \times n$ matrices with entries in { $0, \dots, q-1$ } have non-zero determinant over $\mathbb{R}$?
|
See Sloane, A046747 for the number of singular (0,1)-matrices. It doesn't seem like there's an exact formula, but it's conjectured that the probability that a random (0,1)-matrix is singular is asymptotic to $n^2/2^n$. Over $F_2$ the probability that a random matrix is nonsingular, as $n \to \infty$, approaches the product $(1/2)(3/4)(7/8)\cdots = 0.2887880951$, and so the probability that a random large matrix is singular is only around 71 percent. I should note that a matrix is singular over $F_2$ if its real determinant is even, so this tells us that determinants of 0-1 matrices are more likely to be even than odd.
|
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|
18,666 |
Does there exist a continuous function $f:[0,1]\rightarrow [0,1]$ such that $f$ takes every value in $[0,1]$ an infinite number of times?
|
Yes. In fact, there exists such an $f$ taking every value uncountably many times. Take a continuous surjection $g: [0, 1] \to [0, 1]^2$. (Such things exist: they're space-filling curves.) Then the composite $f$ of $g$ with first projection $[0, 1]^2 \to [0, 1]$ has the required property.
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18,747 |
Well, I don't have any notion of arithmetic geometry , but I would like to understand what arithmetic geometers mean when they say "integer point of a variety/scheme $X$" (like e.g. in "integer points of an elliptic curve"). Is an integer point just defined as a morphism from $Spec\mathbb{Z}$ into $X$? Suppose $X$ is given a structure of variety over, say, $Spec\mathbb{C}$; does the notion of integer point interact with this structure? How is all of this related to finding the integer solutions of a concrete polynomial equation (perhaps with integer coefficients)? Now a very, very , naif question. Suppose you have the plane $\mathbb{A}^{2}_{\mathbb{C}}$ and draw two lines on it: $X=\{x=0\}$ and $X'=\{x=\pi\}$ (with the obvious reduced induced closed subscheme structure). Well, $X$ and $X'$ are clearly isomorphic as schemes over $\mathbb{C}$ (hence as schemes). But, if having integer points is somehow related to finding integer solutions to equations, how do you explain that $X$ (as embedded in $\mathbb{A}^{2}$) has plenty of points with integer coordinates , while $X'$ has none?
This to mean: how can the intrinsic notion of a morphism from $Spec\mathbb{Z}$ be possibly related to finding solutions to concrete (coordinate dependent) equations?
|
To make sense of the notion of integer points, your scheme should be defined over $\mathbb{Z}$. What do we mean by that? Of course we should not ask for a structure map tp $Spec(\mathbb{Z})$, since every scheme has one such map. The right notion is the following. Let $X$ be a scheme over $\mathbb{C}$; so by definition we have a structure map $X \to \mathop{Spec} \mathbb{C}$. Then we say that $X$ is defined over $\mathbb{Z}$ is there exists a scheme $X_{\mathbb{Z}}$ over $\mathbb{Z}$ such that $X$ is the base change of $X_{\mathbb{Z}}$ to $\mathbb{C}$, i. e. $X \cong X_{\mathbb{Z}} \times_{\mathbb{Z}} \mathop{Spec} \mathbb{C}$. Now for such a scheme an integral point is a map $\mathop{Spec}\mathbb{Z} \to X_{\mathbb{Z}}$ such that the composition with the structure map is the identity. Note that the same can be done for every ring $A$ in place of $\mathbb{Z}$. With this definition, the line $\{ x = 0 \}$ is defined over $\mathbb{Z}$, but the line $\{ x = \pi \}$ is not, basically because there is no way to generate its ideal with equations having integer coefficients. So your problem does not arise anymore. EDIT: Abstractly of course the two lines are isomorphic over $\mathbb{C}$, so the line $r = \{x = \pi \}$ actually has a model over $\mathbb{Z}$. The problem is that this model is not compatible with the inclusion in $\mathbb{A}^2$, that is, there will be no map $r_{\mathbb{Z}} \to \mathbb{A}^2_\mathbb{Z}$ whose base change is the inclusion of $r$ into $\mathbb{A}^2$. In order to have this, you would have to ask that the ideal of $r$ in $\mathbb{A}^2$ should be generated by polynomials with integer coefficients. As for your second question, there can be different models, that is, nonisomorphic schemes over $\mathbb{Z}$ which become isomorphic after base change to $\mathbb{C}$. So before discussing the existence of integral points, you have to FIX a model, and the points will in general depend on the model. For instance take the two conics $\{ x^2 + y^2 = 2 \}$ and $\{ x^2 + y^2 = 3 \}$. Both have an obvious choice of a model, given by the inclusion in $\mathbb{A}^2$; moreover they are isomorphic over $\mathbb{C}$. But the integral points on the first one are $(\pm 1, \pm 1)$, while the second has none. Finally you consider the possibility that the structure over $\mathbb{C}$ is not relevant. This is false: the base change $X_\mathbb{Z} \times_{\mathbb{Z}} \mathop{Spec} \mathbb{C}$ is endowed with a natural map to $\mathop{Spec} \mathbb{C}$, and we ask for the isomorphism with $X$ to be over $\mathbb{C}$.
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18,748 |
Let's say we want to define a choice function for certain particular subsets $S \subset2^{\mathbb{R}}$, i.e. we want a function $c:S \rightarrow \mathbb{R}$ such that $c(X)\in X$ for every $X\in S$. We don't want to invoke the axiom of choice. Clearly we require $\emptyset\notin S$. For example, if $S$ is the set of non-empty open sets for the usual topology, then we can fix an enumeration of the rationals and for every open $A$ pick the first rational (in this particular enumeration) lying in $A$. If $S$ is the set of non-empty closed sets, then for any $A\in S$ we can consider the least $n$ such that $\[-n,n\]\cap A$ is non empty and then pick the infimum of this non empty compact set. The question is the following: can you define a choice function for, say, $S=F_{\sigma}\setminus \{\emptyset\}$ or $G_{\delta}\setminus\{\emptyset\}$ or maybe for the higher levels of the Borel hierarchy? Is it possible to prove that such choice function exists for such $S$ without using the axiom of choice?
|
To make sense of the notion of integer points, your scheme should be defined over $\mathbb{Z}$. What do we mean by that? Of course we should not ask for a structure map tp $Spec(\mathbb{Z})$, since every scheme has one such map. The right notion is the following. Let $X$ be a scheme over $\mathbb{C}$; so by definition we have a structure map $X \to \mathop{Spec} \mathbb{C}$. Then we say that $X$ is defined over $\mathbb{Z}$ is there exists a scheme $X_{\mathbb{Z}}$ over $\mathbb{Z}$ such that $X$ is the base change of $X_{\mathbb{Z}}$ to $\mathbb{C}$, i. e. $X \cong X_{\mathbb{Z}} \times_{\mathbb{Z}} \mathop{Spec} \mathbb{C}$. Now for such a scheme an integral point is a map $\mathop{Spec}\mathbb{Z} \to X_{\mathbb{Z}}$ such that the composition with the structure map is the identity. Note that the same can be done for every ring $A$ in place of $\mathbb{Z}$. With this definition, the line $\{ x = 0 \}$ is defined over $\mathbb{Z}$, but the line $\{ x = \pi \}$ is not, basically because there is no way to generate its ideal with equations having integer coefficients. So your problem does not arise anymore. EDIT: Abstractly of course the two lines are isomorphic over $\mathbb{C}$, so the line $r = \{x = \pi \}$ actually has a model over $\mathbb{Z}$. The problem is that this model is not compatible with the inclusion in $\mathbb{A}^2$, that is, there will be no map $r_{\mathbb{Z}} \to \mathbb{A}^2_\mathbb{Z}$ whose base change is the inclusion of $r$ into $\mathbb{A}^2$. In order to have this, you would have to ask that the ideal of $r$ in $\mathbb{A}^2$ should be generated by polynomials with integer coefficients. As for your second question, there can be different models, that is, nonisomorphic schemes over $\mathbb{Z}$ which become isomorphic after base change to $\mathbb{C}$. So before discussing the existence of integral points, you have to FIX a model, and the points will in general depend on the model. For instance take the two conics $\{ x^2 + y^2 = 2 \}$ and $\{ x^2 + y^2 = 3 \}$. Both have an obvious choice of a model, given by the inclusion in $\mathbb{A}^2$; moreover they are isomorphic over $\mathbb{C}$. But the integral points on the first one are $(\pm 1, \pm 1)$, while the second has none. Finally you consider the possibility that the structure over $\mathbb{C}$ is not relevant. This is false: the base change $X_\mathbb{Z} \times_{\mathbb{Z}} \mathop{Spec} \mathbb{C}$ is endowed with a natural map to $\mathop{Spec} \mathbb{C}$, and we ask for the isomorphism with $X$ to be over $\mathbb{C}$.
|
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18,752 |
I am studying the homotopy type of a space,and i hope it would be a $K(\pi,1)$ space.
now i have find its covering,once we can say the covering is $K(\pi,1)$,so is the space
itself.and the covering is $\mathbb{R}^4-M$ where $M=M_1\cup M_2\cup M_3\cup M_4$, $M_1=\{(x,y,z,w)|x,y \in \mathbb R,z,w \in\mathbb Z\}$ $M_2=\{(x,y,z,w)|y,z \in \mathbb R,x,w \in\mathbb Z\}$ $M_3=\{(x,y,z,w)|x,w \in \mathbb R,y,z \in\mathbb Z\}$ $M_4=\{(x,y,z,w)|z,w \in \mathbb R,x,y \in\mathbb Z\}$ I guess $\mathbb{R}^4-M$ is $K(\pi,1)$ space,can someone help prove this?
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To make sense of the notion of integer points, your scheme should be defined over $\mathbb{Z}$. What do we mean by that? Of course we should not ask for a structure map tp $Spec(\mathbb{Z})$, since every scheme has one such map. The right notion is the following. Let $X$ be a scheme over $\mathbb{C}$; so by definition we have a structure map $X \to \mathop{Spec} \mathbb{C}$. Then we say that $X$ is defined over $\mathbb{Z}$ is there exists a scheme $X_{\mathbb{Z}}$ over $\mathbb{Z}$ such that $X$ is the base change of $X_{\mathbb{Z}}$ to $\mathbb{C}$, i. e. $X \cong X_{\mathbb{Z}} \times_{\mathbb{Z}} \mathop{Spec} \mathbb{C}$. Now for such a scheme an integral point is a map $\mathop{Spec}\mathbb{Z} \to X_{\mathbb{Z}}$ such that the composition with the structure map is the identity. Note that the same can be done for every ring $A$ in place of $\mathbb{Z}$. With this definition, the line $\{ x = 0 \}$ is defined over $\mathbb{Z}$, but the line $\{ x = \pi \}$ is not, basically because there is no way to generate its ideal with equations having integer coefficients. So your problem does not arise anymore. EDIT: Abstractly of course the two lines are isomorphic over $\mathbb{C}$, so the line $r = \{x = \pi \}$ actually has a model over $\mathbb{Z}$. The problem is that this model is not compatible with the inclusion in $\mathbb{A}^2$, that is, there will be no map $r_{\mathbb{Z}} \to \mathbb{A}^2_\mathbb{Z}$ whose base change is the inclusion of $r$ into $\mathbb{A}^2$. In order to have this, you would have to ask that the ideal of $r$ in $\mathbb{A}^2$ should be generated by polynomials with integer coefficients. As for your second question, there can be different models, that is, nonisomorphic schemes over $\mathbb{Z}$ which become isomorphic after base change to $\mathbb{C}$. So before discussing the existence of integral points, you have to FIX a model, and the points will in general depend on the model. For instance take the two conics $\{ x^2 + y^2 = 2 \}$ and $\{ x^2 + y^2 = 3 \}$. Both have an obvious choice of a model, given by the inclusion in $\mathbb{A}^2$; moreover they are isomorphic over $\mathbb{C}$. But the integral points on the first one are $(\pm 1, \pm 1)$, while the second has none. Finally you consider the possibility that the structure over $\mathbb{C}$ is not relevant. This is false: the base change $X_\mathbb{Z} \times_{\mathbb{Z}} \mathop{Spec} \mathbb{C}$ is endowed with a natural map to $\mathop{Spec} \mathbb{C}$, and we ask for the isomorphism with $X$ to be over $\mathbb{C}$.
|
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|
18,787 |
Here's a question I can't answer by myself: The Reflection Principle in Set Theory states for each formula $\phi(v_{1},...,v_{n})$ and for each set M there exists a set N which extends M such that the following holds $\phi^{N} (x_{1},...,x_{n})$ iff $\phi (x_{1},...,x_{n})$ for all $x_{1},...x_{n} \in N$ Thus if $\sigma$ is a true sentence then the RFP yields a model of it and as a consequence any finite set of axioms of ZFC has a model (as a consequence ZFC is not finitely axiomatizable by Gödel's Second Incompleteness Theorem) But why can't I just use now the Compactness Theorem (stating that each infinte set of formulas such that each finite subset has a model, has a model itself) to obtain a model of ZFC (which is actually impossible)??
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For any finite set of axioms K of ZFC, ZFC proves "K has a model", via the reflection principle as you note. However, ZFC does not prove "for any finite set of axioms K of ZFC, K has a model". The distinction between these two is what prevents ZFC from proving that ZFC has a model. (That is, even though, as you note, ZFC proves "if every finite set of axioms K of ZFC has a model, then ZFC has a model", as ZFC proves compactness, it does not follow that ZFC proves the consequent of this implication, as in fact ZFC does not prove the antecedent; ZFC only proves each particular instance of the antecedent, but not the universal statement itself.)
|
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|
18,813 |
This question may end up [closed], but I'm going to ask and let the people decide. It's certainly the kind of question that I'd ask people at tea, and it's not one whose answer I've been able find with Google. TeX, I have heard, is Turing complete. In theory, this means that we can do modular arithmetic with LaTeX programs. I'd like to know how this can be done in practice. Background: I've been using the \foreach command in TikZ to draw NxN arrays of nodes, indexed by pairs of integers (m,n). I'd like to be able to use modular arithmetic and an ifthenelse statement to put different decorations on the nodes, depending on the value of (m+n) mod p. Obviously, one can just do this by hand. But that's not the world I want to live in.
|
Get a current version of Ti k Z and use \pgfmathmod !
|
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|
18,847 |
A first course in algebraic topology, at least the ones I'm familiar with, generally gets students to a point where they can calculate homology right away. Building the theory behind it is generally then left for the bulk of the course, in terms of defining singular homology, proof of the harder Eilenberg-Steenrod axioms, cellular chains, and everything else necessary to show that the result is essentially independent of the definitions. A second course then usually takes up the subject of homotopy theory itself, which is harder to learn and often harder to motivate. This has some disadvantages, e.g. it leaves a discussion of Eilenberg-Maclane spaces and the corresponding study of cohomology operations far in the distance. However, it gets useful machinery directly to people who are consumers of the theory rather than looking to research it long-term. Many of the more recent references (e.g. tom Dieck's new text) seem to take the point of view that from a strictly logical standpoint a solid foundation in homotopy theory comes first. I've never seen a course taught this way and I'm not really sure if I know anyone who has, but I've often wondered. So the question is: Has anyone taught, or been taught, a graduate course in algebraic topology that studied homotopy theory first? What parts of it have been successful or unsuccessful?
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I was a heavily involved TA for such a graduate course in 2006 at UC Berkeley. We started with a little bit of point-set topology introducing the category of compactly generated spaces. Then we moved into homotopy theory proper. We covered CW-complexes and all the fundamental groups, Van-Kampen's Theorem, etc. From this you can prove some nice classical theorems, like the Fundamental Theorem of Algebra, the Brauwer Fixed Point Theorem, the Borsuk-Ulam Theorem, and that $R^n \neq R^m$ for $n \neq m$. I felt like this part of the course went fairly well and is sufficiently geometric to be suitable for a first level graduate course (you can draw lots of pictures!). At this point you can take the course in a couple different directions which all seem to have their own disadvantages and problems. The main problem is lack of time. A very natural direction is to discuss obstruction theory, since it is based off of the same ideas and constructions covered so far. However this is not really possible since the students haven't seen homology or cohomology at this point! Instead, for a bit we discussed the long exact sequences you get from fibrations and cofibrations. You could then try to lead into the definition of cohomology as homotopy classes of maps into a $K(A,n)$. But this definition is fairly abstract and doesn't show one of the main feature of homology/cohomology: It is extremely computable. Still, I could imagine a course trying to develop homology and cohomology from this point of view and leading into CW homology and the Eilenberg-Steenrod axioms. Another direction you can go is into the theory of fiber bundles (this is what we tried). The part on covering space theory works fairly well and you have all the tools at your disposal. However when you want to do general fiber bundle theory it can be difficult. A natural goal is the construction of classifying spaces and Brown's representability theorem.
The problem is that the homotopy invariance of fiber bundles is non-trivial to prove. You should expect to have to spend fair amount of time on this. It is really more suited for a second course on algebraic topology. The main problem with all of these approaches is that it is difficult to cover the homotopy theory section and still have enough time to cover homology/cohomology properly. You know this has to be the case since it is hard to do the reverse: cover homology and cohomology, and still have enough time to cover homotopy theory properly. What this means is that you'll be in the slightly distasteful situation of having bunch of students who have taken a first course on algebraic topology, but don't really know about homology or cohomology. This is fine if you know that these students will be taking a second semester of algebraic topology. Then any gaps can be fixed. However, in my experience this is not a realistic expectation. As you well know, you will typically have some students who end up not being interested in algebraic topology and go into analysis or algebraic geometry or some such. Or you might have some students who are second or third year students in other math fields and are taking your course to learn more about homology and cohomology. They would be done a particular disservice by a course focusing on "homotopy first".
|
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|
18,877 |
Does ${\mathbb R}$ have proper, countable index subrings? By countable I mean finite or countably infinite. By subring I mean any additive subgroup which is closed under multiplication (I don't care if it contains $1$.) By index, I mean index as an additive subgroup. Given some real number $x$, when is it possible to find a countable index subring of ${\mathbb R}$ which does not contain $x$?
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Perhaps surprisingly, it turns out that such subrings do exist. This was proved in Section 2 of my paper: Simon Thomas, Infinite products of finite simple groups II,
J. Group Theory 2 (1999), 401--434. The basic idea of the proof is quite simple. Clearly the ring of $p$-adic integers has countable index in the field of $p$-adic numbers. Now the $p$-adic integers are the valuation ring of the obvious valuation on the field of $p$-adic numbers ... and it turns out to be enough to show that $\mathbb{C}$ has an analogous valuation. This is true because $\mathbb{C}$ is isomorphic to the field of Puiseux series over the algebraic closure of $\mathbb{Q}$, which has an appropriate valuation.
|
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|
18,964 |
Using p-adic analysis, Dwork was the first to prove the rationality of the zeta function of a variety over a finite field. From what I have seen, in algebraic geometry, this method is not used much and Grothendieck's methods are used instead. Is this because it is felt that Dwork's method is not general or powerful enough; for example, Deligne proved the Riemann Hypothesis for these zeta functions with Grothendieck's methods, is it felt that Dwork's method can't do this?
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The premise of the question is not correct. Dwork's methods (and modern descendants of them) are a major part of modern arithmetic geometry over $p$-adic fields, and of $p$-adic analysis.
You could look at the two volume collection Geometric aspects of Dwork theory to get
some sense of these developments. Just to say a little bit here: Dwork's approach led to Monsky--Washnitzer cohomology,
which in turn was combined with the theory of crystalline cohomology by Bertheolt to develop the modern theory of rigid cohomology. The $p$-adic analysis of Frobenius actions is also
at the heart of the $p$-adic theory of modular and automorphic forms, and of much of the machinery underlying $p$-adic Hodge theory. The theory of $F$-isocrystals (the $p$-adic analogue of variations of Hodge structure) also grew (at least in part) out of Dworks
ideas. To get a sense for some of Dwork's techniques, you can look at the Bourbaki report Travaux de Dwork , by Nick Katz, and also at Dwork's difficult masterpiece $p$-adic cycles , which has been a source of insipiration for many arithmetic geometers. In some sense the $p$-adic theory is more difficult than the $\ell$-adic theory, which is
why it took longer to develop. (It is like Hodge theory as compared to singular cohomology.
The latter is already a magnificent theory, but the former is more difficult in the sense that it has more elaborate foundations and a more elaborate formalism, and (related to this) has ties to analysis (elliptic operators, differential equations) that are not present
in the same way in the latter.) [For the experts: I am alluding to $p$-adic Hodge theory,
syntomic cohomology, $p$-adic regulators, Serre--Tate theory, and the like.]
|
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|
18,987 |
Using Alexander duality, you can show that the Klein bottle does not embed in $\mathbb{R}^3$. (See for example Hatcher's book Chapter 3 page 256.) Is there a more elementary proof, that say could be understood by an undergraduate who doesn't know homology yet?
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If you are willing to assume that the embedded surface $S$ is polyhedral, you can prove that it is orientable by an elementary argument similar to the proof of polygonal Jordan Theorem. Of course the proof is translation of a homology/transversality/separation argument. Fix a direction (nonzero vector) which is not parallel to any of the faces. For every point $p$ in the complement of $S$, consider the ray starting at $p$ and goint to the chosen direction. If this ray does not intersect edges of $S$, count the number of intersection points of the ray and the surface. If this number is even, you say that $p$ is black, otherwise $p$ is white. If the ray intersects an edge of $S$, you paint $p$ the same color as some nearby point whose ray does not intersect edges. It is easy to see that the color does not change along any path in the complement of $S$ (it suffices to consider only polygonal paths avoiding points whose rays contain vertices of $S$). Now take points $p$ and $q$ near the surface such that the segment $pq$ is parallel to the chosen direction. Then they are of different colors. But if the surface is non-orientable, you can go from $p$ to $q$ along a Mobius strip contained in the surface. This contradicts the above fact about paths in the complement of $S$.
|
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|
19,046 |
I want some recommendation on which software I should install on my computer. I'm looking for an open source program for general abstract mathematical purposes (as opposed to applied mathematics). I would likely use it for group theory, number theory, algebraic geometry and probably polytopes. The kind of program I have in mind is Mathematica or Matlab. Although probably those are not designed for abstract mathematics. Any suggestions?
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Here are some links. Axiom and Maxima are good general purpose computer algebra systems. DataMelt is a free Java-based math software with a lot of examples GAP is a system for computational discrete algebra (with particular emphasis on computational group theory). PARI/GP is a CAS for fast computations in number theory. SAGE is a kind of unified framework for several systems, including GAP, PARI, and Maxima. Octave is a system for numerical computations (it is close to Matlab). Cadabra is a computer algebra system designed for the solution of the field theory problems. CoCoA stands for "Computations in commutative algebra". KANT / KASH stands for "Computational Algebraic Number Theory". Macaulay 2 is a system for research in commutative algebra and algebraic geometry. Snap is a computer program for studying arithmetic invariants of hyperbolic 3-manifolds. Symmetrica is an object oriented computer algebra system for representations, combinatorics and applications of symmetric groups.
|
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|
19,076 |
Some MOers have been skeptic whether something like natural number graphs can be defined coherently such that every finite graph is isomorphic to such a graph. (See my previous questions [ 1 ], [ 2 ], [ 3 ], [ 4 ]) Without attempting to give a general definition of natural number graphs , I invite you to consider the following DEFINITION A natural number $d$ may be called demi-prime iff there is a prime number $p$ such
that $d = (p+1)/2$. The demi-primes' distribution is exactly
like the primes, only shrinked by the
factor $2$: $$2, 3, 4, 6, 7, 9, 10, 12, 15, 16, 19, 21, 22, 24, 27, 30, 31, 34, 36, 37, 40, 42, 45, 49, ...$$ Let D ($k,n$) be the set which consists
of the $k$-th up to the $(k+n-1)$-th
demi-prime number. After some - mildly exhaustive - calculations I feel quite confident to make the following CONJECTURE For every finite graph $G$ there is a $k$ and a bijection $d$ from the vertex set
$V(G)$ to D ($k,|G|$) such that $x,y$ are adjacent if and only if $d(x),d(y)$ are coprime . I managed to show this rigorously for all graphs of order $n\leq $ 5 by brut force calculation, having to take into account all (demi-)primes $d$ up to the 1,265,487 th one for graphs of order 5. For graphs of order 4, the first 1,233 primes did suffice, for graphs of order 3 the first 18 ones. Looking at some generated statistics for $n \leq$ 9 reveals interesting facts (1)(2) , correlations, and lack of correlations, and let it seem probable (at least to me) that the above conjecture also holds for graphs of order $n >$ 5. Having boiled down my initial intuition to a concrete predicate, I would like to pose the following QUESTION Has anyone a clue how to prove or disprove the above conjecture? My impression is that the question is about the randomness of prime numbers: Are they distributed and their corresponding demi-primes composed randomly enough to mimick – via D ($k,n$) and coprimeness – all (random) graphs? (1) E.g., there is one graph of order 5 - quite unimpressive in graph theoretic terms - that is very hard to find compared to all the others: it takes 1,265,487 primes to find this guy, opposed to only 21,239 primes for the second hardest one. (Lesson learned: Never stop searching too early!) It's – to whom it is of interest – $K_2 \cup K_3$: 0 1 0 0 0
1 0 0 0 0
0 0 0 1 1
0 0 1 0 1
0 0 1 1 0 (2) Added: This table shows the position of the smallest prime (among all primes) needed to mimick the named graphs of order $n$. All values not shown are greater than $\approx 2,000,000$ order | 3 4 5 6 7 8
-------------------------------------------------
empty | 14 45 89 89 89 3874
complete | 5 64 336 1040 10864 96515
path | 1 6 3063 21814
cycle | 5 112 21235 49957
|
Theorem: Schinzel's hypothesis H implies the conjecture. Proof: Choose distinct primes $q_S > 100|G|$ indexed by the 2-element subsets $S$ of $G$. For each $i \in G$, let $Q_i$ be the set of $q_S$ for $S$ such that $i \in S$ and the edge $S$ is not part of $G$. Let $P_i$ be the product of the primes in $Q_i$. Let $P = 4 \prod_S q_S^2$. By the Chinese remainder theorem, for each $i$ we can find a positive integer $a_i$ such that $a_i \equiv 1 \bmod{\ell^2}$ for each prime $\ell \le 10|G|$, $a_i \equiv q-1 \bmod{q^2}$ for each $q \in Q_i$, and $a_i \equiv 1 \bmod{q_S}$ for each $q_S \notin Q_i$. Moreover, we can choose the $a_i$ to be distinct. Let $J$ be the set of positive integers up to $\operatorname{max} a_i$, but excluding all of the $a$'s themselves (i.e., $J$ consists of the numbers in the gaps). For each $j \in J$ choose a prime $s_j$ much larger than all the $a_i$ and all the $q_S$. Consider the linear polynomials $P n + a_i$ and $(P n + a_i + 1)/(2P_i)$ In $\mathbf{Z}[n]$. For each prime $\ell \le 10|G|$ and each $\ell$ of the form $q_S$, all these $2|G|$ polynomials are nonzero mod $\ell$ at $n=0$. For each other prime $\ell$, there exists $n$ such that all these polynomials are nonzero mod $\ell$, since $n$ needs to avoid no more than $2|G|$ residue classes mod $\ell$. Furthermore, we can impose the condition that $P n+j$ is divisible by $s_j^2$ for each $j \in J$, and still find $n$ as above. Therefore Schinzel's hypothesis H implies that there exist arbitrarily large positive integers $n$ such that the numbers $P n+a_i$ and $(P n + a_i + 1)/(2P_i)$ are all prime, and such that $P n+j$ is not prime for $j \in J$. This makes the numbers $p_i:=P n + a_i$ consecutive primes such that $(p_i+1)/2 = P_i r_i$ for some prime $r_i$. If $n$ is sufficiently large, then these primes $r_i$ are all distinct and larger than all of the $q_S$. So the greatest common factor of $(p_i+1)/2$ and $(p_j+1)/2$ for $i \ne j$ equals $1$ if there is an edge between $i$ and $j$, and $q_{\{i,j\}}$ otherwise. $\square$ Remark: Given how little is known about consecutive primes, it seems unlikely that the conjecture can be proved unconditionally. But at least now we can be confident that it's true!
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|
19,127 |
I heard the following two questions recently from Carl Mummert , who encouraged me to spread them around. Part of his motivation for the questions was to give the subject of computable model theory some traction on complete metric spaces, by considering the countable objects as stand-ins for the full spaces, to the extent that they are able to do so. Question 1. Is there a countable subset $D$ of the real plane $\mathbb{R}^2$ that is dense and has the property that the Euclidean distance $d(x,y)$ is a rational number for all $x,y\in D$ ? The one dimensional analogue of this question has an easy affirmative answer, since the rationals $\mathbb{Q}$ sits densely in $\mathbb{R}$ and the distance between any two rationals is rational. Question 2. More generally, does every separable complete metric space have a countable dense set $D$ with all distances between elements of $D$ rational? [Edit: Tom Leinster has pointed out that if the space has only two points, at irrational distance, this fails. So let us consider the case of connected spaces, generalizing the situation of Question 1.] If one is willing to change to an equivalent metric (giving rise to the same topology), then the answer to Question 1 is Yes, since the rational plane $\mathbb{Q}\times\mathbb{Q}$ is dense in the real plane $\mathbb{R}\times\mathbb{R}$ , and has all rational distances under the Manhattan metric, which gives rise to the same topology. Is the answer to the correspondingly weakened version of Question 2 also affirmative, if one is willing to change the metric? Note that one cannot find an equivalent metric such that all distances in $\mathbb{R}^2$ become rational, since omitted values in the distance function lead to disconnectivity in the space. This is why the questions only seek to find a dense subset with the rational condition. The question seems related to the question of whether it is possible to find large non-linear arrangements of points in the plane with all pair-wise distances being integers. For example, this is true of the integers $\mathbb{Z}$ sitting inside $\mathbb{R}$ , but can one find a 2 dimensional analogue of this? Clearly, some small arrangements (triangles, etc.) are possible, but I am given to understand that there is a finite upper bound on the size of such arrangements. What is the precise statement about this that is known?
|
Let me answer Question 2. Strong version: no. Consider $[0,1]$ with distance $d(x,y)=|x-y|^{1/3}$. There is no even a triple of points with rational distances - otherwise there would be a nonzero rational solution of $x^3+y^3=z^3$. Weak version: yes. Let $(X,d)$ be the space in question. Construct sets $S_1\subset S_2\subset\dots$ such that each $S_k$ is a maximal $(2^{-k})$-separated net in $X$. Let $S$ be the union of these nets; then $S$ is countable and dense in $X$. Now construct the following metric graph on $S$. For every $k$, connect every pair of points $x,y\in S_k$ by an edge whose length is $(1-10^{-k})d(x,y)$ rounded down to a multiple of $10^{-2k}$. The new distance $d'$ on $S$ is the induced length distance in this graph. It is easy to see that the edges outside $S_k$ do not affect the distances in $S_k$, hence all these distances are rational (multiples of $10^{-2k}$). The new metric $d'$ on $S$ satisfies $\frac12d\le d'\le d$, hence the completion of $(S,d')$ is the same set $X$ with an equivalent metric. UPDATE .
Here is a more detailed description without the term "metric graph". For each $k$, define a function $f_k:\mathbb R_+\to\mathbb R_+$ by
$$
f_k(t) = 10^{-2k}\left\lfloor 10^{2k}(1-10^{-k})t \right\rfloor .
$$
The actual form of $f_k$ does not matter, we only need the following properties: $f_k$ takes only rational values with bounded denominators (by $10^{-k}$). Let $a_k$ and $b_k$ denote the infimum and the supremum of $f_k(t)/t$ over the set $\{t\ge 2^{-k}\}$. Then $\frac12\le a_k\le b_k\le a_{k+1}\le 1$ for all $k$. (Indeed, we have $1-2\cdot10^k\le a_k\le b_k\le 1-10^k$.) For every $x,y\in S_k$, define $\ell(x,y)=f_k(d(x,y))$ where $k=k(x,y)$ is the minimum number such that $x,y\in S_k$. Note that
$$
a_k d(x,y) \le \ell(x,y) \le b_k d(x,y)
$$
for all such pairs $x,y$, since $S_k$ is a $(2^{-k})$-separated set. For a finite sequence $x_0,x_1,\dots,x_n\in S$ define
$$
\ell(x_0,x_1,\dots,x_n) = \sum_{i=1}^n \ell(x_{i-1},x_i) .
$$
I will refer to this expression as the $\ell$- length of the sequence $x_0,\dots,x_n$. Define
$$
d'(x,y) = \inf\{ \ell(x_0,x_1,\dots,x_n) \}
$$
where the infimum is taken over all finite sequences $x_0,x_1,\dots,x_n$ in $S$ such that $x_0=x$ and $x_n=y$. Clearly $d'$ is a metric and $\frac12d\le d'\le d$. It remains to show that $d'$ takes only rational values. Lemma : If $x,y\in S_k$, then $d'(x,y)$ equals the infimum of $\ell$-lengths of sequences contained in $S_k$. Proof : Consider any sequence $x_0,\dots,x_n$ in $S$ such that $x_0=x$ and $y_0=y$. Remove all points that do not belong to $S_k$ from this sequence. I claim that the $\ell$-length became shorter. Indeed, it suffices to prove that
$$
\ell(x_r,x_s) \le \ell(x_r,x_{r+1},\dots,x_{s-1},x_s)
$$
if $x_r$ and $x_s$ are in $S_k$ and the intermediate points are not. By the second property of the functions $f_k$, the left-hand side is bounded above by $b_k d(x_r,x_s)$ and every term $\ell(x_i,x_{i+1})$ in the right-hand side is bounded below by $b_k d(x_i,x_{i+1})$. So it suffices to prove that
$$
b_k d(x_r,x_s) \le b_k\sum_{i=r}^{s-1} d(x_i,x_{i+1}),
$$
and this is a triangle inequality multiplied by $b_k$. Q.E.D. All $\ell$-lengths of sequences in $S_k$ are multiples of some fixed rational number (namely $10^{-2k}$). Hence $d'(x,y)$ is a multiple of the same number if $x,y\in S_k$. Thus all values of $d'$ are rational.
|
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|
19,152 |
I'm ashamed to admit it, but I don't think I've ever been able to genuinely motivate the definition of a topological space in an undergraduate course. Clearly, the definition distills the essence of many examples, but it's never been obvious to me how it came about, compared, for example, to the rather intuitive definition of a metric space. In some ways, the sparseness of the definition is startling as it tries to capture, apparently successfully, the barest notion of 'space' imaginable. I can try to make this question more precise if necessary, but I'd prefer to leave it slightly vague, and hope that someone who has discussed this successfully in a first course, perhaps using a better understanding of history, might be able to help me out. Added 24 March: I'm grateful to everyone for their thoughtful answers so far. I'll have to think over them a bit before I can get a sense of the 'right' answer for myself. In the meanwhile, I thought I'd emphasize again the obvious fact that the standard concise definition has been tremendously successful. For example, when you classify two-manifolds with it, you get equivalence classes that agree exactly with intuition. Then in as divergent a direction as the study of equations over finite fields, there is the etale topology*, which explains very clearly surprising and intricate patterns in the behaviour of solution sets. *If someone objects that the etale topology goes beyond the usual definition, I would argue that the logical essence is the same. It is notable that the standard definition admits such a generalization so naturally, whereas some of the others do not. (At least not in any obvious way.) For those who haven't encountered one before, a Grothendieck topology just replaces subsets of a set $X$ by maps $$Y\rightarrow X.$$ The collection of maps that defines the topology on $X$ is required to satisfy some obvious axioms generalizing the usual ones. Added 25 March: I hope people aren't too annoyed if I admit I don't quite see a satisfactory answer yet. But thank you for all your efforts. Even though Sigfpe's answer is undoubtedly interesting, invoking the notion of measurement, even a fuzzy one, just doesn't seem to be the best approach. As Qiaochu has pointed out, a topological space is genuinely supposed to be more general than a metric space. If we leave aside the pedagogical issue for a moment and speak as working mathematicians, a general concept is most naturally justified in terms of its consequences. As pointed out earlier, topologies that have no trace of a metric interpretation have been consequential indeed. When topologies were naturally generalized by Grothendieck, a good deal of emphasis was put on the notion of an open covering , and not just the open sets themselves. I wonder if this was true for Hausdorff as well. (Thanks for the historical information, Donu!) We can see the reason as we visualize a two-manifold. Any sufficiently fine open covering captures a combinatorial skeleton of the space by way of the intersections. Note that this is not true for a closed covering. In fact, I'm not sure what a sensible condition might be on a closed covering of a reasonable space that would allow us to compute homology with it. (Other than just saying they have to be the simplices of a triangulation. Which also reminds me to point out that homology can be computed for ordinary objects without any notion of topology.) To summarize, a topology relates to analysis with its emphasis on functions and their continuity, and to metric geometry , with its measurements and distances. However, it also interpolates between these and something like combinatorial geometry , where continuous functions and measurements play very minor roles indeed. For myself, I'm still confused. Another afterthought: I see what I was trying to say above is that open sets in topology provide an abstract framework for describing local properties of functions. However, an open cover is also able to encode global properties of spaces. It seems the finite intersection property is important for this, but I'm not able to say for sure. And then, when I try to return to the pedagogical question with all this, I'm totally at a loss. There are very few basic concepts that trouble me as much in the classroom.
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Topology is the art of reasoning about imprecise measurements, in a sense I'll try to make precise. In a perfect world you could imagine rulers that measure lengths exactly. If you wanted to prove that an object had a length of $l$ you could grab your ruler marked $l$, hold it up next to the object, and demonstrate that they are the same length. In an imperfect world however you have rulers with tolerance. Associated to any ruler is a set $U$ with the property that if your length $l$ lies in $U$, the ruler can tell you it does. Call such a ruler $R_U$. Given two rulers $R_U$ and $R_V$ you can easily prove a length lies in $U\cup V$. You just hold both rulers up to the length and the length is in $U\cup V$ if one or the other ruler shows a positive match. You can think of $R_{U\cup V}$ as being a kind of virtual ruler. Similarly you can easily prove that a point lies in $U\cap V$ using two rulers. If you have an infinite family of rulers, $R_{U_i}$, then you can also prove that a length lies in $\bigcup_i U_i$. The length must lie in one of the $U_i$ and you simply exhibit the ruler $R_{U_i}$ matching for the appropriate $i$. But you can't always do the same for $\bigcap_i U_i$. To do so might require an infinitely long proof showing that all of the $R_{U_i}$ match your length. A topology is a (generalised) set of rulers that fits this description. Your notion of 'measurement' in whatever problem you have might not match the notion that the above description tries to capture. But to the extent that it does, topology will work as a way to reason about your problem.
|
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|
19,170 |
In Number of digits in n! , now closed, there was a mention of Dmitry Kamenetsky's formula, $[\bigl(\log(2\pi n)/2+n(\log n-\log e)\bigr)/\log 10]+1$, for the number of decimal digits in $n$-factorial. Here, $[x]$ is the integer part of $x$. The formula appears at A034886 in the Online Encyclopedia of Integer Sequences, http://oeis.org/A034886 . My question is whether this formula is exact for all $n$, or is it occasionally off. No proof of exactness is given at the OEIS, no paper of Kamenetsky appears in Math Reviews. In the other thread, I mentioned the discussion in the Usenet newsgroup sci.math in January-February, Subject: Number of digits in factorial. Although neither proof nor counterexample was found, I'd recommend looking over that discussion before starting in on this question. EDIT 11 Aug 2011: I note that the question also came up at m.se: question 8323, 30 Oct 2010.
|
A counterexample is $n_1 := 6561101970383$, with
$$
\log_{10} \left( (n_1/e)^{n_1} \sqrt{2\pi n_1} \right)
= 81244041273652.999999999999995102483 - \phantom; ,
$$
but
$$
\log_{10} (n_1!)
= 81244041273653.000000000000000618508 + \phantom;.
$$
If I computed correctly, $n_1$ is the first counterexample, and the only one up to $10^{13}$. The computation should reach $10^{15}$ sometime next week, with a probability of about $1 - \exp(-\frac16) \sim 15\%$ of finding an $n_2$. The computation (in gp/pari ) took about 40 CPU hours here, compressed to 4 hours by running in parallel on 10 of the 12 heads of alhambra.math.harvard.edu . This was not done by calculating $\log_{10} (n!)$ to enough precision for every $n \leq 10^{13}$, which would have taken hundreds of times longer. The problem of finding nearly integral values of $\log_{10} (n!)$ is a special case of the "table maker's dilemma" (Wikipedia attributes this felicitous coinage to William Kahan); in this case, the linear-approximation technique suggested by Lefèvre at the bottom of page 15 of his slides takes time $\tilde O(N^{2/3})$ to find all examples with $n < N$. That's what's running on alhambra now. Along the way a few more terms of sequence A177901 turned up:
$252544447$,
$1430841730$,
$5042264463$,
$31774693500$,
$40752166709$,
$46787073630$,
$129532358256$,
$421559495894$,
$2418277169072$,
$6105111564681$,
and then $n_1 = 6561101970383$, which might even turn out to be the last term up to $10^{15}$ because $\log_{10} (n_1!)$ is so close to an integer (about $9$ times closer than necessary for our purpose). [ EDIT It's the last term $<10^{14}$ but not $10^{15}$, see below.]
The term $252544447$ was reported on math.se #8323 by Byron Schmuland [ EDIT and a few months earlier by David Cantrell on sci.math], though it has not been posted to OEIS yet. The further ones seem to be new, and I'll post them on OEIS soon. Kamenetsky was right to suggest that the approximation should fail sometimes: in base 10, we expect $n$ to be a counterexample with probability about $1/cn$ with $c = 12 \log 10$, so on average each range $[N, 10^{12}N]$ should have about one. Thus it is not surprising that the first one (past $n=1!$) turns out to have $13$ digits. This heuristic is also the source of the estimate $1-\exp(-\frac16)$ for the probability of another counterexample in $ [10^{13}, 10^{15}]$. UPDATE The calculation has now passed $10^{14}$, finding no new counterexample. It did, however, find a new term for the OEIS sequence a bit beyond $10^{14}$: $n=125291661119688$, with $\log_{10}(n!)$ close to but just below the nearest integer $1711938609606982$ (where a counterexample must be a bit above), and also not quite as close as $1/(12n)$ — the difference is about $1/(8.4n)$. While I'm at it: I should have mentioned that the gp/pari computation also found (in a minute or two) all the terms in $[10^4,10^8]$ listed by OEIS, which lends the new results some credibility; and I thank Gerry Myerson for drawing my attention to this question with his edit of about two weeks ago.
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|
19,264 |
This is related to the previous question of how to define a conductor of an elliptic curve or a Galois representation. What motivated the use of the word "conductor" in the first place? A friend of mine once pointed out the amusing idea that one can think of the conductor of an elliptic curves as "someone" driving a train which lets you off at the level of the associated modular form. A similar statement can be made concerning Szpiro's conjecture, which provides asymptotic bounds on several invariants of an elliptic curve in terms of its conductor. Here one might think of the conductor as "someone" who controls this symphony of invariants consisting of the minimal discriminant, the real period, the modular degree, and the order of the Shafarevich-Tate group (assuming BSD). Was there some statement of this sort which motivated Artin's original definition of the conductor? Does anyone have a reference for the first appearance of the word conductor in this context? I apologize if this question is inappropriate for MO.
|
It is a translation from the German Führer (which also is the reason that
in older literature, as well as a fair bit of current literature, the conductor
is denoted as f in various fonts). Originally the term conductor appeared in
complex multiplication and class field theory: the conductor of an abelian extension is a certain ideal that controls the situation. Then it drifted off into other
areas of number theory to describe parameters that control other situations. Of course in English we tend not to think of conductor as a leader
in the strong sense of Führer, but more in a musical sense, so it seems like a weird translation. But back in the
1930s the English translation was leader rather than conductor, at least once: see the review of Fueter’s book on complex multiplication in the 1931 Bulletin of the Amer. Math.
Society, page 655. The reviewer writes in the second paragraph "First there is a careful treatment of those ray class fields whose leaders are multiples of the ideal..."
You can find the review yourself at http://www.ams.org/bull/1931-37-09/S0002-9904-1931-05214-9/S0002-9904-1931-05214-9.pdf . I stumbled onto that reference quite by chance (a couple of years ago). If anyone knows other places in older papers in English where conductors were called leaders, please post them as comments below. Thanks! Concerning Artin's conductor, he was generalizing to non-abelian Galois extensions the parameter already defined for abelian extensions and called the conductor. So it was natural to use the same name for it in the general case. Edit: I just did a google search on "leader conductor abelian" and the first hit
is this answer. Incredible: it was posted less than 15 minutes ago!
|
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|
19,269 |
What are the examples of some mathematicians coming very close to a very promising theory or a correct proof of a big conjecture but not making or missing the last step?
|
Freeman Dyson discusses a few examples of this in his article Missed Opportunities . One that I thought was particularly striking was that mathematicians could have discovered special relativity decades before Einstein just by staring at Maxwell's equations hard enough, and also on the basis that the representation theory of the Poincare group is simpler than the representation theory of the Galilean group.
|
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|
19,285 |
Let $X$ be a set and let ${\mathcal N}$ be a collection of nets on $X.$ I've been told by several different people that ${\mathcal N}$ is the collection of convergent nets on $X$ with respect to some topology if and only if it satisfies some axioms. I've also been told these axioms are not very pretty. Once or twice I've tried to figure out what these axioms might be but never came up with anything very satisfying. Of course one could just recode the usual axioms regarding open sets as statements about nets and then claim to have done the job. But, come on, that's nothing to be proud of. Has anyone seen topology axiomatized this way? Does anyone remember the rules?
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Yes. This is given in Kelley's General Topology . (Kelley was one of the main mathematicians who developed the theory of nets so that it would be useful in topology generally rather than just certain applications in analysis.) In the section "Convergence Classes" at the end of Chapter 2 of his book, Kelley lists the following axioms for convergent nets in a topological space $X$ a) If $S$ is a net such that $S_n = s$ for each $n$ [i.e., a constant net], then $S$ converges to $s$ . b) If $S$ converges to $s$ , so does each subnet. c) If $S$ does not converge to $s$ , then there is a subnet of $S$ , no subnet of which converges to $s$ . d) (Theorem on iterated limits): Let $D$ be a directed set. For each $m \in D$ , let $E_m$ be a directed set, let $F$ be the product $D \times \prod_{m \in D} E_m$ and for $(m,f)$ in $F$ let $R(m,f) = (m,f(m))$ . If $S(m,n)$ is an element of $X$ for each $m \in D$ and $n \in E_m$ and $\lim_m \lim_n S(m,n) = s$ , then $S \circ R$ converges to $s$ . He has previously shown that in any topological space, convergence of nets satisfies a) through d). (The first three are easy; part d) is, I believe, an original result of his.) In this section he proves the converse: given a set $S$ and a set $\mathcal{C}$ of pairs (net,point) satisfying the four axioms above, there exists a unique topology on $S$ such that a net $N$ converges to $s \in X$ iff $(N,s) \in \mathcal{C}$ . I have always found property d) to be unappealing bordering on completely opaque, but that's a purely personal statement. Addendum : I would be very interested to know if anyone has ever put this characterization to any useful purpose. A couple of years ago I decided to relearn general topology and write notes this time. The flower of my efforts was an essay on convergence in topological spaces that seems to cover all the bases (especially, comparing nets and filters) more solidly than in any text I have seen. http://alpha.math.uga.edu/~pete/convergence.pdf But "even" in these notes I didn't talk about either the theorem on iterated limits or (consequently) Kelley's theorem above: I honestly just couldn't internalize it without putting a lot more thought into it. But I've always felt/worried that there must be some insight and content there...
|
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|
19,308 |
I am not an expert, but there seems to be an enormous technical difference between algebraic geometry and differential/metric geometry stemming from the fact that there is apparently no such thing as curvature in the former context while curvature is everywhere in the latter (indeed, it is hard to produce nontrivial results in Riemannian geometry that DON'T involve curvature). Of course it seems unreasonable to just port definitions of curvature into an algebraic context, but maybe there are constructions that play the same role in algebraic geometry that curvature does in other kinds of geometry. Here are two specific ways the notion of curvature shows up which I can imagine making sense in more general contexts. Algebraic Chern - Weil Theory? In differential geometry one uses the curvature of a connection on a vector bundle to produce explicit cohomology classes. Does this have an algebraic analogue? Algebraic Curvature Bounds? One supremely important theme in modern geometry involves proving theorems that depend only on the large scale geometry of a space, and the main strategy is to compare the space to a simpler one with the same large scale properties. This reminds me a little bit of tropical geometry wherein one replaces an algebraic variety with a simple combinatorial proxy, but from what little I know the analogy seems to stop there. Any thoughts? I hope this question is not too vague, but it seems worthwhile and part of the problem is that I can't formulate a precise question along these lines. Thanks in advance!
|
An algebraic analog of Chern-Weil theory (explicitly taking symmetric polynomials of curvature) is given by the Atiyah class.
Given a vector bundle $E$ on a smooth variety we can consider the short exact sequence
$$ 0\to End(E) \to A(E) \to T_X\to 0$$
where $T_X$ is the tangent sheaf and $A(E)$ is the "Atiyah algebroid" --- differential operators of order at most one acting on sections of $E$, whose symbol is a scalar first order diffop (hence the map to the tangent sheaf). A (holomorphic or algebraic) connection is precisely a splitting of this sequence, and a flat connection is a Lie algebra splitting. Now algebraically such splittings will often not exist (having a holomorphic connection forces your characteristic classes to have type $(p,0)$ rather than the $(p,p)$ you want..) but nonetheless we can define the extension class, which is the Atiyah class
$$a_E\in H^1(X, End(E)\otimes \Omega^1_X).$$
This is the analog of the curvature form in the Riemannian world -- we now can take symmetric polynomials in the $End(E)$ factor to get the characteristic classes of $E$ in $H^p(X,\Omega_X^p)$ as desired. This answer and Mariano's agree of course in the sense that Atiyah classes can be interpreted via Hochschild and cyclic (co)homology and generalized to arbitrary coherent sheaves (or complexes) on varieties (or stacks) (let me stick to characteristic zero to be safe). Namely the Atiyah class of the tangent sheaf can be used to define a Lie algebra structure (or more precisely $L_\infty$) on the shifted tangent sheaf $T_X[-1]$, and Hochschild cohomology is its enveloping algebra. This Lie algebra acts as endomorphisms of any coherent sheaf (which is another way to say Hochschild cohomology is endomorphisms of the identity functor on the derived category), and one can take characters for these modules, recovering the characteristic classes defined concretely above. (In fact the notion of characters is insanely general... for example an object of any category - with reasonable finiteness - defines a class (or "Chern character") in the Hochschild homology of that category, which is cyclic and so descends to cyclic homology. An example of this is the category of representations of a finite group, whose HH is class functions, recovering usual characters, or coherent sheaves on a variety, recovering usual Chern character. or one can go more general.)
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|
19,312 |
I want to know exactly how derived functor cohomology and Cech cohomology can fail to be the same. I started worrying about this from this answer to an MO question, and Brian Conrad's comments to another MO question . Let $\mathcal{F}$ be a sheaf of abelian groups on a space X. (Here I want to be a little vague about what a "space" means. I'm thinking of either a scheme or a topological space). Then Cech cohomology of X with respect to a cover $U \to X$ can be defined as cohomology of the complex $$ \mathcal{F}(U) \to \mathcal{F}(U^{[2]}) \to \mathcal{F}(U^{[3]}) \to \cdots $$ Where $U^{[ n ]} = U \times_X U \times_X \cdots \times_X U$ . The total Cech cohomology of $X$ , $\check H^{ * }(X, \mathcal{F}) $ , is then given by taking the colimit over all covers $U$ of X. Now if the following condition is satisfied: Condition 1 : For sufficiently many covers $U$ , the sheaf $\mathcal{F}|_{U^{[ n ]}}$ is an acyclic sheaf for each n then this cohomology will agree with the derived functor version of sheaf cohomology. We have, $$\check H^{ * }(X, \mathcal{F}) \cong H^*(X; \mathcal{F}).$$ I am told, however, that even if $\mathcal{F}|_U$ is acyclic this doesn't imply that it is acyclic on the intersections. It is still okay if this condition fails for some covers as long as it is satisfied for enough covers. However I am also told that there are spaces for which there is no cover satisfying condition 1. Instead you can replace your covers by hypercovers. Basically this is an augmented simplicial object $$V_\bullet \to X$$ which you use instead of the simplicial object $U^{[ \bullet +1 ]} \to X$ . There are some conditions which a simplicial object must satisfy in order to be a hypercover, but I don't want to get into it here. You can then define cohomology with respect to a hypercover analogously to Cech cohomology with respect to a cover, and then take a colimit. This seems to always reproduce derived functor sheaf cohomology. So my question is when is this really necessary? Question 1 : What is the easiest example of a scheme and a sheaf of abelian groups (specifically representable ones such as $\mathbb{G}_m$ ) for which Cech cohomology of that sheaf and derived functor cohomology disagree? Question 2 : What is the easiest example of a (Hausdorff) topological space and a reasonable sheaf for which Cech cohomology and derived functor cohomology disagree? I also want to be a little flexible about what a "cover" is supposed to be. I definitely want to allow interesting Grothendieck topologies, and would be interested in knowing if passing to a different Grothendieck topology changes the answer. It changes both the notion of sheaf and the notion of Cech cohomology, so I don't really know what to expect. Also, I edited question 1 slightly from the original version, which just asked about quasi-coherent sheaves. Brian Conrad kindly pointed out to me that for any quasi-coherent sheaf the Cech cohomology and the sheaf cohomology will agree (at least with reasonable assumptions on our scheme, like quasi-compact quasi-separated?) and that the really interesting case is for more general sheaves of groups.
|
Q1: A very simple example is given in Grothendieck's Tohoku paper "Sur quelques points d'algebre homologiquie", sec. 3.8. Edit: The space is the plane, and the sheaf is constructed by using a union of two irreducible curves intersecting at two points. Q2: Cech cohomology and derived functor cohomology coincide on a Hausdorff paracompact space (the proof is given in Godement's "Topologie algébrique et théorie des faisceaux"). I don't know of an example on a non paracompact space where they differ.
|
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|
19,325 |
What are the most striking applications of category theory? I'm trying to motivate deeper study of category theory and I have only come across the following significant examples: Joyal's Combinatorial Species Grothendieck's Galois Theory Programming (unification as computing a coequalizer, Tatsuya Hagino's categorical construction of functional programming) I am sure that these only touch on the surface so I would be most grateful to hear of more examples, thank you! edit: To try and be more precise, "application" in the context of this question means that it makes use of slightly deeper results from category theory in a natural way. So we are not just trying to make a list of 'maths that uses category theory' but some of the results which exemplify it best, and might not have been possible without it.
|
For a while, my answer to this question was algebraic K-theory; what little I know of it, I learned from Quillen's paper, and it was a relief to finally see an example of category theory being used in an essential way to do something that was not just linguistic. Quillen defines the higher K-groups of an exact category by forming a quite different category in some combinatorial manner that seems to strip away any vestige of a connection to something non-categorical, and then taking its geometric realization and homotopy groups. The whole process: ring to module category to Q-construction to geometric realization, was the first argument I'd seen that category theory could do more than just rephrase perfectly good theorems confusingly. (Now my answer would be "perverse sheaves", though.)
|
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|
19,326 |
Given a set of points $X\in\Re^D$, they have labels $Y\in${$-1,+1$}. I would like to separate the data labeled +1 and the data labeled -1 by a polyhedron. $min_w \sum_i \xi_i + \frac{1}{2}\|w\|_2^2$ subject to: $ \xi_i > max_{j=1}^K[1-(w_j^Tx_i+b_j)]$, for $y_i=+1$ and $ \xi_i > min_{j=1}^K[1+(w_j^Tx_i+b_j)]$, for $y_i=-1$
and $ \xi_i > 0 $, for all $i$. Where K is the number of faces of the polyhedron, i represents each sample, j represents each face of the polyhedron. I assume that all positive data go inside the polyhedron while negative data are outside. Following the max-margin principle, we let the distance of the point to the face offset by a margin 1. Optimizing with the first constraint is straightforward. But the second one seems difficult. Is there anyway to optimize them in a fast way to the optimal?
|
For a while, my answer to this question was algebraic K-theory; what little I know of it, I learned from Quillen's paper, and it was a relief to finally see an example of category theory being used in an essential way to do something that was not just linguistic. Quillen defines the higher K-groups of an exact category by forming a quite different category in some combinatorial manner that seems to strip away any vestige of a connection to something non-categorical, and then taking its geometric realization and homotopy groups. The whole process: ring to module category to Q-construction to geometric realization, was the first argument I'd seen that category theory could do more than just rephrase perfectly good theorems confusingly. (Now my answer would be "perverse sheaves", though.)
|
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19,356 |
So I was wondering: are there any general differences in the nature of "what every mathematician should know" over the last 50-60 years? I'm not just talking of small changes where new results are added on to old ones, but fundamental shifts in the nature of the knowledge and skills that people are expected to acquire during or before graduate school. To give an example (which others may disagree with), one secular (here, secular means "trend over time") change seems to be that mathematicians today are expected to feel a lot more comfortable with picking up a new abstraction, or a new abstract formulation of an existing idea, even if the process of abstraction lies outside that person's domain of expertise. For example, even somebody who knows little of category theory would not be expected to bolt if confronted with an interpretation of a subject in his/her field in terms of some new categories, replete with objects, morphisms, functors, and natural transformations. Similarly, people would not blink much at a new algebraic structure that behaves like groups or rings but is a little different. My sense would be that the expectations and abilities in this regard have improved over the last 50-60 years, partly because of the development of "abstract nonsense" subjects including category theory, first-order logic, model theory, universal algebra etc., and partly because of the increasing level of abstraction and the need for connecting frameworks and ideas even in the rest of mathematics. I don't really know much about how mathematics was taught thirty years ago, but I surmised the above by comparing highly accomplished professional mathematicians who probably went to graduate school thirty years ago against today's graduate students. Some other guesses: Today, people are expected to have a lot more of a quick idea of a larger number of subjects, and less of an in-depth understanding of "Big Proofs" in areas outside their subdomain of expertise. Basically, the Great Books or Great Proofs approach to learning may be declining. The rapid increase in availability of books, journals, and information via the Internet (along with the existence of tools such as Math Overflow) may be making it more profitable to know a bit of everything rather than master big theorems outside one's area of specialization. Also, probably a thorough grasp of multiple languages may be becoming less necessary, particularly for people who are using English as their primary research language. Two reasons: first, a lot of materials earlier available only in non-English languages are now available as English translations, and second, translation tools are much more widely available and easy-to-use, reducing the gains from mastery of multiple languages. These are all just conjectures. Contradictory information and ideas about other possible secular trends would be much appreciated. NOTE: This might be too soft for Math Overflow! Moderators, please feel free to close it if so.
|
Many, many things have changed in the last 60 years. A mathematician of the fifties (in Europe) was required to know descriptive geometry, rational mechanics, maybe some astronomy, and a lot of physics. He (yes!) was supposed to know how to calculate rather difficult primitives and have many tricks at his fingertips for checking the convergence of a series. Masterful use of logarithms tables and slide-rules went without saying. Nomography, the graphical representation of mathematical relationships (I guess even the word is forgotten), was a popular option, etc...
|
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|
19,390 |
The modular curve $X_0(N)$ has good reduction at all primes $p$ not dividing $N$. At such a prime, the Eichler-Shimura relation expresses the Hecke operator $T_p$ (as an element of the ring of correspondences on the points of $X_0(N)$ in $\overline{ \mathbb{F}_p }$) in terms of the geometric Frobenius map. This is already weird enough; the definitions of the Hecke operators with which I'm familiar give no indication that such a relation should be true, and the sources I've read so far give no intuition as to why such a relation should be true. (In fact, I still don't feel very comfortable with the Hecke operators themselves; I like the definition given in Milne the best so far, but any enlightening alternate definitions are welcome if they shed light on the question.) What's much weirder, to me anyway, is an important corollary of the Eichler-Shimura relation, which says that given a cusp eigenform $f$ of weight $2$ with respect to $\Gamma_0(N)$ it is possible to construct an elliptic curve $E_f$ whose L-function is (more or less) the Mellin transform of $f$. There are several reasons this is weird to me, but here's a specific one. The modular form $f$ satisfies a functional equation more or less by definition. Its Mellin transform therefore satisfies a corresponding functional equation (part of which has been absorbed into the fact that $f$ is written in terms of its Fourier coefficients), again more or less by definition. The zeta function of an elliptic curve, however, satisfies a functional equation because ( or so I'm told ) of Poincaré duality in the étale cohomology. So: What on Earth does Poincaré duality have to do with modular symmetry? (Ignore the above; I seem to have mixed up the local and global functional equations again.) One of the many things that's weird about the above is that the L-function of an elliptic curve naturally has an Euler product, but for modular forms the Euler product for the Mellin transform comes about because of certain properties of the Hecke operators (which, again, I don't really understand conceptually). What do these properties have to do with multiplying local zeta functions together? I guess I should also clarify what I mean by "intuition": For the first part of the question, if something in the definition of the Hecke operators suggests that they should be related to the Frobenius map if certain natural things were true, and the proof of Eichler-Shimura (which I haven't really looked at yet...) consists of verifying these natural things are true, that would be great intuition. I would appreciate an answer telling me whether or not this was the case in terms of "first principles." For the second part of the question, intuition might more naturally come from a more sophisticated perspective. Here I would appreciate an answer about the "big picture" instead, giving some vague high-level sense of how this all fits into more general things people know about automorphic forms and so forth.
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(1) Short answer to first question: $T_p$ is about $p$-isogenies, and in char. $p$ there is a canonical $p$-isogeny, namely Frobenius. Details: The Hecke correspondence $T_p$ has the following definition, in modular terms:
Let $(E,C)$ be a point of $X_0(N)$, i.e. a modular curve together with a cyclic subgroup
of order $N$. Now $T_p$ (for $p$ not dividing $N$) is a correspondence (multi-valued function) which maps
$(E,C)$ to $\sum_D (E/D, (C+D)/D)$, where $D$ runs over all subgroups of $E$ of degree $p$.
(There are $p+1$ of these.) Here is another way to write this, which will work better in char. $p$:
map $(E,C)$ to $\sum_{\phi:E \rightarrow E'}(E',\phi(C)),$
where the sum is over all degree $p$ isogenies $\phi:E\rightarrow E'.$ Giving a degree
$p$ isogeny in char. 0 is the same as choosing a subgroup of order $D$ of $E$ (its kernel),
but in char. $p$ the kernel of an isogeny can be a subgroup scheme which is non-reduced,
and so has no points, and hence can't be described simply in terms of subgroups of points.
Thus this latter description is the better one to use to compute the reduction of the
correspondence $T_p$ mod $p$. Now if $E$ is an elliptic curve in char. $p$, any $p$-isogeny $E \to E'$ is either
Frobenius $Fr$, or the dual isogeny to Frobenius (often called Vershiebung).
Now Frobenius takes an elliptic curve $E$ with $j$-invariant $j$ to the elliptic curve
$E^{(p)}$ with $j$-invariant
$j^p$. So the correspondence on $X_0(N)$ in char. $p$ which maps $(E,C)$ to $(E^{(p)},
Fr(C))$ is itself the Frobenius correspondence on $X_0(N)$. And the correspondence
which maps $(E,C)$ to its image under the dual to Frobenius is the transpose
to Frobenius (domain and codomain are switched). Since there are no other $p$-isogenies in char. $p$
we see that $T_p$ mod $p = Fr + Fr'$ as correspondences on $X_0(N)$ in char. $p$;
this is the Eichler--Shimura relation. (2) Note that only weight 2 eigenforms with rational Hecke eigenvalues give elliptic curves;
more general eigenforms give abelian varieties. An easy computation shows that if $f$ is a Hecke eigenform, than the $L$-funcion
$L(f,s)$, obtained by Mellin transform, has a degree 2 Euler product. A more conceptual answer would probably involve describing how automorphic representations
factor as a tensor product of local factors, but that it a very different topic from Eichler--Shimura, and I won't say more here.
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|
19,420 |
It's "well-known" that the 19th century Italian school of algebraic geometry made great progress but also started to flounder due to lack of rigour, possibly in part due to the fact that foundations (comm alg etc) were only just being laid, and possibly (as far as I know) due to the fact that in the 19th century not everyone had come round to the axiomatic way of doing things (perhaps in those days one could use geometric plausibility arguments and they would not be shouted down as non-rigorous and hence invalid? I have no real idea about how maths was done then). But someone asked me for an explicit example of a false result "proved" by this school, and I was at a loss. Can anyone point me to an explicit example? Preferably a published paper that contained arguments which were at the time at least partially accepted by the community as being OK but in fact have holes in? Actually, to be honest I'd probably prefer some sort of English historical summary of such things, but I do have access to (living and rigorous) Italian algebraic geometers if necessary ;-) EDIT: A few people have posted solutions which hang upon the Italian-ness or otherwise of the person making the mathematical mistake. It was not my intention to bring the Italian-ness or otherwise of mathematicians into the question! Let me clarify the underlying issue: a friend of mine, interested in logic, asked me about (a) Grothendieck's point of view of set theory and (b) a precise way that one could formulate the statement that he "made algebraic geometry rigorous". My question stemmed from a desire to answer his.
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As for a result that was not simply incorrectly proved, but actually false, there is the case of the Severi bound (*) for the maximum number of singular double points of a surface in P^3. The prediction implies that there are no surfaces in P^3 of degree 6 with more than 52 nodes, but in fact there are such surfaces in P^3 with 65 nodes such as the Barth sextic (and this is optimal by Jaffe--Ruberman ). (*) Francesco Severi; "Sul massimo numero di nodi di una superficie di dato ordine dello spazio ordinario o di una forma di un iperspazio." Ann. Mat. Pura Appl. (4) 25, (1946). 1--41, MR0025179 , doi:10.1007/bf02418077 .
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|
19,435 |
Hello Suppose given a polynomial $P=Q_1\cdots Q_k$ of degree $n$, where each $Q_i$ is irreducible. Suppose also that I know the Galois group $G_i$ (over the rationals) of each irreducible factor $Q_i$. Is there an easy correlation between the Galois group of $P$, and the $G_i$?
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As for a result that was not simply incorrectly proved, but actually false, there is the case of the Severi bound (*) for the maximum number of singular double points of a surface in P^3. The prediction implies that there are no surfaces in P^3 of degree 6 with more than 52 nodes, but in fact there are such surfaces in P^3 with 65 nodes such as the Barth sextic (and this is optimal by Jaffe--Ruberman ). (*) Francesco Severi; "Sul massimo numero di nodi di una superficie di dato ordine dello spazio ordinario o di una forma di un iperspazio." Ann. Mat. Pura Appl. (4) 25, (1946). 1--41, MR0025179 , doi:10.1007/bf02418077 .
|
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|
19,471 |
Is the sum of two measurable set measurable? I think it is not...
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Evidently, there are measure zero sets with a non measurable sum . The article begins as follows: Krzysztof Ciesielski,
Hajrudin Fejzi´c, Chris Freiling, Measure zero sets with non-measurable sum Abstract For any C ⊆ R there is a subset A ⊆ C such that A + A has inner
measure zero and outer measure the same as C + C. Also, there is a
subset A of the Cantor middle third set such that A+A is Bernstein in
[0, 2]. On the other hand there is a perfect set C such that C + C is an
interval I and there is no subset A ⊆ C with A + A Bernstein in I. 1 Introduction. It is not at all surprising that there should be measure zero sets, A, whose sum
A+A = {x+y : x ∈ A, y ∈ A} is non-measurable. Ask a typical mathematician
why this should be so and you are likely to get the following response: The Cantor middle-third set, when added to itself gives an entire
interval, [0, 2]. So certainly there exists a measure zero set that
when added to itself gives a non-measurable set. The intuition being that an interval has much more content than is needed for
a non-measurable set.
Indeed such sets do exist (in ZFC). Sierpi´nski (1920) seems to be the first
to address this issue. Actually, he shows the existence of measure zero sets
X, Y such that X+Y is non-measurable (see [7]). The paper by Rubel (see [6])
in 1963 contains the first proof that we could find for the case X = Y (see also
[5]). Ciesielski [3] extends these results to much greater generality, showing
that A can be a measure zero Hamel basis, or it can be a (non-measurable)
Bernstein set and that A+A can also be Bernstein. He also establishes similar
results for multiple sums, A + A + A etc. This paper is mainly about the statement above and the intuition behind
it. Below we list four conjectures, each of which seems justified by extending
this line of reasoning. Not only does such a set exist, but it can be taken to be a subset of the
Cantor middle-third set, C. (This does not seem to immediately follow
from any of the above proofs. Thomson [9, p. 136] claims this to be
true, but without proof.) The intuition really has nothing to do with the precise structure of the
Cantor set, which might lead one to conjecture the following. Suppose
C is any set with the property that C + C contains a set of positive
measure. Then there must exist a subset A ⊆ C such that A + A is
non-measurable. The intuition relies on the fact that non-measurable sets can have far
less content than an entire interval. Therefore, the claim should also
hold when non-measurable is replaced by other similar qualities. Recall
that if I is a set then a set S is called Bernstein in I if and only if
both S and its complement intersect every non-empty perfect subset
of I. Constructing a set that is Bernstein in an interval is one of the
standard ways of establishing non-measurability. Certainly, any set that
is Bernstein in an interval has far less content than the interval itself.
Therefore, we might conjecture that there is a subset A ⊆ C
with A+A
Bernstein in [0,2]. Combining the reasoning behind the Conjectures 2 and 3, let C be any
set with the property that C + C contains an interval, I. We might
conjecture that there must exist a subset A ⊆ C such that A + A is
Bernstein in I. We will settle these four conjectures in the next four sections. The paper goes on to show that conjectures 1, 2 and 3 are true, but 4 is false.
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|
19,505 |
I have studied differential geometry, and am looking for basic introductory texts on Riemannian geometry. My target is eventually Kähler geometry, but certain topics like geodesics, curvature, connections and transport belong more firmly in Riemmanian geometry. I am aware of earlier questions that ask for basic texts on differential geometry (or topology). However, these questions address mainly differential geometry. I'm more interested in Riemannian geometry here.
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Personally, for the basics, I can't recommend John M. Lee's "Riemannian Manifolds: An Introduction to Curvature" highly enough. If you already know a lot though, then it might be too basic, because it is a genuine 'introduction' (as opposed to some textbooks which just seem to almost randomly put the word on the cover). However, right from the first line: "If you've just completed an introductory course on differential geometry, you might be wondering where the geometry went", I was hooked. It introduces geodesics and curvature beautifully and is very readable. I think the first chapter might be available on the author's website.
|
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|
19,530 |
There seems to be some confusion over what the tangent space to a singular point of an orbifold is. On the one hand there is the obvious notion that smooth structures on orbifolds lift to smooth $G$-invariant structures on $\mathbb R^n$ ($G$ being the finite group so that the orbifold is locally (about some specific point $x$) the quotient of $\mathbb R^n$ by the action of $G$). One might be tempted to consider cone points as differentiable spaces (that is, subsets of some $\mathbb R^k$, inheriting their differential structure by restriction), however, we are told, for example, that $\mathbb R^2/\mathbb Z_3$ and $\mathbb R^2/\mathbb Z_4$ are distinct as orbifolds, so it is not the case that cone points can be modeled merely with cone-like subsets of some $\mathbb R^k$. The definition in which 'smooth' means 'lifts to $G$-invariant smooth' distinguishes these two cones, as the set of functions with 3-fold symmetry and the set of functions with 4-fold symmetry, in $\mathbb R^2$, are distinct. The third item in Satake's seminal paper [On a Generalization of the Notion of Manifold] corroborates this, giving $C^\infty$ forms of degree $p$ at a singularity $x$ as those $C^\infty$ $p$-forms in $\mathbb R^n$ which are invariant under $G_x$. If we require the same property of vectors, that is, that they lift to $G$-invariant vectors in $\mathbb R^n$, then we have that the dimension of the tangent space of an orbifold is the dimension of the invariant subspace upstairs. In particular the dimension tends to drop at the singular points. For example, the dimension of the tangent space at the singularity in $\mathbb R^2/\mathbb Z_3$ is 0. This notion of vector agrees with the notion of vector as derivation on the germ of smooth functions. In this case smooth functions in $\mathbb R^2$ which have 3-fold symmetry necessarily have vanishing derivatives at the origin. On the other hand, one finds descriptions of smooth orbifolds as objects which have tangent bundle-like structures, which are locally $\mathbb R^n/G$. It is not clear what this means as far as smooth structures go, but the explanation above of $\mathbb R^2/\mathbb Z_3$ having a 0 dimensional tangent space at the cone point seems to contradict the notion that the tangent-like space at the singularity in $\mathbb R^2/\mathbb Z_3$ is $\mathbb R^2/\mathbb Z_3$, whatever that means. It is also said that manifolds with boundary can be viewed as orbifolds, which have isotropy group reflection by $\mathbb Z_2$ along their boundaries. It would be nice to include the note that the differentiable structures are different. Specifically, smooth manifolds with boundaries have tangent spaces along their boundaries which are the same dimension as the manifold. In contrast, the same topological space as an orbifold with $\mathbb Z_2$ structure group along the boundary should have a tangent space which is one dimension less than the dimension at a generic point, if the definition of tangent space follows Satake's guideline. Indeed, smooth functions in $\mathbb R^n$ which locally have symmetry by reflection through a codimension 1 hyperplane, have vanishing partial derivatives in the normal direction. I am asking for concurrence or correction and clarification, since I am still not certain I have the correct notion of tangent space to an orbifold, although I'm fairly confident in the first given here.
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Disclaimer: I don't talk to people about orbifolds. This answer may not represent the opinions of orbifolders. As I understand it, the orbifold $\mathbb R^n/G$ is characterized by how manifolds map to it, † not by how it maps to manifolds. In particular, the orbifold is not determined by the ring of smooth functions on it (i.e. the ring of $G$-invariant smooth functions on $\mathbb R^n$). In particular, the ring of 3-fold symmetric smooth functions on $\mathbb R^2$ is isomorphic to the ring of 4-fold symmetric functions on $\mathbb R^2$. Given that the smooth functions on an orbifold don't "remember" everything about it, it seems unreasonable to define the tangent space at a point in terms of derivations of smooth functions. However, we have another construction of the tangent space at a point: equivalence classes of smooth curves through that point. Since this definition has to do with maps into the space rather than maps out of it, we expect it to play well with orbifolds. Any smooth curve through the cone point of $\mathbb R^n/G$ lifts to a curve in $\mathbb R^n$, and any curve in $\mathbb R^n$ induces a curve in $\mathbb R^n/G$, so the tangent space to $\mathbb R^n/G$ should be the same as the tangent space to $\mathbb R^n$. Well, not exactly, since a curve in $\mathbb R^n/G$ can lift to a curve in $\mathbb R^n$ in $G$ different ways. So the tangent space to $\mathbb R^n/G$ at the cone point should really be the quotient of the tangent space of $\mathbb R^n$ by the action of $G$. So far, it seems like I'm arguing that the tangent space to $\mathbb R^n/G$ at the cone point should be the orbifold $\mathbb R^n/G$. But I actually want to say that the tangent space should be the tangent space of $\mathbb R^n$, together with the action of $G$. The reasoning is that the vector space together with the action of the residual group is independent of how you express $\mathbb R^n/G$ as a quotient by a finite group. In other words, even though an isomorphism of orbifolds $\mathbb R^n/G\cong M/G'$ does not induce isomorphisms $\mathbb R^n\cong M$ or $G\cong G'$, it does induce isomorphisms of tangent spaces and residual groups in such a way that respects the actions of the residual groups on the tangent spaces. Once your definition of tangent space is canonically related to the orbifold, you should be welcome to think of it however you like. † Roughly, a map from a manifold $M$ to $\mathbb R^n/G$ should be the same thing as a map from $M$ to $\mathbb R^n$, except that maps that differ by the action of $G$ should be regarded as the same map. More precisely, I think a map from $M$ to $\mathbb R^n/G$ should consist of a $G$-fold covering space of $M$ with a $G$-equivariant map to $\mathbb R^n$.
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|
19,584 |
One of the more misleadingly difficult theorems in mathematics is that all finitely generated projective modules over a polynomial ring are free. It involves some of the most basic notions in commutative algebra, and really sounds as though it should be easy (the graded case, for example, is easy), but it's not. The question at least goes as far back as Serre's FAC, but it wasn't proved until 1976, by Quillen in Projective modules over polynomial rings EDIT: and also independently by Suslin. I decided that this is the sort of fact that I should know a rough outline of how to prove, but the paper was not very helpful. Usually when someone kills off a famous conjecture in 5 pages, it's because they've developed some fantastic new piece of machinery people didn't have before. And, indeed, Quillen is famous for inventing some fancy and wonderful machinery, and the paper is only 5 pages long, but as far as I can tell, none of that fancy machinery actually appears in the proof. So, what was it that Quillen saw, that Serre missed?
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Here is a summary of what I learned from a nice expository account by Eisenbud (written in French), can be found as number 27 here . 1 First, one studies a more general problem: Let $A$ be a Noetherian ring, $M$ a finite presented projective $A[T]$ -module. When is $M$ extended from $A$ , meaning there is $A$ -module $N$ such that $M = A[T]\otimes_AN$ ? The proof can be broken down to 2 punches: Theorem 1 (Horrocks) If $A$ is local and there is a monic $f \in A[T]$ such that $M_f$ is free over $A_f$ , then $M$ is $A$ -free (this statement is much more elementary than what was stated in Quillen's paper). Theorem 2 (Quillen) If for each maximal ideal $m \subset A$ , $M_m$ is extended from $A_m[T]$ , then $M$ is extended from $A$ (on $A$ , locally extended implies globally extended). So the proof of Serre's conjecture goes as follows: Let $A=k[x_1,\dotsc,x_{n-1}]$ , $T=x_n$ , $M$ projective over $A[T]$ . Induction (invert all monic polynomials in $k[T]$ to reduce the dimension) + further localizing at maximal ideals of $A$ + Theorem 1 show that $M$ is locally extended. Theorem 2 shows that $M$ is actually extended from $A$ , so by induction must be free. Eisenbud's note also provides a very elementary proof of Horrocks's Theorem, basically using linear algebra, due to Swan and Lindel (Horrocks's original proof was quite a bit more geometric). As Lieven wrote , the key contribution by Quillen was Theorem 2: patching. Actually the proof is fairly natural, there is only one candidate for $N$ , namely $N=M/TM$ , so let $M'=A[T]\otimes_AN$ and build an isomorphism $M \to M'$ from the known isomorphism locally. It is hard to answer your question: what did Serre miss (-:? I don't know what he tried. Anyone knows? 1 Eisenbud, David , Solution du problème de Serre par Quillen–Suslin, Semin. d’Algebre, Paul Dubreil, Paris 1975-76 (29eme Annee), Lect. Notes Math. 586, 9-19 (1977). DOI: 10.1007/BFb008711 , PDF on the MSRI website , ZBL0352.13005 , MR568878 .
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|
19,644 |
I just received a referee report criticizing that I would too often use the word "canonical". I have a certain understanding of what "canonical" should stand for, but the report shows me that other people might think differently. So I am asking: Is there a definition of "canonical"? What are examples where the use of "canonical" is undoubtedly correct? What are examples where the use of "canonical" is undoubtedly incorrect? VERY LATE EDIT: I just came across this wonderful passage written by André Weil (Oeuvres, vol. 2, page 558): I can assure you, at any rate, that [...] my results are invariant, probably canonical, perhaps even functorial.
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I think there is a multi-level classification associated to "canonicalness," which explains why some clashes of definition occur. Arbitrary — No requirements. Uniform — There may be a few options but these options can be selected by making a few global choices. Canonical — As in the uniform case, but there is only one natural choice of options which applies globally. Canonical examples à la Russell: Choose one sock from each pair in a collection of sock pairs — There is no way to make a uniform choice. Choose one shoe from each pair in a collection of shoe pairs — There are two obvious global solutions, left shoe or right shoe, but no way to prefer one over the other. Choose one object from each set in a collection of sets each consisting of a bowtie and possibly other items — There is only one obvious global solution. I think the main point of contention is distinguishing uniform and canonical . Some will argue that it's not canonical if there is a choice to be made, while some will argue that a finite number of global choices is still canonical. There is yet another use of canonical to mean something like 'universally sanctioned' (this is closer to the religious term). The second occurrence of canonical above is of this type.
|
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|
19,684 |
In the study of number theory (and in other branches of mathematics) presence of Hecke Algebra and Hecke Operator is very prominent. One of the many ways to define the Hecke Operator $T(p)$ is in terms of double coset operator corresponding to the matrix $ \begin{bmatrix} 1 & 0 \\ 0 & p \end{bmatrix}$ . On the other hand Hecke Algebra $\mathcal{H}(G,K)$ associated to a group $G$ of td-type ( topological group, such that every neighborhood of unity contains a compact open subgroup), where $K$ is a compact open subgroup of $G$ is defined as the space of locally constant compactly supported $K$ bi-invariant functions on $G$. Convolution product makes it an associative algebra. I was told that the hecke algebra $\mathcal{H}(Gl(2,\mathbb{Q}_p) , Gl(2,\mathbb(Z)_p))$ corresponds to the classical algebra of hecke operators attached to $p$ via Satake Isomorphism Theorem. Using Satake Isomorphism theorem I can show $\mathcal{H}(Gl(2,\mathbb{Q}_p) ,Gl(2,\mathbb(Z)_p))$ is commutative and finitely generated over $\mathbb{C}$. So my question is how one uses Satake Isomorphism Theorem (or otherwise) to see this?
And secondly in general what is the relation between hecke operators and hecke algebra?
|
The fact that Hecke operators (double coset stuff coming from $SL_2(\mathbf{Z})$ acting on modular forms) and Hecke algebras (locally constant functions on $GL_2(\mathbf{Q}_p)$) are related has nothing really to do with the Satake isomorphism. The crucial observation is that instead of thinking of modular forms as functions on the upper half plane, you can think of them as functions on $GL_2(\mathbf{R})$ which transform in a certain way under a subgroup of $GL_2(\mathbf{Z})$, and then as functions on $GL_2(\mathbf{A})$ ($\mathbf{A}$ the adeles) which are left invariant under $GL_2(\mathbf{Q})$ and right invariant under some compact open subgroup of $GL_2(\widehat{\mathbf{Z}})$. Now there's just some general algebra yoga which says that if $H$ is a subgroup of $G$ and $f$ is a function on $G/H$, and $g\in G$ such that the $HgH$ is a finite union of cosets $g_iH$, then you can define a Hecke operator $T=[HgH]$ acting on the functions on $G/H$, by $Tf(g)=\sum_i f(gg_i)$; the lemma is that this is still $H$-invariant. Next you do the tedious but entirely elementary check that if you consider modular forms not as functions on the upper half plane but as functions on $GL_2(\mathbf{A})$, then the classical Hecke operators have interpretations as operators $T=[HgH]$ as above, with $T_p$ corresponding to the function supported at $p$ and with $g=(p,0;0,1)$. Because the action is "all going on locally" you may as well compute the double coset space locally, that is, if $H=H^pH_p$ with $H_p$ a compact open subgroup of $GL_2(\mathbf{Q}_p)$, then you can do all your coset decompositions and actions locally at $p$. Now finally you have your link, because you can think of $T$ as being the characteristic function of the double coset space $HgH$ which is precisely the sort of Hecke operator in your Hecke algebra of locally constant functions. Furthermore the sum $f(gg_i)$ is just an explicit way of writing convolution, so everything is consistent. I don't know a book that explains how to get from the classical to the adelic point of view in a nice low-level way, but I am sure there will be some out there by now. Oh---maybe Bump?
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|
19,840 |
My question is fairly simple, and may at first glance seem a bit silly, but stick with me. If we are given the rationals, and we pick an element, how do we recognize whether or not what we picked is an integer? Some obvious answers that we might think of are: A. Write it in lowest terms, and check the denominator is 1. B. Check that the p-adic valuation is non-negative, for all p. C. Decide whether the number is positive (or negative) and add 1 to itself (or -1 to itself) until it is bigger than the rational you picked. (If these multiples ever equaled your rational, then you picked an integers.) Each of these methods has pluses and minuses. For example, in option A we presuppose we know how to write an arbitrary rational number q as a quotient of integers and reduce. In C, we have issues with stopping times. etc.. To provide some context for my question: We know, due to the work of Davis, Putnam, Robinson, and Matijasevic, that the positive existential theory of $\mathbb{Z}$ is undecidable. The same question for $\mathbb{Q}$ is not entirely answered. One approach to this new question is to show that that, using very few quantifiers, one can describe the set of integers inside the rationals; and then reduce to the integer case. For example, see Bjorn Poonen's paper "Characterizing integers among rational numbers with a universal-existential formula." There, he finds a way to describe the p-valuation of a rational number (i.e. he finds a way to encode option B in the language of quantifiers and polynomials on the rationals). I'm wondering if there are other characterizations of the integers which would follow suit.
|
Here's one way: Show that the rational number is an algebraic integer. This may sound like a silly idea, but it has non-trivial applications. A rational number is an integer if it has an expression as a sum of products of algebraic integers. See for example, Prop. 5 in the appendix of Groups and Representations by J.L. Alperin, and Rowen B. Bell, where this is used to prove that, for an irreducible character $\chi$ of a finite group $G$, $\chi(1)$ divides $|G|$.
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|
19,857 |
Note: I have modified the question to make it clearer and more relevant. That makes some of references to the old version no longer hold. I hope the victims won't be furious over this. Motivation: Recently Pace Nielsen asked the question "How do we recognize an integer inside the rationals?". That reminds me of this question I had in the past but did not have chance to ask since I did not know of MO. There seems to be a few evidence which suggest some possible relationship between decidability and prime numbers: 1) Tameness and wildness of structures One of the slogan of modern model theory is " Model theory = the geography of tame mathematics". Tame structure are structures in which a model of arithmetic can not be defined and hence we do not have incompleteness theorem. A structure which is not tame is wild. The following structures are tame: Algebraic closed fields. Proved by Tarski. Real closed fields e.g $\mathbb{R}$. Proved by Tarski. p-adic closed fields e.g $ \mathbb{Q}_p$. Proved by Ax and Kochen. Tame structures often behave nicely. Tame structures often admits quantifier elimination i.e. every formula are equivalent to some quantifier free formula, thus the definable sets has simple description. Tame structures are decidable i.e there is a program which tell us which statements are true in these structure. The following structures are wilds; Natural number (Godel incompleteness theorem) Rational number ( Julia Robinson) Wild structure behaves badly (interestingly). There is no program telling us which statements are true in these structures. One of the difference between the tame structure and wild structure is the presence of prime in the later. The suggestion is strongest for the case of p-adic field, we can see this as getting rid of all except for one prime. 2) The use of prime number in proof of incompleteness theorem The proof of the incompleteness theorems has some fancy parts and some boring parts. The fancy parts involves Godel's Fixed point lemma and other things. The boring parts involves the proof that proofs can be coded using natural number. I am kind of convinced that the boring part is in fact deeper. I remember that at some place in the proof we need to use the Chinese Remainder theorem, and thus invoke something related to primes. 3) Decidability of Presburger arithmetic and Skolem arithmetic ( extracted from the answer of Grant Olney Passmore) Presburger arithmetic is arithmetic of natural number with only addition. Skolem arithmetic is arithmetic of natural number with only multiplication. Wishful thinking: The condition that primes or something alike definable in the theory will implies incompleteness. Conversely If a theory is incomplete, the incompleteness come from something like primes. Questions: (following suggestion by François G. Dorais) Forward direction: Consider a bounded system of arithmetic, suppose the primes are definable in the system. Does it implies incompleteness. Backward direction: Consider a bounded system of arithmetic, suppose the system can prove incompleteness theorem, is primes definable in the system? is the enumeration of prime definable? is the prime factoring function definable? Status of the answer: For the forward direction: A weak theory of prime does not implies incompleteness. For more details, see the answer of Grant Olney Passmore and answer of Neel Krishnaswami For backward direction: The incompleteness does not necessary come from prime. It is not yet clear whether it must come from something alike prime. For more details, see the answer of Joel David Hamkins. Since perhaps this is as much information I can get, I accept the first answer by Joel David Hamkins. Great thanks to Grant Olney Passmore and Neel Krishnaswami who also point out important aspects. Recently, Francois G. Dorais also post a new and interesting answer.
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Goedel did indeed use the Chinese remainder theorem in his proof of the Incompleteness theorem. This was used in what you describe as the `boring' part of the proof, the arithmetization of syntax. Contemporary researchers often agree with your later assessment, however, that the arithmetization of syntax is profound. This is the part of the proof that reveals the self-referential nature of elementary number theory, for example, since in talking about numbers we can in effect talk about statements involving numbers. Ultimately, we arrive in this way at a sentence that asserts its own unprovability, and this gives the Incompleteness Theorem straight away. But there are other coding methods besides the Chinese Remainder theorem, and not all of them involve primes directly. For example, the only reason Goedel needed CRT was that he worked in a very limited formal language, just the ring theory language. But one can just as easily work in a richer language, with all primitive recursive functions, and the proof proceeds mostly as before, with a somewhat easier time for the coding part, involving no primes. Other proofs formalize the theory in the hereditary finite sets HF, which is mutually interpreted with the natural numbers N, and then the coding is fundamentally set-theoretic, also involving no primes numbers especially.
|
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|
19,930 |
I wonder how people wrote papers in the pre-LaTeX era? I mean, when typewriters and simple computers were (60th-70th?). Did they indeed put formulas by hand in the already printed articles?
|
To elaborate on Felipe's answer: My father was a mathematicians, and, before the IBM Selectric typewriter, I can tell you that he bought an expensive manual German typewriter along with boxes of these plastic sticks, each with a metal head containing a math symbol. Very slow and very painful. Later, department secretaries (there used to be more of them) would have IBM selectric typewriters, which would use a metal coated plastic ball, and you would switch the ball to get different symbols. This normally worked pretty well, except sometimes the teeth on the bottom of the ball (which was serrated for some reason) would break. This would mess up the typing action, so it would no longer type the symbol or character properly. And even after TeX came along, there were only mainframes back then with ASCII terminals and line printers. I was one of the first people to type my thesis on TeX, and it required the following conditions:
a) MIT AI lab was freely accessible by anyone (I was at Harvard, not MIT) 24 hours a day
b) They had a machine running LISP and MACSYMA and wanted people to test it, so they gave out free accounts, usable only outside regular business hours, to anyone who asked
c) TeX was installed on this machine
d) People who worked in the MIT AI Lab would leave their offices open or unlocked, so if you walked in there in the middle of the night, you could go in there and use their Symbolics LISP machine. The last remaining challenge was getting access to the laser printer (which was the size of a room) that was inside the locked machine room. Luckily, I had a friend (MIT math graduate student who has been mentioned and cited often on MathOverflow) whose girlfriend (now wife) is the daughter of an MIT CS professor. I asked if I could borrow the father's key to the machine room. Miraculously, the answer was yes. I then gave up TeX for many years. At Courant they had an amazing typist named Caroline Connie, who could type up long difficult papers at record speed.
|
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|
19,932 |
Hi,
to completely describe a classical mechanical system, you need to do three things: -Specify a manifold $X$, the phase space. Intuitively this is the space of all possible states of your system. -Specify a hamilton function $H:X\rightarrow \mathbb{R}$, intuitivly it assigns to each state its energy. -Specify a symplectic form $\omega$ on $X$. What is $\omega$ intuitively? What kind of information about physics does it capture?
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To elaborate on a comment of Steve Huntsman: the symplectic form turns a form $d H$ into a flow $X_H$ with a number of properties, but other types of forms can do a similar job. Indeed, there are a number of situations in physics where the relevant $\omega$ is not symplectic, for example for the following reasons: $\omega$ might be degenerate in the sense that $i_X \omega = 0$ for certain $X \ne 0$. This occurs for instance when you pull back $\omega$ to a constraint surface in phase space. Or you might be working on the Lagrangian side, taking the pull-back of $\omega$ along a non-invertible Lagrangian. In non-holonomic mechanics, $\omega$ is sometimes not closed, with the derivative $d \omega$ being related to the non-integrability of the constraint distribution. The point is that such forms all lead to valid generalizations of Hamilton's equations, so using a symplectic form to write down Hamilton's equations is to a large extent motivated by the fact that it "just works". The physical properties offered by using a symplectic form instead of an arbitrary two-form are the following: Non-degeneracy: the evolution vector field $X_H$ is determined uniquely by the Hamiltonian $H$. By contrast, if you have gauge freedom, there will typically be constraints in the phase space, hence a degenerate symplectic form (see above), leading to a non-unique evolution (which is what gauge freedom is --- several mathematically distinct evolutions being physically the same). Closedness: the system preserves the symplectic form
$$
L_{X_H} \omega = d i_{X_H} \omega + i_{X_H} d \omega = 0
$$
if $\omega$ is closed. In the classical literature, this gives rise to a series of conservation laws called the "Poincare invariants". Again, non-holonomic systems typically don't exhibit this property, leading to all sort of weirdness.
|
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|
19,957 |
My son is one year old, so it is perhaps a bit too early to worry about his mathematical education, but I do. I would like to hear from mathematicians that have older children: What do you wish you'd have known early? What do you think you did particularly well? What do you think would be particularly bad? Is there a book (for children or parents) that you recommend? (This a community wiki, so please give one advice per answer, as usual.) Background I ask here because I believe that the challenges a mathematician faces in educating a child are special. For example, at least some websites and books address the parents' fear of not knowing how to solve homework, which keeps them from becoming involved. On the contrary, I fear I might get too involved and either bore my son or make him think he likes math when in fact his skills are elsewhere. Christos Papadimitriou said in an interview that, even though his father was teaching math in high-school, they never discussed math. I wonder if that means his father didn't teach him how to count and I wonder if it's a good strategy. (It certainly turned out well in one case.) Timothy Gowers (in Mathematics, a very short introduction ) says that it was inappropriate to explain to his son, who was six, the concept 'zero' using the group axioms. (Or something to this effect, I don't have the book near to check.) That was surprising to me, because I wouldn't have thought that I need to restrain myself from mentioning abstract concepts. ( Update. Here's the quote: "[The non-abstract] way of thinking makes it hard to answer questions such as the one asked by my son John (when six): how can nought times nought be nought, since nought times nought means that you have no noughts? A good answer, though not one that was suitable at the time, is that it can be deduced from the [field axioms] as follows. [...]") There is a somewhat related Mathoverflow question . This one is different, because I'm looking for advice (rather than statistics/anecdotes) and because my goal is to give my son a good math education (rather than to make him a mathematician). I also found an online book that seems to give particularly good generic advice. Here I'm looking more for advice geared towards parents that are mathematicians. In short, I'm looking for specific advice on how a mathematician should approach his/her child's math education, especially for the 1 to 10 age range.
|
The story I heard from a senior colleague when I was at this stage was: "Twenty years ago I had no children and five theories on how to bring up children. Now I have four grown-up children and no theories."
|
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|
19,987 |
It seems in writing math papers collaborators put their names in the alphabetical order of their last name. Is this a universal accepted norm? I could not find a place putting this down formally.
|
This tendency of mathematicians is so well-known and universal that it has been taken as an axiom. See Andrew Appel's seminal work establishing whether different computer science conferences are mathematics or science .
|
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|
20,020 |
I apologize that this is vague, but I'm trying to understand a little bit of the historical context in which the zoo of quantum invariants emerged. For some reason, I have in my head the folklore: The discovery in the 80s by Jones of his new knot polynomial was a shock because people thought that the Alexander polynomial was the only knot invariant of its kind (involving a skein relation, taking values in a polynomial ring, ??). Before Jones, there were independent discoveries of invariants that each boiled down to the Alexander polynomial, possibly after some normalization. Is there any truth to this? Where is this written?
|
The skein relation approach to knot invariants was not very popular before the Jones polynomial. The Alexander polynomial was thought of as coming from homology (of the cyclic branched cover); Conway had found the skein relation, but it was not well-known. Of course once you start investigating skein relations systematically, you rapidly find the Jones, Kauffman, and HOMFLY relations. Basically, people had been looking for invariants using their standard tools like homology, and had trouble constructing interesting invariants that way. The idea of just looking for a skein relation was new. The notion of "polynomial invariants" by itself is too vague to give a place to look.
|
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|
20,045 |
When I was teaching calculus recently, a freshman asked me the conditions of the Riemann integrability of composite functions. For the composite function $f \circ g$ , He presented three cases:
1) both $f$ and $g$ are Riemann integrable;
2) $f$ is continuous and $g$ is Riemann integrable;
3) $f$ is Riemann integrable and $g$ is continuous. For case 1 there is a counterexample using Riemann function. For case 2 the proof of the integrability is straight forward. However, for case 3, I can neither give a proof nor construct any counterexample. Even under the condition that $g$ is differentiable, I cannot work out anything. How to reply my student?
|
Let $f$ be bounded and discontinuous on exactly the Cantor set $C$ (for example, the characteristic function of $C$). Let $g$ be continuous increasing on $[0,1]$ and map a set of positive measure (for example a fat Cantor set) onto $C$. Then $f \circ g$ is discontinuous on a set of positive measure. So $f$ is Riemann integrable, $g$ is continuous, and $f \circ g$ is not Riemann integrable. Of course, a Freshman calculus student wont know about "measure zero" so this example is not good for an elementary course.
|
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|
20,057 |
Chas and Sullivan constructed in 1999 a Batalin-Vilkovisky algebra structure on the shifted homology of the loop space of a manifold: $\mathbb{H}_*(LM) := H_{*+d}(LM;\mathbb{Q})$. This structure includes a product which combines the intersection product and Pontryagin product and a BV operater $\Delta: \mathbb{H}_*(LM) \to \mathbb{H}_{*+1}(LM)$. I was wondering about the applications of this structure. Has it even been used to prove theorems in other parts of mathematics? A more concrete question is the following: Usually, considering a more complicated structure on topological invariants of a space allows you to prove certain non-existince results. For example, the cup product in cohomology allows you to distinguish between $ S^2 \vee S^1 \vee S^1 $ and
$T^2$. Is there an example of this type for string topology?
|
Hossein Abbaspour gave an interesting connection between 3-manifold topology and the string topology algebraic structure in arXiv:0310112 . The map $M \to LM$ given by sending a point $x$ to the constant loop at $x$ allows one to split $\mathbb{H}_*(LM)$ as $H_*(M) \oplus A_M$ . He showed essentially that the restriction of
the string product to the $A_M$ summand is nontrivial if and only if $M$ is hyperbolic. There are some technical details in the statements in his paper, but it was written pre-Perelman and I believe the statements can be made a bit more elegant in light of the Geometrization Theorem. Philosophically, Sullivan has said that he his goal in inventing string topology was to try to find new invariants of smooth structures on manifolds. His original idea was that if you have to use the smooth structure to smoothly put chains into transversal positions to intersect them then you might hope that the answer will depend on the smooth structure. Unfortunately, we now know that the string topology BV algebra depends only on the underlying homotopy type of the manifold (there are now quite a few different proofs of various parts of this statement). The string topology BV algebra is only a piece of a potentially much richer algebraic structure. Roughly speaking, $\mathbb{H}_*(LM)$ is a homological conformal field theory. This was believed to be true for quite some time but it took a while before it was finally produced by Veronique Godin arxiv:0711.4859 . She constructed an action of the PROP made from the homology of moduli spaces of Riemann surfaces with boundary. Restricting this action to pairs of pants recovers the original Chas-Sullivan structure. Unfortunately, for degree reasons, nearly all of the higher operations vanish. In particular, any operation given by a class in the Harer stable range of the homology of the moduli space must act by zero. Hirotaka Tamanoi has a paper that spells out the details, but it is nothing deep. Furthermore, it seems that the higher operations are homotopy invariant as well. For instance Lurie gets this as a corollary of his work on the classification of topological field theories. Last I heard, Sullivan, ever the optimist, believes that there is still hope for string topology to detect smooth structures. He says that one should be able to extend from the moduli spaces of Riemann surfaces to a certain piece of the boundary of the Deligne-Mumford compactification. I've heard that the partial compactification here is meant to be that one allows nodes to form, but only so long as the nodes collectively do not separate the incoming boundary components from the outgoing boundary. Sullivan now has some reasons to hope that operations coming from homology classes related to the boundary of these moduli spaces might see some information about the underlying smooth structure of the manifold.
|
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|
20,071 |
I am very interested in reading some and skimming through the list of invited talks at the International Congress of Mathematicians. Since the proceedings contain talks supposedly by top experts in each area, even the list of invited talks would hopefully provide some picture of how mathematics changed throughout the last century or so. I looked it up but wikipedia only provides links to the proceedings of ICM since 1998. So it excludes many talks I really want to read, like those by Serre, Grothendieck, Auslander, Quillen, etc. Does anyone know how to find the rest of the ICM proceedings, hopefully online? Thanks. UPDATE: The whole collection of all the proceedings of the ICMs is available here !
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Update: (Oct. 2018) For the first time, all ICM 2018 lectures (plenary, invited and special) as well as panels and special events are presented (by good-quality videos) on the ICM 2018 You tube channel . Update: (Dec 2017) The ICM launched a new website. All previous ICM proceedings are available here . However, I cannot find the old page with access to individual papers and search options. Just recently the International Mathematical Union (IMU) put online all the previous proceedings on the ICM's! Here : (update:broken). This webpage is based on joint work by R. Keith Dennis (Ithaca) and Ulf Rehmann (Bielefeld). So you can read for free all the articles (including Hilbert's famous problem paper; Martin Grötschel demonstrated it at the opening ceremony of ICM 2010). I don't know when the proceedings of ICM2010 will be added. UPDATE: the 2010 articles have now been added! Update The ICM2014 papers are available here (you can download each of 4 volumes; the IMU site does not contain these papers yet.) As for the talks themselves, there is a page with all the ICM 2010 plenary talks here ; links for videos from earlier ICM's (plenary talks and other events) can be found here . Update (August, 6 2014) Many (46 for now) of the ICM 2014 proceedings contributions are already available on arXiv, via this search . (I got it from Peter Woit's blog.) Videos of lectures can be found here . Update (May 2014): Starting 1992 there is also every four years the European Congress of Mathematics (ECM) that the European Mathematical Society (EMS) is running.
The proceedings of the first three ECMs are now freely open. These volumes are available here . (Digitising the proceedings of the first three ECMs, published by Birkhäuser was a task carried out by the EMS Electronic Publishing Committee.) Starting with the 4ECM (2004), the Proceedings are published by the EMS Publishing House. The EMS decided to make them freely available online too. I expect that this will happen soon and I will keep you posted. Further update (June 2014) The ECM Proceedings are now available here ! I was told that in a few months, the EMS will put also the 6ECM volume.
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20,112 |
On another thread I asked how I could encourage my final year undergraduate colleagues to take an algebraic geometry or complex analysis courses during their graduate studies. Willie Wong proposed me following idea - to show them some interesting results in this fields with relatively simply proofs and some consequences in other fields.
Thus by 'interesting' result in algebraic geometry I here mean the result which may convince 3rd year undergraduate student to study algebraic geometry. In fact I'm supposed to give some talk during the seminar dedicated to final year undergraduates, and I can propose my own topic, so I thought that it could work.
But my problem is that I'm just wanna-be student of algebraic geometry and I don't have enough insight and knowledge to find a topic which 'could work'. Also I doubt whether it is possible to present some intriguing ideas of algebraic geometry to audience without any preparation in this field. So in short my first question is as above: Is it possible to present some intriguing ideas of algebraic geometry to audience without any preparation in this field? To be more precise - the talk should take a one or two meetings, 90 minutes each one.
The audience will be, as I said 3rd year undergraduate students, all of them after two semester course in algebra, some of them after one or two semesters in commutative algebra. All taking the course in one complex variable, and after one semester introductory course in differential geometry.
Some of them may be interested in number theory. The second, related question is as follows: If You think that the answer to the previuos question is positive, please try to give an example of idea/theorem/result which would be accessible in such time for such audience, and which You find interesting enough to make them consider possibility of studying algebraic geometry. Try to think about the results which shows connections of AG with some other fields of mathematics.
|
If you want to teach something intriguing, you should do something that introduces a new geometric idea while also involving algebra in an essential way. I recommend that you give an introduction to the projective plane, showing the other students that it is a natural extension of ordinary space which makes some geometric properties more uniform (such as intersection properties of curves), gives a fruitful new way to think about old topics (like asymptotes), and lets you do things that are impossible to conceive without it (reducing rational points mod $p$). There should be substantial interplay between algebra and geometry, but make sure to draw pictures to emphasize the geometric aspects. In algebra, we can conceive of the quadratic formula in a uniform manner, but the ancient Greeks [Edit: Babylonians, not Greeks] couldn't do this because they didn't have the idea of negative numbers. So they had several quadratic formulas on account of not being able to write something as simple as $ax^2 + bx + c = 0$ at one stroke (for any signs on $a, b$, and $c$, with $a$ nonzero). Our extended skill at algebra lets us work with one case where the ancients had to take multiple cases. We can also say with complex numbers that any quadratic equation has two roots, allowing for a double root to count as one root with multiplicity two.
The thrust of what comes next is to extend the plane so that geometric properties become nicer in a similar way the algebra is becoming nicer when we use more general number systems. Consider the intersection properties of lines in the plane. There is a dichotomy: usually two lines in the plane meet in one point, but some pairs of lines (the parallel ones) meet in no points. Let's see what this looks like under stereographic projection. Lines in the plane become circles through the north pole, but not including the north pole itself. It's natural to close up the image and take that whole circle as a substitute for the original line. So we can see that lines in the plane naturally close up into circles through the north pole. Under stereographic projection, the old dichotomy between parallel and non-parallel lines takes on a new appearance: a pair of non-parallel lines corresponds under stereographic projection to a pair of circles intersecting in two different points, one of which is the north pole, while a pair of parallel lines corresponds under stereographic projection to a pair of circles which are tangent at the north pole. It is natural to think of two tangent circles as having their point of tangency be an intersection point of multiplicity two, much like a quadratic polynomial can have a double root. So after stereographic projection we can "see" two points of intersection for any pair of lines. This geometric construction is something like the algebraic use of more general number systems to find roots to all quadratic equations. The moral to take from this example is that in a larger space, curves that used to not intersect may now intersect (or rather, their natural closures in the new space intersect) with a uniform count of the number of intersection points. If the students agree that enlarging number systems to create solutions to polynomial equations is good, they should agree that enlarging space to make intersection properties more uniform is good too. Another important feature is that the sphere, like the plane, is a homogeneous object: we can transform (rotate) the space to carry one point to any other point. On the sphere as a space in its own right, there is truly nothing special about the north pole. An even better geometric extension of the plane is the projective plane, although at first it will feel unfamiliar and strange because you can't see it all at once.
You should introduce it in a uniform manner as points described with homogeneous coordinates $[x,y,z]$ where $x$, $y$, and $z$ are not all 0 and, say,
$$
[3,6,2] = [1,2,2/3] = [1/2,1,1/3] = [3/2,3,1] \text{ and } [0,5,0] = [0,1,0].
$$
Although it is impossible to see the whole projective plane at once, we can get glimpses of large parts of it using three different charts: $U_0$ is the points where $x \not= 0$, $U_1$ is the points where $y \not= 0$ and $U_2$ is the points where $z \not= 0$. These three charts together cover the projective plane. Any nonzero coordinate can be scaled to 1 and that fixes the other two homogeneous coordinates of the point, e.g., $[x,y,1] = [x',y',1]$ if and only if $x = x'$ and $y = y'$. This means we can identify each of $U_0$, $U_1$, and $U_2$ with the usual plane (e.g., identify $U_2$ with ${\mathbf R}^2$ by identifying $[x,y,1]$ with $(x,y)$). This means the projective plane locally looks like the plane, much like the sphere does, except we can't see all of it at the same time as we can with the sphere. (In case you want to show students that the projective plane is a really natural model of something they have known in another context, think about nonzero ideals in ${\mathbf R}[x]$. Any ideal has a generator, but the polynomial generator is only defined up to a nonzero scaling factor. Usually we normalize the generator to be monic, but if we don't want to insist on a particular choice of generator then the right model for the generator is a point in projective space. In particular, for any nonzero ideal $(f(x))$ where $\deg f(x) \leq 2$, write $f(x) = ax^2 + bx + c$; the coefficients $a, b, c$ are only defined up to an overall scaling factor, so the point $[a,b,c]$ is one way to think about that ideal.) Next introduce curves in the projective plane as solutions to homogeneous polynomial equations in $x$, $y$, and $z$ and explain what the algebraic process of homogenization and dehomogenization of polynomials is, e.g., it makes $y = 2x + 1$ into $y = 2x + z$ or $x^2 - y^2 = x+ 1$ into $x^2 - y^2 = xz + z^2$. In particular a line in the projective plane is the solution set to any equation $ax + by + cz = 0$ where the coefficients are not all 0. Now let's look at what a point on a specific curve in the projective plane looks like in each of the three standard charts, carry out the same kind of calculus computation in each chart, and compare the results with each other. We will use the curve $C : x^2 + y^2 = z^2$ in the projective plane (not to be confused with a surface in 3-space given by the same equation) and the points $P = [3,4,5]$ and $Q = [1,0,1]$ which lie on $C$. How do $C$, $P$, and $Q$ appear in each of the charts $U_0$, $U_1$, and $U_2$? a) In $U_0$, which is identified with the plane by $[x,y,z] \mapsto (y/x,z/x)$, $C$ becomes the hyperbola $z^2 - y^2 = 1$, $P$ becomes $(4/3,5/3)$, and $Q$ becomes $(0,1)$. Here we identify $U_0$ with the usual $yz$-plane. By calculus, the tangent line to $z^2 - y^2 = 1$ at the point $(4/3,5/3)$ is $z = (4/5)y + 3/5$ and the tangent line at $(0,1)$ is $z = 1$. Note that we actually miss two points from $C$ when we look at the intersection of it with $U_0$: $[0,1,\pm 1]$. b) In $U_1$, $C$ becomes the hyperbola $z^2 - x^2 = 1$ in the $xz$-plane, $P$ becomes the point $(3/4,5/4)$ with tangent line $z = (3/5)x + 4/5$, and $Q$ doesn't actually live in this chart (kind of like the north pole under stereographic projection not going to anything the in the plane). Here two points from $C$ are missing: $[1,0,\pm 1]$. c) In $U_2$, $C$ becomes the circle $x^2 + y^2 = 1$, $P$ becomes $(3/5,4/5)$ with tangent line $y = (-3/4)x + 5/4$, and $Q$ becomes $(1,0)$ with tangent line $x = 1$. Every point from $C$ lies in $U_2$, so no points are missing here. We see the "complete" curve in this chart. It is essential to draw three pictures here (of the $yz$-plane, $xz$-plane, and $xy$-plane) and mark $P$ and $Q$ in each (except you don't see $Q$ in the $xz$-plane). Now comes the beautiful comparison step: in all three charts the homogenization of the tangent line at $P$ is exactly the same equation: $3x + 4y = 5z$. The tangent line at $Q$ in $U_0$ and $U_2$ homogenizes in both cases back to $x = z$. This suggests there should be an intrinsic concept of tangent line in the projective plane to the curve $C$ at the points $P$ and $Q$, and you can compute the tangent line by looking at any chart containing the relevant point of interest, doing calculus there, and then homogenizing back. The homogenization of your ordinary linear equation to a homogenuous linear equation will always be the same, and its solutions in the projective plane define the tangent line to the projective curve at that point. As further evidence of the consistency of this new space and the geometry in it, look at the intersections of the two tangent lines at $P$ and $Q$: in $U_0$ -- the $yz$-plane -- the tangent lines meet in $(1/2,1)$ while in $U_2$ -- the $xy$-plane -- the tangent lines meet in $(1,1/2)$. These points both homogenize back to the same point $[2,1,2]$, which is the unique (!) point in the projective plane satisfying $3x + 4y = 5z$ and $x = z$. Remember that $Q$ went missing in the chart $U_1$? Well, its tangent line did not go missing: the projective line $x = z$ in the projective plane meets the chart $U_1$ in the ordinary line $x = z$ of the $xz$-plane, which is an asymptote to the piece of $C$ we can see in $U_1$. This is really amazing: asymptotes to (algebraic) curves in the usual plane are "really" the tangent lines to missing points on the complete version of that curve in the projective plane. To see this from another point of view, move around $C$ clockwise in the chart $U_2$ (where it's a circle) and figure out the corresponding motion along the piece of $C$ in the chart $U_0$ (where it's a hyperbola): as you pass through the point $Q = (1,0)$ in $U_2$, what happens in the chart $U_0$ is that you jump off one branch of the hyperbola onto the other branch by skipping through an asymptote, sort of. (There is a second point on $C$ in $U_2$ that you don't see in $U_0$ -- the point $R = [-1,0,1]$ is $(-1,0)$ in $U_2$ -- and paying attention to that point may help here.) The conic sections -- parabolas, hyperbolas, and ellipses -- which look quite different in ${\mathbf R}^2$, simplify in the projective plane because they all look like the same kind of curve (once you close them up): $y = x^2$ becomes $yz = x^2$, $xy = 1$ becomes $xy = z^2$, and $x^2 + y^2 = 1$ becomes $x^2 + y^2 = z^2$, which is the same as $x^2 = (z-y)(x+y) = z'y'$, where $z' = z-y$ and $y' = z+y$. I like to think about this as a fancy analogue of the Greek [Edit: Babylonian] use of many forms of the quadratic formula because they didn't have the right algebraic technique to realize there is one quadratic formula. Using the projective plane we see there is really one conic section. You might want to show by examples the nicer intersection properties of lines in the projective plane: any two lines in the projective plane meet in exactly one point. This is just a glimpse of the fact that curves in the projective plane have nicer intersection properties than in the ordinary plane, but to get the correct theorem in that direction for curves other than lines, you need to (a) work over the complex numbers and (b) introduce an appropriate concept of intersection multiplicity for intersection points of curves, somewhat like the idea of tangent circles intersecting in a point of multiplicity two which I mentioned earlier. The relevant theorem here is Bezout's theorem, but to state it correctly is complicated precisely because it is technical to give a good definition of what the intersection multiplicity is for two curves meeting at a common point. For the student who wants to be a number theorist, compare reduction mod $p$ in the usual plane and the projective plane. In the study of Diophantine equations (e.g., to show $y^2 = x^3 - 5$ has no integral solutions), it is very useful to reduce mod $p$, and there is a natural way to reduce a point in ${\mathbf Z}^2$ modulo $p$ However, there's no reasonable way to reduce all points in ${\mathbf Q}^2$ modulo $p$: when the rational numbers have denominator divisible by $p$, you can't make sense of them mod $p$: we can reduce $(-7/4,51/8)$ mod 5, for example, but not mod 2. In the projective plane, however, we can reduce rational points mod $p$ by the idea of choosing a set of primitive integral coordinates, where the homogeneous coordinates are relatively prime. For example, $[-7/4,51/8,1] = [-14,51,4]$ in ${\mathbf P}^2({\mathbf Q})$, and this can be reduced mod $p$ for any $p$ at all. For example, in ${\mathbf P}^2({\mathbf F}_2)$ it becomes $[0,1,0]$. (There is another primitive set of homogeneous coordinates for the point, namely $[14,-51,-4]$, but that reduces mod $p$ to the same thing as before, so this reduction mod $p$ process is well-defined.) This suggests that the projective plane has better mapping properties than the usual plane, in some sense.
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20,275 |
Today my fellow grad student asked me a question, given a map f from X to Y, assume $f_*(\pi_i(X))=0$ in Y, when is f null-homotopic? I search the literature a little bit, D.W.Kahn Link And M.Sternstein has worked on this, and Sternstein even got a necessary and sufficient condition, for suitable spaces. http://www.jstor.org/stable/pdfplus/2037939.pdf However, his condition is a little complicated for me as a beginner. Right now I just wanted a counter example of a such a map. Kahn in his paper said one can have many such examples using Eilenberg Maclance spaces. Well, we can certainly show a lot of map between E-M spaces induce zero map on homopoty groups just by pure group theoretic reasons, but I can not think of a easy example when you can show that map, if it exists, is not null-homotopic. Could someone give me some hint? or, maybe even some examples arising from manifolds?
|
Consider ordinary singular cohomology with varying coefficients. You can look at the short exact sequence of abelian groups: $$0 \to \mathbb{Z}/2 \to \mathbb{Z}/4 \to \mathbb{Z}/2 \to 0$$ This gives rise, for any space X, to a short exact sequence of chain complexes: $$0 \to C^i(X;\mathbb{Z}/2) \to C^i(X;\mathbb{Z}/4) \to C^i(X;\mathbb{Z}/2) \to 0$$ and hence you get a long exact sequence in cohomology. Thus we get an interesting boundary map known as the Bockstein $$H^i(X; \mathbb{Z}/2) \to H^{i+1}(X; \mathbb{Z}/2).$$ This is natural in X and so is represented by a (homotopy class of) map(s) of Eilenberg-Maclane spaces: $$K(i, \mathbb{Z}/2) \to K(i+1, \mathbb{Z}/2)$$ This map is necessarily zero on homotopy groups. To show that this map is not null-homotopy, you just need to find a space for which the Bockstein is non-trivial. There are lots of examples of this. Rather then explain one, I suggest you look up "Bockstein homomorphism" in a standard algebraic topology reference, e.g. Hatcher's book.
|
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|
20,314 |
Hi all.
I'm looking for english books with a good coverage of distribution theory.
I'm a fan of Folland's Real analysis, but it only gives elementary notions on distributions.
Thanks in advance.
|
Grubb's recent Distributions And Operators is supposed to be quite good. There's also the recommended reference work, Strichartz, R. (1994), A Guide to Distribution Theory and Fourier Transforms The comprehensive treatise on the subject-although quite old now-is Gel'fand, I.M.; Shilov, G.E. (1966–1968), Generalized functions, 1–5, . A very good,though quite advanced,source that's now available in Dover is Trèves, François (1967), Topological Vector Spaces, Distributions and Kernels That book is one of the classic texts on functional analysis and if you're an analyst or aspire to be,there's no reason not to have it now. But as I said,it's quite challenging. That should be enough to get you started.And of course,if you read French,you really should go back and read Schwartz's original treatise.
|
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|
20,386 |
I would like to know if practicing mathematics, constituting a hobby for some of you who are neither academics nor (advanced) mathematics, is an important part of your career. How do you go and learn a new mathematical field on your own? Do you just pick up a book and go over all proofs and do all exercises on your own? Is there any technique would you like to share?
|
I'm no longer in academia and while my job has some mathematical challenges, they aren't as interesting as the challenges I find in other branches of mathematics. That means I do mathematics on my own as a hobby. As I see it, the main challenge is this: how do you know you understand? You can try doing the exercises in textbooks. But how do you know you have the correct solution? And if the book has solutions, how can you trust the similarity metric you use to decide whether your solution is the same one? It seems to me that from time to time you need external calibration to make sure you're on the right track. In a sense you have to always test yourself, in the sense of falsifiability. If you work in academia, your colleagues will keep doing this for you. I know two ways to achieve this on your own: (1) Blog about what you have learnt. Because of the "someone is wrong on the Internet" syndrome , you're likely to get a response if you say something that is incorrect. You can write page after page of insightful material that will (apparently) be completely ignored, but if you make a mistake you'll get corrected (at least if you can get a following of some sort). As a side effect, communicating stuff to other people, even " rubber ducks ", can really deepen your own understanding. (2) Try to turn what you have learnt into computer programs that solve a problem. Computer programs don't provide proofs (well, that's not always true), but they will give you confidence and sanity checks. This doesn't just limit you to numerical analysis. Whether you're doing group theory, or algebraic geometry, or logic, or algebraic topology, or combinatorics, or even analysis, there are often subdomains of those disciplines for which you can write computer programs that will likely fail if you don't understand the theory.
|
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|
20,471 |
Why is it that every nontrivial word in a free group (it's easy to reduce to the case of, say, two generators) has a nontrivial image in some finite group? Equivalently, why is the natural map from a group to its profinite completion injective if the group is free? Apparently, this follows from a result of Malcev's that finitely generated matrix groups over an arbitrary commutative ring are residually finite, but is there a more easily accessible proof if we only want the result for free groups?
|
Here is a direct proof for free groups. Let $x_1,\dots,x_m$ be the generators of our group. Consider a word $x_{i_n}^{e_n}\dots x_{i_2}^{e_2}x_{i_1}^{e_1}$ where $e_i\in\{\pm 1\}$ and there are no cancellations (that is, $e_k=e_{k+1}$ if $i_k=i_{k+1}$). I'm going to map this word to a nontrivial element of $S_{n+1}$, the group of permutations of $M:=\{1,\dots,n+1\}$. It suffices to construct permutations $f_1,\dots,f_m\in S_{n+1}$ such that $f_{i_n}^{e_n}\dots f_{i_2}^{e_2}f_{i_1}^{e_1}\ne id_M$. For each $k=1,\dots,n$, assign $f_{i_k}(k)=k+1$ if $e_k=1$, or $f_{i_k}(k+1)=k$ if $e_k=-1$. This gives us injective maps $f_1,\dots,f_m$ defined on subsets of $M$. Assign yet unassigned values of $f_i$'s arbitrarily (the only requirement is that they are bijections). The resulting permutations satisfy $f_{i_n}^{e_n}\dots f_{i_2}^{e_2}f_{i_1}^{e_1}(1)=n+1$. Edit : As pointed out by Steve D in comments, this proof can be found in a book by Daniel E. Cohen, "Combinatorial group theory: a topological approach" (1989). The book can be found on the net if you are determined; the proof in on page 7 and in Proposition 5 on page 11. Edit [DZ]: I have a hard time reading multiple subscripts, so here is an example of Sergei Ivanov's construction. Take the word $cca^{-1}bc^{-1}a$. This has length $6$, so we will find a homomorphism to $S_7$ whose image of this word is nontrivial because it sends $1$ to $7$. We'll choose values of permutations so that the $k$th suffix sends $1$ to $k+1$: Suffix 1: $a$ $\alpha=\bigg(\begin{array}{} 1&2 &3 &4& 5& 6& 7 \\\ 2&?&?&?&?&?&? \end{array} \bigg)$ Suffix 2: $c^{-1}a$ $\gamma=\bigg(\begin{array}{} 1&2 &3 &4& 5& 6& 7 \\\ ?&?&2&?&?&?&? \end{array} \bigg)$ Suffix 3: $bc^{-1}a$ $\beta=\bigg(\begin{array}{} 1&2 &3 &4& 5& 6& 7 \\\ ?&?&4&?&?&?&? \end{array} \bigg)$ Suffix 4: $a^{-1}bc^{-1}a$ $\alpha=\bigg(\begin{array}{} 1&2 &3 &4& 5& 6& 7 \\\ 2&?&?&?&4&?&? \end{array} \bigg)$ Suffix 5: $ca^{-1}bc^{-1}a$ $\gamma=\bigg(\begin{array}{} 1&2 &3 &4& 5& 6& 7 \\\ ?&?&2&?&6&?&? \end{array} \bigg)$ Suffix 6: $cca^{-1}bc^{-1}a$ $\gamma=\bigg(\begin{array}{} 1&2 &3 &4& 5& 6& 7 \\\ ?&?&2&?&6&7&? \end{array} \bigg)$ These conditions on $\alpha, \beta,$ and $\gamma$ don't conflict, they can be extended to permutations, and then $\gamma\gamma\alpha^{-1}\beta\gamma^{-1}\alpha(1) = 7$.
|
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|
20,493 |
Hi,
given a connection on the tangent space of a manifold, one can define its torsion:
$$T(X,Y):=\triangledown_X Y - \triangledown_Y X - [X,Y]$$
What is the geometric picture behind this definition—what does torsion measure intuitively?
|
The torsion is a notoriously slippery concept. Personally I think the best way to understand it is to generalize past the place people first learn about torsion, which is usually in the context of Riemannian manifolds. Then you can see that the torsion can be understood as a sort of obstruction to integrability. Let me explain a little bit first. The torsion really makes sense in the context of general G-structures . Here $G \subseteq GL_n(\mathbb{R}) = GL(V)$ is some fixed Lie group. Typical examples are $G = O(n)$ and $G = GL_n(\mathbb{C})$. We'll see that these will correspond to Riemannian metrics and complex structures respectively. Now given this data, we have an exact sequence of vector spaces, $$0 \to K \to \mathfrak{g} \otimes V^\ast \stackrel{\sigma}{\to} V \otimes\wedge^2 V^\ast \to C \to 0 $$ Here $\sigma$ is the inclusion $\mathfrak{g} \subseteq V \otimes V^\ast$ together with anti-symmetrization. K and C are the kernel and cokernel of $\sigma$. If we are given a manifold with $G$-structure, we then get four associated bundles, which fit into an exact sequence: $$ 0 \to \rho_1P \to ad(P) \otimes T^*M \to \rho_3P \to \rho_4P \to 0$$ Now the difference of two connections which are both compatible with the G-structure is a tensor which is a section of the second space $\rho_2P = ad(P) \otimes T^*M$. This means that we can write any connection as $$\nabla + A$$
where $A$ is a section of $\rho_2(P)$. Now the torsion of any G-compatible connection is a section of this third space. Suppose that we have two compatible connections. Then their torsions are sections of this third space. However since we can write the connections as $\nabla$ and $\nabla + A$, the torsion differ by $\sigma(A)$. Thus they have the same image in the fourth space $\rho_4(P)$. The section of this fourth space is the intrinsic torsion of the G-structure. It measures the failure of our ability to find a torsion free connection. If this obstruction vanishes, then the torsion free connections form a torsor over sections of the smaller bundle $\rho_1P$. Now some examples: $G = O(n)$. This is the case of a Riemannian structure. In this case $\sigma$ is an isomorphism so that the there is always a unique torsion free connection. The Levi-Civita connection. $G = GL_m(\mathbb{C})$. This is the case of a complex structure. More precisely a $GL_m(\mathbb{C})$-structure is the same as an almost complex structure. In this case the intrinsic torsion can be identified with the Nijenhuis tensor. So it vanishes precisely when the almost-complex structure is integrable (i.e. a ordinary complex structure). $G = Sp(n)$. Having an $Sp(n)$-structure on a manifold for which the intrinsic torsion vanishes is equivalent to having a symplectic manifold. From these examples you can see that the vanishing of torsion can be viewed as a sort of integrability condition. In these latter two cases the space of torsion free connections consists of more then a single point. There are many such connections. That's one reason why we don't see them popping up more often.
|
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|
20,497 |
I have just learned here that we know numbers that are not periods; is it known meanwhile that the ring of periods is not a field? I know that it is conjectured that $1/\pi$ is not a period, but the existence of a period whose inverse is not a period seems to be still open. Is this correct? More generally: is it believed that the unit group of the ring of periods is bigger than the nonzero algebraic numbers?
|
I think the questions were about unconditional proofs or counter examples. I don't have an answer to any of those questions but I think it is still interesting to understand how the yoga of motives suggests natural answers to theses questions. Even though this may seem trivial to people familiar with the subject. Let's work in the setting of Voevodsky's $\otimes$-triangulated categories $DM^{eff}(\mathbb{Q}):= DM_{gm}^{eff}(Spec(\mathbb{Q});\mathbb{Q}) \subset DM_{gm}(Spec(\mathbb{Q});\mathbb{Q}) =: DM(\mathbb{Q})$. Remember that the latter is obtained by formally inverting $\mathbb{Q}(1)$ and that it is a rigid $\otimes$-triangulated category. Question 1 : Is the ring of (effective) periods a field? Following Beilinson's "Remarks on Grothendieck's standard conjecctures", let's assume Motivic conjecture: There exists a non degenerate t-structure on Voevodsky's category $DM(\mathbb{Q}) := DM(Spec(\mathbb{Q});\mathbb{Q})$ and such that the Betti realization function $\omega_B: DM(\mathbb{Q}) \to D^bMod_f(\mathbb{Q})$ is a $t$-exact $\otimes$-functor. This is an extremely strong conjecture as it implies the standard conjectures in characteristic 0. Under this conjecture, the heart of the motvitic $t$-structure is a tannakian category $MM(\mathbb{Q})$. We have Betti and Rham realization functors $\omega_B,\omega_{dR}: MM(\mathbb{Q}) \rightrightarrows Mod_f(\mathbb{Q})$. And we can define
$$
Per := Isom^\otimes(\omega_{dR},\omega_{B})
$$
This is a fpqc-torsor under the motivic Galois group $G_B := Aut^\otimes(\omega_B)$. Define the algebra of motivic periods as the ring of regular functions on the Betti/de Rham torsor:
$$
P_{mot} := \mathcal{O}(Per)
$$
Integration of differential forms (or more generally the Riemann-Hilbert correspondance) defines an $\mathbb{C}$-point
$$
Spec(\mathbb{C}) \longrightarrow Per
$$
The image of the corresponding morphism $P_{mot} \to \mathbb{C}$ is the ring of periods $P$. Note: I think this whole part is actually known unconditionnally in the setting of Nori's motives (see arXiv:1105.0865v4). Period conjecture: The morphism $P_{mot} \to P$ is an isomorphism. Now based on these tiny little conjectures we can say Prop: $P_{mot}$ is not a field so $P$ isn't either. Proof: Indeed in his comment G-torsor whose ring of regular functions is a field. @quasi-coherent explained how faithfull flatness would imply that if $P_{mot}$ were a field then it would be algebraic over $\mathbb{Q}$ which contradicts the fact that $2\pi i$ belongs to the image of $P_{mot}\to \mathbb{C}$. Question 2 Is it true that $(P_{mot}^{eff})^\times = \overline{\mathbb{Q}}^\times$? This post is getting too long already so I'll try and write down the rest later but the basic idea is that invertible effective motives are Artin motives. This can be proved in terms of weights or niveau (level).
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|
20,551 |
Does anyone know of a good place to find already-done BibTeX entries for standard books in advanced math? Or is this impossible because the citation should include items specific to your copy? (I am seeing the latter as potentially problematic because the only date I can find in my copy of Hartshorne is 2006, whereas the citations I can find all put the publication date at 1977.)
|
I recommend using the AMS website MREF, located here . EDIT : Another remark about your question. Don't worry too much about getting things like the printing date for a book correct (it changes every time they make a new printing run). Just make sure that you have the author, title, and edition in some standard format (and for papers, the journal, date, and page number). Every journal will reformat things into their house style and verify that your bibliographic entries are correct. Make those copy-editors work for their money!
|
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20,590 |
Here's the setup: you have a first-order theory T, in a countable language L for simplicity. Let k be a cardinal and suppose T is k-categorical. This means that, for any two models M,N |= T of cardinality k, there is an isomorphism f : M --> N. Supposing all this happens inside of ZFC, let's say I change the underlying model of ZFC, e.g by restricting to the constructible sets, or by forcing new sets in. I would like to understand what happens to the k-categoricity of T. I'll assume the set theory doesn't change so drastically that we lose L or T. Then, a priori, a bunch of things may happen: (i) We may lose all isomorphisms between a pair of models M,N of cardinality k;
(ii) Some models that used to be of cardinality k may no longer have bijections with k;
(iii) k may become a different cardinal, meaning new cardinals may appear below it, or others may disappear by the introduction of new bijections;
(iv) some models M, or k itself, may disappear as sets, leading to a new set being seen as "the new k". Overall, nearly every aspect of the phrase "T is k-categorical" may be affected. How likely is it to still be true? Do some among (i)-(iv) not matter, or is there some cancellation of effects? (Say, maybe all isomorphisms M-->N disappear, but so do all bijections between N and k?)
|
Categoricity is absolute. By the Ryll-Nardzewski theorem, for a countable language, $\aleph_0$-categoricity of a complete theory $T$ is equivalent to $T$ proving for each natural number $n$ that there are only finitely many inequivalent formulas in $n$ variables. This property is evidently arithmetic and, thus, absolute. Likewise, again in a countable language, it follows from the Baldwin-Lachlan theorem that a theory is categorical in some (hence, by Morley's theorem, all) uncountable cardinality just in case every model is prime and minimal over a strongly minimal set. Moreover, the strongly minimal formula may be taken to be defined over the prime model and the primality and minimality of every model over this strongly minimal formula is something which will be witnessed by an explicit analysis, hence, something arithmetic and absolute. For uncountable languages, the situation is a little more complicated, but again categoricity is equivalent to an absolute property. Shelah shows that either the theory is totally transcendental and Morley's analysis in the case of countable languages applies, or the theory is strictly superstable though unidimensional.
|
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|
20,671 |
This question makes sense for any topological group $G$ , but I'd particularly like to know the answer for $G$ a compact, connected Lie group. $G$ acts on itself by conjugation. One has the equivariant singular cohomology $H^{\ast}_G(G) = H^{\ast}( (G\times EG)/G )$ , with integer coefficients, say. What is $H^{\ast}_G(G)$ ? Concretely, $H^{\ast}_G(G)$ is the target of the Serre spectral sequence for the fibration $$G \hookrightarrow (G\times EG)/G \to BG.$$ When $G$ is path-connected, so that $BG$ is simply connected, this spectral sequence has $E_2^{p,q}=H^p(BG; H^q(G))$ (trivial local system). Does the spectral sequence always degenerate at $E_2$ ? It does when $G$ is abelian, because then the conjugation action is trivial.
|
I asked Dan Freed, who gave a very clean general solution to this problem (as expected).
Here it is (all mistakes in the transcription are mine of course). The claim is that the equivariant cohomology of G acting on G is indeed the tensor product of cohomology of BG with cohomology of G - in other words the Leray spectral sequence for the fibration $G=\Omega BG \to G/G=LBG \to BG$ degenerates at E_2. To see this we will use the Leray-Hirsch theorem -- i.e. if we can show that every class on the fiber (G) extends to a class on the total space (LBG) then we will be done. Now the cohomology of G is generated (as an exterior algebra) by its primitive classes, and these all come from the generators of the cohomology of BG by transgression. So we just need to show that these transgressed classes actually lift to LBG. But there is a nice direct construction of these classes on LBG. Namely we use the tautological correspondence $$LBG \leftarrow S^1\times LBG \rightarrow BG$$
where the right arrow is the evaluation map. Thus given a class on BG we can lift it to $S^1 \times LBG$ and then integrate along the circle to get a class on $LBG$. When we restrict these classes to a fiber, i.e. $G=\Omega BG$, we recover the usual transgression construction. (This can be seen very explicitly with differential forms.. the transgression involves the path fibration $$\Omega BG \to P(BG) \to BG$$
and is given by the same kind of tautological/evaluation construction for the map from the interval times the path space to BG, integrating over an interval.. when restricted to $\Omega BG$, ie closed paths, this integration becomes the integration over the circle we had above.) So we've explicitly lifted all the generators of the cohomology of G to equivariant classes, ie to G/G, hence we're done. (Presumably this can also be seen concretely in the Cartan model, that the generators of H^*G lift to conjugation-equivariant classes..)
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20,675 |
Katzarkov-Kontsevich-Pantev define a smooth dg ($\mathbb{C}$-)algebra $A$ to be a dg algebra which is a perfect $A \otimes A^{op}$-module. They say that an $A$-module $M$ is perfect if the functor $Hom(M,-)$ from ($A$-modules) to ($\mathbb{C}$-modules) preserves small homotopy colimits. (Definition 2.23 of KKP ) * Kontsevich-Soibelman define smooth dg (or $A_\infty$) algebras in the same way; but they define perfect $A$-modules to be ones which are quasi-isomorphic to a direct summand of an extension of a sequence of modules each of which is quasi-isomorphic to a shift of $A$. (Definition 8.1.1 of KS ) ** First question: I have trouble wrapping my head around what it means for a functor to preserve small homotopy colimits. I don't understand the definition from Kontsevich-Soibelman, either. What are the "moral" meanings of these conditions? Second question: Presumably these two conditions are equivalent. How do you prove this? Third question: If $A$ is a commutative $(\mathbb{C}$-)algebra, then presumably smoothness of $A$ as a dg algebra is equivalent to smoothness of $\operatorname{Spec} A$ as a ($\mathbb{C}$-)scheme. (Maybe you need $A$ to be finite type?) How do you prove this? Furthermore, Example 8.1.4 of KS gives some more examples of dg algebras that are allegedly smooth (for instance free algebras $k\langle x_1, \dots, x_n \rangle$). Again, I don't know how to prove that these are in fact smooth, and Kontsevich-Soibelman don't seem to provide proofs; or maybe I overlooked something. *The answers below indicate that this is the standard definition of compact (also known as small ) module. **The answers below indicate that this is the standard definition of perfect module.
|
The condition of Hom(M,-) being a continuous functor, i.e. preserving (small homotopy) colimits is equivalent (in the present stable setting) to the maybe more concrete condition of preserving arbitrary direct sums. The issue is not finite direct sums, that's automatic (since the derived Hom is an exact functor, it automatically commutes with FINITE colimits).
This is better known as a compact object of the dg category (see n-lab for lots of discussion).
This is a strong finiteness condition on a module. Its importance in algebraic geometry was realized by Thomason (in a dream form of his friend Trobaugh), who proved some amazing fundamental results about the behavior (and abundance) of compact/perfect objects on schemes. One can think of Hom(M,-) as the functional on the category defined by M (Hom being a kind of inner product), and this is saying the functional is "continuous"..
There's a discussion of the different common notions of finiteness for modules and their properties (including the foundational results of Thomason and Neeman) in Section 3.1 of this paper (sorry for the self-referencing -- none of this is in the least original, but it's a convenient discussion with plentiful references). This includes an explanation of why compactness is the same as being in the thick subcategory category generated by the free module, which is your definition of perfect from K-S (ie built out of the free module by taking sums, cones, summands), and also the same as being a dualizable object with respect to tensor product in case your category is the derived category of quasicoherent sheaves (ie this is a "commutative" notion not applicable in the NC setting you're discussing). By the way the definition from K-S is the standard definition of perfection, that from K-K-P is the standard definition of compactness.. there are settings where the two notions don't agree (for example for sheaves on the classifying space of a finite group in a modular characteristic, or on BS^1) [hence our terminology of "perfect stack" in the paper with Francis and Nadler, which is a stack where these notions for sheaves agree and these nice finite objects generate]. The fact that perfection of the diagonal (ie A as an A-bimodule) is equivalent to smoothness in the case of a scheme is a reformulation of the homological criterion of Serre for smoothness of a point in terms of finite Tor amplitude of the skyscraper at the point - ie we're saying all points are smooth all at once.
|
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20,696 |
In his Huygens and Barrow, Newton and Hooke , Arnold mentions a notorious teaser that, in his opinion, "modern" mathematicians are not capable of solving quickly. Then, he adds that the exception that proved the rule in this case of his was the German mathematician Gerd Faltings . My question is whether any of you knows the complete story behind those lines in Arnold's book. I mean, did Arnold pose the problem somewhere (maybe Квант?) and Faltings was the only one that submitted a solution after Arnold's own heart? Is the previous conjecture totally unrelated to the actual development of things? I thank you in advance for your insightful replies. P.S. It seems that this teaser of Arnold eventually became a cult thingy in certain branches of the Russian mathematical community. Below you can find a photograph taken by a fellow of mine of one of the walls of IUM's cafeteria (where IUM stands for Independent University of Moscow ). As the Hindu mathematician Bhāskara would say (or so the legend has it): BEHOLD!
|
Here is a problem which I heard Arnold give in an ODE lecture when I was an undergrad. Arnold indeed talked about Barrow, Newton and Hooke that day, and about how modern mathematicians can not calculate quickly but for Barrow this would be a one-minute exercise. He then dared anybody in the audience to do it in 10 minutes and offered immediate monetary reward, which was not collected. I admit that it took me more than 10 minutes to do this by computing Taylor series. This is consistent with what Angelo is describing. But for all I know, this could have been a lucky guess on Faltings' part, even though he is well known to be very quick and razor sharp. The problem was to find the limit $$ \lim_{x\to 0} \frac
{ \sin(\tan x) - \tan(\sin x) }
{ \arcsin(\arctan x) - \arctan(\arcsin x) }
$$ The answer is the same for
$$ \lim_{x\to 0} \frac
{ f(x) - g(x) }
{ g^{-1}(x) - f^{-1}(x) }
$$
for any two analytic around 0 functions $f,g$ with $f(0)=g(0)=0$ and $f'(0)=g'(0)=1$, which you can easily prove by looking at the power expansions of $f$ and $f^{-1}$ or, in the case of Barrow, by looking at the graph. End of Apr 8 2010 edit Beg of Apr 9 2012 edit Here is a computation for the inverse functions. Suppose
$$
f(x) = x + a_2 x^2 + a_3 x^3 + \dots
\quad \text{and} \quad
f^{-1}(x) = x + A_2 x^2 + A_3 x^3 + \dots
$$ Computing recursively, one sees that for $n\ge2$ one has
$$ A_n = -a_n + P_n(a_2, \dotsc, a_{n-1} ) $$
for some universal polynomial $P_n$. Now, let
$$
g(x) = x + b_2 x^2 + b_3 x^3 + \dots
\quad \text{and} \quad
g^{-1}(x) = x + B_2 x^2 + B_3 x^3 + \dots
$$ and suppose that $b_i=a_i$ for $i\le n-1$ but $b_n\ne a_n$. Then by induction one has $B_i=A_i$ for $i\le n-1$, $A_n=-a_n+ P_n(a_2,\dotsc,a_{n-1})$ and $B_n=-b_n+ P_n(a_2,\dotsc,a_{n-1})$. Thus, the power expansion for $f(x)-g(x)$ starts with $(a_n-b_n)x^n$, and the power expansion for $g^{-1}(x)-f^{-1}(x)$ starts with $(B_n-A_n)x^n = (a_n-b_n)x^n$. So the limit is 1.
|
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20,740 |
The title really is the question, but allow me to explain. I am a pure mathematician working outside of probability theory, but the concepts and techniques of probability theory (in the sense of Kolmogorov, i.e., probability measures) are appealing and potentially useful to me. It seems to me that, perhaps more than most other areas of mathematics, there are many, many nice introductory (as well as not so introductory) texts on this subject. However, I haven't found any that are written from what it is arguably the dominant school of thought of contemporary mainstream mathematics, i.e., from a structuralist (think Bourbaki) sensibility. E.g., when I started writing notes on the texts I was reading, I soon found that I was asking questions and setting things up in a somewhat different way. Here are some basic questions I couldn't stop from asking myself: [0) Define a Borel space to be a set $X$ equipped with a $\sigma$ -algebra of subsets of $X$ . This is already not universally done (explicitly) in standard texts, but from a structuralist approach one should gain some understanding of such spaces before one considers the richer structure of a probability space.] What is the category of Borel spaces, i.e., what are the morphisms? Does it have products, coproducts, initial/final objects, etc? As a significant example here I found the notion of the product Borel space -- which is exactly what you think if you know about the product topology -- but seemed underemphasized in the standard treatments. What is the category of probability spaces, or is this not a fruitful concept (and why?)? For instance, a subspace of a probability space is, apparently, not a probability space: is that a problem? Is the right notion of morphism of probability spaces a measure-preserving function? What are the functorial properties of probability measures? E.g., what are basic results on pushing them forward, pulling them back, passing to products and quotients, etc. Here again I will mention that product of an arbitrary family of probability spaces -- which is a very useful-looking concept! -- seems not to be treated in most texts. Not that it's hard to do: see e.g. http://alpha.math.uga.edu/~pete/saeki.pdf I am not a category theorist, and my taste for how much categorical language to use is probably towards the middle of the spectrum: that is, I like to use a very small categorical vocabulary (morphisms, functors, products, coproducts, etc.) as often as seems relevant (which is very often!). It would be a somewhat different question to develop a truly categorical take on probability theory. There is definitely some nice mathematics here, e.g. I recall an arxiv article (unfortunately I cannot put my hands on it at this moment) which discussed independence of events in terms of tensor categories in a very persuasive way. So answers which are more explicitly categorical are also welcome, although I wish to be clear that I'm not asking for a categorification of probability theory per se (at least, not so far as I am aware!).
|
$\def\Spec{\mathop{\rm Spec}}
\def\R{{\bf R}}
\def\Ep{{\rm E}^+}
\def\L{{\rm L}}
\def\EpL{\Ep\L}$ One can argue that an object of the right category of spaces in measure theory is not a set equipped with a σ-algebra of measurable sets,
but rather a set $S$ equipped with a σ-algebra $M$ of measurable sets and a σ-ideal $N$ of negligible sets , i.e., sets of measure 0.
The reason for this is that you can hardly state any theorem of measure theory or probability theory without referring to sets of measure 0.
However, objects of this category contain less data than the usual measured spaces, because they are not equipped with a measure.
Therefore I prefer to call them enhanced measurable spaces , since they are measurable spaces enhanced with a σ-ideal of negligible sets.
A morphism of enhanced measurable spaces $(S,M,N)→(T,P,Q)$ is a map $S\to T$ such that
the preimage of every element of $P$ is a union of an element of $M$ and a subset of an element of $N$ and the preimage of every element of $Q$ is a subset of an element of $N$ . Irving Segal proved in “ Equivalences of measure spaces ” (see also Kelley's “ Decomposition and representation theorems in measure theory ”)
that for an enhanced measurable space $(S,M,N)$ that admits a faithful measure (meaning $μ(A)=0$ if and only if $A∈N$ ) the following properties are equivalent. The Boolean algebra $M/N$ of equivalence classes of measurable sets is complete; The space of equivalence classes of all bounded (or unbounded) real-valued functions on $S$ modulo equality almost everywhere is Dedekind-complete; The Radon-Nikodym theorem is true for $(S,M,N)$ ; The Riesz representation theorem is true for $(S,M,N)$ (the dual of $\L^1$ is isomorphic to $\L^∞$ ); Equivalence classes of bounded functions on $S$ form a von Neumann algebra (alias W*-algebra). An enhanced measurable space that satisfies these conditions (including the existence of a faithful measure) is called localizable .
This theorem tells us that if we want to prove anything nontrivial about measurable spaces, we better restrict ourselves to localizable enhanced measurable spaces.
We also have a nice illustration of the claim I made in the first paragraph:
none of these statements would be true without identifying objects that differ on a set of measure 0.
For example, take a nonmeasurable set $G$ and a family of singleton subsets of $G$ indexed by themselves.
This family of measurable sets does not have a supremum in the Boolean algebra of measurable sets, thus disproving a naive version of (1). But restricting to localizable enhanced measurable spaces does not eliminate all the pathologies:
one must further restrict to the so-called compact and strictly localizable enhanced measurable spaces,
and use a coarser equivalence relation on measurable maps: $f$ and $g$ are weakly equal almost everywhere if for any measurable subset $B$ of the codomain the symmetric difference $f^*B⊕g^*B$ of preimages of $B$ under $f$ and $g$ is a negligible subset of the domain.
(For codomains like real numbers this equivalence relation coincides with equality almost everywhere.) An enhanced measurable space is strictly localizable if it splits as a coproduct (disjoint union) of σ-finite (meaning there is a faithful finite measure)
enhanced measurable spaces.
An enhanced measurable space $(X,M,N)$ is (Marczewski) compact if there is a compact class $K⊂M$ such that for any $m∈M∖N$ there is $k∈K∖N$ such that $k⊂m$ .
Here a compact class is a collection $K⊂2^X$ of subsets of $X$ such that for any $K'⊂K$ the following finite intersection property holds:
if for any finite $K''⊂K'$ we have $⋂K''≠∅$ , then also $⋂K'≠∅$ . The best argument for such restrictions is the following Gelfand-type duality theorem for commutative von Neumann algebras . Theorem. The following 5 categories are equivalent. The category of compact strictly localizable enhanced measurable spaces with measurable maps modulo weak equality almost everywhere. The category of hyperstonean topological spaces and open continuous maps. The category of hyperstonean locales and open maps. The category of measurable locales (and arbitrary maps of locales). The opposite category of commutative von Neumann algebras and normal (alias ultraweakly continuous) unital *-homomorphisms. I actually prefer to work with the opposite category of the category of commutative von Neumann algebras,
or with the category of measurable locales.
The reason for this is that the point-set definition of a measurable space
exhibits immediate connections only (perhaps) to descriptive set theory, and with additional effort to Boolean algebras,
whereas the description in terms of operator algebras or locales immediately connects measure theory to other areas of the central core of mathematics
(noncommutative geometry, algebraic geometry, complex geometry, differential geometry, topos theory, etc.). Additionally, note how the fourth category (measurable locales) is a full subcategory of the category of locales.
Roughly, the latter can be seen as a slight enlargement of the usual category of topological spaces,
for which all the usual theorems of general topology continue to hold (e.g., Tychonoff, Urysohn, Tietze, various results about paracompact and uniform spaces, etc.).
In particular, there is a fully faithful functor from sober topological spaces (which includes all Hausdorff spaces) to locales.
This functor is not surjective, i.e., there are nonspatial locales that do not come from topological spaces.
As it turns out, all measurable locales (excluding discrete ones) are nonspatial.
Thus, measure theory is part of (pointfree) general topology, in the strictest sense possible. The non-point-set languages (2–5) are also easier to use in practice.
Let me illustrate this statement with just one example: when we try to define measurable bundles of Hilbert spaces
on a compact strictly localizable enhanced measurable space in a point-set way, we run into all sorts of problems
if the fibers can be nonseparable, and I do not know how to fix this problem in the point-set framework.
On the other hand, in the algebraic framework we can simply say that a bundle of Hilbert spaces is a Hilbert W*-module over the corresponding von Neumann algebra. Categorical properties of von Neumann algebras (hence of compact strictly localizable enhanced measurable spaces)
were investigated by Guichardet in “ Sur la catégorie des algèbres de von Neumann ”.
Let me mention some of his results, translated in the language of enhanced measurable spaces.
The category of compact strictly localizable enhanced measurable spaces admits equalizers and coequalizers, arbitrary coproducts, hence also arbitrary colimits.
It also admits products (and hence arbitrary limits), although they are quite different from what one might think.
For example, the product of two real lines is not $\R^2$ with the two obvious projections.
The product contains $\R^2$ , but it also has a lot of other stuff, for example, the diagonal of $\R^2$ ,
which is needed to satisfy the universal property for the two identity maps on $\R$ .
The more intuitive product of measurable spaces ( $\R\times\R=\R^2$ ) corresponds to the spatial
tensor product of von Neumann algebras and forms a part of a symmetric monoidal structure on the category of measurable spaces.
See Guichardet's paper for other categorical properties (monoidal structures on measurable spaces, flatness, existence of filtered limits, etc.). Another property worthy of mentioning is that the category of commutative von Neumann algebras
is a locally presentable category, which immediately allows one to use the adjoint functor theorem to construct commutative
von Neumann algebras (hence enhanced measurable spaces) via their representable functors. Finally let me mention pushforward and pullback properties of measures on enhanced measurable spaces.
I will talk about more general case of $\L^p$ -spaces instead of just measures (i.e., $\L^1$ -spaces).
For the sake of convenience, denote $\L_p(M)=\L^{1/p}(M)$ , where $M$ is an enhanced measurable space.
Here $p$ can be an arbitrary complex number with a nonnegative real part.
We do not need a measure on $M$ to define $\L_p(M)$ .
For instance, $\L_0$ is the space of all bounded functions (i.e., the commutative von Neumann algebra corresponding to $M$ ), $\L_1$ is the space of finite complex-valued measures (the dual of $\L_0$ in the ultraweak topology),
and $\L_{1/2}$ is the Hilbert space of half-densities.
I will also talk about extended positive part $\EpL_p$ of $\L_p$ for real $p$ .
In particular, $\EpL_1$ is the space of all (not necessarily finite) positive measures on $M$ . Pushforward for $\L_p$ -spaces. Suppose we have a morphism of enhanced measurable spaces $M\to N$ .
If $p=1$ , then we have a canonical map $\L_1(M)\to\L_1(N)$ , which just the dual of $\L_0(N)→\L_0(M)$ in the ultraweak topology.
Geometrically, this is the fiberwise integration map.
If $p≠1$ , then we only have a pushforward map of the extended positive parts, namely, $\EpL_p(M)→\EpL_p(N)$ , which is nonadditive unless $p=1$ .
Geometrically, this is the fiberwise $\L_p$ -norm.
Thus $\L_1$ is a functor from the category of enhanced measurable spaces to the category of Banach spaces
and $\EpL_p$ is a functor to the category of “positive homogeneous $p$ -cones”.
The pushforward map preserves the trace on $\L_1$ and hence sends a probability measure to a probability measure. To define pullbacks of $\L_p$ -spaces (in particular, $\L_1$ -spaces) one needs to pass to a different category of enhanced measurable spaces.
In the algebraic language, if we have two commutative von Neumann algebras $A$ and $B$ ,
then a morphism from $A$ to $B$ is a usual morphism of commutative von Neumann algebras $f\colon A\to B$ together with an operator valued weight $T\colon\Ep(B)\to\Ep(A)$ associated to $f$ .
Here $\Ep(A)$ denotes the extended positive part of $A$ .
(Think of positive functions on $\Spec A$ that can take infinite values.)
Geometrically, this is a morphism $\Spec f\colon\Spec B\to\Spec A$ between the corresponding enhanced measurable spaces and a choice of measure on each fiber of $\Spec f$ .
Now we have a canonical additive map $\EpL_p(\Spec A)\to\EpL_p(\Spec B)$ ,
which makes $\EpL_p$ into a contravariant functor from the category of enhanced measurable spaces
and measurable maps equipped with a fiberwise measure to the category of “positive homogeneous additive cones”. If we want to have a pullback of $\L_p$ -spaces themselves and not just their extended positive parts,
we need to replace operator valued weights in the above definition
by finite complex-valued operator valued weights $T\colon B\to A$ (think of a fiberwise finite complex-valued measure).
Then $\L_p$ becomes a functor from the category of enhanced measurable spaces to the category of Banach spaces (if the real part of $p$ is at most $1$ )
or quasi-Banach spaces (if the real part of $p$ is greater than $1$ ).
Here $p$ is an arbitrary complex number with a nonnegative real part.
Notice that for $p=0$ we get the original map $f\colon A\to B$ and in this (and only this) case we do not need $T$ . Finally, if we restrict ourselves to an even smaller subcategory defined by the additional condition $T(1)=1$ (i.e., $T$ is a conditional expectation; think of a fiberwise probability measure),
then the pullback map preserves the trace on $\L_1$ and in this case the pullback of a probability measure is a probability measure. There is also a smooth analog of the theory described above.
The category of enhanced measurable spaces and their morphisms is replaced by the category of smooth manifolds and submersions, $\L_p$ -spaces are replaced by bundles of $p$ -densities,
operator valued weights are replaced by sections of the bundle of relative 1-densities,
the integration map on 1-densities is defined via Poincaré duality (to avoid any dependence on measure theory) etc.
There is a forgetful functor that sends a smooth manifold to its underlying enhanced measurable space. Of course, the story does not end here, there are many
other interesting topics to consider: products of measurable spaces,
the difference between Borel and Lebesgue measurability, conditional expectations, etc.
An index of my writings on this topic is available.
|
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|
20,782 |
Background Recall that, given two commutative rings $A$ and $B$ , the set of morphisms of rings $A\to B$ is in bijection with the set of morphisms of schemes $\mathrm{Spec}(B)\to\mathrm{Spec}(A)$ . Furthermore, we know that Spec $(A)$ has a base of open affine subsets, the "basic" or "principal" open affines $D(f)$ for all $f\in A$ . Furthermore, $D(f)\cong \mathrm{Spec}(A_f)$ as schemes, and the inclusion $D(f)\hookrightarrow\mathrm{Spec}(A)$ corresponds to the localization map $A\to A_f$ . But answers to a recent MathOverflow question show that open affine subschemes of affine schemes can arise in other ways. Question In order to try to make sense of the situation above, I'd like to know the following. Given a commutative ring $A$ , is there a "ring-theoretic" characterization of the ring homomorphisms $A\to B$ that realize $\mathrm{Spec}(B)$ as an open affine subscheme of $\mathrm{Spec}(A)$ (more precisely, those morphisms such that the induced map $\mathrm{Spec}(B)\to\mathrm{Spec}(A)$ is an open immersion)? Of course, "ring-theoretic" is a bit vague. Let's certainly avoid any tautological characterizations. I would prefer if an answer didn't make any reference to the Zariski topology (for instance, the morphisms $A\to A_f$ make perfect sense without the Zariski topology), but I'm not sure whether that's reasonable. Update: I received two great answers, thank you both! I chose the one that was closer to the kind of condition that I had in mind. But Dan Petersen's answer was also very interesting and unexpected.
|
There is the following characterisation. I don't think it's too tautological. Let $T \subseteq A$ be the set of f such that the induced map $A[f^{-1}] \to B[f^{-1}]$ is an isomorphism. Then $\mathrm{Spec}(B) \to \mathrm{Spec}(A)$ is an open immersion if and only if the image of $T$ in $B$ generates the unit ideal.
|
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20,877 |
In intuitive terms, what is the main difference? We know that homology is essentially the number of $n$-cycles that are not $n$-boundaries in some simplicial complex $X$. This is, more or less, the number of holes in the complex. But what is the geometrical interpretation of cohomology?
|
As to what cohomology actually measures, I think a general theme is "the failure of locally trivial things to be globally trivial", or perhaps "the failure of local solutions to glue together to form a global solution". In the de Rham cohomology of a smooth manifold, any closed form ω is "locally trivial" in that you can cover the manifold by contractible charts, over each of which a solution to dα=ω exists by the Poincaré lemma. The cohomology class [ω] measures the failure of existence of a global solution of this equation. Similar remarks can be made about simplicial, singular, and (especially) Čech cohomology.
|
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|
20,882 |
It is well-known that the axiom of choice is equivalent to many other assumptions, such as the well-ordering principle , Tychonoff's theorem , and the fact that every vector space has a basis. Even though all these formulations are equivalent, I have heard many people say that they 'believe' the axiom of choice, but they don't 'believe' the well-ordering principle. So, my question is what do you consider to be the most unintuitive application of choice? Here is the sort of answer that I have in mind. An infinite number of people are about to play the following game. In a moment, they will go into a room and each put on a different hat. On each hat there will be a real number. Each player will be able to see the real numbers on all the hats (except their own). After all the hats are placed on, the players have to simultaneously shout out what real number they think is on their own hat. The players win if only a finite number of them guess incorrectly. Otherwise, they are all executed. They are not allowed to communicate once they enter the room, but beforehand they are allowed to talk and come up with a strategy (with infinite resources). The very unintuitive fact is that the players have a strategy whereby they can always win. Indeed, it is hard to come up with a strategy where at least one player is guaranteed to answer correctly, let alone a co-finite set. Hint : the solution uses the axiom of choice.
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I have enjoyed the other answers very much. But perhaps it
would be desirable to balance the discussion somewhat with
a counterpoint, by mentioning a few of the
counter-intuitive situations that can occur when the axiom
of choice fails. For although mathematicians often point to
what are perceived as strange consequences of AC, many of
the situations that can arise when one drops the axiom are
also quite bizarre. There can be a nonempty tree $T$, with no leaves, but which has no infinite
path. That is, every finite path in the tree can be extended one more step, but there is no
path that goes forever. A real number can be in the closure of a set $X\subset\mathbb{R}$, but
not the limit of any sequence from $X$. A function $f:\mathbb{R}\to\mathbb{R}$ can be continuous
in the sense that $x_n\to x\Rightarrow f(x_n)\to f(x)$,
but not in the $\epsilon\ \delta$ sense. A set can be infinite, but have no countably infinite subset. Thus, it can be incorrect to say that $\aleph_0$ is the smallest infinite
cardinality, since there can be infinite sets of
incomparable size with $\aleph_0$. (see this MO
answer .) There can be an equivalence relation on $\mathbb{R}$, such that the number of equivalence classes is strictly greater than the size of $\mathbb{R}$. (See François's excellent answer .) This is a consequence of AD, and thus relatively consistent with DC and countable AC. There can be a field with no algebraic closure. The rational field $\mathbb{Q}$ can have different nonisomorphic algebraic closures (due to Läuchli, see Timothy Chow's comment below). Indeed, $\mathbb{Q}$ can have an uncountable algebraic closure, as well as a countable one. There can be a vector space with no basis. There can be a vector space with bases of different
cardinalities. The reals can be a countable union of countable
sets. Consequently, the theory of Lebesgue measure can fail totally. The first uncountable ordinal $\omega_1$ can be
singular. More generally, it can be that every uncountable $\aleph_\alpha$ is
singular. Hence, there are no infinite regular uncountable
well-ordered cardinals. See the Wikipedia
page for additional examples.
|
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20,946 |
Given a polynomial equation $x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0=0$, where $n$ is even and all the coefficients $a_i$ are real, what is the best way to determine whether it has a real root or not? I know Sturm's theorem , but I am wondering if it's possible to determine this by the sign of some form of resolvent ( e.g. , discriminant works for $n=2$, but not 4 or beyond)? Please point me to some reference if this has already been studied. Thanks for your time! I apologize for not searching hard enough - I just realized there is a similar question discussed here before, and has already got lots of useful comments. My hope is to find a function of these real coefficients and whether the polynomial has real root or not is determined by its sign.
|
There is indeed an easy way to check if a univariate poly with real coefficients has a real root, without computing the roots. Note that the answer for odd degree polynomials is always yes. For an even degree polynomial $p(x)$ do the following: Compute the Hermite form of the polynomial. This is a symmetric matrix defined e.g. here : (on page 4, near the bottom denoted by $H_1(p)$ ). The entries of this matrix can be filled using Newton identities . The number of real roots of $p(x)$ is equal to the signature of the Hermite matrix $H_1(p)$ , i.e., the number of positive eigenvalues minus the number of negative eigenvalues. Since the Hermite form is symmetric, its characteristic poly has
only real roots. Therefore, we can apply Descartes' rule of signs to its char.
polynomial, which would give us exactly the number of positive and negative
eigenvalues of the Hermite form. This process gives you the exact number of real roots of $p(x)$ without computing the roots, and in particular you can see if the polynomial has a real root. Hope this helps. -Amirali Ahmadi
|
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20,960 |
I've asked this question in every math class where the teacher has introduced the Gamma function, and never gotten a satisfactory answer. Not only does it seem more natural to extend the factorial directly, but the integral definition $\Gamma(z) = \int_0^\infty t^{z-1} e^{-t}\,dt$, makes more sense as $\Pi(z) = \int_0^\infty t^{z} e^{-t}\,dt$. Indeed Wikipedia says that this function was introduced by Gauss , but doesn't explain why it was supplanted by the Gamma function. As that section of the Wikipedia article demonstrates, it also makes its functional equations simpler: we get $$\Pi(z) \; \Pi(-z) = \frac{\pi z}{\sin( \pi z)} = \frac{1}{\operatorname{sinc}(z)}$$ instead of $$\Gamma(1-z) \; \Gamma(z) = \frac{\pi}{\sin{(\pi z)}}\;;$$ the multiplication formula is simpler: we have $$\Pi\left(\frac{z}{m}\right) \, \Pi\left(\frac{z-1}{m}\right) \cdots \Pi\left(\frac{z-m+1}{m}\right) = \left(\frac{(2 \pi)^m}{2 \pi m}\right)^{1/2} \, m^{-z} \, \Pi(z)$$
instead of
$$\Gamma\left(\frac{z}{m}\right) \, \Gamma\left(\frac{z-1}{m}\right) \cdots \Gamma\left(\frac{z-m+1}{m}\right) = (2 \pi)^{(m-1)/2} \; m^{1/2 - z} \; \Gamma(z);$$ the infinite product definitions reduce from
$$\begin{align}
\Gamma(z) &= \lim_{n \to \infty} \frac{n! \; n^z}{z \; (z+1)\cdots(z+n)}
= \frac{1}{z} \prod_{n=1}^\infty \frac{\left(1+\frac{1}{n}\right)^z}{1+\frac{z}{n}}
\\
\Gamma(z) &= \frac{e^{-\gamma z}}{z} \prod_{n=1}^\infty \left(1 + \frac{z}{n}\right)^{-1} e^{z/n} \\
\end{align}$$
to
$$\begin{align}
\Pi(z) &= \lim_{n \to \infty} \frac{n! \; n^z}{(z+1)\cdots(z+n)}
= \prod_{n=1}^\infty \frac{\left(1+\frac{1}{n}\right)^z}{1+\frac{z}{n}}
\\
\Pi(z) &= e^{-\gamma z} \prod_{n=1}^\infty \left(1 + \frac{z}{n}\right)^{-1} e^{z/n}; \\
\end{align}$$
and the Riemann zeta functional equation reduces from $$\zeta(s) = 2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Gamma(1-s)\ \zeta(1-s)$$ to $$\zeta(s) = 2^s\pi^{s-1}\ \sin\left(\frac{\pi s}{2}\right)\ \Pi(-s)\ \zeta(1-s).$$ I suspect that it's just a historical coincidence, in the same way $\pi$ is defined as circumference/diameter instead of the much more natural circumference/radius. Does anyone have an actual reason why it's better to use $\Gamma(z)$ instead of $\Pi(z)$?
|
From Riemann's Zeta Function, by H. M. Edwards, available as a Dover paperback, footnote on page 8: "Unfortunately, Legendre subsequently introduced the notation $\Gamma(s)$ for $\Pi(s-1).$Legendre's reasons for considering $(n-1)!$ instead of $n!$ are obscure (perhaps he felt it was more natural to have the first pole at $s=0$ rather than at $s = -1$) but, whatever the reason, this notation prevailed in France and, by the end of the nineteenth century, in the rest of the world as well. Gauss's original notation appears to me to be much more natural and Riemann's use of it gives me a welcome opportunity to reintroduce it."
|
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|
21,003 |
Is there any polynomial $f(x,y)\in{\mathbb Q}[x,y]{}$ such that $f\colon\mathbb{Q}\times\mathbb{Q} \rightarrow\mathbb{Q}$ is a bijection?
|
Jonas Meyer's comment: Quote from arxiv.org/abs/0902.3961 , Bjorn Poonen, Feb. 2009: "Harvey Friedman asked whether there exists a polynomial $f(x,y)\in Q[x,y]$ such that the induced map $Q × Q\to Q$ is injective. Heuristics suggest that most sufficiently complicated polynomials should do the trick. Don Zagier has speculated that a polynomial as simple as $x^7+3y^7$ might already be an example. But it seems very difficult to prove that any polynomial works. Our theorem gives a positive answer conditional on a small part of a well-known conjecture." – Jonas Meyer Added June 2019 Poonen's paper is published as: Bjorn Poonen, Multivariable polynomial injections on rational numbers , Acta Arith. 145 (2010), no. 2, pp 123-127, doi: 10.4064/aa145-2-2 , arXiv: 0902.3961 .
|
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|
21,010 |
Let $Y$ be a normal projective surface, let $X$ be a smooth projective surface and let $\pi:Y\longrightarrow X$ be a finite morphism. Why are all singularities of $Y$ cyclic quotient singularities? And what does this mean? Furthermore, why are these singularities rational? And again, what does that mean? (I edited the question. So the example of Ekedahl doesn't work anymore.) Take a minimal resolution of singularities $\rho:Y^\prime\longrightarrow Y$. Then the above apparently shows that $R^0 \rho_\ast \mathcal{O}_{Y^\prime} = \mathcal{O}_Y$ and $R^i \rho_\ast \mathcal{O}_{Y^\prime} = 0$ for $i>0$. Is this something special for surfaces? The reason I ask this question is the following. If $Y$ is a normal variety (say over the field of complex numbers) with the above data, do we still have $R^0 \rho_\ast \mathcal{O}_{Y^\prime} = \mathcal{O}_Y$ and $R^i \rho_\ast \mathcal{O}_{Y^\prime} = 0$ for $i>0$. Note . The case of a surface over the complex numbers is dealt with in Compact complex surfaces by Barth, Hulek, Peters and van de Ven. I believe they show that cyclic quotient singularities are rational in this case.
|
Jonas Meyer's comment: Quote from arxiv.org/abs/0902.3961 , Bjorn Poonen, Feb. 2009: "Harvey Friedman asked whether there exists a polynomial $f(x,y)\in Q[x,y]$ such that the induced map $Q × Q\to Q$ is injective. Heuristics suggest that most sufficiently complicated polynomials should do the trick. Don Zagier has speculated that a polynomial as simple as $x^7+3y^7$ might already be an example. But it seems very difficult to prove that any polynomial works. Our theorem gives a positive answer conditional on a small part of a well-known conjecture." – Jonas Meyer Added June 2019 Poonen's paper is published as: Bjorn Poonen, Multivariable polynomial injections on rational numbers , Acta Arith. 145 (2010), no. 2, pp 123-127, doi: 10.4064/aa145-2-2 , arXiv: 0902.3961 .
|
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|
21,024 |
Actually I have several related questions, not worth opening different threads: What is the exterior derivative intuitively? What is its geometric meaning?
A possible answer I know is, that it is dual to the boundary operator of singular homology. However I would prefer a more direct interpretation. What is a conceptually nice definition of the exterior derivative?
|
Many years back I wrote something about an intuitive way to look at differential forms here . In particular, figure 4 illustrates Stokes' theorem in a way that generalises to higher dimensions. Note that these are just sketches for intuition, and I've found them useful for illustrating various fields arising in physics, but they're not anything rigorous. They're also, in some sense, dual to the diagrams in Misner, Thorne and Wheeler. (There are some errors in my document, but I lost the source code many years ago...)
|
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|
21,110 |
I only know of one proof of Hilbert 90, which is very smart if not magical. See for example http://hilbertthm90.wordpress.com/2008/12/11/hilberts-theorem-90the-math/ Does anyone know of a more intuitive proof or know a good way to view the proof? I have accepted the answer by Emerton, great thanks as well to David Speyer and Brian Conrad.
|
Here is a proof of Hilbert's Theorem 90 in the case of cyclic extensions which
I think is fairly conceptual. The key point (which is also at the heart of Grothendieck's
very general version in terms of flat descent) is that if we want to verify that
a linear transformation has a certain eigenvalue (in our particular case, the eigenvalue
of interest will be 1), we can do so after extending scalars. The set-up: we have a cyclic extension $L/K$, with Galois group generated by $\sigma$,
and an element $a \in L$ of norm 1. We want to find $b \in L$ such that $a = b/\sigma(b)$.
As in David Speyer's answer, rewrite this as the equation $a\sigma(b) = b$. The map $b \mapsto a\sigma(b)$ is a $K$-linear transformation of the $K$-vector space
$L$, and we want to show that it has a fixed point, i.e. that it has $1$ as an eigenvector.
Well, we can verify this after extending scalars (the eigenvectors of a matrix don't
change if we enlarge the ground field), and so we tensor up with $L$ over $K$. Now $L\otimes_K L \cong L\times\cdots \times L$, an isomorphism of $L$-algebras,
and under this isomorphism the action of $\sigma$ on the left just becomes the cyclic permutation of factors on the right. (To see the isomorphism, write $L = K(\alpha),$
as we may by the primitive element theorem. If $f(X)$ is a minimal polynomial of
$\alpha$ over $K$, then $L \cong K[X]/f(X),$ and so $L\otimes_K L \cong L[X]/f(X).$
But over $L$, the polynomial $f(X)$ splits as $f(X) = (X-\alpha_1)\cdots (X-\alpha_n),$
where the $\alpha_i$ are all the conjugates of $\alpha$. Choosing the labelling
appropriately, we may assume that $\alpha_i = \sigma^{i-1}(\alpha)$. Then
$L[X]/f(X) = L[X]/(X-\alpha_1)\cdots (X-\alpha_n) \cong L\times\cdots \times L,$
and $\sigma$ does indeed just permute the factors.) Under the isomorphism $L\otimes_K L \cong L\times\cdots \times L,$
the base-change of our linear transformation $b \mapsto a \sigma(b)$ is given by
$(b_1,\ldots,b_n) \mapsto (a b_n, \sigma(a) b_1, \ldots, \sigma^{n-1}(a) b_{n-1}).$
This transformation has the obvious non-zero fixed vector
$(1,\sigma(a),\sigma(a)\sigma^2(a),\ldots,\sigma(a)\ldots\sigma^{n-1}(a)).$
(Remember that Norm$(a) = 1$, and so the last entry is also just $a^{-1}$.) Thus our original linear transformation (before extending scalars) has a non-zero fixed vector as well,
as required. How does this relate to Brian Conrad's comment? Well, the preceding argument
generalizes massively to Grothendieck's theory of faithfully flat descent, which
in particular shows that any quasi-coherent sheaf in the flat topology in fact
arises from a Zariski sheaf. That may sound quite complicated, but what the argument
amounts to is precisely what we used in the preceding argument: If $A \rightarrow B$
is a faithfully flat map of rings, and we want to study the "spectral theory"
of a linear operator on an $A$-module, we can do so after extending scalars to $B$.
(Of course, one has to be precise about what "spectral theory" means when we are
working over rings that aren't fields. This is where faithfully flat comes in:
it is the condition that extending scalars from $A$ to $B$ is exact, and takes
non-zero modules to non-zero modules; this turns out to be exactly the right
generalization of the more naive notion we used above, that extending scalars
preserves the eigenvalues of a matrix.) Finally, here is an aside about the relation with Galois cohomology: In cohomological language, Hilbert's Theorem 90 is the statement that $H^1(Gal(L/K), L^{\times}) = 0$
for any finite Galois extension of fields $L/K$. To recover the statement involving
norms, one proceeds as follows: if $Gal(L/K)$ is cyclic, with generator $\sigma$,
and the norm of $a \in L$ equals 1,
then $\sigma \mapsto a$ determines a $1$-cocyle on $Gal(L/K)$ with values in $L^{\times}$.
By the vanishing of $H^1$, this must be a coboundary, which means that there exists $b$
such that $a = \sigma(b)/b.$ The cohomological statement (which, as Brian Conrad pointed out, is still a very special
case of Grothendieck's general theory) can be proved by the same extension of scalars argument as above.
|
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|
21,171 |
Possible Duplicate: Cohomology and fundamental classes Given an oriented manifold $M$ and an oriented submanifold $\phi:N\to M$ we can obtain a homology class $\phi_*[N]\in H_*(M)$ where $[N]$ is the fundamental class of $N$. In general, it is not true that every homology class of $M$ can be represented by a submanifold in this manner, however for some special cases it is. For example, for $M$ an oriented (and closed maybe?) 4-manifold every homology class can be represented by a submanifold. Another example is when $M$ an Euclidean configuration space. My questions are: 1) Under what circumstances can every homology class of $M$ be represented by a submanifold and 2) What are some examples of manifolds who have homology classes not representable in this manner?
|
Here are a few simple answers to the question you asked: Every class in $H_{n-1}(M;Z)$ for $M$ orientable is represented by a submanifold: choose a smooth map $f:M\to S^1$ representing the Poincare dual in $H^1(M;Z)=[M,S^1]$ and take the preimage of a point. In dimensions>2 it can be taken connected. Similarly, every class in $H_{n-2}(M;Z)$ for $M$ orientable is represented by a submanifold: choose a smooth map $f:M\to CP^\infty$ representing the Poincare dual in $H^2(M;Z)=[M,CP^\infty]$, homotop $f$ into a finite skeleton, say $CP^N$, and take the preimage of $CP^{N-1}$. Transversality says that if you can represent $x\in H_k(M)$ by a map of a smooth manifold (e.g. elements in the image of the Hurewicz map, or by Thom) , and $2k < n$, then you can represent it by an embedded submanifold (as Andy mentions above). For example, any class in $H_1(M)$ for $dim(M)\ge 3$. With care you can also make this work for $2k=n$, and there are techniques available in the "metastable" range (no triple points) involving generalizations of Whitney's trick and other ways to replace double points.
|
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|
21,267 |
A number field $K$ is said to be monogenic when $\mathcal{O}_K=\mathbb{Z}[\alpha]$ for some $\alpha\in\mathcal{O}_K$. What is currently known about which $K$ are monogenic? Which are not? From Marcus's Number Fields , I'm familiar with the proof that the cyclotomic fields are monogenic, and for example that $\mathbb{Q}(\sqrt{7},\sqrt{10})$ is not monogenic (it is exercise 30 of chapter 2), but because Marcus eschews anything local, I haven't seen any of the perhaps more natural proofs of these results. If $K$ is monogenic, is there an effective method of determining those $\alpha\in\mathcal{O}_K$ for which $\mathcal{O}_K=\mathbb{Z}[\alpha]$? More generally, what is known about the minimal number of generators of $\mathcal{O}_K$ as a $\mathbb{Z}$-algebra? That is, can we determine, or at least put non-trivial bounds on, the minimal $m$ such that $\mathcal{O}_K=\mathbb{Z}[\alpha_1,\ldots,\alpha_m]$ for some $\alpha_i\in\mathcal{O}_K$? We know that any $\mathcal{O}_K$ has an integral basis of $n=[K:\mathbb{Q}]$ elements, so certainly $m\leq n$ (I'm considering that trivial).
|
Zev, when $[K:{\mathbf Q}] > 2$ , finding all $\alpha$ which are ring generators for ${\mathcal O}_K$ is a hard problem in general: there are only finitely many choices modulo the obvious condition that if $\alpha$ works then so does $a + \alpha$ for any integer $a$ . In other words, up to adding an integer there are only finitely many possible choices -- which could of course mean there are no choices. Here is a nice example: what are the possible ring generators for the integers of ${\mathbf Q}(\sqrt[3]{2})$ ? We know a basis for the ring of integers is $1, \sqrt[3]{2}, \sqrt[3]{4}$ , so a ring generator over $\mathbf Z$ would, up to addition by an integer, have the form $\alpha_{x,y} = x\sqrt[3]{2} + y\sqrt[3]{4}$ for some integers $x$ and $y$ which are not both 0. The index of the ring ${\mathbf Z}[\alpha_{x,y}]$ in the full ring of integers is the absolute value of the determinant of the matrix expressing $1, \alpha_{x,y}, \alpha_{x,y}^2$ in terms of $1, \sqrt[3]{2},\sqrt[3]{4}$ , and after a computation that turns out to be $|x^3 - 2y^3|$ . We want this to be 1 in order to have a ring generator, which means we have to find all the integral solutions to the equation $x^3 - 2y^3 = \pm 1$ . Well, that's a pretty famous example of an equation with only finitely many integral solutions. Up to sign the only solutions are $(1,0)$ and $(1,1)$ , so $\alpha_{x,y}$ is $\sqrt[3]{2}$ or $\sqrt[3]{2} + \sqrt[3]{4}$ up to sign (and then addition by an integer). Here's a more general cubic exercise, just to put the previous example in some perspective (among concrete examples). Let ${\mathbf Q}(\alpha)$ be a cubic field where $\alpha^3 + b\alpha + c = 0$ for integers $b$ and $c$ . a) Show for $x, y \in {\mathbf Z}$ not both 0 that $[{\mathbf Z}[\alpha]:{\mathbf Z}[x\alpha + y\alpha^2]] = |x^3 + bxy^2 + cy^3|$ . Therefore if $1,\alpha,\alpha^2$ is known to be a ${\mathbf Z}$ -basis of the ring of integers, finding all other ring generators besides $\alpha$ , up to addition by integers, amounts to solving $x^3 + bxy^2 + cy^3 = \pm 1$ in integers. b) It is natural to guess from part a that if $\alpha^3 + a\alpha^2 + b\alpha + c = 0$ and $x, y \in {\mathbf Z}$ are not both 0 the index $[{\mathbf Z}[\alpha]:{\mathbf Z}[x\alpha + y\alpha^2]]$ should be $|x^3 + ax^2y + bxy^2 + cy^3|$ . Decide if that natural guess is right! In general, finding all possible ring generators (modulo addition by an integer) for the ring of integers in a number field amounts to solving some norm-form equation equal to $\pm 1$ , and beyond the quadratic case that kind of equation will have just a finite number of integral solutions. A place to look for further discussion is Narkiewicz's massive tome on algebraic number theory: pp. 64--65 and especially p. 80. It turns out the question of finiteness of the number of possible ring generators up to addition by an integer goes back to Nagell. The general case was settled by Gyory in 1973; see MathSciNet MR0437489. There's actually a whole book on this theme: Diophantine Equations and Power Integral Bases by István Gaál, Birkhauser, 2002. Update in 2018: to address your question about finding a ring of integers needing many generators as a $\mathbf Z$ -algebra (not just as a $\mathbf Z$ -module), see my answer at Explicit family of number rings $\mathcal{O}_{K_n}$ requiring $n$ generators? .
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|
21,290 |
Let $k$ be a field. What is an explicit power series $f \in k[[t]]$ that is transcendental over $k[t]$? I am looking for elementary example (so there should be a proof of transcendence that does not use any big machinery).
|
If $k$ has characteristic zero, then $\displaystyle e^t = \sum_{n \ge 0} \frac{t^n}{n!}$ is certainly transcendental over $k[t]$; the proof is essentially by repeated formal differentiation of any purported algebraic relation satisfied by $e^t$. Edit: Let me fill in a few details. Given a polynomial $P$ in $e^t$ of degree $d$ where each coefficient is a polynomial in $k[t]$ of degree at most $m$, the possible terms that appear in any formal derivative of $P$ lie in a vector space of dimension $(m+1)(d+1)$, so by taking at least $(m+1)(d+1)$ formal derivatives we obtain too many linear relationships between the terms $t^k e^{nt}$. The coefficient of $e^{dt}$ in particular eventually dominates all other coefficients.
|
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|
21,295 |
Let $(\mathcal{M},g)$ be a $C^{\infty}$-Riemannian manifold. A basic fact is that $g$ endows the manifold $\mathcal{M}$ with a metric space structure, that is, we can define a distance function $d:\mathcal{M}\times\mathcal{M}\longrightarrow\mathbb{R}$ (the distance between two points will be the infimum of the lengths of the curves which join the points)
which is compatible with the topology of $\mathcal{M}$. Of course $d$ is continuous function, but what can we say about the differentiability of $d$?, is it smooth?. If not, Is there some criterion to know when it is? Thanks in advance.
|
As others mentioned, you have to remove the diagonal of $M\times M$ or square the distance function. Then, for a complete $M$, the answer is the following. The distance function is differentiable at $(p,q)\in M\times M$ if and only if there is a unique length-minimizing geodesic from $p$ to $q$. Furthermore, the distance function is $C^\infty$ in a neighborhood of $(p,q)$ if and only if $p$ and $q$ are not conjugate points along this minimizing geodesic. Thus, the function is smooth everywhere if and only if $M$ is simply connected and the geodesics have no conjugate points. This property has numerous equivalent reformulations, including the following for every pair of points, there is a unique minimizing geodesic between them; for every pair of points, there is a unique geodesic between them; every geodesic is minimizing; the exponential map at every point $p\in M$ is a diffeomorphism from $T_pM$ to $M$. In general, the distance function has one-sided directional derivatives everywhere. This derivative has a nice description in the case when you fix $p\in M$ and study the function $f=d(p,\cdot)$. Namely let $q\in M$, $q\ne p$, and denote by $\vec{qp}$ the set of initial velocity vectors (in $T_qM$) of unit-speed minimizing geodesics from $q$ to $p$. Then, for a vector $v\in T_qM$, the one-sided derivative $f'_v$ of $f$ in the direction of $v$ is
$$ f'_v=\min\{-\langle v,\xi\rangle:\xi\in \vec{qp}\} . $$
This follows from the first variation formula and holds not only in Riemannian manifolds but also in Alexandrov spaces. It is not hard to derive the above differentiablity properties from this. I don't have a textbook reference for this precise formulation in the Riemannian case, but any book that covers Berger's lemma about geodesics realizing the diameter probably has directional derivatives as a sublemma. For Alexandrov spaces, the standard reference is Burago-Gromov-Perelman's paper. An intro-level proof (not in a full generality) can be found in (a shameless advertisement follows) "A course in metric geometry" by Burago, Burago and myself, section 4.5.
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|
21,367 |
I just taught the classical impossible constructions for the first time, and in finding my class a reference for the transcendence of pi, I found a dearth of distinct proofs. In particular, those that I read all require the existence of infinitely many primes, which strikes me as extraneous. Is there a known proof that requires only knowledge that I would "expect", namely, integral calculus to get your hands on the actual constant and algebraic properties of polynomials in connection with the assumption that the constant is algebraic?
|
The infinitude of primes (more precisely, the existence of arbitrarily large primes) might actually be necessary to prove the transcendence of $\pi$. As I explained in an earlier answer , there are structures which satisfy many axioms of arithmetic but fail to prove the unboundedness of primes or the existence of irrational numbers. Shepherdson presented a simple method for constructing such models, I will present such a model where $\pi$ is rational! The Shepherdson integers $S$ consist of all Puiseux polynomials of the form
$$a = a_0 + a_1T^{q_1} + \cdots + a_kT^{q_k}$$
where $0 < q_1 < \cdots < q_k$ are rationals, $a_0 \in \mathbb{Z}$, and $a_1,\dots,a_k \in \mathbb{R}$. This is a discrete ordered domain, where $a < b$ iff the most significant term of $b-a$ is positive; this corresponds to making $T$ infinitely large. This ring $S$ satisfies open induction axioms
$$\phi(0) \land \forall x(\phi(x) \to \phi(x+1)) \to \forall x(x \geq 0 \to \phi(x))$$
where $\phi(x)$ is a quantifier free formula (possibly with parameters). So the ring $S$ satisfies the same basic axioms as $\mathbb{Z}$, but only a very limited amount of induction. In the field of fractions of $S$, $\pi$ is equal to the ratio $\pi T/T$. In other words, $\pi$ is a rational number! Is $\pi T/T$ really $\pi$? The integers form a subring of $S$, and if $p,q \in \mathbb{Z}$ then $p/q < \pi T/T$ in $S$ if and only if $p/q < \pi$ in $\mathbb{R}$. So $\pi T/T$ defines the same Dedekind cut as $\pi$ does, which is a very accurate description of $\pi$. Indeed, any proof of the transcendence of $\pi$ must ultimately be based on the comparison of $\pi$ and its powers with certain rational numbers, which $\pi T/T$ will accomplish just as well as the real number $\pi$. However, the usual definitions of $\pi$ are not easily formalizable in this basic theory, so there is much room for debate here and I wouldn't claim that $\pi T/T$ satisfies all reasonable definitions of $\pi$. Shepherdson only presented this argument for real algebraic numbers like $\sqrt{2}$, which have a finitary description in this theory and leave little room for debate. In any case, the conclusion to draw from this is that basic arithmetic with open induction does not suffice to prove that $\pi$, or any other real number, is irrational (never mind transcendental). What about primes? In the ring $S$, the only primes are the ones from $\mathbb{Z}$. Although there are infinitely many primes in $S$, it is not true that there are arbitrarily large primes. For example, there are no primes larger than $T$. Thus $S$ is a model where the unboundedness of primes fails and so does the irrationality of $\pi$. This only shows that basic arithmetic with open induction does not suffice to prove either result. A possible line of attack to show that the unboundedness of primes is necessary to prove the transcendence of $\pi$ would be to show that the minimum amount of induction necessary to prove that $\pi$ is transcendental also suffices to prove the unboundedness of primes. Unfortunately, I do not know how much induction is necessary to prove the transcendence of $\pi$. (And the minimum amount of induction necessary to prove the unboundedness of primes is still an open problem.) Well, here is a partial answer, which is a bit of a bummer. There is another Shepherdson domain $S_0$ similar to the above where $\pi$ is transcendental over $S_0$ and $S_0$ does not have arbitrarily large primes. This shows that the transcendence of $\pi$ does not imply the unboundedness of primes over basic arithmetic with open induction. The ring $S_0$ is the subring of $S$ where the coefficients of the Puiseux polynomial are restricted to be algebraic numbers. The unboundedness of primes fails in $S_0$ because the real algebraic numbers form a real closed field just like $\mathbb{R}$. The number $\pi$ is transcendental over $S_0$ because it is transcendental over the field of real algebraic numbers. This is not entirely surprising since open induction is a very weak base theory and the Shepherdson type rings are very pathological. To constrain such pathologies Van Den Dries suggested requiring that the domain is integrally closed in its field of fractions; he called such domains normal but I don't know if this is standard terminology. Neither $S$ nor $S_0$ are normal. More convincing examples would be normal discrete ordered domains. The methods of Macintyre and Marker ( Primes and their residue rings in models of open induction , MR1001418 ) suggest that normal analogues of $S$ and $S_0$ might exist. The conclusion that I draw from this is that open induction is probably too weak a base theory to study this question. Stronger base theories run into the difficulty that it is still not known just how little induction is necessary to prove the unboundedness of primes. The next reasonable candidate is bounded-quantifier induction (IΔ 0 ), which is not known to imply the unboundedness of primes. Using the Euler product $\pi^2/6 = \prod_p (1-p^{-2})^{-1}$ looks promising, but so far I can only make sense of this product in IΔ 0 + Exp which is known to prove the unboundedness of primes.
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21,375 |
I was hoping to see this pop up on the recent big list question about etymology or terms and symbols. Since it has not, and I can't find an answer, I will ask: What is the reason for the $L$ in $L$-function? I've read that the general use of the term cames from Dirichlet's $L$-functions $L(s,\chi).$ Was there any motivation behind Dirichlet's use or was it just an arbitary choice? If so, is there any compelling reason that we keep this name other than tradition?
|
It is not known why Dirichlet denoted his functions with an $L$. Perhaps he chose $L$ for Legendre (I am not serious). The reason may be alphabetical. Just before $L$-functions are introduced in his 1837 paper on primes in arithmetic progression (Math. Werke vol. 1, 313--342), there are certain functions $G$ and $H$, and the letters $I, J$, and $K$ may not have seemed appropriate labels for a function. While $L(s,\chi)$ and $L(\chi,s)$ are common notations for the $L$-function of a character $\chi$, neither decorated notation is due to Dirichlet; he simply wrote different $L$-functions as $L_0, L_1, L_2,\dots$. Update (Jan. 12, 2016): I learned a few days ago from Ellen Eischen that the Kubota Tractor Corporation has a model called the "(compact) Standard L-Series," and today I saw a Kubota L-series go past my department building. Here is a photo I took. If you're looking for a modern reinterpretation of what the L stands for in L-series, the webpage https://www.kubota.com/product/tlbseries.aspx gives the answer, and it's not Langlands: L means Loader or Landscaper.
|
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|
21,415 |
Let $\mathfrak g$ be a finite-dimensional Lie algebra over $\mathbb C$, and let $\mathfrak g \text{-rep}$ be its category of finite-dimensional modules. Then $\mathfrak g\text{-rep}$ comes equipped with a faithful exact functor "forget" to the category of finite-dimensional vector spaces over $\mathbb C$. Moreover, $\mathfrak g\text{-rep}$ is symmetric monoidal with duals, and the forgetful functor preserves all this structure. By Tannaka-Krein duality (see in particular the excellent paper André Joyal and Ross Street, An introduction to Tannaka duality and quantum groups, 1991 ), from this data we can reconstruct an affine algebraic group $\mathcal G$ such that $\mathfrak g \text{-rep}$ is equivalent (as a symmetric monoidal category with a faithful exact functor to vector spaces) to the category of finite-dimensional representations of $\mathcal G$. However, it is not true that every finite-dimensional Lie algebra is the Lie algebra of an algebraic group . So it is not true that $\mathcal G$ is, say, necessarily the simply-connected connected Lie group with Lie algebra $\mathfrak g$, or some quotient thereof. So my question is: Given $\mathfrak g$, what is an elementary description of $\mathcal G$ (that avoids the machinery of Tannaka-Krein)? For example, perhaps $\mathcal G$ is some Zariski closure of something...?
|
Rather than an answer, this is more of an anti-answer: I'll try to persuade you that you probably don't want to know the answer to your question. Instead of some exotic nonalgebraic Lie algebra, let's start with the one-dimensional Lie algebra $\mathfrak{g}$ over a field $k$. A representation of $\mathfrak{g}$ is just a finite-dimensional vector space + an endomorphism. I don't know what the affine group scheme attached to this Tannakian category is, but thanks to a 1954 paper of Iwahori, I can tell you that its Lie algebra can be identified with the set of pairs $(\mathfrak{g},c)$ where $\mathfrak{g}$ is a homomorphism of abelian groups $k\to k$ and $c$ is an element of $k$. So if $k$ is big, this is huge; in particular, you don't get an algebraic group. (Added: $k$ is algebraically closed.) By contrast, the affine group scheme attached to the category of representations of a semisimple Lie algebra $\mathfrak{g}$ in characteristic zero is the simply connected algebraic group with Lie algebra $\mathfrak{g}$. In summary: this game works beautifully for semisimple Lie algebras (in characteristic zero), but otherwise appears to be a big mess. See arXiv:0705.1348 for a few more details.
|
{
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|
21,424 |
I am writing an exam for my students, and the topic is intro knots theory. I have no idea how to put knots into the file, but I know many MO users who can draw amazing diagrams in their papers. Can someone please provide some hints on what can be used, preferably with some example codes? I do not need complicated diagrams, just some simple knots and links with few crossings. Thanks in advance.
|
Knotinfo has .png files of all knots of 12 crossings or less. http://www.indiana.edu/~knotinfo
|
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|
21,552 |
I have read Hartshorne's Algebraic Geometry from chapter 1 to chapter 4, so I'd like to find some suggestions about the next step to study arithmetic geometry. I want to know how to use scheme theory and its cohomology to solve arithmetic problems. I also want to learn something about moduli theory. Would you please recommend some books or papers? Thank you very much!
|
My suggestion, if you have really worked through most of Hartshorne, is to begin reading papers, referring to other books as you need them. One place to start is Mazur's "Eisenstein Ideal" paper. The suggestion of Cornell--Silverman is also good. (This gives essentially the complete proof, due to Faltings, of the Tate conjecture for abelian varieties over number fields, and of the Mordell conjecture.) You might also want to look at Tate's original paper on the Tate conjecture for abelian varieties over finite fields,
which is a masterpiece. Another possibility is to learn etale cohomology (which you will have to learn in some form or other if you want to do research in arithemtic geometry). For this, my suggestion is to try to work through Deligne's first Weil conjectures paper (in which he proves the Riemann hypothesis), referring to textbooks on etale cohomology as you need them.
|
{
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|
21,555 |
Modular forms could be defined for arbitrary subgroups of the modular group, and I have read that this is done in some papers, but every definition of a modular form I have seen has been with respect to congruence subgroups.
|
At a certain level, it's mostly a matter of (i) terminology and (ii) reading the right books. Technically the word "modular" in modular forms refers to the "modular group $SL_2(\mathbb{Z})$ ". In Miyake's book Modular Forms , he defines an automorphic form with respect to an arbitrary Fuchsian group $\Gamma$ (i.e., a discrete subgroup of $SL_2(\mathbb{R})$ ). Then he goes on to say (p. 114) that "Automorphic functions and forms for modular groups are called modular functions and modular forms respectively." Despite the title, plenty of the book deals with the general case, or with the special case of Fuchsian groups associated to quaternion algebras, which do not yield modular forms according to his definition. In Shimura's book Introduction to the Arithmetic Theory of Automorphic Functions he defines (pp. 28-29) automorphic functions and forms with respect to an arbitrary Fuchsian group of the first kind (i.e., finite hyperbolic covolume). The phrase "modular forms" is sometimes used in his book, but doesn't appear to get a formal definition. These are, to my mind, the two most standard and authoritative references on "modular forms", and they both entertain the concept of a modular form with respect to a rather general Fuchsian group, whatever they want to call it. On the other hand, there are reasons for restricting to Fuchsian groups which are arithmetic (which is a technical term here) and of congruence type . A theorem of Margulis shows that arithmeticity is equivalent to having a sufficiently rich theory of Hecke operators, which is highly important in number-theoretic applications. Similarly, being arithmetic of congruence type puts you in the realm of Shimura varieties, and gives you a rich theory of models of the Riemann surfaces defined over various abelian number fields. On the other hand, it is a famous consequence of Belyi's theorem that every algebraic curve over $\mathbb{Q}$ can be uniformized by a finite-index subgroup of $SL_2(\mathbb{Z})$ (generally of non-congruence type). So if one is interested in the "special" arithmetic properties of modular curves, it makes sense to restrict to congruence type. Indeed, continuing work of John Voight and myself indicates that the congruence type condition is even more arithmetically significant than the arithmeticity [sic!]. We define congruence subgroups of non-arithmetic Fuchsian triangle groups and derive some of the arithmetic applications (using techniques from group theory and the arithmetic theory of branched coverings) that are parallel to those satisfied by the usual modular curves. See http://alpha.math.uga.edu/~pete/triangle-091309.pdf Note that this work is not yet finished, to my consternation. (Mea culpa. Mea culpa.)
|
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|
21,562 |
Almost any mathematical concept has antecedents; it builds on, or is related to, previously known concepts. But are there concepts that owe little or nothing to previous work? The only example I know is Cantor's theory of sets. Nothing like his concrete manipulations of actual infinite objects had been done before.
|
Shannon's work on Information theory. Maybe the math wasn't new but the ideas (such as positing a qualitative metric of information and identifying its relevance to design of communication systems) definitely were.
|
{
"source": [
"https://mathoverflow.net/questions/21562",
"https://mathoverflow.net",
"https://mathoverflow.net/users/4692/"
]
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|
21,599 |
Let $\beta \mathbb{N}$ be the Stone-Cech compactification of the natural numbers $\mathbb{N}$, and let $x, y \in \beta \mathbb{N} \setminus \mathbb{N}$ be two non-principal elements of this compactification (or equivalently, $x$ and $y$ are two non-principal ultrafilters). I am interested in ways to "model" the ultrafilter $y$ using the ultrafilter $x$. More precisely, Q1. (Existence) Does there necessarily exist a continuous map $f: \beta \mathbb{N} \to \beta \mathbb{N}$ which maps $x$ to $y$, while mapping $\mathbb{N}$ to $\mathbb{N}$? To put it another way: does there exist a function $f: \mathbb{N} \to \mathbb{N}$ such that $\lim_{n \to x} f(n) = y$? Q2. (Uniqueness) Suppose there are two continuous maps $f, g: \beta \mathbb{N} \to \beta\mathbb{N}$ with $f(x)=g(x)=y$, which map $\mathbb{N}$ to $\mathbb{N}$. Is it then true that $f$ and $g$ must then be equal on a neighbourhood of $x$? I suspect the answer to both questions is either "no" or "undecidable in ZFC", though perhaps there exist "universal" ultrafilters $x$ for which the answers become yes. But I do not have enough intuition regarding the topology of $\beta \mathbb{N}$ (other than that it is somewhat pathological) to make this more precise. (The fact that $\beta\mathbb{N}$ is not first countable does seem to indicate that the answers should be negative, though.)
|
The answer to Q1 is no. This has been well studied in set theory; you're basically asking whether any two non-principal ultrafilters on $\mathbb{N}$ are comparable under the Rudin-Keisler ordering. Variations on your question have led to many, many interesting developments in set theory, but your question Q1 is easy to answer by a cardinality argument. First note that every $f:\mathbb{N}\to\mathbb{N}$ has a unique extension to a continuous function $\bar{f}:\beta\mathbb{N}\to\beta\mathbb{N}$. Any $x \in \beta\mathbb{N}$ has at most $2^{\aleph_0}$ images through such $\bar{f}$, but there are $2^{2^{\aleph_0}}$ ultrafilters on $\mathbb{N}$, so there are very many $y \in \beta\mathbb{N}$ which are not images of $x$ through such $\bar{f}$. The answer to Q2 is also no. Let $y$ be a nonprincipal ultrafilter on $\mathbb{N}$. The sets $A \times A\setminus\Delta$ where $A \in y$ and $\Delta = \{(n,n) : n \in \mathbb{N}\}$ form a filter base on $\mathbb{N}\times\mathbb{N}$. Let $x$ be an ultrafilter on $\mathbb{N}\times\mathbb{N}$ that contains all these sets. The left and right projections $\pi_1,\pi_2:\mathbb{N}\times\mathbb{N}\to\mathbb{N}$ both send $x$ to $y$, but they are not equal on any neighborhood of $x$. However, the answer to Q2 is yes when $x$ is a selective ultrafilter. Recall that $x$ is selective if for every $h:\mathbb{N}\to\mathbb{N}$ there is a set $A \in x$ on which $h$ is either constant or one-to-one. Given $f,g:\mathbb{N}\to\mathbb{N}$ such that $\bar{f}(x) = \bar{g}(x)$ is nonprincipal, then we can find $A \in x$ on which $f$ and $g$ are both one-to-one. In that case, $f\circ g^{-1}$ must be well-defined on some $A' \in x$. Any extension of $f\circ g^{-1}$ to the complement of $A'$ must map $x$ to $x$, which means that $f \circ g^{-1}$ is the identity on some $A'' \in x$. Thus $f$ and $g$ are equal on the neighborhood of $x$ defined by $A''$.
|
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|
21,709 |
For example, I find the first group isomorphism theorem to be vastly more opaque when presented in terms of commutative diagrams and I've had similar experiences with other elementary results being expressed in terms of exact sequences. What are the benefits that I am not seeing?
|
Holy cow, go beyond the first homomorphism theorem! For example, if you have a long exact sequence of vector spaces and linear maps
$$
0 \rightarrow V_1 \rightarrow V_2 \rightarrow \cdots \rightarrow V_n \rightarrow 0
$$
then exactness implies that the alternating sum of the dimensions is 0.
This generalizes the "rank-nullity theorem" that $\dim(V/W) = \dim V - \dim W$, which is the special case of $0 \rightarrow W \rightarrow V \rightarrow V/W \rightarrow 0$. Replace vector spaces and linear maps by finite abelian groups and group homomorphisms and instead you find the alternating product of the sizes of the groups has to be 1. The purpose of this general machinery is not the small cases like the first homomorphism theorem. Exact sequences and commutative diagrams are the only way to think about or formulate large chunks of modern mathematics. For instance, you need commutative diagrams to make sense of universal mapping properties (which is the way many concepts are defined or at least most clearly understood) and to understand the opening scene in the movie "It's My Turn". Here is a nice exercise. When $a$ and $b$ are relatively prime,
$\varphi(ab) = \varphi(a)\varphi(b)$, where $\varphi(n)$ is Euler's $\varphi$-function from number theory. Question: Is there a formula for $\varphi(ab)$ in terms of $\varphi(a)$ and $\varphi(b)$ when $(a,b) > 1$? Yes:
$$
\varphi(ab) = \varphi(a)\varphi(b)\frac{(a,b)}{\varphi((a,b))}.
$$
You could prove that by the formula for $\varphi(n)$ in terms of prime factorizations, but it wouldn't really explain what is going on because it doesn't provide any meaning to the formula. That's kind of like the proofs by induction which don't really give any insight into what is going on. But it turns out there is a nice 4-term short exact sequence of abelian groups (involving units groups mod $a$, mod $b$, and mod $ab$) such that, when you apply the above "alternating product is 1" result, the general $\varphi$-formula above falls right out. Searching for an explanation of that formula in terms of exact sequences forces you to try to really figure out conceptually what is going on in the formula.
|
{
"source": [
"https://mathoverflow.net/questions/21709",
"https://mathoverflow.net",
"https://mathoverflow.net/users/4692/"
]
}
|
21,793 |
An entertaining topological party trick that I have seen performed is to turn your pants inside-out while having your feet tied together by a piece of string. For a demonstration, check out this video . I have heard some testimonial evidence that it is also possible to turn your pants backwards , again with the constraint of having your feet tied together. This second claim seems pretty dubious to me. Question. Is it indeed possible to turn your pants backwards, while having your feet tied together by a piece of string? A set of instructions or a video demonstration would suffice for a yes answer. A precise mathematical formulation of the problem together with a proof of impossibility would suffice for a no answer.
|
I think that the answer is no, by consideration of linking numbers. First simplify the human body plus cord joining the ankles to a circle, and
assign it an orientation. Also assign an orientation to each pant cuff.
This can be done, e.g., so that each cuff has linking number +1 with the
"body" (in which case the two cuffs are oppositely oriented). Now suppose that there is an isotopy of the pants that turns them backwards.
This means the left cuff is now on the right ankle and vice versa. But this
also reverses the linking numbers, which is impossible.
|
{
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"https://mathoverflow.net/questions/21793",
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"https://mathoverflow.net/users/2233/"
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|
21,854 |
I have a question about vector bundles on the algebraic surface $\mathbb{P}^1\times\mathbb{P}^1$. My motivation is the splitting theorem of Grothendieck, which says that every algebraic vector bundle $F$ on the projective line $\mathbb{P}^1$ is a direct sum of $r$ line bundles, where $r$ is the rank of $F$. My question is: to what extent is this true for $\mathbb{P}^1\times\mathbb{P}^1$, if at all? At first glance I imagine this is classical, but I haven't had luck working it out, nor do I have a good reference for where this might be done. Irrespective of the content of the answer, would it follow that the answer would be the same if the question were asked for $(\mathbb{P}^1)^k$? Thanks!
|
The splitting theorem is most certainly false for vector bundles on $\mathbb{P}^1\times\mathbb{P}^1$ . In fact, the theory of vector bundles on quadric surfaces is probably as complicated as the theory of vector bundles on $\mathbb{P}^2$ (that is, very complicated). Here is a simple example of an indecomposable rank 2 bundle. By the Künneth formula, we see that $\mathrm H^1(\mathcal O(1,-2))$ is not 0; hence there is a non-split extension $$
0 \to \mathcal O(1,-2) \to E \to \mathcal O \to 0.
$$ I claim that $E$ is not split. Suppose that it splits as the sum $L_1 \oplus L_2$ of two line bundles. If $L_1$ were to map to 0 in $\mathcal O$ , the map $E \to \mathcal O$ would factor through $L_2$ , and then the map $L_2 \to \mathcal O$ , being surjective, would have to be an isomorphism, and the sequence would split. Hence both $L_1$ and $L_2$ admit a non-zero map to $\mathcal O$ ; this means that they have to be of the form $\mathcal O(m,n)$ with $m,n \leq 0$ . But $L_1\otimes L_2$ , which is the determinant of $E$ , is $\mathcal O(1,-2)$ , and this gives a contradiction. I know that there have been people studying the moduli theory of vector bundles on $\mathbb{P}^1\times\mathbb{P}^1$ (see for example arXiv:0810.4392 ; but there must be lots of earlier papers).
|
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