source_id
int64 1
4.64M
| question
stringlengths 0
28.4k
| response
stringlengths 0
28.8k
| metadata
dict |
|---|---|---|---|
4,953 |
Do we know any problem in NP which has a super-linear time complexity lower bound? Ideally, we would like to show that 3SAT has super-polynomial lower bounds, but I guess we're far away from that. I'd just like to know any examples of super-linear lower bounds. I know that the time hierarchy theorem gives us problems which can be solved in O(n^3) but not in O(n^2), etc. Thus I put the word "natural" in the question. I ask for problems in NP, because otherwise someone would give examples of EXP-complete problems. I know there are time-space tradeoffs for some problems in NP. I don't know if any of them imply a super-linear time complexity lower bound though. (To address a question below about machine models, consider either multitape Turing machines or the RAM model.)
|
Sorry I am so late to the discussion, but I just registered... There are non-linear time lower bounds on multitape Turing machines for NP-complete problems. These lower bounds follow from the fact that the class of problems solvable in nondeterministic linear time is not equal to the class of those solvable in (deterministic) linear time, in the multitape Turing machine setting. This is proved in: Wolfgang J. Paul, Nicholas Pippenger, Endre Szemerédi, William T. Trotter: On Determinism versus Non-Determinism and Related Problems (Preliminary Version) FOCS 1983: 429-438 In fact, unraveling the proof shows that there must be some problem solvable in nondeterministic linear time that is not solvable in $o(n \cdot (\log^* n)^{1/4})$ time (again, on a multitape Turing machine). Note the * in the logarithm; this is just "barely" above linear. One known application of this result is that a natural NP-complete problem in automata theory cannot be solved in $o(n \cdot (\log^* n)^{1/4})$ time: Etienne Grandjean: A Nontrivial Lower Bound for an NP Problem on Automata. SIAM J. Comput. 19(3): 438-451 (1990) Unfortunately the lower bound of Paul et al. relies crucially on the geometry that arises from accessing one-dimensional tapes. We don't know how to prove a non-linear lower bound even if you allow the Turing machine to have a constant number of two -dimensional tapes. We can prove time lower bounds for NP problems on general computational models if you severely restrict the extra workspace used by the machine. (This is getting into my own work so I won't say more unless you're truly interested.) As for the comment above me: the sorting lower bound holds only in a comparison-based model, which is extremely restricted. The claim that sorting requires Omega(n log n) time on general computational models is false. There are faster algorithms for sorting integers. See for example: Yijie Han: Deterministic sorting in O(n log logn) time and linear space. J. Algorithms 50(1): 96-105 (2004)
|
{
"source": [
"https://mathoverflow.net/questions/4953",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1042/"
]
}
|
4,994 |
It is not unusual that a single example or a very few shape an entire mathematical discipline. Can you give examples for such examples? (One example, or few, per post, please) I'd love to learn about further basic or central examples and I think such examples serve as good invitations to various areas. (Which is why a bounty was offered.) Related MO questions: What-are-your-favorite-instructional-counterexamples , Cannonical examples of algebraic structures , Counterexamples-in-algebra , individual-mathematical-objects-whose-study-amounts-to-a-subdiscipline , most-intricate-and-most-beautiful-structures-in-mathematics , counterexamples-in-algebraic-topology , algebraic-geometry-examples , what-could-be-some-potentially-useful-mathematical-databases , what-is-your-favorite-strange-function ; Examples of eventual counterexamples ; To make this question and the various examples a more useful source there is a designated answer to point out connections between the various examples we collected. In order to make it a more useful source, I list all the answers in categories, and added (for most) a date and (for 2/5) a link to the answer which often offers more details. (~year means approximate year, *year means a year when an older example becomes central in view of some discovery, year? means that I am not sure if this is the correct year and ? means that I do not know the date. Please edit and correct.) Of course, if you see some important example missing, add it! Logic and foundations: $\aleph_\omega$ (~1890), Russell's paradox (1901), Halting problem (1936), Goedel constructible universe L (1938), McKinsey formula in modal logic (~1941), 3SAT (*1970), The theory of Algebraically closed fields (ACF) (?), Physics: Brachistochrone problem (1696), Ising model (1925), The harmonic oscillator, (?) Dirac's delta function (1927), Heisenberg model of 1-D chain of spin 1/2 atoms, (~1928), Feynman path integral (1948), Real and Complex Analysis: Harmonic series (14th Cen.) {and Riemann zeta function (1859)}, the Gamma function (1720), li(x), The elliptic integral that launched Riemann surfaces (*1854?), Chebyshev polynomials (?1854) punctured open set in C^n (Hartog's theorem *1906 ?) Partial differential equations : Laplace equation (1773), the heat equation, wave equation, Navier-Stokes equation (1822), KdV equations (1877), Functional analysis: Unilateral shift, The spaces $\ell_p$, $L_p$ and $C(k)$, Tsirelson spaces (1974), Cuntz algebra, Algebra: Polynomials (ancient?), Z (ancient?) and Z/6Z (Middle Ages?), symmetric and alternating groups (*1832), Gaussian integers ($Z[\sqrt -1]$) (1832), $Z[\sqrt(-5)]$ , $su_3$ ($su_2)$ , full matrix ring over a ring , $\operatorname{SL}_2(\mathbb{Z})$ and SU(2), quaternions (1843), p-adic numbers (1897), Young tableaux (1900) and Schur polynomials, cyclotomic fields, Hopf algebras (1941) Fischer-Griess monster (1973), Heisenberg group, ADE-classification (and Dynkin diagrams), Prufer p-groups, Number Theory: conics and pythagorean triples (ancient), Fermat equation (1637), Riemann zeta function (1859) elliptic curves , transcendental numbers, Fermat hypersurfaces, Probability: Normal distribution (1733), Brownian motion (1827), The percolation model (1957), The Gaussian Orthogonal Ensemble, the Gaussian Unitary Ensemble, and the Gaussian Symplectic Ensemble, SLE (1999), Dynamics: Logistic map (1845?), Smale's horseshoe map (1960). Mandelbrot set (1978/80) (Julia set), cat map , (Anosov diffeomorphism) Geometry: Platonic solids (ancient), the Euclidean ball (ancient), The configuration of 27 lines on a cubic surface, The configurations of Desargues and Pappus, construction of regular heptadecagon (*1796), Hyperbolic geometry (1830), Reuleaux triangle (19th century), Fano plane (early 20th century ??), cyclic polytopes (1902), Delaunay triangulation (1934) Leech lattice (1965), Penrose tiling (1974), noncommutative torus , cone of positive semidefinite matrices, the associahedron (1961) Topology: Spheres, Figure-eight knot (ancient), trefoil knot (ancient?) (Borromean rings (ancient?)), the torus (ancient?), Mobius strip (1858), Cantor set (1883), Projective spaces (complex, real, quanterionic..), Poincare dodecahedral sphere (1904), Homotopy group of spheres , Alexander polynomial (1923), Hopf fibration (1931), The standard embedding of the torus in R^3 (*1934 in Morse theory), pseudo-arcs (1948), Discrete metric spaces, Sorgenfrey line, Complex projective space, the cotangent bundle (?), The Grassmannian variety, homotopy group of spheres (*1951), Milnor exotic spheres (1965) Graph theory: The seven bridges of Koenigsberg (1735), Petersen Graph (1886), two edge-colorings of K_6 (Ramsey's theorem 1930), K_33 and K_5 (Kuratowski's theorem 1930), Tutte graph (1946), Margulis's expanders (1973) and Ramanujan graphs (1986), Combinatorics: tic-tac-toe (ancient Egypt(?)) (The game of nim (ancient China(?))), Pascal's triangle (China and Europe 17th), Catalan numbers (18th century), (Fibonacci sequence (12th century; probably ancient), Kirkman's schoolgirl problem (1850), surreal numbers (1969), alternating sign matrices (1982) Algorithms and Computer Science: Newton Raphson method (17th century), Turing machine (1937), RSA (1977), universal quantum computer (1985) Social Science: Prisoner's dilemma (1950) (and also the chicken game, chain store game, and centipede game), the model of exchange economy, second price auction (1961) Statistics: the Lady Tasting Tea (?1920), Agricultural Field Experiments (Randomized Block Design, Analysis of Variance) (?1920), Neyman-Pearson lemma (?1930), Decision Theory (?1940), the Likelihood Function (?1920), Bootstrapping (?1975)
|
The harmonic oscillator is a fundamental example in both classical and quantum mechanics.
|
{
"source": [
"https://mathoverflow.net/questions/4994",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1532/"
]
}
|
5,045 |
Why the word "hidden" present in hidden markov model? What exactly is hidden.
Whatever is hidden in HMM isn't it hidden in normal Markov Models?
|
The unobserved state. Let's consider a hidden Markov model for my cat's behavior. Bella can be in five states: hungry, tired, playful, cuddly, bored. She can respond to these states with six behaviors: whining, scratching, cuddling, pouncing, sleeping and stalking. A hidden Markov model would consist of two matrices, one 5x5 and the other 5x6. The 5x5 matrix gives the probabilities that, if she is hungry at time $t$, she will be tired at time $t+1$, and so forth. So we can compute the probability that she is in different emotional states by taking powers of this matrix. However, we can't observe her emotions -- they are hidden. The 5x6 matrix gives the probability that, if she is hungry at time $t$, she will whine at time $t+1$. (Very close to $1$.) These are the behaviors we observe. In an ordinary Markov model, there would just be a single 6x6 matrix, which directly described the probability of transitions like whining ---> clawing. As you can see, an ordinary Markov model is less able to reflect the complexity of my cat's inner life. See the wikipedia article for much more information.
|
{
"source": [
"https://mathoverflow.net/questions/5045",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1692/"
]
}
|
5,139 |
Many of us have -- or at some point want to have -- children, and wonder how we can do our best to fulfill the "nurture" component of helping them develop mathematical talent... not because we want them to be mathematical professionals, but we'd like them to have the choice, like we did. However, this is not a question about how one "should" raise children, because of how fiercely debatable even the meaningfulness of such a question would be. Instead, What reputable statistics are known about how mathematicians or highly mathematically talented people have been raised, especially during their early, formative years when "potential" is thought to be more of a variable? (This question is community wiki, so everyone can edit it.) Edit: I changed the title from "How are mathematicians raised?", since I personally only intended to ask for descriptive, not prescriptive, answers. The "However" paragraph above was inserted to further emphasize this.
|
I was raised by a pack of wild mathematicians. We roamed the great planes proving theorems and conjecturing.
|
{
"source": [
"https://mathoverflow.net/questions/5139",
"https://mathoverflow.net",
"https://mathoverflow.net/users/84526/"
]
}
|
5,159 |
In Mumford's "The red book of varieties and schemes" one of the examples (G on pg 74) is the space Spec $(\prod_{i=1}^\infty k)$, where $k$ is a field. He mentions that "Logicians assure us that we can prove more theorems if we use these outrageous spaces". Are the any examples of theorems proved using such spaces, or any references to logicians saying such a thing?
|
At (about) the time Mumford was giving his lectures at Harvard, Ax was lecturing on his work with Kochen in which they proved a conjecture of Artin for almost all p by using ultrafilters. This is clearly what Mumford was thinking of. The reference for the Ax-Kochen work is: MR0184930 Ax, James; Kochen, Simon Diophantine problems over local fields. I. Amer. J. Math. 87 1965 605--630. Ib. 87 1965 631--648.
|
{
"source": [
"https://mathoverflow.net/questions/5159",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1709/"
]
}
|
5,211 |
Is there any geometric way to understand the exact sequence in Example 8.20.1 in Ch II of Hartshorne (p. 182), or its dual from theorem 8.13? Here is the sequence: $0\to O_{\mathbb{P}^n}\to O_{\mathbb{P}^n}(1)^{n+1}\to T_{\mathbb{P}^n}\to 0$
|
Yes! The geometric picture is very nice and very easy. It is explained on pages 408-409 of Griffiths-Harris. Here is roughly how it works: Let's work over $\mathbb{C}$ for simplicity. Think of $\mathbb{P}^n$ as being the quotient of $X := \mathbb{C}^{n+1} - 0$ by the action of $\mathbb{C}^\ast$. On $X$ we have the vector fields $d/dx_i$, where the $x_i$ are the standard coordinates on $\mathbb{C}^{n+1}$. Check that if $v_i$ are linear functionals on $\mathbb{C}^{n+1}$, then the vector field $\sum v_i d/dx_i$ on $X$ descends to a vector field on $\mathbb{P}^n$. The surjection $\mathcal{O}(1)^{n+1} \to \mathcal{T}$ corresponds to taking $n+1$ linear functionals $v_i$ and projecting the vector field $\sum v_i d/dx_i$ down to $\mathbb{P}^n$.
The kernel $\mathcal{O}$ corresponds to the vector field $E = \sum_i x_i d/dx_i$. Intuitively, $E$ is a "radial" vector field on $X$, and if you pretend that $\mathbb{P}^{n}$ is a "sphere" in $X$, then $E$ is "normal" to this "sphere", so it vanishes when we project it down. Jonathan Wise gives a nice (and rigorous) explanation of this below. Aside: I think the reason why this is called the Euler sequence is because the vector field $E$ is known as the Euler vector field. And perhaps the reason why $E$ is called the Euler vector field is because its flow is exponential, and $e = 2.718\dots$ is also known as Euler's number. But I'm not sure, and someone should correct me if I'm wrong about this. Edit: Today somebody told me that the relation $E f = d f$ for $f$ a homogeneous degree $d$ polynomial (as in Charles' answer) was discovered by Euler and is known as "Euler's relation".
|
{
"source": [
"https://mathoverflow.net/questions/5211",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1724/"
]
}
|
5,243 |
This is related to another question of mine . Suppose you met someone who was well-acquainted with the basic properties of rings, but who had never heard of a module. You tell him that modules generalize ideals and quotients, but he remains unimpressed. How do you convince him that studying modules of a ring is a good way to understand that ring? (In other words, why does one have to work "external" to the ring?) Your answer should also explain why it is a good idea to study a group by studying its representations.
|
In short, I'd tell your friend: "If you believe a ring can be understood geometrically as functions its spectrum, then modules help you by providing more functions with which to measure and characterize its spectrum." Elements of a module over a ring $R$ are like generalized functions on $Spec(R)$. We can talk about the support of a module element, or its vanishing set. More concretely, think of how global sections of a line bundle can act as functions you can use to define map into projective space. When you glue together a module on open sets of a spectrum or a scheme, you get to glue using maps which are module isomorphisms, which are more flexible than the ring isomorphisms required to glue together a scheme. Borrowing intuition from smooth manifold land, the twist in the Moebius band (as a line bundle on the circle) is formed by gluing a copy of the reals to itself via multiplication by $-1$, a module map, not a ring map. This allows us to think of functions like $\cos(\theta/2)$ as being globally defined: as a map to the Moebius band. In the same vein, when you have a representation $V$ of a group $G$, each element $v\in V$ gives you a nice evaluation map from $G$ into $V$, so lurking everywhere we've got these morphisms from our object of interest into a known object, which are nicely related to each other via the group laws. A fortiori, this certainly doesn't capture the full utility of group representations, but a priori I think it's a decent justification.
|
{
"source": [
"https://mathoverflow.net/questions/5243",
"https://mathoverflow.net",
"https://mathoverflow.net/users/290/"
]
}
|
5,249 |
Everybody knows that there are $D_n=n! \left( 1-\frac1{2!}+\frac1{3!}-\cdots+(-1)^{n}\frac1{n!} \right)$ derangements of $\{1,2,\dots,n\}$ and that there are $D_n(q)=(n)_q! \left( 1-\frac{1}{(1)_q!}+\frac1{(2)_q!}-\frac1{(3)_q!}+\cdots+(-1)^{n}\frac1{(n)_q!} \right)$ elements in $\mathrm{GL}(q,n)$ which do not have $1$ as an eigenvalue; here $q$ is a prime-power, $(k)_q!=(1)_q(2)_q\cdots(k)_q$ are the $q$-factorials, and $(k)_q=1+q+q^2+\cdots+q^{k-1}$ are the $q$-numbers. Now, there are $D_n^+=\tfrac12\bigl(|D_n|-(-1)^n(n-1)\bigr)$ and $D_n^-=\tfrac12\bigl(|D_n|+(-1)^n(n-1))\bigr)$ even and odd derangements of $\{1,2,\dots,n\}$, as one can see, for example, by computing the determinant $\left| \begin{array}{cccccc} 0 & 1 & 1 & \cdots & 1 & 1 \\ 1 & 0 & 1 & \cdots & 1 & 1 \\ 1 & 1 & 0 & \cdots & 1 & 1 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 1 & 1 & 1 & \cdots & 0 & 1 \\ 1 & 1 & 1 & \cdots & 1 & 0 \end{array} \right|$ and looking at the result. How should one define $D_n^+(q)$ and $D_n^-(q)$?
|
In short, I'd tell your friend: "If you believe a ring can be understood geometrically as functions its spectrum, then modules help you by providing more functions with which to measure and characterize its spectrum." Elements of a module over a ring $R$ are like generalized functions on $Spec(R)$. We can talk about the support of a module element, or its vanishing set. More concretely, think of how global sections of a line bundle can act as functions you can use to define map into projective space. When you glue together a module on open sets of a spectrum or a scheme, you get to glue using maps which are module isomorphisms, which are more flexible than the ring isomorphisms required to glue together a scheme. Borrowing intuition from smooth manifold land, the twist in the Moebius band (as a line bundle on the circle) is formed by gluing a copy of the reals to itself via multiplication by $-1$, a module map, not a ring map. This allows us to think of functions like $\cos(\theta/2)$ as being globally defined: as a map to the Moebius band. In the same vein, when you have a representation $V$ of a group $G$, each element $v\in V$ gives you a nice evaluation map from $G$ into $V$, so lurking everywhere we've got these morphisms from our object of interest into a known object, which are nicely related to each other via the group laws. A fortiori, this certainly doesn't capture the full utility of group representations, but a priori I think it's a decent justification.
|
{
"source": [
"https://mathoverflow.net/questions/5249",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1409/"
]
}
|
5,257 |
I think it's pretty intuitive how singular/simplicial cohomology detects "holes" in a space. How can we directly visualize how and in what sense the Cech cohomology of a cover does this? In case it's of any interest, here are two examples I've looked at with the constant sheaf $\mathbb{Z}$ : (1) The disk, covered "Venn diagram style" with three open patches $U_1, U_2, U_3$ overlapping near the center (like this , but with overlaps), and (2) The restriction of this cover to the boundary circle of the disk: three opens $U_1, U_2, U_3$ with 3 double intersections $U_{12}, U_{13}, U_{23}$ and no triple intersection . If you look at the Cech complex in (2), the $H^1=\mathbb{Z}$ "comes from" the fact that you can write down a triple of elements $(1,0,0)$ on $U_{12}, U_{13}$ , $U_{23}$ which "would" disagree on the triple overlap in (1), but since it's "missing", $(1,0,0)$ gets counted as a cocycle, which is not a coboundary. Even better, the presentation of this $H^1$ you get from the Cech complex is $\mathbb{Z}^3/\{(a,b,c)=(b,c,a)\}$ , which is isomorphic to $\mathbb{Z}$ because you can "rotate" all the coordinates "around the missing intersection" into the first component. I think a like minded analysis of higher dimensional analogues provides similar intuition. Are there any formulations of the Cech complex to really make precise how this intuition should work? What's going on here? Follow up: Following Mariano's answer below, I started reading about Abstract simplicial complexes and their cohomology , which seem like just what I was looking for. What helped me most were the ideas that The (constant sheaf) Cech cohomology of a cover $\cal{U}$ of $X$ "is" the simplicial cohomology of its nerve $N(\cal{U})$ , an abstract simplicial complex , The simplicial cohomology of an abstract simplicial complex "is" the singular cohomology of its geometric realization, and The geometric realization of the nerve of a covering of $X$ is a "simple approximation" of $X$ , So in this sense, we can say precisely that Cech (constant sheaf) cohomology on a cover detects holes in a "simple approximation" to $X$ defined by that cover. In particular, seeing the faces of a simplicial complex encoded as formal wedge products of its vertices totally made my day :)
|
The complex which computes Cech cohomology for a covering is the "same" one as the one that computes the cohomology of the nerve of the covering. It is not hard to see that the geometric realization of that nerve is, in some sense, an approximation to the original space.
Since you apparently find it intuitive that simplicial cohomology detects the geometry, then this should convince you that Cech cohomology also does :P
|
{
"source": [
"https://mathoverflow.net/questions/5257",
"https://mathoverflow.net",
"https://mathoverflow.net/users/84526/"
]
}
|
5,268 |
The Whitehead tower of a (pointed) space is a tower of spaces which successively kills the bottom homotopy groups. The first two spaces can be constructed functorially (at least for suitably nice spaces) as the connected component and the universal cover. Can the remaining spaces be constructed functorially? For the dual situation the answer is yes. I.e. for the Postnikov tower where we have a tower of spaces where the bottom homotopy groups are intact, but where we have killed off all the higher homotopy groups does have a functorial construction (again for nice spaces). The construction I know passes through simplicial sets. I'm wondering if something similar exists for the Whitehead tower?
|
The nth stage of the Whitehead tower of X is the homotopy fiber of the map from X to the nth (or so) stage of its Postnikov tower, so you can use your functorial construction of the Postnikov tower plus a functorial construction of the homotopy fiber (such as the usual one using the path space of the target). The nth stage of the Whitehead tower of X is also the cofibrant replacement for X in the right Bousfield localization of Top with respect to the object S n (or so). Since Top is right proper and cellular this localization exists by the result of chapter 5 of Hirschhorn's book on localizations of model categories. You might look there to see how the cofibrant replacement functor is constructed. With some care you should be able to define functorially the maps in the tower as well. (BTW, the Postnikov tower can similarly be obtained functorially by a left Bousfield localization of Top.)
|
{
"source": [
"https://mathoverflow.net/questions/5268",
"https://mathoverflow.net",
"https://mathoverflow.net/users/184/"
]
}
|
5,323 |
At the time of writing, question 5191 is closed with the accusation of homework. But I don't have a clue about what is going on in that question (other than part 3) [Edit: Anton's comments at 5191 clarify at least some of the things going on and are well worth reading] [Edit: FC's excellent answers shows that my lack of clueness is merely due to ignorance on my own part] so I'll ask a related one. My impression is that it's generally believed that there are infinitely many Mersenne primes, that is, primes of the form $2^n-1$. My impression is also that it's suspected that there are only finitely many Fermat primes, that is, primes of the form $2^n+1$ (a heuristic argument is on the wikipedia page for Fermat primes). [EDIT: on the Wikipedia page there is also a heuristic argument that there are infinitely many Fermat primes!] So I'm going to basically re-ask some parts of Q5191, because I don't know how to ask that a question be re-opened in any other way, plus some generalisations. 1) For which odd integers $c$ is it generally conjectured that there are infinitely many primes of the form $2^n+c$? For which $c$ is it generally conjectured that there are only finitely many? For which $c$ don't we have a clue what to conjecture? [Edit: FC has shown us that there will be loads of $c$'s for which $2^n+c$ is (provably) prime only finitely often. Do we still only have one $c$ (namely $c=-1$) for which it's generally believed that $2^n+c$ is prime infinitely often?] 2) Are there any odd $c$ for which it is a sensible conjecture that there are infinitely many $n$ such that $2^n+c$ and $2^{n+1}+c$ are simultaneously prime? Same question for "finitely many $n$". 3) Are there any pairs $c,d$ of odd integers for which it's a sensible conjecture that $2^n+c$ and $2^n+d$ are simultaneously prime infinitely often? Same for "finitely often".
|
Buzzard is correct to be skeptical of the most naive arguments: Erdos observed that $2^n + 9262111$ is never prime. [ edit Jan 2017 by Buzzard: the 9262111 has sat here for 7 years but there's a slip in Pomerance's slides where he calculates the CRT solution. The correct conclusion from Pomerance's arguments is that $2^n+1518781$ is never prime. Thanks to Robert Israel for pointing out that $2^{104}+9262111$ is prime.] Question one is an incredibly classical problem, of course. Observe that the proof that $2^n + 3$ and $2^n + 5$ are both prime finitely often can plausibly work for a single expression $2^n + c$ for certain $c$. It suffices to find a finite set of pairs $(a,p)$ where $p$ are distinct primes such that every integer is congruent to $a$ modulo $p - 1$ for at least one pair $(a,p)$. Then take $-c$ to be congruent to $2^{a}$ modulo $p$. (Key phrase: covering congruences).
I could write some more, but I can't really do any better than the following very nice elementary talk by Carl Pomerance: www.math.dartmouth.edu/~carlp/PDF/covertalkunder.pdf Apparently the collective number theory brain of mathoverflow is remaking 150 year old conjectures that have been known to be false for over 50 years! I was going to let this post consist of the first line, but I guess I'm feeling generous today. On the other hand, I'm increasingly doubtful that I'm going to get an answer to question 2339 .
|
{
"source": [
"https://mathoverflow.net/questions/5323",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1384/"
]
}
|
5,344 |
Given a compact smooth manifold $M$, it's relatively well known that $C^\infty(M)$ determines $M$ up to diffeomorphism. That is, if $M$ and $N$ are two smooth manifolds and there is an $\mathbb{R}$-algebra isomorphism between $C^\infty(M)$ and $C^\infty(N)$, then $M$ and $N$ are diffeomorphic. (See, for example, Milnor and Stasheff's "Characteristic Classes" book where an exercise walks one through the proof of this fact). Thus, in some sense, all the information about the manifold is contained in $C^\infty(M)$. Further, the tools of logic/set theory/model theory etc. have clearly been applied with greater success to purely algebraic structures than to, say, differential or Riemannian geometry. This is partly do the fact that many interesting algebraic structures can be defined via first-order formulas, whereas in the geometric setting, one often uses (needs?) second-order formulas. So, my question is two-fold: First, is there a known characterization of when a given (commutive, unital) $\mathbb{R}$-algebra is isomorphic to $C^\infty(M)$ for some compact smooth manifold $M$? I imagine the answer is either known, very difficult, or both. Second, has anyone applied the machinery of logic/etc to, say, prove an independence result in differential or Riemannian geometry? What are the references?
|
There is a very cool answer to your question, and it goes by the name well-adapted models for synthetic differential geometry . Andrew Stacey already indicated it in his reply, but maybe I can expand a bit more on this. Synthetic differential geometry is an axiom system that characterizes those categories whose objects may sensibly be regarded as spaces on which differential calculus makes sense. These categories are called smooth toposes . A model for this is a particular such category with these properties. A well-adapted model is one which has a full and faithful embedding of the category of smooth manifolds. (This is "well adapted" from the point of view of ordinary differential geoemtry: ordinary differential geometry embeds into these more powerful theories of smooth structures). The striking insight is that this perspective in particular usefully unifies the ideas of algebraic geometry with that of differential geoemtry to a grander whole. Indeed, the category of presheaves on the opposite of (finitely generated) commutative rings is a model for the axioms, and of course this is the context in which algebraic geometry takes place. But we are entitled to take probe categories considerably richer than just that of duals of commutative rings. In particular, we may consider a category of commutative rings that have a notion of being "smooth" the way a ring of smooth functions is "smooth". These are the C-infinity rings or generalized smooth algebra s. Every ring of smooth functions on a smooth manifold is an example, but there are more. The formal dual of these rings are spaces called smooth loci . This is a smooth analog of the notion of affine scheme. (Notice that the notion of "smooth" as used here is that of differential geometry, not quite that of algebraic geometry, which is more like "singularity free". But they are not unrelated). The main theorem going in the direction of an answer to your question is that the category of manifolds embeds fully and faithfully into that of smooth loci. See at the link smooth loci for the details. But inside the category SmoothLoci, manifolds are characterized as the formal dual to their smooth rings of functions, so that's one way to answer your question. There is a grand story developing from this point on, but for the moment this much is maybe sufficient as a reply.
|
{
"source": [
"https://mathoverflow.net/questions/5344",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1708/"
]
}
|
5,351 |
The Hahn-Banach theorem is rightly seen as one of the Big Theorems in functional analysis. Indeed, it can be said to be where functional analysis really starts. But as it's one of those "there exists ..." theorems that doesn't give you any information as to how to find it; indeed, it's quite usual when teaching it to introduce the separable case first (which is reasonably constructive) before going on to the full theorem. So it's real use is in situations where just knowing the functional exists is enough - if you can write down a functional that does the job then there's no need for the Hahn-Banach Theorem. So my question is: what's a good example of a space where you need the Hahn-Banach theorem? Ideally the space itself shouldn't be too difficult to express, and normed vector spaces are preferable to non-normed ones (a good non-normed vector space would still be nice to know but would be of less use pedagogically). Edit: It seems wrong to accept one of these answers as "the" answer so I'm not going to do that. If forced, I would say that $\ell^\infty$ is the best example: it's probably the easiest non-separable space to think about and, as I've learnt, it does need the Hahn-Banach theorem. Incidentally, one thing that wasn't said, and which I forgot about when asking the question, was that such an example is by necessity going to be non-separable since countable Hahn-Banach is provable merely with induction.
|
I'd like to summarize the answer that has developed from Eric Shechter's book, via Mark Meckes, plus the remark from Gerald Edgar. Since it's not really my answer, I'm making this a community answer. The Hahn-Banach theorem is really the Hahn-Banach axiom. Like the axiom of choice, Hahn-Banach cannot be proved from ZF. What Hahn and Banach proved is that AC implies HB. The converse is not true: Logicians have constructed axiom sets that contradict HB, and they have constructed reasonable axioms strictly between AC and HB. So a version of Andrew's question is, is there a natural Banach space that requires the HB axiom? For the question, let's take HB to say that every Banach space $X$ embeds in its second dual $X^{**}$. As Shechter explains, Shelah showed the relative consistency of ZF + DC + BP (dependent choice plus Baire property). As he also explains, these axioms imply that $(\ell^\infty)^* = \ell^1$. This is contrary to the Hahn-Banach theorem as explained in the next point. A striking way to phrase the conclusion is that $\ell^1$ and its dual $\ell^\infty$ become reflexive Banach spaces. $c_0$ is the closed subspace of $\ell^\infty$ consisting of sequences that converge to 0. The quotient $\ell^\infty/c_0$ is an eminently natural Banach space in which the norm of a sequence is $\max(\lim \sup,-\lim \inf)$. (Another example is $c$, the subspace of convergent sequences. In $\ell^\infty/c$, the norm is half of $\lim \sup - \lim \inf$.) The inner product between $\ell^1$ and $c_0$ is non-degenerate, so in Shelah's axiom system, $(\ell^\infty/c_0)^* = 0$. Without the Hahn-Banach axiom, the Banach space $\ell^\infty/c_0$ need not have any non-zero bounded functionals at all.
|
{
"source": [
"https://mathoverflow.net/questions/5351",
"https://mathoverflow.net",
"https://mathoverflow.net/users/45/"
]
}
|
5,353 |
We've all heard it. I even got it in Norwegian recently. It's number 1 on the list of responses to the statement "I'm a mathematician.". Does anyone have any good comebacks? What other responses have you heard? Standard community wiki rules (not that anyone seems to take any notice of them): one answer per post please. Also, no snide comebacks, please - keep it nice. Extra kudos for answers that would actually educate the other person.
|
The tricky thing about that response is that there's often some pride in not being good at math, perhaps because they've gotten so far with out having to know any math (so they must be pretty good, right?). More often than not, what the person is getting at is that they were right when they said to themselves in grade school "I'm never going to need this." I size the person up, and if I think the answer will be an emphatic "YES", I respond with something like, "but you can read, right?" Then I try to draw a parallel between math and reading. Neither of them (in much quantity) is strictly necessary for getting by, or even to be successful, but if you read (or think mathematically) often, it enriches your life. At this point, their curiosity is usually piqued, so I'll try to steer the discussion to how your understanding of the world might be improved by thinking about probability, symmetry, or abstraction (depending on what I think will best engage the person). I try to get across that doing mathematics can be for pure enjoyment or curiosity with examples of interesting or counterintuitive elementary results.
|
{
"source": [
"https://mathoverflow.net/questions/5353",
"https://mathoverflow.net",
"https://mathoverflow.net/users/45/"
]
}
|
5,357 |
Some theorems give far more than you feel they ought to: a weak hypothesis is enough to prove a strong result. Of course, there's almost always a lot of machinery hidden below the waterline. Such theorems can be excellent starting-points for someone to get to grips with a new(ish) subject: when the surprising result is no longer surprising then you can feel that you've gotten it. Let's have some examples.
|
Every compact metric space is (unless it's empty) a topological quotient of the Cantor set. What, every compact metric space? Yes, every compact metric space.
|
{
"source": [
"https://mathoverflow.net/questions/5357",
"https://mathoverflow.net",
"https://mathoverflow.net/users/45/"
]
}
|
5,372 |
Many mathematical areas have a notion of "dimension", either rigorously or naively, and different dimensions can exhibit wildly different behaviour. Often, the behaviour is similar for "nearby" dimensions, with occasional "dimension leaps" marking the boundary from one type of behaviour to another. Sometimes there is just one dimension that has is markedly different from others. Examples of this behaviour can be good provokers of the "That's so weird, why does that happen?" reaction that can get people hooked on mathematics. I want to know examples of this behaviour. My instinct would be that as "dimension" increases, there's more room for strange behaviour so I'm more surprised when the opposite happens. But I don't want to limit answers so jumps where things get remarkably more different at a certain point are also perfectly valid.
|
My favourite of these is that there is precisely one differentiable structure on $\mathbb{R}^n$ up to diffeomorphism for all $n$, except when $n=4$, when there are uncountably many.
|
{
"source": [
"https://mathoverflow.net/questions/5372",
"https://mathoverflow.net",
"https://mathoverflow.net/users/45/"
]
}
|
5,450 |
Ok, hotshots. You're at a party, and you're chatting with some non-mathematicians. You tell them that you're a mathematician, and then they ask you to elaborate a bit on what you study, or they ask you to explain why you like math so much. What are some engaging ways to do this in general? What are some nice elementary results, accessible to people with any mathematical background or lack thereof, that can be used to illustrate why math is interesting, and its depth, breadth, and beauty? Note the scenario: you're at a party so it should be a relatively quick and snappy kind of thing that doesn't require a blackboard or paper to explain. Easy Mode: How do you explain what your particular sub-field of math is about in an accurate but still understandable and engaging way? Hard Mode: Assuming you work in a less applied area, how can you do this without mentioning any real-world applications ? Again, note the scenario. This question is inspired by this one , in particular Anton's answer .
|
In parties where people are eating pizza, it is quite nice to see people taking their slice and curving the edge so that the slice stays straight. Then you can tell them that this is effectively Gauss's "Theorema Egregium": the initial Gaussian curvature is zero, so by curving the slice in one direction they force the slice to stay straight in the perpendicular direction. You can probably continue the discussion talking about rolling pieces of paper into cylinders but not into tori ("doughnuts"), or maybe talking about soap bubbles!
|
{
"source": [
"https://mathoverflow.net/questions/5450",
"https://mathoverflow.net",
"https://mathoverflow.net/users/83/"
]
}
|
5,485 |
Although we are not so numerous as other respected professionals, like for example lawyers, I wonder if we could come up with a reasonable estimate of our population. Needless to say, the question more or less amounts to the definition of"mathematician". Since I should like to count only research mathematicians (and not, say, high-school teachers) some criterion of publishing should be applied. But it should not be too strict in order not to exclude Grothendieck, for example, who has not published any mathematics for a long time. An excuse for asking a question so soft as to verge on the flabby is that it might be considered an exercise in Fermi-type order of magnitude estimation.
|
Current count of Mathematics Genealogy Project is 137672 (I am assuming that the PhD students that graduated are ranked as "research mathematicians"). But the problem is.. Mathematics Genealogy is mostly for universities of developed countries. There could be some really good university in Russia, China or Korea out there that doesn't give us the correct statistics. Another problem is.. Mathematics Genealogy Project counts even the dead mathematicians (like Hilbert, Hasse, Kepler and so on).. and I am assuming you want a report of living mathematicians.. but hey, I'm quite surprised by the number even 200k is pretty low for the living!
|
{
"source": [
"https://mathoverflow.net/questions/5485",
"https://mathoverflow.net",
"https://mathoverflow.net/users/450/"
]
}
|
5,499 |
There are mathematicians whose creativity, insight and taste have the power of driving anyone into a world of beautiful ideas, which can inspire the desire, even the need for doing mathematics, or can make one to confront some kind of problems, dedicate his life to a branch of math, or choose an specific research topic. I think that this kind of force must not be underestimated; on the contrary, we have the duty to take advantage of it in order to improve the mathematical education of those who may come after us, using the work of those gifted mathematicians (and even their own words) to inspire them as they inspired ourselves. So, I'm interested on knowing who (the mathematician), when (in which moment of your career), where (which specific work) and why this person had an impact on your way of looking at math. Like this, we will have an statistic about which mathematicians are more akin to appeal to our students at any moment of their development. Please, keep one mathematician for post, so that votes are really representative.
|
Thurston . When I was a graduate student, Thurston's work really inspired me to appreciate the role of imagination and visualization in geometry/topology. A prominent mathematician once remarked to me that Thurston was the most underappreciated mathematician alive today. When I pointed out that Thurston had a Fields medal and innumerable other accolades, he replied that this was not incompatible with his thesis.
|
{
"source": [
"https://mathoverflow.net/questions/5499",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1234/"
]
}
|
5,518 |
The standard approach to proving that $H^n(X; G)$ is represented by $K(G, n)$ seems to be to prove that $\text{Hom}(X, K(G, n))$ defines a cohomology theory and then use Eilenberg-Steenrod uniqueness. This is utterly spiffing, but as far as I can see gives little geometric intuition. In his treatment, Hatcher mentions that there is a more direct cell-by-cell proof, albeit a somewhat messy and tedious one. I haven't been able to find any such proof, but I'd really like to see one; I think it would help me solidify my mental picture of Eilenberg-MacLane spaces. Does anyone have a reference?
|
I'd suggest looking up some basic material on obstruction theory. There, you generally find classification of maps $X \to Y$ with domain a CW-complex in terms of cohomology groups $H^s(X;\pi_t(Y))$. The proofs are often very cellular indeed. In the case where the range is an Eilenberg-Maclane space (for an abelian group), the dirty proof is something like: Any map from $X$ is homotopic to one where the (n-1)-skeleton $X^{(n-1)}$ maps to the basepoint of $Y$. A map on the n-skeleton $X^{(n)}$ sending the (n-1)-skeleton to the basepoint is determined, up to homotopy, by a choice of element of $G$ for each n-cell of $X$, essentially by definition of homotopy. This is an element in the n'th CW-chain group $C^n_{CW}(X;G)$. Such a map extends to all higher skeleta if and only if the attaching maps for all the (n+1)-cells become nullhomotopic in $Y$. Thus the map extends if and only if it's represented by a cocycle, i.e. an element of $Z^n_{CW}(X;G)$. This is a complete invariant, up to homotopy, of maps that are trivial on the (n-1)-skeleton. (Higher cells have basically unique maps up to homotopy.) Any homotopy between two such maps can be pushed to a homotopy that's trivial on the (n-2)-skeleton of $X$. Such a homotopy is determined, up to a "track" (a homotopy between homotopies), by a choice of element of $G$ for each (n-1)-cell of $X$. Such a homotopy alters the map on the n-skeleton (as an element of $C^n_{CW}(X;G)$) by adding a coboundary element, something in $B^n_{CW}(X;G)$. Therefore, the full mapping space mod homotopy is $H^n_{CW}(X;G)$. This is a little messy. Often it's nice to use the filtration of $X$ by subcomplexes $X^{(n)}$ and use that each inclusion in the filtration induces a fibration of mapping spaces
$$F(X^{(n)}/X^{(n-1)},Y) \to F(X^{(n)},Y) \to F(X^{(n-1)},Y)$$
to clean this homotopical analysis up a little into something slightly more systematic. This leads to a spectral sequence for the homotopy groups of the mapping spaces in terms of the cohomology of $X$ with coefficients in the homotopy groups of $Y$, but you have to be a little careful because there is a "fringe" that exhibits some non-abelian-group-like behavior.
|
{
"source": [
"https://mathoverflow.net/questions/5518",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1202/"
]
}
|
5,553 |
This is a follow-up to a previous question . What graphs have incidence matrices of full rank? Obvious members of the class: complete graphs. Obvious counterexamples: Graph with more than two vertices but only one edge. I'm tempted to guess that the answer is graphs that contain spanning trees as subgraphs. However, I haven't put much thought into this.
|
The first answer identifies "incidence matrix" with "adjacency matrix". The latter is the
vertices-by-vertices matrix that Sciriha writes about. But the original question
appears to concern the incidence matrix, which is vertices-by-edges. The precise
answer is as follows. Theorem: The rows of the incidence matrix of a graph are linearly independent over the reals if and only if no connected component is bipartite. Proof. Some steps are left for the reader :-) Note first that the sum of rows indexed by the vertices
in one color class of a bipartite component is equal to the sum of the rows indexed by
the other color class. Hence if some component is bipartite, the rows of the incidence matrix are linearly dependent. For the converse, we have to show that the incidence matrix of a connected non-bipartite
graph has full rank. Select a spanning tree $T$ of our graph $G$. Since $G$ is not bipartite, there is an edge $e$ of $G$ such that the subgraph $H$ formed by $T$ and $e$ is not bipartite. The trick is to show that the columns of the incidence matrix indexed by the edges of H
form an invertible matrix. We see that $H$ is built by "planting trees on an odd cycle".
We complete the proof by induction on the number of edges not in the cycle. The base case is when $H$ is an odd cycle. It is easy to show that its incidence matrix is invertible. Otherwise there is a vertex of valency one, $x$ say, such that $H \setminus x$ is connected and not bipartite. Then the incidence matrix of $H \setminus x$ is invertible and again it is easy to see this implies that the incidence matrix of $H$ is invertible. Remark: I do not know who first wrote this result down. It is old, and is rediscovered
at regular intervals.
|
{
"source": [
"https://mathoverflow.net/questions/5553",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1674/"
]
}
|
5,635 |
Purely for fun, I was playing around with iteratively applying $\DeclareMathOperator{\Aut}{Aut}\Aut$ to a group $G$; that is, studying groups of the form $$ {\Aut}^n(G):= \Aut(\Aut(...\Aut(G)...)) $$ Some quick results: For finitely-generated abelian groups, it isn't hard to see that this sequence eventually arrives at trivial group. For $S_n$, $n\neq 2,6$, the group has no center and no outer automorphisms, and so the conjugation action provides an isomorphism $G\simeq \Aut(G)$. Furthermore, I believe that if $G$ is a non-abelian finite simple group, then $\Aut(\Aut(G))\simeq \Aut(G)$, though this is based on hearsay. If one considers topological automorphisms of topological groups, then $\Aut(\mathbb{R})\simeq \mathbb{Z}/2\mathbb{Z}\times \mathbb{R}$, and so $$\Aut(\Aut(\mathbb{R}))\simeq \Aut(\mathbb{Z}/2\mathbb{Z}\times \mathbb{R})\simeq\mathbb{Z}/2\mathbb{Z}\times \mathbb{R}.$$ When the sequence $\Aut^n(G)$ is constant for sufficiently large $n$, we will say the sequence stabilizes . Despite my best efforts, I have been unable to find a group $G$ such that the sequence $\Aut^n(G)$ is provably non-stabilizing. Is this possible? A slightly deeper question is whether there are groups $G$ such that the sequence becomes periodic after some amount of time. That is, $\Aut^n(G)\simeq \Aut^{n+p}(G)$ for some $p$ and for $n$ large enough, but $\Aut^n(G)\not\simeq \Aut^{n+m}(G)$ for $m$ between 0 and $p$. A simple way to produce such an example would be to give two groups $G \neq H$ such that $\Aut(G)\simeq H$ and $\Aut(H)\simeq G$. Does anyone know an example of such a pair?
|
I remember that my old grad classmate from Berkeley, Joel Hamkins, worked on the transfinite version of this problem. The Automorphism Tower Problem , by Simon Thomas, is an entire book on this subject. The beginning of the book gives the example of the infinite dihedral group $D_\infty$, in the sense of $\mathbb{Z}/2 \ltimes \mathbb{Z}$. It says that the automorphism tower of this group has height $\omega+1$. It also treats Joel's theorem, which says that every automorphism tower does stabilize, transfinitely. A Proceedings paper with the same author and title says that Wielandt showed that every finite centerless group has a finite automorphism tower. An improved answer: Simon's book later shows that the automorphism tower of the finite group $D_8$ has height $\omega+1$, and that for general finite groups no one even knows a good transfinite bound. (The $8$ may look like a typo for $\infty$, but it's not :-).) Apparently the centerless condition is essential in Wielandt's condition. Also, to clarify what these references mean by the automorphism tower, they specifically use the direct limit of the conjugation homomorphisms $G \to \mbox{Aut}(G)$. $D_8$ is abstractly isomorphic to its automorphism group. This is a different version of the question that I suppose does not have a transfinite extension. Section 5 of Thomas' book implies that it's an open problem whether the tower terminates in this weaker sense, for finite groups. Finally an arXiv link to Joel Hamkins' charming paper, Every group has a terminating transfinite automorphism tower . As other people in this thread have pointed out, it's unsatisfying to make an automorphism tower that only stabilizes transfinitely as a direct limit, when all of the finite terms of the tower are abstractly isomorphic to the base group $G$. I Googled around a bit more and came back to the same two sources, Thomas' book, and this time a joint result of Hamkins and Thomas which is in chapter 8 of the book. If an automorphism tower stabilizes after exactly $n \in \mathbb{N}$ steps in the direct limit sense, then it also stabilizes after exactly $n$ steps in the weaker abstract isomorphism sense. (Otherwise the direct limit "wouldn't know to stop".) Hamkins and Thomas do better than that. For any two ordinals $\alpha$ and $\beta$, which may or may not be finite numbers, they find one group $G$ whose automorphism tower has height $\alpha$ and $\beta$ in two different models of ZFC set theory. (Whether it's really the "same" group in different worlds is unclear to me, but their models are built to argue that it is so.) I would suppose that it is possible to make a tower without isomorphic terms by taking a product of these groups, even without the two-for-one property. Other than one paper on the Grigorchuk group by Bartholdi and Sidki, I haven't found anything on automorphism towers of finitely generated groups. The Grigorchuk group has a countably infinite tower, but I'd have to learn more to know whether the terms are abstractly isomorphic.
|
{
"source": [
"https://mathoverflow.net/questions/5635",
"https://mathoverflow.net",
"https://mathoverflow.net/users/750/"
]
}
|
5,751 |
I've just been asked for a good example of a situation in maths where using infinity can greatly shorten an argument. The person who wants the example wants it as part of a presentation to the general public. The best I could think of was Goodstein sequences: if you take a particular instance of Goodstein's theorem, then the shortest proof in Peano arithmetic will be absurdly long unless the instance is very very small, but using ordinals one has a lovely short proof. My question is this: does anyone have a more down-to-earth example? It doesn't have to be one where you can rigorously prove that using infinity hugely shortens the shortest proof. Just something where using infinity is very convenient even though the problem itself is finite. (This is related to the question asked earlier about whether finite mathematics needs the axiom of infinity, but it is not quite the same.) A quick meta-question to add: when I finally got round to registering for this site, I lost the hard-earned reputation I had gained as a non-registered user. I am now disgraced, so to speak. Is that just my tough luck?
|
A simple instance where infinity makes things simple is when proving that there exist transcendental numbers. Instead of having to come up with ways to tell a transcendental number from an algebraic one, you simply say «well, there are too many numbers for all of them to be algebraic, so there you have it».
|
{
"source": [
"https://mathoverflow.net/questions/5751",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1459/"
]
}
|
5,760 |
In particular, suppose I have an affine algebraic variety over $\mathbb{R}^n$ described by generators of a radical ideal $I$ and I want to find (perhaps not all of the) points in the variety. Several important questions come up in practice: are there versions of Buchberger's algorithm that work with inexact data? For instance, suppose that the coefficients of the polynomials generating $I$ are known only to floating point precision. Some CAS will try to find solutions assuming that these coefficients are exact . Are there CAS that do something more intelligent (e.g., make certain guarantees given that the numerical coefficients are the truncation of exact coefficients)? does a sparse system of polynomial equations yield a Gröbner basis with sparse elements? In other words, if each polynomial in the original system has a small number of non-zero coefficients relative to $n$, do the basis elements also have this property? what bounds are known for the size of a Gröbner basis in terms of size and sparsity of the original system? are there more appropriate algorithms (than Buchberger's) if we just want to find a single point in the variety? (Suppose that any such point is sufficient.) More generally, which algorithms are better suited to address the kinds of issues mentioned above?
|
First, the Gröbner basis is not sparse. I am speaking a little off-the-cuff, but empirically when I ask SAGE for the Gröbner basis of $(y^n-1,xy+x+1)$ in the ring $\mathbb{Q}[x,y]$, it gets worse and worse as $n$ increases. Any bound would have to be in terms of the degrees of the original generators as well as their sparseness, and I suspect that the overall picture is bad. Overall your questions play to the weaknesses of Gröbner bases. You would need new ideas to make not just the computations of the bases, but also the actual answer numerically stable. You would also need new ideas to make Gröbner bases sparse. You are probably better off with three standard ideas from numerical analysis: Divide and conquer, chasing zeroes with an ODE, and Newton's method. If you have the generators for the variety in an explicit polynomial form, then you are actually much better off than many uses for these methods that involve messy transcendental functions. Because you can use standard analysis bounds, specifically bounding the norms of derivatives, to rigorously establish a scale to switch between divide-and-conquer and Newton's method, for instance. Moreover you can subdivide space adaptively; the derivative norms might let you stop much faster when you are far away from the variety. To explain what I mean by derivative bounds, imagine for simplicity finding a zero of one polynomial in the unit interval $[0,1]$. If the polynomial is $100-x-x^5$, then a simple derivative bound shows that it has no zeroes. If the polynomial is $40-100x+x^5$, then a simple derivative bound shows that it has a unique zero and Newton's method must converge everywhere within the interval. If the polynomial is something much more complicated like $1-x+x^5$, then you can subdivide the interval and eventually the derivative bounds become true. Also, with polynomials, you can make bounds to know that there are no zeroes in an infinite interval under suitable conditions. You can do something similar in higher dimensions. You can divide space into rectangular boxes, and just median subdivisions. It's not very elegant, but it works well enough in low dimensions. In high dimensions, the whole problem can be intractable; you need to say something about why you think that the solution locus is well-behaved to know what algorithm is suitable.
|
{
"source": [
"https://mathoverflow.net/questions/5760",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1557/"
]
}
|
5,772 |
What is the exact relationship between principal bundles, representations, and vector bundles?
|
Let G be an algebraic group (or, since the question was tagged as differential geometry, a Lie group). Then if we're given a principal G-bundle $E_G$ and a representation V of G, we get a vector bundle out of this through the associated bundle construction: $(E_G \times V)/G$ is a vector bundle with generic fiber V. Here, G acts on $E_G \times V$ as $g(x,v) = (xg^{-1},gv).$ This shows that fixing a G-bundle determines an exact tensor functor from the category of representations of G to the category of vector bundles. There's a converse to this which says that giving an exact tensor functor from representations of G to vector bundles is equivalent to a G-bundle.
|
{
"source": [
"https://mathoverflow.net/questions/5772",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1648/"
]
}
|
5,786 |
Because adjoint functors are just cool, and knowing that a pair of functors is an adjoint pair gives you a bunch of information from generalized abstract nonsense, I often find myself saying, "Hey, cool functor. Wonder if it has an adjoint?" The problem is, I don't know enough category theory to be able to check this for myself, which means I can either run and ask MO or someone else who might know, or give up. I know a couple of necessary conditions for a functor to have a left/right adjoint. If it doesn't preserve either limits or colimits, for example, I know I can give up. Is there an easy-to-check sufficient condition to know when a functor's half of an adjoint pair? Are there at least heuristics, other than "this looks like it might work?"
|
The adjoint functor theorem as stated here and the special adjoint functor theorem (which can also both be found in Mac Lane) are both very handy for showing the existence of adjoint functors. First here is the statement of the special adjoint functor theorem: Theorem Let $G\colon D\to C$ be a functor and suppose that the following conditions are satisfied: (i) $D$ and $C$ have small hom-sets (ii) $D$ has small limits (iii) $D$ is well-powered i.e., every object has a set of subobjects (where by a subobject we mean an equivalence class of monics) (iv) $D$ has a small cogenerating set $S$ (v) $G$ preserves limits Then $G$ has a left adjoint. Example I think this is a pretty standard example. Consider the inclusion CHaus $\to$ Top of the category of compact Hausdorff spaces into the category of all topological spaces. Both of these categories have small hom-sets, it follows from Tychonoff's Theorem that CHaus has all small products and it is not so hard to check it has equalizers so it has all small limits and the the inclusion preserves these. CHaus is well-powered since monics are just injective continuous maps and there are only a small collection of topologies making any subspace compact and Hausdorff. Finally, one can check that $[0,1]$ is a cogenerator for CHaus . So $G$ has a left adjoint $F$ and we have just proved that the Stone-Čech compactification exists. If you have a candidate for an adjoint (say the pair $(F,G)$) and you want to check directly it is often easiest to try and cook up a unit and/or a counit and verify that there is an adjunction that way - either by using them to give an explicit bijection of hom-sets or by checking that the composites
$$G \stackrel{\eta G}{\to} GFG \stackrel{G \epsilon}{\to} G$$
and
$$F \stackrel{F \eta}{\to} FGF \stackrel{\epsilon F}{\to} F$$
are identities of $G$ and $F$ respectively. I thought (although I am at the risk of this getting excessively long) that I would add another approach. One can often use existing formalism to produce adjoints (although this is secretly using one of the adjoint functor theorems in most cases so in some sense is only psychologically different). For instance as in Reid Barton's nice answer if one can interpret the situation in terms of categories of presheaves or sheaves it is immediate that certain pairs of adjoints exist. Andrew's great answer gives another large class of examples where the content of the special adjoint functor theorem is working behind the scenes to make verifying the existence of adjoints very easy. Another class of examples is given by torsion theories where one can produce adjoints to the inclusions of certain subcategories of abelian (more generally pre-triangulated) categories by checking that certain orthogonality/decomposition properties hold. I can't help remarking that one instance where it is very easy to produce adjoints is in the setting of compactly generated (and well generated) triangulated categories. In the land of compactly generated triangulated categories one can wave the magic wand of Brown representability and (provided the target has small hom-sets) the only obstruction for a triangulated functor to have a right/left adjoint is preserving coproducts/products (and the adjoint is automatically triangulated).
|
{
"source": [
"https://mathoverflow.net/questions/5786",
"https://mathoverflow.net",
"https://mathoverflow.net/users/382/"
]
}
|
5,790 |
I seem to recall that there is a straightforward subfactor construction that yields fusion categories given by G-graded vector spaces and representations of G, for finite groups G. Is there an analogous construction for 2-groups? Some background: A 2-group is a monoidal groupoid, for which the isomorphism classes of objects form a group. Sinh showed that up to monoidal equivalence, these are classified by a group G (isom. classes of objects), a G-module H (automorphisms of identity), and an element of H 3 (G,H). In the context of this discussion, we can limit our attention to G finite, H=C x . One notable feature is that when the action of G on H is trivial, the three-cocycle twists the associator in the G-graded vector space category. I'm mostly curious about how to tell when two elements of H 3 (G,H) yield Morita-equivalent fusion categories, and am wondering if subfactors or planar algebras make it easy to detect this.
|
The adjoint functor theorem as stated here and the special adjoint functor theorem (which can also both be found in Mac Lane) are both very handy for showing the existence of adjoint functors. First here is the statement of the special adjoint functor theorem: Theorem Let $G\colon D\to C$ be a functor and suppose that the following conditions are satisfied: (i) $D$ and $C$ have small hom-sets (ii) $D$ has small limits (iii) $D$ is well-powered i.e., every object has a set of subobjects (where by a subobject we mean an equivalence class of monics) (iv) $D$ has a small cogenerating set $S$ (v) $G$ preserves limits Then $G$ has a left adjoint. Example I think this is a pretty standard example. Consider the inclusion CHaus $\to$ Top of the category of compact Hausdorff spaces into the category of all topological spaces. Both of these categories have small hom-sets, it follows from Tychonoff's Theorem that CHaus has all small products and it is not so hard to check it has equalizers so it has all small limits and the the inclusion preserves these. CHaus is well-powered since monics are just injective continuous maps and there are only a small collection of topologies making any subspace compact and Hausdorff. Finally, one can check that $[0,1]$ is a cogenerator for CHaus . So $G$ has a left adjoint $F$ and we have just proved that the Stone-Čech compactification exists. If you have a candidate for an adjoint (say the pair $(F,G)$) and you want to check directly it is often easiest to try and cook up a unit and/or a counit and verify that there is an adjunction that way - either by using them to give an explicit bijection of hom-sets or by checking that the composites
$$G \stackrel{\eta G}{\to} GFG \stackrel{G \epsilon}{\to} G$$
and
$$F \stackrel{F \eta}{\to} FGF \stackrel{\epsilon F}{\to} F$$
are identities of $G$ and $F$ respectively. I thought (although I am at the risk of this getting excessively long) that I would add another approach. One can often use existing formalism to produce adjoints (although this is secretly using one of the adjoint functor theorems in most cases so in some sense is only psychologically different). For instance as in Reid Barton's nice answer if one can interpret the situation in terms of categories of presheaves or sheaves it is immediate that certain pairs of adjoints exist. Andrew's great answer gives another large class of examples where the content of the special adjoint functor theorem is working behind the scenes to make verifying the existence of adjoints very easy. Another class of examples is given by torsion theories where one can produce adjoints to the inclusions of certain subcategories of abelian (more generally pre-triangulated) categories by checking that certain orthogonality/decomposition properties hold. I can't help remarking that one instance where it is very easy to produce adjoints is in the setting of compactly generated (and well generated) triangulated categories. In the land of compactly generated triangulated categories one can wave the magic wand of Brown representability and (provided the target has small hom-sets) the only obstruction for a triangulated functor to have a right/left adjoint is preserving coproducts/products (and the adjoint is automatically triangulated).
|
{
"source": [
"https://mathoverflow.net/questions/5790",
"https://mathoverflow.net",
"https://mathoverflow.net/users/121/"
]
}
|
5,800 |
The classic puzzle goes something like this: "You are standing in front of a lake with a 3 gallon bucket and a 5 gallon bucket, how can you get 4 gallons of water?" Is there an easy way to generate the triple (A,B,C) where you can get C gallons of water using buckets of size A and B?
|
Simon's answer points out that the Euclidean algorithm shows that gcd(A,B) divides C is necessary, but the lack of large container makes the problem more difficult, because obviously you can't get $C$ if $C \gt A+B$. However, the following modification of the algorithm seems to work. Let's assume $A \lt B$ and gcd$(A, B)=1$ for simplicity. By pouring from B into A and dumping A, you can get any positive integer $B-nA$ left in B. Do this until the answer is less than $A$. Then by transferring the contents to bucket A and filling it from B into it, we get $2B-(n+1)A$. Then subtract $A$ again until you have $0 \lt 2B-kA \lt A$, and we can iterate this process to get $3B-(k+1)A$, etc. This gives any integer linear combination $rB-sA$ up to the size of $A+B$ because once you get the right multiple of $B$ into the combination, you can always add bucketsful of $A$. You just need to get $rB\equiv C$ (mod A) in order to find a combination for $C$, which happens if gcd$(A, B)=1$. With this algorithm run in the general case you can get any multiple of gcd($A, B$) up to $A+B$ (this holds trivially when $A=B$). (Sorry, I had to edit this answer several times because parts disappeared, until I realized that my inequality signs were being parsed as HTML tag starts even after dollar signs.)
|
{
"source": [
"https://mathoverflow.net/questions/5800",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1857/"
]
}
|
5,847 |
Faa di Bruno's formula ( MathWorld , Wikipedia ) gives the kth derivative of f(g(t)) as a sum over set partitions. (I'm not sure how well-known the combinatorial interpretation is; for example see a 2006 paper of Michael Hardy ). Is there a similar formula for the kth derivative of f(g(h(t))), i. e. of a threefold composition? One has, I think, ${d^2 \over dt^2} f(g(h(t))) = f^{\prime\prime}(g(h(t))) g^\prime(h(t))^2 (h^\prime(t))^2 + f^\prime(g(h(t))) g^{\prime\prime}(h(t)) h^{\prime}(t)^2$
$ + f^\prime(g(h(t))) g^\prime(h(t)) h^{\prime\prime}(t)$ but I am not even sure this is right (computation is tricky!) and I do not know how to generalize this.
|
The Faa di Bruno formula can be well reinterpreted in terms of trees, and then the generalization is obvious. Here's how: First of all, the $k$th derivative of $f\circ g \circ h$ depends only on the first $k$ derivatives of $f,g,h$, evaluated at the appropriate spots. So we can without loss of generality identify smooth functions with their power series --- these should be appropriately centered, so we might as well assume that $f(0) = g(0) = h(0) = 0$. So the data of the "function" $f$ is the sequence $f^{(1)},f^{(2)},f^{(3)},\dots$ of derivatives at $0$. (Any combinatorial fact about derivatives can be checked by composing polynomials.) Second, rather than trying to guess formulas by thinking only in terms of functions of a single variable, moving to vector land. So $f,g,h$ are functions (rather, formal power series) between different vector spaces. For the Faa di Bruno formula in this generality, see M. Hardy, "Combinatorics of Partial Derivatives", the electronic journal of combinatorics, vol. 13 (R1), 2006. Third, as soon as you're working in vectors, you realize that if you thought $f$ was a smooth vector-valued function $V \to W$, then $f^{(n)}$ is a linear function $V^{\otimes n} \to W$. And then it's natural to draw it in the Penrose notation (R. Penrose, "Applications of negative dimensional tensors". In D.J.A. Welsh, editor, Combinatorial mathematics and its applications , pages 221–244, London, 1971. Mathematical Institute, Oxford, Academic Press) as a directed tree. You should label the $n$ incoming edges with the vector space $V$, the one outgoing edge with $W$, and the vertex with $f$ --- you don't need to label it with $f^{(n)}$ because the $n$ is counted by the number of incoming edges. Fourth, work a la Feynman. Declare that a vertex with $n$ incoming and one outgoing edge is $f^{(n)}$. But when you evaluate it, you should divide by a symmetry factor, which counts the number of automorphisms of the diagram ($n!$). Put an $x$ label on each incoming edge. Then:
$$ f(x) = \sum_\Gamma \frac{\Gamma}{|{\rm Aut} \Gamma|} $$
where the sum ranges over all diagrams $\Gamma$ that can be drawn with a single vertex, which is labeled by $f$: x x x x x x
| | | | | |
| | | \ | /
1 | 1 \ / 1 \|/
f(x) = - * f + - * f + - * f + ...
1 | 2 | 6 |
| | |
| | | Ok, then the Faa di Bruno formula says the following: $$ f(g(x)) = \sum_\Gamma \frac{\Gamma}{|{\rm Aut} \Gamma|} $$
where the sum ranges over all diagrams with two "rows": an $f$ on the bottom, and some $g$s on the top: x x x x x
| | | | |
| \ / | |
1 * g 1 * g 1 * *
f(g(x)) = - | + - | + - | | +
1 | 2 | 2 \ /
* f * f * f
| | |
x x x x x x x x x
| | | | | | | | |
\|/ \ / / | | |
1 * g 1 *g *g 1 *g *g *g
+ - | + - | | + - \ | / +
6 | 2 \ / 6 \|/
* f *f *f
| | |
x x x x x x x x x x x x x x x x x x x x
\ | | / | | | | | | | | | | | | | | | |
\\ // | | | | | | | | | | | | | | | |
\V/ \|/ / \| |/ \| | | | | | |
1 *g 1 *g *g 1 *g *g 1 *g *g *g 1 *g *g *g *g
+ -- | + - | | + - | | + - \ | / + -- \ \ / / +
24 | 6 \ / 8 \ / 4 \|/ 24 \ V /
*f *f *f *f *f
| | | | |
+ ... Here I've sorted the diagrams into the degree in $x$. The numerical factor counts the number of combinatorial automorphisms of each diagram (see e.g. the 1/8 and the 1/4), and I list every diagram once up to isomorphism. Which is to say that I'm really considering the whole groupoid of diagrams, and computing a "groupoid integral" in the sense of Baez and Dolan. Anyway, the composition of three functions can again be understood as counting labeled graphs with symmetry. x x x x x x x
| \ / | | | |
*h *h *h *h *h *h
1 | 1 | 1 \ / 1 | |
f(g(h(x))) = - *g + - *g + - *g + - *g *g + ...
1 | 2 | 2 | 2 \ /
*f *f *f *f
| | | | I should mention one more thing. The above sums of diagrams compute the whole power series, and so they're computing the $n$th derivatives of the composition divided by $n!$. For example,
$$ \frac1{4!} (f\circ g)^{(4)} = \frac1{24} f^{(1)}g^{(4)} + \frac16 f^{(2)} g^{(3)} g^{(1)} + \frac18 f^{(2)} (g^{(2)})^2 + \frac14 f^{(3)} g^{(2)} (f^{(1)})^2 + \frac1{24} f^{(4)} (g^{(1)})^4 $$
If you want to evaluate the honest derivative, that's fine too: label the $x$s so that you can distinguish between them, and demand that isomorphisms of diagrams preserve the labellings. Then no diagrams have nontrivial automorphisms (the automorphism group of any diagram is a subgroup of $S_n$ acting on the $n$ $x$s), and it's still true that you should think in terms of the sum over all isomoprhism classes of diagrams.
|
{
"source": [
"https://mathoverflow.net/questions/5847",
"https://mathoverflow.net",
"https://mathoverflow.net/users/143/"
]
}
|
5,853 |
I learned most of my math font rendering from watching others (for example, I draw ζ terribly). In most cases it is passable, but I'm often uncomfortable using fonts like Fraktur on the board. Does anyone know good resources for learning to write these passably?
|
I often give this to Freshmen. Learning the right way to start with is easier than trying to change later.
|
{
"source": [
"https://mathoverflow.net/questions/5853",
"https://mathoverflow.net",
"https://mathoverflow.net/users/360/"
]
}
|
5,892 |
If random variable $X$ has a probability distribution of $f(x)$ and random variable $Y$ has a probability distribution $g(x)$ then $(f*g)(x)$, the convolution of $f$ and $g$, is the probability distribution of $X+Y$. This is the only intuition I have for what convolution means. Are there any other intuitive models for the process of convolution?
|
I remember as a graduate student that Ingrid Daubechies frequently referred to convolution by a bump function as "blurring" - its effect on images is similar to what a short-sighted person experiences when taking off his or her glasses (and, indeed, if one works through the geometric optics, convolution is not a bad first approximation for this effect). I found this to be very helpful, not just for understanding convolution per se, but as a lesson that one should try to use physical intuition to model mathematical concepts whenever one can. More generally, if one thinks of functions as fuzzy versions of points, then convolution is the fuzzy version of addition (or sometimes multiplication, depending on the context). The probabilistic interpretation is one example of this (where the fuzz is a a probability distribution), but one can also have signed, complex-valued, or vector-valued fuzz, of course.
|
{
"source": [
"https://mathoverflow.net/questions/5892",
"https://mathoverflow.net",
"https://mathoverflow.net/users/812/"
]
}
|
5,936 |
Many mathematicians seem to think that the only way to give a mathematics talk is by using chalk on a blackboard. To some, even using a whiteboard is heresy. And we Don't Talk About Computers. I'd like to know people's opinions on this. In a - probably useless - attempt to forestall a flame war, let me try to narrow down the question a little. I'm interested primarily in the effect that the medium has on the talk. This will make comparisons difficult since saying "X gave a rubbish talk using a computer" leaves open the possibility that X would have given an even worse talk using a blackboard. So please try to analyse why you have a preferred medium for presentations. This isn't meant to be a collection of tips for making good presentations (though I can see the value in that). It may be just my impression, but it seems that this attitude is peculiar to mathematicians. Is this true, in your opinion? Is there something special about mathematics that makes it so right for blackboard talks? I'm interested in opinions both as speakers and listeners. Your answers are allowed to be different from the different perspectives! It may be that the type of talk has a bearing on the medium. Try to take this into account as well. I'm possibly completely out of date on this - maybe everyone has now fully Embraced Their Inner Beamer. Please be nice . By all means, praise good behaviour ("X gave a really nice talk using a combination of rope, blancmange, and a small aubergine; it wouldn't have worked so well with only chalk") but let bad behaviour rot in the dungeon of obscurity.
|
The fact that giving a presentation at the black board slows down the speed of presentation and helps the audience to digest the stuff is related to the fact that mathematicians use a language that has a high information density (formulas, diagrams) compared to other fields. Moreover the symbols and other structures used in that language CAN be written on a black board. This is different for a biologist say: using a black board for him/her can be very inefficient, because of the necessity to write down more or less complete sentences or long phrases. This is too much of slowing down. Information dense structures used in biology are pictures or visualizations of numerical data that are hard to create at the black board.
|
{
"source": [
"https://mathoverflow.net/questions/5936",
"https://mathoverflow.net",
"https://mathoverflow.net/users/45/"
]
}
|
5,957 |
According to categorical lore, objects in a category are just a way of separating morphisms. The objects themselves are considered slightly disparagingly. In particular, if I can't distinguish between two objects by using morphisms, then I should consider them equivalent (not equal, that would be evil ). In this view, then, metric spaces with continuous functions are just plain wrong. The category of metric spaces is equivalent to the full subcategory of topological spaces consisting of metrisable spaces. Choosing a metric is evil. So what's the right view of metric spaces, that makes the metric both worth having and not (so much of) an arbitrary choice? I put the "so much of" there because the obvious answer is having isometries as morphisms, but then the category becomes too rigid to be of any conceivable use. So what's the best middle ground?
|
A classic category-theoretic view of metric spaces says that the "correct" maps are the distance-decreasing ones:
$$
d(f(a), f(a')) \leq d(a, a')
$$
where $A$ and $B$ are metric spaces, $f: A \to B$, and $a, a' \in A$. Then all maps are continuous, and the isomorphisms are the isometries onto. This comes from viewing metric spaces as enriched categories, as proposed by Lawvere . The enriched functors are then exactly the distance-decreasing maps. Edit Let me add some detail. Consider the set $V=[0, \infty]$ of non-negative reals. (The inclusion of $\infty$ isn't important here.) It's ordered by $\geq$, and so can be regarded as a category: there's one map $x \to y$ if $x \geq y$, and there are no maps $x \to y$ otherwise. It becomes a monoidal category under $+$ and $0$. A $V$-enriched category is then a set $A$ of objects (or points) together with, for each pair $(a, b)$ of points, an object $A(a, b)$ of $V$ --- that is, a non-negative real, which you might prefer to call $d(a, b)$. Composition then becomes the triangle inequality, and identities the assertion that the distance from a point to itself is $0$. So, a $V$-enriched category is a "generalized metric space": there's no requirement of symmetry (so you could take distance to be the work done in moving between points of a mountainous region) or that points distance $0$ apart are equal (which is just like not asking for isomorphic objects of a category to be equal). You should then be able to see that $V$-enriched functors are what I said they were. Edit re Lipschitz maps I don't want to evangelize this point
of view too much. But it's a matter of fact that Lipschitz maps do arise naturally in this framework. To explain this I first need to explain a little about 'change of
base' for enriched categories. Any lax monoidal functor $\Phi:
\mathcal{V} \to \mathcal{W}$ induces a functor $\Phi_*:
\mathcal{V}-\mathbf{Cat} \to \mathcal{W}-\mathbf{Cat}$, in an obvious
way. For example, if $\Phi: \mathbf{Vect} \to \mathbf{Set}$ is the
forgetful functor, then $\Phi_*$ sends a linear category to its
underlying ordinary category. This means that given a lax monoidal $\Phi: \mathcal{V} \to \mathcal{W}$,
a $\mathbf{V}$-enriched category $\mathbf{A}$, and a
$\mathbf{W}$-enriched category $\mathbf{B}$, we can define a
$\Phi$ -enriched functor $\mathbf{A} \to \mathbf{B}$ to be a
$\mathcal{W}$-enriched functor $\Phi_*(\mathbf{A}) \to \mathbf{B}$. One
might also call this a 'functor over $\Phi$'. That's completely general enriched category theory. Now let's
apply it to $\mathcal{V} = \mathcal{W} = [0, \infty]$. For any $M
\geq 0$, multiplication by $M$ defines a (strict) monoidal functor
$M\cdot -: [0, \infty] \to [0, \infty]$. Let $A$ and $B$ be metric
spaces. Then an $(M\cdot -)$-enriched functor from $A$ to $B$ is
precisely a function $f: A \to B$ such that
$$
d(f(a), f(a')) \leq M\cdot d(a, a')
$$
for all $a, a' \in A$. In other words, it's a Lipschitz map. A bit more can be squeezed out of this. The maps $M\cdot -$ are the strict monoidal endofunctors of $[0, \infty]$. But we can talk
about $\phi$-enriched maps of metric spaces for any lax monoidal endofunctor of $[0, \infty]$. `Lax monoidal' means that
$$
\phi(0) = 0,
\ \ \ \ \ \
\phi(x + y) \leq \phi(x) + \phi(y),
$$
which is a kind of concavity property (satisfied by $\phi(x) = \sqrt{x}$,
for instance). Then a $\phi$-enriched map from $A$ to $B$ is a
function $f: A \to B$ such that
$$
d(f(a), f(a')) \leq \phi(d(a, a'))
$$
for all $a, a' \in A$. Is that kind of map found useful?
|
{
"source": [
"https://mathoverflow.net/questions/5957",
"https://mathoverflow.net",
"https://mathoverflow.net/users/45/"
]
}
|
6,019 |
How would you calculate the order of a list of reviews sorted by "Most Helpful" to "Least Helpful"? Here's an example inspired by product reviews on Amazon: Say a product has 8 total reviews and they are sorted by "Most Helpful" to "Least Helpful" based on the part that says "x of y people found this review helpful". Here is how the reviews are sorted starting with "Most Helpful" and ending with "Least Helpful": 7 of 7 21 of 26 9 of 10 6 of 6 8 of 9 5 of 5 7 of 8 12 of 15 What equation do I need to use to calculate this sort order correctly? I thought I had it a few times but the "7 of 7" and "6 of 6" and "5 of 5" always throw me off. What am I missing?
|
See How not to sort by average ranking .
|
{
"source": [
"https://mathoverflow.net/questions/6019",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1901/"
]
}
|
6,070 |
I know (at least I think I know) that some of the main motivating problems in the development of etale cohomology were the Weil conjectures. I'd like to know what other problems one can solve using the machinery of etale cohomology. I know a little bit about how etale cohomology groups appear in algebraic number theory but I'd like to know about ways that these things show up in other mathematical subjects as well. Is there anything that an algebraic topologist should really know about etale cohomology? What about a differential geometer?
|
$\DeclareMathOperator{\gal}{Gal}$
Here's a comment which one can make to differential geometers which at least explains what etale cohomology "does". Given an algebraic variety over the reals, say a smooth one, its complex points are a complex manifold but with a little extra structure: the complex points admit an automorphism coming from complex conjugation. Hence the singular cohomology groups inherit an induced automorphism, which is extra information that is sometimes worth carrying around. In short: the cohomology of an algebraic variety defined over the reals inherits an action of $\gal(\mathbb{C}/\mathbb{R})$. The great thing about etale cohomology is that a number theorist can now do the same trick with algebraic varieties defined over $\mathbb{Q}$. The etale cohomology groups of this variety will have the same dimension as the singular cohomology groups (and are indeed isomorphic to them via a comparison theorem, once the coefficient ring is big enough) but the advantage is that that they inherit a structure of the amazingly rich and complicated group $\gal(\bar{\mathbb{Q}}/\mathbb{Q})$. I've often found that this comment sees off differential geometers, with the thought "well at least I sort-of know the point of it now". A differential geometer probably doesn't want to study $\gal(\bar{\mathbb{Q}}/\mathbb{Q})$ though. If I put my Langlands-philosophy hat on though, I can see a huge motivation for etale cohomology: Langlands says that automorphic forms should give rise to representations of Galois groups, and etale cohomology is a very powerful machine for constructing representations of Galois groups, so that's why I might be interested in it even if I'm not an algebraic geometer. Finally, I guess a much simpler motivating good reason for etale cohomology is that geometry is definitely facilitated when you have cohomology theories around. That much is clear. But if you're doing algebraic geometry over a field that isn't $\mathbb C$ or $\mathbb R$ then classical cohomology theories aren't going to cut it, and the Zariski topology is so awful that you can't use it alone to do geometry---you're going to need some help. Hence etale cohomology, which gives the right answers: e.g. a smooth projective curve over any field has a genus, and etale cohomology is a theory which assigns to it an $H^1$ of dimension $2g$ (<pedant> at least if you use $\ell$-adic cohomology for $\ell$ not zero in the field <\pedant>).
|
{
"source": [
"https://mathoverflow.net/questions/6070",
"https://mathoverflow.net",
"https://mathoverflow.net/users/493/"
]
}
|
6,074 |
What's the relationship between Kahler differentials and ordinary differential forms?
|
Let $M$ be a differentiable manifold, $A=C^\infty (M)$ its ring of global differentiable functions and $\Omega^1 (M)$ the A-module of global differential forms of class $C^\infty$.
The A-module of Kähler differentials $\Omega_k(A)$ is the free A-module over the symbols $db$ ($b \in A$) divided out by the relations $d(a+b)=da+db,\quad d(ab)= adb+bda,\quad d\lambda=0 \quad(a,b\in A, \quad \lambda \in k)$ There is an obvious surjective map $\quad \Omega_k(A) \to \Omega^1 (M)$ because the relations displayed above are valid in the classical interpretation of the calculus (Leibniz rule).
However, I do not believe at all that it is injective. For example, if $\: M=\mathbb R$, I see absolutely no reason why $\mathrm{d}\sin(x)=\cos(x)\mathrm{d}x$ should be true
in $\Omega_k(A) $ (beware the sirens of calculus). Things would be worse if we considered $C^\infty$
functions which, unlike the sine, are not analytic.
The same sort of reasoning applies to holomorphic manifolds and also to local rings of differentiable or holomorphic functions on manifolds. To sum up: the differentials used in differentiable or holomorphic manifold theory are a quotient of the corresponding Kähler differentials but are not isomorphic to them.
(And I think David's claim that they are isomorphic is mistaken.)
|
{
"source": [
"https://mathoverflow.net/questions/6074",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1867/"
]
}
|
6,079 |
I would like to study/understand the (complete) classification of compact lie groups. I know there are a lot of books on this subject, but I'd like to hear what's the best route I can follow (in your opinion, obviously), since there are a lot of different ideas involved.
|
First, here's a rough outline of how the classification works: Prove that if G and H are simply connected and have the same Lie algebra, then G and H are isomorphic as Lie groups. Prove that if G is any Lie group, its universal cover $\tilde{G}$ inherits a natural Lie group structure for which G = $\tilde{G}/Z$ where $Z\subseteq Z(\tilde{G})$. This reduces classification to a) understanding the Lie algebras and b) understanding the centers of simply connected Lie groups. 3. Classify (simple) Lie algebras. This is done via root diagrams (Dynkin diagrams). 4. For each simply connected compact Lie group, compute its center. For references, I'd check out Fulton and Harris' book "Representation Theory". I'm not sure if it actually does 4., but that's a fairly easy exercise afterwards (except for perhaps the exceptional groups).
|
{
"source": [
"https://mathoverflow.net/questions/6079",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1049/"
]
}
|
6,132 |
Is it true that any finitely presented group can be realized as fundamental group of compact 3-manifold with boundary ?
|
A couple of extra points. Any compact 3-manifold with boundary $M$ can be doubled to give a closed 3-manifold $D$ . As $M$ is a retract of $D$ , it follows that $\pi_1(M)$ injects into $\pi_1(D)$ . Therefore, any "poison subgroup" (such as the Baumslag--Solitar groups that Autumn mentions above) applies just as well to compact 3-manifolds as closed 3-manifolds. Other classes of poison subgroups can be constructed from cohomological conditions. The Kneser--Milnor Theorem implies that any closed, irreducible 3-manifold with infinite fundamental group is aspherical. It follows that any freely indecomposable infinite group with cohomologial dimension greater than 3 cannot be a subgroup of a closed 3-manifold (and hence of a compact 3-manifold, by the previous paragraph). EDIT: Oh, and yet another source of poison subgroups comes from Scott's theorem that 3-manifold groups are coherent , meaning that every finitely generated subgroup is finitely presented. This rules out subgroups like $F\times F$ (where $F$ is a free group), which is not coherent.
|
{
"source": [
"https://mathoverflow.net/questions/6132",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1441/"
]
}
|
6,180 |
In the same vein as Kate and Scott 's questions, why are fusion categories interesting? I know that given a "suitably nice" fusion category (which probably means adding adjectives such as "unitary," "spherical," and "pivotal"), we get a subfactor planar algebra which, in turn, gives us a subfactor. Also, I vaguely understand that these categories give us Turaev Viro TQFTs. What else do fusion categories do?
What's a good reference for the Turaev Viro stuff?
|
Fusion categories (over $\mathbb{C}$) are a natural generalization of finite groups and their behavior over $\mathbb{C}$. The complex representation theory of a finite group is a fusion category, but there are many others. In fact, you can think of a fusion category as a non-commutative, non-cocommutative generalization of a finite group. A finite-dimensional Hopf algebra is that too, but they don't have to be semisimple, while the semisimple ones give you many fusion categories, but again not by any means all of them. Many of the basic results about the structure and representation theory of finite groups generalize, or seem like they could generalize, to fusion categories. This principle has been worked out to a very incomplete but interesting extent by Etingof and others. For instance there is an analogue of the theorem that the dimension of complex irrep of a finite group $G$ divides $|G|$. (Addendum: A qualified analogue, as Scott and Noah point out. If the category is braided, it is a strict analogue; otherwise it is an analogue of dividing $|G|^2$.) There are also semisimple Hopf algebras and other fusion categories that look a lot like $p$-groups. You can think of the whole theory as a rebooted theory of finite groups. However, we are miles and miles away from any fusion category equivalent of the classification of finite simple groups. It is a struggle to make fusion categories that are not derived very closely from finite groups, or do not come from quantum groups at roots of unity. Only a few types of examples are known, and who knows what else is out there. One enticing thing that does change is that dimensions of irreducible objects in a fusion category don't have to be integers. For instance, one of the simplest fusion categories is the Fibonacci category. It has two irreducible objects, the trivial one $I$ and the other object $F$. The dimension of $F$ is the golden ratio, as you can infer from the branching equation $F \otimes F \cong F \oplus I$. (But the dimensions are algebraic integers, and even cyclotomic algebraic integers. Hence divisibility is still a sensible question.) You could also ask, why the semisimple case. As you learn in undergraduate or basic graduate representation theory, the semisimple representation theory of a finite group is much cleaner than the modular representation theory in positive characteristic. And yes, you also get 3-manifold invariants and subfactors. For references: Really Turaev and Viro's original paper, state sum invariants of 3-manifolds and 6j symbols , is pretty good. The generalization to spherical categories is due to Barrett and Westbury, Invariants of Piecewise-Linear 3-manifolds . And there is a discussion in Turaev's book. A sketch: Recall that a basis-independent expression in tensor calculus has the structure of a graph with vertices labelled by tensors and edges labelled by vector spaces. A monoidal category allows the evaluation of similar expressions, except that the graph must be planar and acyclic. In a rigid pivotal category, there are good duals and the graph just needs to be planar. In a spherical category, left trace equals right trace, so a closed graph can be drawn on a sphere. If it is spherical, rigid, and semisimple, then you can use the graph of a tetrahedron to make a local interaction on the tetrahedra of a triangulated 3-manifold, and the result up to normalization is the Turaev-Viro 3-manifold invariant. (In this setting you should dualize the tetrahedra, so that a tensor morphism in the category is associated to a face of the tetrahedron.)
|
{
"source": [
"https://mathoverflow.net/questions/6180",
"https://mathoverflow.net",
"https://mathoverflow.net/users/351/"
]
}
|
6,200 |
I would like to know what quantization is, I mean I would like to have some elementary examples, some soft nontechnical definition, some explanation about what do mathematicians quantize?, can we quantize a function?, a set?, a theorem?, a definition?, a theory?
|
As I'm sure you'll see from the many answers you'll get, there are lots of notions of "quantization". Here's another perspective. Recall the primary motivation of, say, algebraic geometry: a geometric space is determined by its algebra of functions. Well, actually, this isn't quite true --- a complex manifold, for example, tends to have very few entire functions (any bounded entire function on C is constant, and so there are no nonconstant entire functions on a torus, say), so in algebraic geometry, they use "sheaves", which are a way of talking about local functions. In real geometry, though (e.g. topology, or differential geometry), there are partitions of unity, and it is more-or-less true that a space is determined by its algebra of total functions. Some examples: two smooth manifolds are diffeomorphic if and only if the algebras of smooth real-valued functions on them are isomorphic. Two locally compact Hausdorff spaces are homeomorphic if and only if their algebras of continuous real-valued functions that vanish at infinity (i.e. for any epsilon there is a compact set so that the function is less than epsilon outside the compact set) are isomorphic. (From a physics point of view, it should be taken as a definition of "space" that it depends only on its algebra of functions. Said functions are the possible "observables" or "measurements" --- if you can't measure the difference between two systems, you have no right to treat them as different.) So anyway, it can be useful to recast geometric ideas into algebraic language. Algebra is somehow more "finite" or "computable" than geometry. But not every algebra arises as the algebra of functions on a geometric space. In particular, by definition the multiplication in the algebra is "pointwise multiplication", which is necessarily commutative (the functions are valued in R or C, usually). So from this point of view, "quantum mathematics" is when you try to take geometric facts, written algebraically, and interpret them in a noncommutative algebra. For example, a space is locally compact Hausdorff iff its algebra of continuous functions is commutative c-star algebra, and any commutative c-star algebra is the algebra of continuous functions on some space (in fact, on its spectrum). So a "quantum locally compact Hausdorff space" is a non -commutative c-star algebra. Similarly, "quantum algebraic space" is a non -commutative polynomial algebra. Anyway, I've explained "quantum", but not "quantization". That's because so far there's just geometry ("kinetics"), and no physics ("dynamics"). Well, a noncommutative algebra has, along with addition and multiplication, an important operation called the "commutator", defined by $[a,b]=ab-ba$. Noncommutativity says precisely that this operation is nontrivial. Let's pick a distinguished function H, and consider the operation $[H,-]$. This is necessarily a differential operator on the algebra, in the sense that it is linear and satisfies the Leibniz product rule. If the algebra were commutative, then differential operators would be the same as vector fields on the corresponding geometric space, and thus are the same as differential equations on the space. In fact, that's still true for noncommutative algebras: we define the "time evolution" by saying that for any function (=algebra element) f, it changes in time with differential [H,f]. (Using this rule on coordinate functions defines the geometric differential equation; in noncommutative land, there does not exist a complete set of coordinate functions, as any set of coordinate functions would define a commutative algebra.) Ok, so it might happen that for the functions you care about, $[a,b]$ is very small. To make this mathematically precise, let's say that (for the subalgebra of functions that do not have very large values) there is some central algebra element $\hbar$, such that $[a,b]$ is always divisible by $\hbar$. Let $A$ be the algebra, and consider the $A/\hbar A$. If $\hbar$ is supposed to be a "very small number", then taking this quotient should only throw away fine-grained information, but some sort of "classical" geometry should still survive (notice that since $[a,b]$ is divisible by $\hbar$, it goes to $0$ in the quotient, so the quotient is commutative and corresponds to a classical geometric space). We can make this precise by demanding that there is a vector-space lift $(A/\hbar A) \to A$, and that $A$ is generated by the image of this lift along with the element $\hbar$. Anyway, so with this whole set up, the quotient $A/\hbar A$ actually has a little more structure than just being a commutative algebra. In particular, since $[a,b]$ is divisible by $\hbar$, let's consider the element $\{a,b\} = \hbar^{-1} [a,b]$. (Let's suppose that $\hbar$ is not a zero-divisor, so that this element is well-defined.) Probably, $\{a,b\}$ is not small, because we have divided a small thing by a small thing, so that it does have a nonzero image in the quotient. This defines on the quotient the structure of a Poisson algebra . In particular, you can check that $\{H,-\}$ is a differential operator for any (distinguished) element $H$, and so still defines a "mechanics", now on a classical space. Then quantization is the process of reversing the above quotient. In particular, lots of spaces that we care about come with canonical Poisson structures. For example, for any manifold, the algebra of functions on its cotangent bundle has a Possion bracket. "Quantizing a manifold" normally means finding a noncommutative algebra so that some quotient (like the one above) gives the original algebra of functions on the cotangent bundle. The standard way to do this is to use Hilbert spaces and bounded operators, as I think another answerer described.
|
{
"source": [
"https://mathoverflow.net/questions/6200",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1651/"
]
}
|
6,250 |
Are there any good examples of theorems in reasonably expressive theories (like Peano arithmetic) for which it is substantially easier to prove (in a metatheory) that a proof exists than it is actually to find the proof? When I say "substantially easier," the tediousness of formalizing an informal proof should not be considered. In other words, a rigorous but informal proof doesn't count as a nonconstructive demonstration that a formal proof exists, since there is no essential difficulty in producing a formal proof besides workload. If there is some essential barrier, then there's something wrong either with the proof or with the definition of "formal proof." (Although the concept of "informal proof" is by nature vague, it seems reasonable to say that the procedure transforming a typical informal proof to a formal proof is a primitive recursive computation, even though it's probably impossible to make that precise enough to prove.) I don't know what such a nonconstructive meta-proof could look like, but I also don't see why one couldn't exist. The only near-example I can think of right now is in the propositional calculus: you can prove that a propositional formula is a theorem by checking its truth table, which does not explicitly provide a deduction from the axioms; however, in most interesting cases, that probably isn't much easier (if at all) than exhibiting a proof, just maybe more mechanical. (Of course, estimating the difficulty of proving or disproving a candidate for a propositional theorem is a huge open problem.) Also, I think there is actually a simple way to convert a truth-table proof into an actual deduction from the axioms of propositional calculus, although I don't remember and didn't retrace all of the details. (Anyway, I'm not even so interested in the propositional calculus for this question, since it's not very expressive.) Another thing that seems vaguely relevant is Godel's completeness theorem, which states that a formula is a theorem of a first-order theory if and only if it is true in every model of that theory. I don't know in what cases it would be easier, though, to show that some formula were true in every model than it would be just to prove the theorem in the theory.
|
Set theory provides a good example. It is often convenient in set theory to work with the concept of "classes" and treat them as mathematical objects of their own kind. The standard axiomatization of set theory with classes is called Goedel-Bernays set theory, denoted GBC, whereas the usual ZFC axioms have only set objects. But there is a general theorem that any statement purely about sets that is proved by using classes in GBC can be proved without them purely in ZFC. This is what it means to say that GBC is a conservative extension of ZFC. To prove this general theorem, it suffices to observe that any ZFC model can be expanded to a model of GBC, by adding only the classes definable from parameters. There are many other examples of conservative extensions in mathematics, and all of them would seem to be examples of the type that you seek. For example, PA has a conservative extension to the analogous theory true in the collection HF of hereditarily finite sets. Thus, to prove a theorem about numbers in PA, one can freely use heredtiarily finite sets (e.g. sequences of numbers, sequences of sets of numbers, etc.), not just as coded by numbers via Goedel coding, but as actual mathematical objects. And surely this makes proofs much easier. Perhaps another type of example arises in the absoluteness phenomenon in set theory. For example, the Shoenfield Absoluteness theorem states that any Sigma^1_2 statement has the same truth value in any two models of set theory with the same ordinals. In particular, to prove that a particular Sigma^1_2 statement is true in ZFC, it suffices to prove it under the assumption also that V=L, where one also has all kinds of additional structure available. The Absoluteness Theorems (and there are many) can therefore be viewed under the rubric you mention. But of course, in all these cases, we have an actual proof in the weaker theory. To prove that there is a proof, is a proof, so I believe ultimately there will be no way to avoid the quibbling over whether it is easy or hard to translate the high-level proof into a low level proof, since in principle it will always be possible to do this.
|
{
"source": [
"https://mathoverflow.net/questions/6250",
"https://mathoverflow.net",
"https://mathoverflow.net/users/302/"
]
}
|
6,281 |
When I think of a "simplicial complex", I think of the geometric realization of a simplicial set (a simplicial object in the category of sets). I'll refer to this as "the first definition". However, there is another definition of "simplicial complex", e.g. the one on wikipedia : it's a collection $K$ of simplices such that any face of any simplex in $K$ is also in $K$, and the intersection of two simplices of $K$ is a face of both of the two simplices. There is also the notion of " abstract simplicial complex ", which is a collection of subsets of $\{ 1, \dots, n \}$ which is closed under the operation of taking subsets. These kinds of simplicial complexes also have corresponding geometric realizations as topological spaces. I'll refer to both of these definitions as "the second definition". The second definition looks reasonable at first sight, but then you quickly run into some horrible things, like the fact that triangulating even something simple like a torus requires some ridiculous number of simplices (more than 20?). On the other hand, you can triangulate the torus much more reasonably using the first definition (or alternatively using the definition of "Delta complex" from Hatcher's algebraic topology book, but this is not too far from the first definition anyway). I believe you can move back and forth between the two definitions without much trouble. (I think you can go from the first to the second by doing some barycentric subdivisions, and going from the second to the first is trivial.) Due to the fact that the second definition is the one that's listed on wikipedia, I get the impression that people still use this definition. My questions are: Are people still using the second definition? If so, in which contexts, and why? What are the advantages of the second definition?
|
Simplicial sets and simplicial complexes lie at two ends of a spectrum, with Delta complexes, which were invented by Eilenberg and Zilber under the name "semi-simplicial complexes", lying somewhere in between. Simplicial sets are much more general than simplicial complexes and have the great advantage of allowing quotients and products to be formed without the necessity of subdivision, as is required for simplicial complexes. In this way simplicial sets are like CW complexes, only more combinatorial or categorical. The price to pay for this is that simplicial sets are perhaps less geometric, or at least not as nicely geometric as simplicial complexes. So the choice of which to use may depend in part on how geometric the context is. In some areas simplicial sets are far more natural and useful than simplicial complexes, in others the reverse is true. If one drew a Venn diagram of the people using one or the other structure, the intersection might be very small. Delta complexes, being something of a compromise, have some of the advantages and disadvantages of each of the other two types of structure. When I wrote my algebraic topology book I had the feeling that Delta complexes had been largely forgotten over the years, so I wanted to re-publicize them, both as a pedagogical tool in introductory algebraic topology courses and as a sort of structure that arises very naturally in many contexts. For example the classifying space of a category is a Delta complex. Incidentally, I've added 5 pages at the end of the Appendix in the online version of my book going into a little more detail about these various types of simplicial structures. (I owe a debt of thanks to Greg Kuperberg for explaining some of this stuff to me a couple years ago.)
|
{
"source": [
"https://mathoverflow.net/questions/6281",
"https://mathoverflow.net",
"https://mathoverflow.net/users/83/"
]
}
|
6,376 |
for forgetful functors, we can usually find their left adjoint as some "free objects", e.g. the forgetful functor: AbGp -> Set, its left adjoint sends a set to the "free ab. gp gen. by it". This happens even in some non-trivial cases. So my question is, why these happen? i.e. why that a functor forgets some structure (in certain cases) implies that they have a left adjoint? Thanks.
|
Forgetful functors usually have a left adjoint because they usually preserve limits. For example, the underlying set of the direct product of two groups is the direct product of the underlying sets, and similarly for equalizers (that gives you all finite limits). However, functors that preserve limits don't have to have left adjoints, because once in a while what you want to do to construct a free object results in a proper class. An example is complete lattices .
Freyd's Adjoint Functor Theorem gives a necessary and sufficient condition for a limit-preserving functor to have a left adjoint. The proof and related results is discussed in section 1.9 of Toposes, Triples and Theories .
|
{
"source": [
"https://mathoverflow.net/questions/6376",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1657/"
]
}
|
6,377 |
I'm studying Steenrod operations from Hatcher's book. Like homology, one can use them only knowing the axioms, without caring for the actual construction. But while there are plenty of intuitive reasons to introduce homology, I cannot find any for the Steenrod operations. I can follow the steps in the proofs given by Hatcher, but I don't understand why one introduces all these spaces like $\Lambda X$, $\Gamma X$ and so on (in Hatcher's notation, I don't know if it's universal). Does anyone know how to get an intuitive grasp of what's going on?
|
Steenrod operations are an example of what's known as a power operation . Power operations result from the fact that cup product is "commutative, but not too commutative". The operations come from a "refinement" of the operation of taking $p$th powers (squares if $p=2$), whose construction rests on this funny version of commutativity. A cohomology class on $X$ amounts to a map $a: X\to R$, where $R = \prod_{n\geq0} K(F_2,n)$. So the cup product of $a$ and $b$ is given by
$$X\times X \to R\times R \xrightarrow{\mu} R.$$
In other words, the space $R$ carries a product, which encodes cup product. (There is another product on $R$ which encodes addition of cohomology classes.) You might expect, since cup product is associative and commutative, that if you take the $n$th power of a cohomology class, you get a cohomology class on the quotient $X^n/\Sigma_n$, where $\Sigma_n$ is the symmetric group, i.e.,
$$X^n \xrightarrow{a^n} R^n \rightarrow R$$
should factor through the quotient $X^n/\Sigma_n$. This isn't quite right, because cup product is really only commutative up to infinitely many homotopies (i.e., it is an "E-infinity structure" on $R$). This means there is a contractible space $E(n)$ with a free action of $\Sigma_n$, and a product map:
$$\mu_n' : E(n)\times R^n\to R$$
which is $\Sigma_n$ invariant, so it factors through $(E(n)\times R^n)/\Sigma_n$. Thus, given $a: X\to R$, you get
$$P'(a): (E(n)\times X^n)/\Sigma_n \to (E(n)\times R^n)/\Sigma_n \to R.$$
If you restrict to the diagonal copy of $X$ in $X^n$, you get a map
$$P(a):E(n)/\Sigma_n \times X\to R.$$ If $n=2$, then $E(2)/\Sigma_2$ is what Hatcher seems to call $L^\infty$; it is the infinite real proj. space $RP^\infty$. So $P(a)$ represents an element
in $H^* RP^\infty \times X \approx H^*X[x]$; the coefficients of this polynomial in $x$ are the Steenrod operations on $a$. Other cohomology theories have power operations (for K-theory, these are the Adams operations). You can also describe the steenrod squares directly on the chain level: the account in the book by Steenrod and Epstein is the best place to find the chain level description.
|
{
"source": [
"https://mathoverflow.net/questions/6377",
"https://mathoverflow.net",
"https://mathoverflow.net/users/828/"
]
}
|
6,379 |
What is an integrable system , and what is the significance of such systems? (Maybe it is easier to explain what a non-integrable system is.) In particular, is there a dichotomy between "integrable" and "chaotic"? (There is an interesting Wikipedia article , but I don't find it completely satisfying.) Update (Dec 2010): Thanks for the many excellent answers. I came across another quote from Nigel Hitchin: "Integrability of a system of differential equations should manifest
itself through some generally recognizable features: the existence of many conserved quantities the presence of algebraic geometry the ability to give explicit solutions. These guidelines would be interpreted in a very broad sense." (If there are some aspects mentioned by Hitchin not addressed by the current answers, additions are welcome...) Closely related questions: What does it mean for a differential equation "to be integrable"? basic questions on quantum integrable systems
|
This is, of course, a very good question. I should preface with the disclaimer that despite having worked on some aspects of integrability, I do not consider myself an expert. However I have thought about this question on and (mostly) off. I will restrict myself to integrability in classical (i.e., hamiltonian) mechanics, since quantum integrability has to my mind a very different flavour. The standard definition, which you can find in the wikipedia article you linked to, is that of Liouville. Given a Poisson manifold $P$ parametrising the states of a mechanical system, a hamiltonian function $H \in C^\infty(P)$ defines a vector field $\lbrace H,-\rbrace$, whose flows are the classical trajectories of the system. A function $f \in C^\infty(P)$ which Poisson-commutes with $H$ is constant along the classical trajectories and hence is called a conserved quantity . The Jacobi identity for the Poisson bracket says that if $f,g \in C^\infty(P)$ are conserved quantities so is their Poisson bracket $\lbrace f,g\rbrace$. Two conserved quantities are said to be in involution if they Poisson-commute. The system is said to be classically integrable if it admits "as many as possible" independent conserved quantities $f_1,f_2,\dots$ in involution. Independence means that the set of points of $P$ where their derivatives $df_1,df_2,\dots$ are linearly independent is dense. I'm being purposefully vague above. If $P$ is a finite-dimensional and symplectic, hence of even dimension $2n$, then "as many as possible" means $n$. (One can include $H$ among the conserved quantities.) However there are interesting infinite-dimensional examples (e.g., KdV hierarchy and its cousins) where $P$ is only Poisson and "as many as possible" means in practice an infinite number of conserved quantities. Also it is not strictly necessary for the conserved quantities to be in involution, but one can allow the Lie subalgebra of $C^\infty(P)$ they span to be solvable but nonabelian. Now the reason that integrability seems to be such a slippery notion is that one can argue that "locally" any reasonable hamiltonian system is integrable in this sense. The hallmark of integrability, according to the practitioners anyway, seems to be coordinate-dependent. I mean this in the sense that $P$ is not usually given abstractly as a manifold, but comes with a given coordinate chart. Integrability then requires the conserved quantities to be written as local expressions (e.g., differential polynomials,...) of the given coordinates.
|
{
"source": [
"https://mathoverflow.net/questions/6379",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1532/"
]
}
|
6,418 |
What is the fastest algorithm that exists to solve a particular NP-Complete problem? For example, a naive implementation of travelling salesman is $O(n!)$, but with dynamic programming it can be done in $O(n^2 2^n)$. Is there any "easier" NP-Complete problem that has a better running time? Note that I'm curious about exact solutions, not approximations.
|
If P is an NP-complete problem, then define P k = instances of P in which the instances have been blown up from size n to size n k by padding them with blanks. Then P k is also NP-complete, but if P takes time exp(p(n)) to solve where p is some polynomial then P k can be solved in time essentially exp(p(n 1/k )) (there's a little more time required to check that the input really does have the right amount of padding but unless the running time is polynomial this is a negligable fraction of the total time). So there is no "easiest" problem: for every problem you name this construction gives another easier but still NP-complete problem. As for non-artificial problems: most hard graph problems like Hamiltonian circuit, that are hard when restricted to planar graphs, can be solved in time exponential in √n or in (√n)(log n) by dynamic programming using a recursive partition by graph separators.
|
{
"source": [
"https://mathoverflow.net/questions/6418",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1646/"
]
}
|
6,551 |
In trying to think of an intuitive answer to a question on adjoints , I realised that I didn't have a nice conceptual understanding of what an adjoint pair actually is. I know the definition (several of them), I've read the nlab page (and any good answers will be added there), I've worked with them , I've found examples of functors with and without adjoints, but I couldn't explain what an adjunction is to a five-year-old, the man on the Clapham omnibus, or even an advanced undergraduate. So how should I intuitively think of adjunctions? For more background: I'm a topologist by trade who's been learning category theory recently (and, for the most part, enjoying it) but haven't truly internalised it yet. I'm fully convinced of the value of adjunctions, but haven't the same intuition into them as I do for, say, the uniqueness of ordinary cohomology.
|
The example I would give a five-year old is the following. Take the category $\mathbb R$ whose objects are the real numbers (or perhaps rational numbers for the five-year old) and a single
morphism $x \to y$ whenever $x \leq y$. Let $\mathbb Z$ be the full subcategory consisting of the integers. The inclusion $i : \mathbb Z \to \mathbb R$ has a right and a left adjoint: the former is the floor function, the other is the ceiling function. I think the five-year old will agree that these are approximations so I would then say that left and right adjoints are just jazzed up versions of approximations.
|
{
"source": [
"https://mathoverflow.net/questions/6551",
"https://mathoverflow.net",
"https://mathoverflow.net/users/45/"
]
}
|
6,554 |
A lot of the terminology of category theory has obvious antecedents in analysis: limits, completeness, adjunctions, continuous (functors), to name but a few. However, analysis and category theory seem to be at opposite poles of the spectrum. Is there anything deep here, or is it a case of "it has wings, so let's call it a duck"? This was partly inspired by the top-rated answer to the question What is a metric space? and by the (slightly unsatisfactory answers to the) question Can adjoint linear transformations be naturally realized as adjoint functors? . In particular, in the first case - the categorical view of metric spaces - there does seem to be an obvious route between the two worlds, do the terminologies correspond there?
|
Names in category theory are often born when someone realizes that a concept in one particular topic can be generalized in a categorical way. The generally-defined concept is then named after the original narrowly-defined one. The case of metric spaces provides a slightly notorious example. As discussed in that other question , metric spaces can be viewed as an example of enriched categories. So, given any concept in metric space theory, you can try to generalize it to the context of enriched categories. This happened with the property of completeness of metric spaces, which one might call Cauchy-completeness since it's about Cauchy sequences. This concept turns out to generalize very smoothly to enriched categories, and to be a useful and important property there. Many people call the property "Cauchy-completeness" in the general context of enriched categories too. But a significant minority disagree with this choice, feeling that it's stretching the terminology too far. For example, when applied to ordinary ( Set -enriched) categories, the property merely says that every idempotent morphism in the category splits. This doesn't "feel" like the completeness condition on metric spaces. So there are other names in currency too, such as "Karoubi complete" (especially popular in the French school). It's true that many pieces of categorical terminology do come from analysis, but maybe all that says is that analysis is an old and venerable subject. Exact is another example. It's used to mean several slightly different things in category theory, confusingly, but the most common usage is that a functor is "left exact" if it preserves finite limits. Now that comes from homological algebra, where one talks about exact sequences; a functor between abelian categories preserves left exact sequences iff it preserves finite limits. And that in turn, I believe, comes from the terminology of differential equations.
|
{
"source": [
"https://mathoverflow.net/questions/6554",
"https://mathoverflow.net",
"https://mathoverflow.net/users/45/"
]
}
|
6,651 |
Gauss's Lemma on irred. polynomial says, Let R be a UFD and F its field of fractions. If a polynomial f(x) in R[x] is reducible in F[x], then it is reducible in R[x]. In particular, an integral coefficient polynomial is irreducible in Z iff it is irreducible in Q. For me this tells me something on how the horizontal divisors in the fibration from the arithmetic plane SpecZ[x] to SpecZ intersects the generic fiber: a prime divisor (the divisor defined by the prime ideal (f(x)) in Z[x]) intersect the generic fiber exactly at one point (i.e. the prime ideal (f(x)) in Q[x]) with multiplicity one. Now here is my question: Give a ring R, with Frac(R)=F, and a polynomial f(x) in R[x] such that f(x) is reducible in F[x], but is irreducible in R[x]. Of course, R should not be a UFD. I'd like to see an example for number fields as well as a geometric example (where R is the affine coordinate ring of an open curve or higher dimensional stuff). Thanks
|
Gauss' Lemma over a domain R is usually taken to be a stronger statement, as follows: If R is a domain with fraction field F, a polynomial f in R[T] is said to be primitive if the ideal generated by its coefficients is not contained in any proper principal ideal. One says that Gauss' Lemma holds in R if the product of two primitive polynomials is primitive. (This implies that a polynomial which is irreducible over R[T] remains irreducible over F[T].) Say that a domain is a GL-domain if Gauss' Lemma holds. It is known that this property is intermediate between being a GCD-domain and having irreducible elements be prime (which I call a EL-domain; this is not standard). Here is a relevant MathSciNet review: MR0371887 (51 #8104)
Arnold, Jimmy T.; Sheldon, Philip B.
Integral domains that satisfy Gauss's lemma.
Michigan Math. J. 22 (1975), 39--51. Let $D$ be an integral domain with identity. For a polynomial $f(x)\in D[X]$ , the content of $f(X)$ , denoted by $A_f$ , is the ideal of $D$ generated by the coefficients of $f(X)$ . The polynomial $f(x)$ is primitive if no nonunit of $D$ divides each coefficient of $f(X)$ (or equivalently, if $D$ is the $v$ -ideal associated with $A_f$ ). On the other hand, $f(X)$ is superprimitive if $A_f{}^{-1}=D$ . The authors study, among other things, the relation between the following four properties on an integral domain: (1) each pair of elements has a greatest common divisor; (2) each primitive polynomial is superprimitive; (3) the product of two primitive polynomials is primitive; (4) each irreducible element is prime. In an integral domain $D$ , the implications (1) $\Rightarrow$ (2) $\Rightarrow$ (3) $\Rightarrow$ (4) hold, while no reverse implication holds in general. On the other hand, the properties (2), (3) and (4) are equivalent in $D[X]$ . On the other hand, when R is Noetherian, all of these conditions are equivalent, and equivalent to being a UFD: see, e.g., Theorem 17 of http://alpha.math.uga.edu/~pete/factorization.pdf Thus a Noetherian domain satisfies Gauss' Lemma iff it is a UFD. In particular, such rings must be integrally closed, but this condition is not sufficient: e.g. take the ring of integers of any number field which is not of class number one (for instance Z[\sqrt{-6}]).
|
{
"source": [
"https://mathoverflow.net/questions/6651",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1657/"
]
}
|
6,675 |
I just realized that I am a barbarian when it comes to writing. But I am not entirely sure, so this might be the right place to ask. When typing display-mode formulae do you guys add a period after the formula ends a sentence? Like: This is the formula for a circle $$x^2 + y^2 = r^2.$$ Therefore blabla... or This is the formula for a circle $$x^2 + y^2 = r^2$$ Therefore blabla... My supervisor has been complaining a lot that I don't use period and commas in my display-mode formulae. But I get uneasy doing that because it doesn't feel natural to me, I took a look at two books at random and both of them so far do the punctuation in their display formulae.. I know this is stupid of me and its amazing I have never noticed that. Edit: This would be a fantastic opportunity to see what people actually like as opposed to what they think they like. Everyone who has an opinion on what the punctuation should be should provide an illustrative example of such so that by the voting it can be seen what is actually preferred. If you do this, make your answer just the example (so provide any general homilies in another answer) so that the voting truly reflects the community view of the example.
|
My meta-guide with respect to that is Tautology 2.3.1 — A mathematical text is, before everything else, a text. from Michèle Audin's Conseils aux auteurs de textes mathématiques , which you can get from her webpage. A corollary is that when one writes a mathematical text one is writing sentences , to which all rules which apply to sentences of course apply. And, say, sentences end in a period.
|
{
"source": [
"https://mathoverflow.net/questions/6675",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1245/"
]
}
|
6,701 |
Is there a canonical definition of the concept of inner products for vector spaces over arbitrary fields, i.e. other fields than $\mathbb R$ or $\mathbb C$?
|
As Qiaochu wrote, the answer to the question is "not really, but..." Let me amplify on the "but" part. Positive-definiteness inherently requires an ordering on your field. Conversely, if you have an ordered field, then the theory of inner products goes through verbatim. (The ordering does not have to be Archimedean, so this indeed gives lots more examples.) Let's assume that you have a field K of characteristic different from 2 which cannot be ordered: by a theorem of Artin-Schreier, this is equivalent to -1 being a sum of squares in the field. Then you don't have "positive definiteness". What is more, the "standard inner product" $$q(x_1,...,x_n) = x_1^2 + ... + x_n^2$$ will be isotropic for sufficiently large $n$ , i.e., there will exist nonzero vectors $v = (x_1,...,x_n)$ such that $q(v) = 0$ . For instance, if K is finite, this occurs as soon as $n \ge 3$ . Let me remark that "isotropic inner products" are not inherently worthless. I have a preliminary version of a wonderful book, "Linear Algebra Methods in Combinatorics" by Laszlo Babai, which indeed makes nice use of the above inner product over finite fields, even in characteristic 2. (See http://www.cs.uchicago.edu/research/publications/combinatorics . Unfortunately it seems that the book never came to fruition. I got my copy more than 10 years ago when I took an undergraduate course in combinatorics from Babai.) On the other hand, to any quadratic form over a field K of characteristic not 2, you can associate a symmetric bilinear form. See (among infinitely other references) p. 2 of http://alpha.math.uga.edu/~pete/quadraticforms.pdf As above, it is plausible that an algebraic substitute for "inner product space" is "vector space endowed with an anisotropic quadratic form", i.e., a regular quadratic form without nonzero vectors v for which $q(v) = 0$ . Witt discovered that you can do a lot of "geometry" in this case: especially, he defined reflection through the hyperplane determined by any anisotropic vector: see (e.g....) pp. 17-18 of the above reference. More is true here than is included in my introductory notes: for instance the orthogonal group of an anistropic quadratic form has the "compactness properties" of the standard real orthogonal group O(n) (that is, it contains no nontrivial split subtorus).
|
{
"source": [
"https://mathoverflow.net/questions/6701",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2058/"
]
}
|
6,704 |
There are a few questions about CM rings and depth. Why would one consider the concept of depth? Is there any geometric meaning associated to that? The consideration of regular sequence is okay to me. (currently I'm regarding it as a generalization of not-a-zero-divisor that's needed to carry out induction argument, e.g. as in $\operatorname{dim} \frac{M}{(a_1,\dotsc,a_n)M} = \operatorname{dim} M - n$ for $M$ -regular sequence $a_1,\dotsc,a_n$ ; correct me if I'm wrong!) But I don't understand why the length of a maximal regular sequence is of interest. Is it merely due to some technical consideration in cohomology that we want many $\operatorname{Ext}$ groups to vanish? What does CM rings mean geometrically? As I read from Eisenbud's book, there doesn't seem to be an exact geometric concept that corresponds to it. Nonetheless I would still like to know about any geometric intuition of CM rings. I know that it should be locally equidimensional. Some examples of CM rings come from complete intersection (I read this from wiki). But what else? Why do we care about CM rings? If I understand it correctly, CM rings ⇔ unmixedness theorem holds for every ideal for a noetherian ring, which should mean every closed subschemes have equidimensional irreducible components (and there's no embedded components). This looks quite restrictive.
|
"Life is really worth living in a Noetherian ring $R$ when all the local rings have the property that every s.o.p. is an R-sequence. Such a ring is called Cohen–Macaulay (C–M for short).": Hochster, "Some applications of the Frobenius in characteristic 0", 1978 . Section 3 of that paper is devoted to explaining what it "really means" to be Cohen–Macaulay. It begins with a long subsection on invariant theory, but then gets to some algebraic geometry that will interest you. In particular, he points out that if $R$ is a standard graded algebra over a field, then it is a module-finite algebra over a polynomial subring $S$ , and that $R$ is Cohen–Macaulay if and only if it is free as an $S$ -module. Equivalently, the scheme-theoretic fibers of the finite morphism $\operatorname{Spec} R \to \operatorname{Spec} S$ all have the same length. At the end of section 3, Hochster explains that the CM condition is exactly what is required to make intersection multiplicity "work correctly": If $X$ and $Y$ are CM, then you can compute the intersection multiplicity of $X$ and $Y$ without all those higher $\operatorname{Tor}$ s that Serre had to add to the definition. He gives lots of examples and explains "where Cohen–Macaulayness comes from" (or doesn't) in each one. The whole thing is eminently readable and highly recommended.
|
{
"source": [
"https://mathoverflow.net/questions/6704",
"https://mathoverflow.net",
"https://mathoverflow.net/users/-1/"
]
}
|
6,711 |
Is there a (preferably simple) example of a function $f:(a,b)\to \mathbb{R}$ which is everywhere differentiable, such that $f'$ is not Riemann integrable? I ask for pedagogical reasons. Results in basic real analysis relating a function and its derivative can generally be proved via the mean value theorem or the fundamental theorem of calculus. Proofs via FTC are often simpler to come up with and explain: you just integrate the hypothesis to get the conclusion. But doing this requires $f'$ (or something) to be integrable; textbooks taking such an approach typically stipulate that $f'$ is continuous. Proofs via MVT can avoid such unnecessary assumptions but may require more creativity. So I'd like an example to show that the extra work does actually pay off. Note that derivatives of everywhere differentiable functions cannot be arbitrarily badly behaved. For example, they satisfy the conclusion of the intermediate value theorem.
|
I believe this answers the question: MR0425042 (54 #13000)
Goffman, Casper
A bounded derivative which is not Riemann integrable.
Amer. Math. Monthly 84 (1977), no. 3, 205--206. In 1881 Volterra constructed a bounded derivative on $[0,1]$ which is not Riemann integrable. Since that time, a number of authors have constructed other such examples. These examples are generally relatively complicated and/or involve nonelementary techniques. The present author provides a simple example of such a derivative $f$ and uses only elementary techniques to show that $f$ has the desired properties. The paper is available here :
|
{
"source": [
"https://mathoverflow.net/questions/6711",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1044/"
]
}
|
6,743 |
Several years ago I attended a colloquium talk of an expert in Galois theory. He motivated some of his work on its relation with the inverse Galois problem . During the talk, a guy from the audience asked: " why should I, as a number theorist, should care about the inverse Galois problem? " I must say that as a young graduate student that works on Galois theory, I was amazed or even perhaps shocked from this question. But later, I realized that I should have asked myself this question long ago. Can you pose reasons to convince a mathematician (not just number theorist) of the importance of the inverse Galois problem? Or maybe why it's unimportant if you want to ruin the party ;)
|
For me, it's one of those questions that would not be so interesting if the answer is Yes but which would probably be very interesting if the answer is No. If not all groups are Galois groups over Q, then there is probably some structure that can be regarded as an obstruction, and then this structure would probably be essential to know about. For instance, not all groups are Galois groups over local fields -- they have to be solvable. This is by basic properties of the higher ramification filtration, which is, surprise, essential to know about if you want to understand local fields. So you could say it's an approach to finding deeper structure in the absolute Galois group. Why not just do that directly? The problem with directly looking for structure is that it's not a yes/no question, and so sometimes you lose track of what exactly you're doing (although in new and fertile subjects often you don't). So the inverse Galois problem has the advantage of being a yes/no question and the advantage that things would be really interesting if the answer is No. Unfortunately, I think the answer is expected to be Yes, though correct me if I'm wrong.
|
{
"source": [
"https://mathoverflow.net/questions/6743",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2042/"
]
}
|
6,749 |
The salamander lemma is a lemma in homological algebra from which a number of theorems quickly drop out, some of the more famous ones include the snake lemma, the five lemma, the sharp 3x3 lemma (generalized nine lemma), etc. However, the only proof I've ever seen of this lemma is by a diagram chase after reducing to R-mod by using mitchell's embedding theorem. Is there an elementary proof of this lemma by universal properties in an abelian category (I don't know if we can weaken the requirements past an abelian category)? If you haven't heard of the salamander lemma, here's the relevant paper: Bergman - Diagram chasing in double complexes . And here's an article on it by our gracious administrator, Anton Geraschenko: The salamander lemma . Also, small side question, but does anyone know a good place to find some worked-out diagram-theoretic proofs that don't use Mitchell and prove everything by universal property? It's not that I have anything against doing it that way (it's certainly much faster), but I'd be interested to see some proofs done without it, just working from the axioms and universal properties. PLEASE NOTE THE EDIT BELOW EDIT: Jonathan Wise posted an edit to his answer where he provided a great proof for the original question (doesn't use any hint of elements!). I noticed that he's only gotten four votes for the answer, so I figured I'd just bring it to everyone's attention, since I didn't know that he'd even added this answer until yesterday. The problem is that he put his edit notice in the middle of the text without bolding it, so I missed it entirely (presumably, so did most other people).
|
There's a proof of the snake lemma without elements (a non-elementary proof?) on my (old) website. Edit : I added a section about the salamander lemma. Much later edit : As Charles Rezk points out below ( 1 2 ), my proof of the salamander lemma is correct only in a special case. I will correct the proof when I find the tex file. What makes working with elements in an abelian category easier than working with objects is that elements of the target of an epimorphism can be lifted to the source. If your abelian category has enough projectives, then a proof with elements can usually be adapted to one without elements by replacing each element by a surjection from a projective object. If you don't have enough projectives, you can still get by without elements. You have to replace the concept of "element" with "epimorphism from something"; then every "element" can still be lifted by passing to a more refined epimorphism. This is just code for working locally in the topology generated by epimorphisms. (That it is a topology is implied by AB2.) Since there are always enough injective sheaves of abelian groups, this gives an exact embedding of any abelian category in an abelian category with enough injectives (or, if we work with "cotopologies", a category with enough projectives). This permits one to apply the simpler approach outlined above (using projectives) rather than "pro-epimorphisms". Once one has an embedding in a Grothendieck abelian category (the category of sheaves of abelian groups always is one), it is not much further to a proof of Mitchell's embedding theorem anyway.
|
{
"source": [
"https://mathoverflow.net/questions/6749",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1353/"
]
}
|
6,762 |
X is a Noetherian scheme, F is an injective object in the category of quasi-coherent sheaves on X. U is an open subset of X. Why F's restriction on U is still an injective object in the category of quasi-coherent sheaves on U?
|
The restriction-by-zero type arguments can actually be made to work, with some effort and an extra hypothesis. Suppose $X$ is locally Noetherian , $j: U \to X$ the inclusion of an open subscheme. Let $Mod(X)$ and $QCoh(X)$ be the categories of $O_X$-modules, and quasi-coherent $O_X$-modules, respectively. The "some effort" is the following Lemma Lemma If $X$ is locally Noetherian, then the injective objects in $QCoh(X)$ are precisely the injective objects of $Mod(X)$ which are quasi-coherent as sheaves of modules. Pf : Any injective object of $Mod(X)$ which is quasi-coherent must certainly be injective in the smaller category $QCoh(X)$. For the converse, it suffices to show that any injective object $I$ of $QCoh(X)$ injects into some $I'$ which is a quasi-coherent injective object of $Mod(X)$, for then $I$ will be a retract of $I'$ and so injective in $Mod(X)$. This seems tricky, but is proved in Theorem 7.18 of Hartshorne's "Residues and duality". Now, let's prove the result using the Lemma: If $J$ is an injective object in $QCoh(X)$, then the hard direction of the Lemma implies that it is injective in $Mod(X)$. The restriction-by-zero argument applies in this category, allowing us to conclude that $j^* J$ is injective in $Mod(U)$. It's clearly quasi-coherent, so applying the easy direction of the Lemma we see that it is injective in $QCoh(U)$ as desired. [Aside: On a Noetherian scheme, any quasi-coherent sheaf is a union of its coherent subsheaves and one can "extend" coherent sheaves on U to coherent sheaves on X (see e.g., Hartshorne Ex. II.5.15). Using these facts, one should be able to give a more direct argument in the Noetherian case.]
|
{
"source": [
"https://mathoverflow.net/questions/6762",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2008/"
]
}
|
6,789 |
Of course "flatness" is a word that evokes a very particular geometric picture, and it seems to me like there should be a reason why this word is used, but nothing I can find gives me a reason! Is there some geometric property corresponding to "flatness" (of morphisms, modules, whatever) that makes the choice of terminology obvious or at least justifiable?
|
A lot of people will tell you that flatness means "continuously varying fibres" in some sense, and that flatness was invented to have correspondingly nice consequences , which is true. But there is a way to expect this (vague) interpretation a priori from an alternative, equivalent definition: An $A$-module $M$ is flat $\iff$ $I \otimes_A M \to IM$ is an isomorphism for every ideal $I$. I would prefer to present this as the definition of flatness, and present the fact that tensoring with $M$ preserves exact sequences as a theorem. Why? Thinking "geometrically", $I$ just corresponds (uniquely) to a closed subscheme $Z=Z(I)=$ $=Spec(A/I)\subseteq Spec(A)$. If we think of $M$ in the usual geometric way as a module of generalized functions on $X$ (like sections of a bundle), and $M/IM \simeq M\otimes_A A/I$ as its restriction to $Z$, then the above definition of flatness can be interepreted directly to mean that $M$ restricts nicely to closed subschemes $Z$ . More precisely, it says that what we lose in this restriction, the submodule $IM$ of elements which "vanish on $Z$", is easy to understand: it's just formal linear combinations of elements $i\otimes m$, with no surprise relations among them, i.e. the tensor product $I \otimes_A M$. In topology, continuous functions "restrict nicely" to points and closed sets (by taking limits), so you can see, without much experience at all, how this definition corresponds in an intuitive way to continuity. Having this motivation in place, the best thing to do is to check out examples along the lines of Dan Erman's answer to see the analogy with continuity and limits at work.
|
{
"source": [
"https://mathoverflow.net/questions/6789",
"https://mathoverflow.net",
"https://mathoverflow.net/users/382/"
]
}
|
6,820 |
Is there an infinite field $k$ together with a polynomial $f \in k[x]$ such that the associated map $f \colon k \to k$ is not surjective but misses only finitely many elements in $k$ (i.e. only finitely many points $y \in k$ do not lie in the image of $f$)? For finite fields $k$, there are such polynomials $f$. If such a poynomial $f$ exists, then $k$ cannot be algebraically closed; the field $\mathbb{R}$ doesn't work either.
|
Since such a polynomial would have to have degree at least 2, its existence implies that
the set of k-rational points of the affine line over k is thin in the sense of Serre's Topics In Galois Theory . It follows from the results presented in that book that this cannot be the case over any Hilbertian field . This includes finite extensions of Q, finite extensions of F(t) for any field F, and many other fields. What about p-adic fields?
|
{
"source": [
"https://mathoverflow.net/questions/6820",
"https://mathoverflow.net",
"https://mathoverflow.net/users/296/"
]
}
|
6,833 |
What is the difference between connected strongly-connected and complete? My understanding is: connected : you can get to every vertex from every other vertex. strongly connected : every vertex has an edge connecting it to every other vertex. complete : same as strongly connected. Is this correct?
|
Connected is usually associated with undirected graphs (two way edges): there is a path between every two nodes. Strongly connected is usually associated with directed graphs (one way edges): there is a route between every two nodes. Complete graphs are undirected graphs where there is an edge between every pair of nodes.
|
{
"source": [
"https://mathoverflow.net/questions/6833",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1983/"
]
}
|
6,870 |
Consider an elliptic curve $y^2=x^3+ax+b$. It is well known that we can (in the generic case) create an addition on this curve turning it into an abelian group: The group law is characterized by the neutral element being the point at infinity and the fact that $w_1+w_2+w_3=0$ if and only if the three points $w_j$ are the intersections (with multiplicity) of a line and the elliptic curve. Groups can be hard to work with, but in most cases proving that the group is in fact a group is easy. The elliptic curve is an obvious exception. Commutativity is easy, but associativity is hard, at least to this non-algebraist: The proof looks like a big calculation, and the associativity seems like an algebraic accident rather than something that ought to be true. So this is my question, then: Why is the group law on an elliptic curve associative? Is there some good reason for it? Is the group perhaps a subgroup or a quotient of some other group that is easier to understand? Or can it be constructed from other groups in some fashion? I gather that historically, the group law was discovered via the addition law for the Weierstrass $\wp$-function . The addition law is itself not totally obviuos, plus this approach seems limited to the case where the base field is $\mathbb{C}$. In any case, I'll elaborate a bit on this shortly, in a (community wiki) answer.
|
Everything I am writing below is carried out explicitly in Chapter III of Silverman's book on elliptic curves. In the earlier chapters, he defines the Picard group. For any curve over any field, algebraic geometers are interested in an associated group called the Picard group. It is a certain quotient of the free abelian group on points of the curve. It consists of formal sums of points on the curve modulo those formal sums that come from looking at the zeroes and poles of rational functions. It is a very important tool in the study of algebraic curves. The very special thing about elliptic curves, as opposed to other curves, is that they turn out to be in natural set-theoretic bijection with their own Picard groups (or actually, the subgroup $\operatorname{Pic}^0(E)$ ). The bijection is as follows: let $O$ be the point at infinity. Then send a point $P$ on the elliptic curve to the formal sum of points $[P] - [O].$ (It is not obvious that this is a bijection, but the work to prove it is all "pure geometric reasoning" with no computations.) So there is automatically a group law on the points of $E.$ Then it requires no messy formulas to show that under this group law, the sum of three collinear points is $O.$ So for free, you also get that this group law is the same as the one you defined in the question and that the one you defined is associative!
|
{
"source": [
"https://mathoverflow.net/questions/6870",
"https://mathoverflow.net",
"https://mathoverflow.net/users/802/"
]
}
|
6,874 |
In the early 1990's, Gil Kalai introduced me to a very interesting generalization of homology theory called intersection homology, which existed for like 10 years back then I believe. Defined initially by Goresky and MacPherson, this is a version of homology which agrees with ordinary homology on manifolds, but also retains crucial properties like Poincare Duality and Hodge Theory on singular (non-)manifolds. The original definition was combinatorial, but it was later re-interpreted in sheaf-theoretic terms (perverse sheaves?). Back then it certainly looked like an exciting new development. So, I'm curious - where does the field stand today? Is it still thriving, or has it been merged with something else, or just faded away?
|
Intersection homology and cohomology are still around, but as a topic they have just substantially been renamed. They are part of the theory of perverse sheaves, which are widely used in the Langlands program, in algebraic geometry approaches to categorification, and elsewhere in algebraic geometry. To the extent that intersection homology was intended for topology, it has stoked relatively less interest than in algebraic geometry. On the one hand, there has been a trend away from homological algebra in geometric topology. On the other hand, singularities are part of the structure of intersection homology. Singularies are more germane to geometric topology than to algebraic topology in the sense of homotopy theory. Both singularities and homological algebra are major aspects of algebraic geometry.
|
{
"source": [
"https://mathoverflow.net/questions/6874",
"https://mathoverflow.net",
"https://mathoverflow.net/users/25/"
]
}
|
6,889 |
I've been learning about Dedekind zeta functions and some basic L-functions in my introductory algebraic number theory class, and I've been wondering why some functions are called L-functions and others are called zeta functions. I know that the zeta function is a product of L-functions, so it seems like an L-function is somehow a component of a zeta function (at least in the case of Artin L-functions, they correspond to specific representations). Is this the idea behind the distinction between "zeta function" and "L-function"? How do things generalize to other kinds of zeta- and L-functions?
|
Let me say first that a Dedekind zeta function is always a product of Artin L-functions. It is the structure of the Galois closure which is relevant here. Let me give a nice example which is indicative of the general case. Let $p(x) \in \mathbb{Z}[x]$ be an irreducible cubic, and let $\alpha$ be a root of $p$. Then $K=\mathbb{Q}(\alpha)$ has trivial automorphism group, and its Galois closure (say $L/\mathbb{Q}$) is an S3-extension. The group S3 has three irreducible representations: the trivial representation, the "sign representation" $\chi$ which is also one-dimensional, and an irreducible two-dimensional representation which we will call $\rho$. Then we have the relations $\zeta_K(s)=\zeta_{\mathbb{Q}}(s)L(s,\rho)$ and $\zeta_L(s)=\zeta_{\mathbb{Q}}(s)L(s,\chi)L(s,\rho)^2$. The proofs of these facts are part of the formalism of Artin L-functions. Generally, the distinction is really a matter of history. Certain objects were named zeta functions - Hasse-Weil, Dedekind - while Dirichlet chose the letter "L" for the functions he made out of characters. However, one feature is that "zeta" functions tend to have poles, and they often "factor" into L-functions. These vagaries are made more precise in various places, for example Iwaniec-Kowalski Ch. 5 and some survey articles on the "Selberg class" of Dirichlet series.
|
{
"source": [
"https://mathoverflow.net/questions/6889",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1355/"
]
}
|
6,928 |
What papers should we read to start? What basic knowledge do we need to understand the question? What is this area really about? And what are people researching on it?
|
Iwasawa theory has its origins in the following counterintuitive insight of Iwasawa: instead of trying to describe the structure of any particular Galois module, it is often easier to describe every Galois module in an infinite tower of fields at once. The specific example that Iwasawa studied was the $p$-Sylow subgroup of the class group of $K_n = \mathbb{Q}(\zeta_{p^n})$. It's naturally a $\mathbb{Z}_p$-module as well as a $G_n$ = Gal$(K_n/K_1)$-module, but the group ring $\mathbb{Z}_p[G_n]$ isn't very nice; it's not a domain, for instance. If we instead look at the inverse limits of the $p$ parts of the class groups of all the fields $K_n$ at once, as modules over $\mathbb{Z}_p[G_n]$, we get a module over the inverse limit $\varprojlim\mathbb{Z}_p[G_n]$. This ring is much easier to understand; it's a complete 2-dimensional regular local ring that is (non-canonically) isomorphic to a power series ring, and there is a strong structure theorem for modules over this ring. Using this structure theorem, Iwasawa proved many theorems about the class numbers of cyclotomic fields. For a simple example: $p$ divides the class number of one of the fields $K_n$ if and only if it divides the class number of all of the fields $K_n$. There's an even bigger payoff to the theory: a profound connection with special values of $L$-functions. In the function field case, Weil had interpreted the Hasse-Weil $L$-function as computing the characteristic polynomial of Frobenius acting on the Jacobian of a curve. Iwasawa's idea was that the analogue for number fields should be the "characteristic ideal" of the ring $\varprojlim\mathbb{Z}_p[G_n]$ acting on ideal class groups. It turns out this characteristic ideal has a generator that is essentially the same as a $p$-adic $L$-function closely related to the ordinary Dirichlet $L$-functions. This was Iwasawa's "main conjecture" and is now a theorem. It implies the Herbrand-Ribet theorem and essentially every classical result relating cyclotomic fields and zeta values. There have been many generalizations since but it's safe to call an area "Iwasawa theory" if it studies some Galois representation ranging over an infinite tower of fields and connects it to $p$-adic $L$-functions. The most fruitful Galois modules from the point of view of $L$-functions seem to be Bloch and Kato's generalized Selmer groups; the ideal class group can be interpreted as a Selmer group, and so can the classical Selmer group of an abelian variety. There's a lot of current research in this area. To start reading, I recommend Washington's book on cyclotomic fields. Chapter 13 is fun and is a good use of some of the main techniques of Iwasawa theory. You don't need anything but the basic background in chapters 1-4 to read sections 1-4 of Chapter 13, which contain the types of theorem I was referring to in the first two paragraphs of this answer. The explicit computations in the first ten chapters also give the link to $p$-adic L-functions. If you know some algebraic number theory, you should be fine to read this book. I also strongly recommend Greenberg's PCMI notes on the Iwasawa theory of elliptic curves, which you can find here: http://www.math.washington.edu/~greenber/Park.ps If you're comfortable with class field theory, and have read the first few sections of Chapter 13 in Washington, then Coates and Sujatha's recent book, Cyclotomic Fields and Zeta Values , is a pleasure to read.
|
{
"source": [
"https://mathoverflow.net/questions/6928",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2110/"
]
}
|
6,932 |
I've noticed that there seem to be two approaches to learning class field theory. The first is to first learn about local fields and local class field theory, and then prove the basic theorems about global class field theory from the corresponding local facts. This often means that global class field theory is given the idelic formulation, as local fields have already been covered. Alternatively, I'm about to take a course on class field theory (which is the sequel to an undergraduate course on algebraic number theory and basic zeta/L-functions) which dives directly into global class field theory and will follow the original (1920s) formulations (ideal-theoretic) and proofs of the basic results. I'm wondering what are people's opinions of the two different approaches to class field theory. Does it make more sense to start local and go global, or is it a better idea to learn the subject more historically? I asked my professor here at Princeton about it, since I was aware that Harvard's CFT course starts with local, he responded that since what we're really interested in are number fields anyway, it's much more relevant to proceed immediately with global class field theory. Thoughts? EDIT/UPDATE: Based on input from this thread and more experience, here is the approach I've decided to follow: Learn global class field theory using more elementary proofs, following something like Janusz (or another source if you don't like Janusz's style) Learn the cohomology-heavy proofs of local class field theory. I particularly like Milne 's notes for this. Continue and learn the proof of global class field theory using cohomology of ideles. You could just continue in Milne, or try the chapters in Cassels-Frohlich
|
I learned class field theory from the Harvard two-semester algebraic number theory sequence that Davidac897 alluded to, so I can really only speak for the "local first" approach (I don't even know what a good book to follow for doing the other approach would be, although I found this interesting book review which seems relevant to the topic at hand.). This is a tough question to answer, partly because local-first/global-first is not the only pedagogical decision that needs to be made when teaching/learning class field theory, but more importantly because the answer depends upon what you want to get out of the experience of learning class field theory (of course, it also depends upon what you already know). Class field theory is a large subject and it is quite easy to lose the forest for the trees (not that this is necessarily a bad thing; the trees are quite interesting in their own right). Here are a number of different things one might want to get out of a course in class field theory, in no particular order (note that this list is probably a bit biased based on my own experience). (a) a working knowledge of the important results of (global) class field theory and ability to apply them to relevant situations. This is more or less independent of the items below, since one doesn't need to understand the proofs of the results in order to apply them. I second Pete Clark's recommendation of Cox's book /Primes of the form x^2 + ny^2/. Now on to stuff involved in the proofs of class field theory: (b) understanding of the structure and basic properties of local fields and adelic/idelic stuff (not class field theory itself, but material that might be taught in a course covering class field theory if it isn't assumed as a prerequisite). (c) knowledge of the machinery and techniques of group cohomology/Galois cohomology, or of the algebraic techniques used in non-cohomology proofs of class field theory. Most of the "modern" local-first presentations of local class field theory use the language of Galois cohomology. (It's not necessary, though; one can do all the algebra involved without cohomology. The cohomology is helpful in organizing the information involved, but may seem like a bit much of a sledgehammer to people with less background in homological algebra.) (d) understanding of local class field theory and the proofs of the results involved (usually via Galois cohomology of local fields) as done, e.g. in Serre's /Local Fields/. (e) understanding of class formations, that is, the underlying algebraic/axiomatic structure that is common to local and global class field theory. (Read the Wikipedia page on "class formations" for a good overview.) In both cases the main results of class field theory follow more or less from the axioms of class formations; the main thing that makes the results of global class field theory harder to prove than the local version is that in the global case it is substantially harder to prove that the class formation axioms are in fact satisfied. (f) understanding the proofs of the "hard parts" of global class field theory. Depending upon one's approach, these proofs may be analytic or algebraic (historically, the analytic proofs came first, which presumably means they were easier to find). If you go the analytic route, you also get: (g) understanding of L-functions and their connection to class field theory (Chebotarev density and its proof may come in here). This is the point I know the least about, so I won't say anything more. There are a couple more topics I can think of that, though not necessary to a course covering class field theory, might come up (and did in the courses I took): (h) connections with the Brauer group (typically done via Galois cohomology). (i) examples of explicit class field theory: in the local case this would be via Lubin-Tate formal groups, and in the global case with an imaginary quadratic base field this would be via the theory of elliptic curves with complex multiplication (j-invariants and elliptic functions; Cox's book mentioned above is a good reference for this). Obviously, this is a lot, and no one is going to master all these in a first course; although in theory my two-semester sequence covered all this, I feel that the main things I got out of it were (c), (d), (e), (h), and (i). (I already knew (b), I acquired (a) more from doing research related to class field theory before and after taking the course, and (f) and (g) I never really learned that well). A more historically-oriented course of the type you mention would probably cover (a), (f), and (g) better, while bypassing (b-e). Which of these one prefers depends a lot on what sort of mathematics one is interested in. If one's main goal is to be able to use class field theory as in (a), one can just read Cox's book or a similar treatment and skip the local class field theory. Algebraically inclined people will find the cohomology in items (c) and (d) worth learning for its own sake, and they will find it simpler to deal with the local case first. Likewise, people who prefer analytic number theory or the study of L-functions in general will probably prefer the insights they get from going via (g). I'm not sure I'm reaching a conclusion here: I guess what I mean to say is -- I took the "modern" local-first, Galois cohomology route (where by "modern" we actually mean "developed by Artin and Tate in the 50's") and, being definitely the algebraic type, I enjoyed what I learned, but still felt like I didn't have a good grip on the big picture. (Note: I learned the material out of Cassels and Frohlich mostly, but if I had to choose a book for someone interested in taking the local-first route I'd probably suggest Neukirch's /Algebraic Number Theory/ instead.) Other approaches may give a better view of the big picture, but it can be hard to keep an eye on the big picture when going through the gory details of proving everything. (PS, directed at the poster, whom I know personally: David, if you're interested in advice geared towards your specific situation, you should of course feel welcome to contact me directly about it.)
|
{
"source": [
"https://mathoverflow.net/questions/6932",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1355/"
]
}
|
7,018 |
One of my favorite results in algebraic geometry is a classical result of AX (see http://terrytao.wordpress.com/2009/03/07/infinite-fields-finite-fields-and-the-ax-grothendieck-theorem/ ) I'll recall the version of the theorem that I learned in an undergraduate class in model theory. An algebraic map $F: \mathbb{C}^{n} \to \mathbb{C}^{n}$ is injective iff it is bijective. The model theoretic proof of this result is very simple. Without discussing details ( truth in TACF_0 is the same as truth in TACF_p for p big enough )one can replace $\mathbb{C}$ by the algebraic closure of $F_p$ and prove the result there. Although the algebraic proof is also "simple" (Hilbert's Nullstellensatz is what one needs ) I personally think that the model theoretic proof is great by its simplicity and elegance. So my first question is: Are there more examples like AX's theorem in which the model theoretic proof is much more simple that the ones given by other areas. (I should mention that some algebraist I know consider quantifier elimination for TACF non a model theoretic fact, so for them the proof that I refer above is an algebraic proof) My second question: One of the most famous model theoretic applications to algebra and number theory is Hrushovski's proof on Mordell-Lang of the function field Mordell-Lang conjecture. I'd like to know what are the research questions that applied model theorists are currently working on, besides continuation to Hrushovski's work . In particular, I'd like to know if there is any model theorist that work in applications to Iwasawa theory.
|
It's hard for me to think of an area of algebra that applied model theorists haven't touched recently. I have not heard of any logicians working on Iwasawa theory, but it wouldn't surprise me if there are some. Diophantine geometry: here is a survey article by Thomas Scanlon on applications of model theory to geometry, including discussions of Mordell-Lang and the postivie-characteristic Manin-Mumford conjecture. Number fields: Bjorn Poonen has shown that there is a first-order sentence in the language of rings which is true in all finitely-generated fields of characteristic 0 but false in all fields of positive characteristic. It was conjectured by Pop that any two nonisomorphic finitely-generated fields have different first-order theories. Polynomial dynamics: see here for a recent preprint by Scanlon and Alice Medvedev. It turns out that first-order theories of algebraically closed difference fields where the automorphism is "generic" are quite nice. Differential algebra: By some abstract model-theoretic nonsense ("uniqueness of prime models in omega-stable theories"), it follows that any differential field has a "differential closure" (in analogy to algebraic closure) which is unique up to isomorphism over the base field. There are much more advanced applications, e.g. here . Geometric group theory: Zlil Sela has recently shown that any two finitely-generated nonabelian free groups are elementarily equivalent (i.e. they have the same first-order theory). According to the wikipedia article, this work is related to his solution of the isomorphism problem for torsion-free hyperbolic groups, but I don't understand this enough to say whether this counts as an "application" of model theory. Exponential fields: Boris Zilber has suggested a model-theoretic approach to attacking Schanuel's Conjecture. His conjecture that the complex numbers form a "pseudo-exponential field" is actually a strengthening of Schanuel's Conjecture, but the picture that it suggests is appealing. See here for more. This is in addition to the work on Tannakian formalism, valued fields, and motivic integration that have already been mentioned in other answers, and I haven't even gotten to all the work by the model theorists studying o-minimality. This was just a pseudo-random list I've come up with spontaneously, and no offense is meant to the areas of applied model theory that I've left off of here!
|
{
"source": [
"https://mathoverflow.net/questions/7018",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2089/"
]
}
|
7,052 |
What would the slice-ribbon conjecture imply for 4-dimensional topology? I've heard people speak of the slice-ribbon conjecture as an approach to the 4-dimensional smooth Poincare conjecture, and to the classification of homology 3-spheres which bound homology 4-balls. But I've never understood what they were talking about.
|
I think of the ribbon-slice conjecture as a wish that would simplify certain 4D questions. Let me explain this in 3 examples. Given an embedded "ribbon disk" in 4-space (where the Morse function has no local maxima) one can push it up into 3-space and obtain an immersed disk (whose boundary is still the given knot) where the singularities are mild: these are the so called "ribbon singularities", arcs of double points such that on one of the sheets, the arc lies in the interior. (Picture...) One would actually call this immersed disk in 3-space a "ribbon" (that is allowed to cut through itself). It contains the information about the embedded disk in 4-space by pushing one sheet of each ribbon singularity into the forth dimension.
There is a fairly obvious algorithm how to create all such ribbons in 3-space, starting from the unlink and adding bands.
No such simple 3D-picture exists for arbitrary slice disks and one may wish that any slice knot is ribbon. A knot K is slice if and only if there is a ribbon knot R such that the connected sum K # R is ribbon. One may wish that one didn't have to stabilize. Consider the monoid M of oriented knots under connected sum. If -K is the reversed mirror image of K then K # (-K) is ribbon. So it's very tempting to try to turn M into a group (where -K would become the inverse of K) by identifying two knots K' and K if K' # (-K) is ribbon. But the wish doesn't come true: this is not an equivalence relation and if we force it to be one then by 2 we end up with the knot concordance group (where two knots K' and K are identified if K' # (-K) is slice). It is amazing that there are no proposed counter examples to this conjecture, not even for links.
|
{
"source": [
"https://mathoverflow.net/questions/7052",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2051/"
]
}
|
7,074 |
Let R be a local, normal domain of dimension 2. Suppose that R contains a finite field. I am interested in knowing when the class group of R is torsion. In characteristic 0, this is known to be related to R being a rational singularity. Lipman showed that if X is a desingularization of Spec(R), then one has an exact sequence: $0 \to Pic^{0}(X) \to Cl(R) \to H $ Here $Pic^{0}(X)$ is the numerically trivial part of the Picard group of $X$, and $H$ is a finite group. Thus the second one is torsion if and only if the first one is. I do not have much understanding of the first group, unfortunately. Does anyone know an answer or reference to this? Does anyone know an example in positive characteristic such that $Cl(R)$ is not torsion? Thanks a lot.
|
As requested in the comments, here's an example of a local, normal $2$-dimensional domain R in positive characteristic such that $\mathrm{Cl}(R)$ is not torsion: choose an elliptic curve $E \subset \mathbf{P}^2$ over a field $k$ such that $E(k)$ is not torsion, and take R to be the local ring at the origin of the affine cone on $E$ (i.e., $R = k[x,y,z]/(f)_{(x,y,z)}$ where $f$ is a homoegenous cubic defining $E$). This can be done over $k = \overline{\mathbf{F}_p(t)}$. Proof : The normality follows from the fact that R is a hypersurface singularity (hence even Gorenstein) and isolated and $2$-dimensional (hence regular in codim 1). Blowing up at the origin defines a map $f:X \to \mathrm{Spec}(R)$. One can then show the following: $X$ is smooth, and $X$ can be identified with the Zariski localisation along the zero section of the total space of the line bundle $L = \mathcal{O}_{\mathbf{P^2}}(-1)|_E$ (these are general facts about cones). By Lipman's theorem, it suffices to show that $\mathrm{Pic}^0(X)$ contains non-torsion elements. As $X$ is fibered over $E$ with a section, the pullback $\mathrm{Pic}^0(E) \to \mathrm{Pic}^0(X)$ is a direct summand. As $\mathrm{Pic}^0(E) \simeq E(k)$ has non-torsion elements by assumption, so does $\mathrm{Pic}^0(X)$. Also, an additional comment: In general, Lipman's theorem tells you that $\mathrm{Cl}(R)$ is torsion if and only if $\mathrm{Pic}^0(X)$ is torsion. Now $\mathrm{Pic}(X) \simeq \lim_n \mathrm{Pic}(X_n)$ where $X_n$ is the $n$-th order thickening of the exceptional fibre $E$. Because we are blowing up a point, the sheaf of ideals $I$ defining $E$ is ample on $E$. The kernel and cokernel of $\mathrm{Pic}(X_n) \to \mathrm{Pic}(X_{n-1})$ are identified with $H^1(E,I|_E^{\otimes n+1})$ and $H^2(E,I|_E^{\otimes n+1})$. As $I|_E$ is ample, it follows that the system "$\lim_n \mathrm{Pic}(X_n)$" is eventually stable. Thus, $\mathrm{Pic}(X) \simeq \mathrm{Pic}(X_n)$ for $n$ sufficiently big. As $X_n$ is a proper variety, it follows that if we are working over a finite field (resp. an algebraic closure of a finite field), then $\mathrm{Pic}^0(X)$ is finite (resp. ind-finite).
|
{
"source": [
"https://mathoverflow.net/questions/7074",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2083/"
]
}
|
7,091 |
Let K/Q be a field, probably not a finite extension. Is it possible for a polynomial to be irreducible over K but have a root in every completion of K? What about all but finitely many completions? This question is related to the question "Can a non-surjective polynomial map from an infinite field to itself miss only finitely many points?", and should help prove that such a polynomial can not exist for any subfield of the algebraic closure of the rationals. The idea is that we make the candidate polynomial monic and have algebraic integers for coefficients, then take any maximal ideal in the ring of integers of the candidate field and complete it using the ideal - since the polynomial must have a root in the residue field, it will have a root in the completion. I'm wondering if this forces the polynomial to have a root in the original field - hence the question. The same question only for function fields is also interesting, in order to prove the above for subfields of the algebraic closure of F p (t)
|
For a finite extension $K/\mathbf{Q}$, the answer is no. Suppose that $f$ is an irreducible polynomial with coefficients in $K$ and splitting field $L$. If $G$ is the Galois group of $L/K$, then the polynomial $f$ gives rise to a faithful transitive permutation representation $G \rightarrow S_d$, where $d$ is the degree of $f$. If $P$ is a prime in $O_K$ that is unramified in $L$, then $f$ has a root over $O_{K,P}$ if and only if the corresponding Frobenius element $\sigma_P \in S_d$ has a fixed point. On the other hand, a theorem of Jordan says that every transitive subgroup of $S_d$ contains an element with no fixed points. By the Cebotarev density theorem, it follows that $f$ fails to have a point modulo $P$ for a positive density of primes $P$. For an infinite extension $K/\mathbf{Q}$, the answer is (often) yes. Let $K$ be the compositum of all cyclotomic extensions. Then $x^5 - x - 1$ is irreducible over $K$, because its splitting field over $\mathbf{Q}$ is $S_5$. On the other hand, it has a root in every completion (easy exercise). Finally, your claim that "since a the polynomial must have a root in the residue field, it will have a root in the completion" is false. Hensel's lemma comes with hypotheses.
|
{
"source": [
"https://mathoverflow.net/questions/7091",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2024/"
]
}
|
7,112 |
This is an extension of this question about symmetric algebras in positive characteristic. The title is also a bit tongue-in-cheek, as I am sure that there are multiple "correct" answers. Let $\mathfrak g$ be a Lie algebra over $k$ . One can define the universal enveloping algebra $U\mathfrak g$ in terms of the adjunction: $$\text{Hom}_{\rm LieAlg}(\mathfrak g, A) = \text{Hom}_{\rm AsAlg}(U\mathfrak g, A)$$ for any associative algebra $A$ . Then it's easy enough to check that $U\mathfrak g$ is the quotient of the free tensor algebra generated by $\mathfrak g$ by the ideal generated by elements of the form $xy - yx - [x,y]$ . (At least, I'm sure of this when the characteristic is not $2$ . I don't have a good grasp in characteristic $2$ , though, because I've heard that the correct notion of "Lie algebra" is different.) But there's another good algebra, which agrees with $U\mathfrak g$ in characteristic $0$ . Namely, if $\mathfrak g$ is the Lie algebra of some algebraic group $G$ , then I think that the algebra of left-invariant differential operators is some sort of "divided-power" version of $U\mathfrak g$ . So, am I correct that this notions diverge in positive characteristic? If so, does the divided-power algebra have a nice generators-and-relations description? More importantly, which rings are used for what?
|
The notions do indeed diverge in positive characteristic: there is the enveloping algebra, and then (in the case that $\mathfrak g$ is the Lie algebra of an algebraic group G) there is also the hyperalgebra of G, which is the divided-power version you mention. In characteristic 0 these two algebras coincide, but in positive characteristic they differ very much. In particular, the hyperalgebra is not finitely-generated in positive characteristic; see Takeuchi's paper "Generators and Relations for the Hyperalgebras of Reductive Groups" for the reductive case. There is also a good exploration of the hyperalgebra in Haboush's paper "Central Differential Operators of Split Semisimple Groups Over Fields of Positive Characteristic." One can obtain the hyperalgebra as follows. I don't know in what generality the following construction holds, so let's say that $\mathfrak g$ is the Lie algebra of an algebraic group G defined over $\mathbb Z$. Then there is a $\mathbb Z$-form of the enveloping algebra of G (the Kostant $\mathbb Z$-form) formed by taking divided powers, and upon base change this algebra becomes the hyperalgebra. Alternatively, one can take an appropriate Hopf-algebra dual of the ring of functions on G (cf Jantzen's book "Representations of Algebraic Groups"). As for their uses, in positive characteristic the hyperalgebra of G captures the representation theory of G in the way that the enveloping algebra does in characteristic 0, i.e. the finite-dimensional representations of G are exactly the same as the finite-dimensional representations of the hyperalgebra. This fails completely for the enveloping algebra: instead, the enveloping algebra sees only the representations of the first Frobenius kernel of G.
|
{
"source": [
"https://mathoverflow.net/questions/7112",
"https://mathoverflow.net",
"https://mathoverflow.net/users/78/"
]
}
|
7,114 |
As compared to classes of graphs embeddable in other surfaces. Some ways in which they're exceptional: Mac Lane's and Whitney's criteria are algebraic characterizations of planar graphs. (Well, mostly algebraic in the former case.) Before writing this question, I didn't know whether generalizations to graphs embedded in other surfaces existed, but some lucky Google-fu turned up some references -- in particular there seems to be a generalization of Whitney's criterion due to Jack Edmonds for general surfaces, although frustratingly I can't find the paper, and the main reference I found implies that there might be a small problem on the Klein bottle. Anyone know if Edmonds' result is as easy to prove as Whitney's? Kuratowski's classic characterization of planar graphs by forbidden minors. Of course this does generalize to other surfaces, but this result is both incredibly deep and difficult (as opposed to the proof of Kuratowski, which is by no means trivial but is obtainable by a sufficiently dedicated undergraduate -- actually my working it as an exercise is largely what motivated the question) and is in some sense "essentially combinatorial" in that it applies to a wider class of families that aren't inherently topologically defined. In the other direction of difficulty, the four-color theorem. It's apparently not difficult to show (except for the plane) that what turns out to be the tight upper bound on the chromatic number of a graph embeddable on a surface (other than, for whatever reason, the Klein bottle) is, in fact, an upper bound -- the problem is showing tightness! Whereas it's pretty much trivial to show that $K_4$ is planar (to be fair, though, tightness is easy to check for surfaces of small genus -- the problem's in the general case), but the four-color theorem requires inhuman amounts of calculation and very different, essentially ad-hoc methods. I realize that the sphere has genus 0, which makes it unique, and has trivial fundamental group, which ditto, but while I imagine this information is related to the exceptionalness of the plane/sphere, it's not really all that satisfying as an answer. So, why is it that methods that work everywhere else tend to fail on the sphere? Related questions: reasons-for-the-importance-of-planarity-and-colorability
|
(I think that the question of why planar graphs are exceptional is important. It can be asked not only in the context of graphs embeddable on other surfaces. Let me edit and elaborate, also borrowing from the remarks.) Duality: Perhaps duality is the crucial property of planar graphs. There is a theorem asserting that the dual of a graphic matroid M is a graphic matroid if and only if M is the matroid of a planar graph. In this case, the dual of M is the matroid of the dual graph of G. (See this wikipedia article ).
This means that the circuits of a planar graph are in one to one correspondence with cuts of the dual graph. One important manifestation of the uniqueness of planar graphs (which I believe is related to duality) is Kasteleyn's formula for the number of perfect matchings and the connection with counting trees. Robust geometric descriptions: Another conceptual difference is that (3-connected or maximal) planar graphs are graphs of convex 3-dimensional polytopes and thus have extra geometric properties that graphs on surfaces do not share. The geometric definition of planar graphs (unlike various generalizations) is very robust. A graph is planar if it can be drawn in the plane such that the edges do not intersect in their interiors and are represented by Jordan curves; The class of planar graphs is also what we get if we replace "Jordan curves" by "line intervals," or if we replace "no intersection" by "even number of crossings". The Koebe-Andreev-Thurston theorem allows to represent every planar graph by the "touching graph" of nonoverlapping circles. Both (related) representations via convex polytopes and by circle packings, can respect the group of automorphisms of the graph and its dual. Simple inductive constructions. Another exceptional property of the class of planar graphs is that planar graphs can be constructed by simple inductive constructions. (In this respect they are similar to the class of trees, although the inductive constructions are not so simple as for trees.) This fails for most generalizations of planar graphs. A related important property of planar graphs, maps, and triangulations (with labeled vertices) is that they can be enumerated very nicely. This is Tutte theory. (It has deep extensions to surfaces.) It is often the case that results about planar graphs extend to other classes. As I mentioned, Tutte theory extends to triangulations of other surfaces. Another example is the fundamental Lipton-Tarjan separator theorem, which extends to all graphs with a forbidden minor. The study of planar graphs have led to important graph theoretic concepts Another reason (of a different nature) why planar graphs are exceptional is that several important graph-theoretic concepts were disvovered by looking at planar graphs (or special planar graphs.) The notion of vertex coloring of graphs came (to the best of my knowledge) from the four color conjecture about planar graphs. Similarly, Hamiltonian paths and cycles were first studied for planar graphs. Graphs on surfaces and other notions generalizing planarity. Considering the class of all graphs that can be embedded in a given surface is a natural and important extension of planarity. But, in fact, for various questions, graphs embeddable on surfaces may not be the right generalization of planar graphs. David Eppstein mentioned another generalization via the colin de Verdier invariant. This describes a hiearachy of graphs where the next class after planar graphs are "linklessly embeddable graphs". Those are graphs that can be embedded in space without having two disjoint cyles geometrically link. As it turned out this is also a very robust notion and it leads to a beautiful class of graphs. (They all have at most 4v-10 edges where v is the number of vertices; The known case of Hadwiger's conjecture for graphs not having K_6 minor implies that they are all 5 colorable.) Further classes in this hierarchy are still very mysterious. Other extensions of planarity are: 3) (not literally) Graphs not having K_r as a minor; 4;5) (Both very problematic) As Joe mentioned, graphs of d-polytopes, and also graphs obtained from sphere packings in higher dimensions; 6) (not graphs) r-dimensional simplicial complexes that cannot be embedded in twice the dimension, 7) [A notion that I promoted over the years:] graphs (and skeleta) of d-polytope with vanishing second (toric) g-number, and many more. Forbidden minors and coloring. As for the second and third items in the question. About coloring I am not sure if we should consider 4-coloring planar graphs and coloring graphs on other surfaces as very related phenomena. Regarding forbidden minors. Kuratowski's theorem on surfaces is a special case (and also an important step of the proof) of a much more general result (Wagner's conjecture proved by Robertson and Seymour) about any minor-closed class of graphs. This result can be regarded as extending Kuratowski theorem and also (and perhaps more importantly) extending Kruskal and Nash-Williams theorem on trees. Indeed Kuratoski's theorem is related nicely to the more general picture of topological obstruction to embeddibility. If you would like to propose a different (perhaps topological) understanding of the extension of Kuratowski's theorem for surfaces, then maybe you should start by the well-quasi ordering theorem for trees.
|
{
"source": [
"https://mathoverflow.net/questions/7114",
"https://mathoverflow.net",
"https://mathoverflow.net/users/382/"
]
}
|
7,120 |
Kind of an odd question, perhaps, so I apologize in advance if it is inappropriate for this forum. I've never taken a mathematics course since high school, and didn't complete college. However, several years ago I was affected by a serious illness and ended up temporarily disabled. I worked in the music business, and to help pass the time during my convalescence I picked up a book on musical acoustics. That book reintroduced me to calculus with which I'd had a fleeting encounter with during high school, so to understand what I was reading I figured I needed to brush up, so I picked up a copy of Stewart's "Calculus". Eventually I spent more time working through that book than on the original text. I also got a copy of "Differential Equations" by Edwards and Penny after I had learned enough calculus to understand that. I've also been learning linear algebra - MIT's lectures and problem sets have really helped in this area. I'm fascinated with the mathematics of the Fourier transform, particularly its application to music in the form of the DFT and DSP - I've enjoyed the lectures that Stanford has available on the topic immensely. I just picked up a little book called "Introduction To Bessel Functions" by Frank Bowman that I'm looking forward to reading. The difficulty is, I'm 30 years old, and I can tell that I'm a lot slower at this than I would have been if I had studied it at age 18. I'm also starting to feel that I'm getting into material that is going to be very difficult to learn without structure or some kind of instruction - like I've picked all the low-hanging fruit and that I'm at the point of diminishing returns. I am fortunate though, that after a lot of time and some great MDs my illness is mostly under control and I now have to decide what to do with "what comes after." I feel a great deal of regret, though, that I didn't discover that I enjoyed this discipline until it was probably too late to make any difference. I am able, however, to return to college now if I so choose. The questions I'd like opinions on are these: is returning to school at my age for science or mathematics possible? Is it worth it? I've had a lot of difficulty finding any examples of people who have gotten their first degrees in science or mathematics at my age. Do such people exist? Or is this avenue essentially forever closed beyond a certain point? If anyone is familiar with older first-time students in mathematics or science - how do they fare?
|
This is indeed not a typical math overflow question, but never mind that. Of course you can learn mathematics at the age of 30 after having stopped studying it at the age of 18! Examples are abundant -- in almost every math department I've ever been in, there are at least one or two older graduate students that took some years off (after high school, after college or both) and did quite well upon their return. Being older than 18 may not be a bad thing. Many 18 year-olds are neither well-prepared nor well-motivated to study mathematics (or something else) at the university level: a lot of them are there because their parents want them to be, and most of them are there because their parents are paying. It is true that essential skills get rusty after years of disuse -- when I teach "freshman calculus", older students often do not do very well, even if "older" means 21 or 22: they've forgotten too much precalculus mathematics. But you have been learning about calculus, differential equations and linear algebra on your own and enjoying it! You're looking forward to reading a book on Bessel functions!! You're well past the point where older, rusty students have trouble. You can do it, for sure, and it sounds like you want to, so you should. By the way, 30 is not remotely old. I am a few years older and I think better and more quickly now than I did when I was your age.
|
{
"source": [
"https://mathoverflow.net/questions/7120",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2160/"
]
}
|
7,153 |
I'm not sure whether this is non-trivial or not, but do there exist simple examples of an affine scheme $X$ having an open affine subscheme $U$ which is not principal in $X$? By a principal open of $X = \mathrm{Spec} \ A$, I mean anything of the form $D(f) = \{\mathfrak p \in \mathrm{Spec} \ A : f \notin \mathfrak p\}$, where $f$ is an element of $A$.
|
Let X be an elliptic curve with the identity element O removed. Let U=X-P where P is a point of infinite order. Then U is affine by a Riemann-Roch argument. Now suppose that U=D(f). Then on the entire elliptic curve, the divisor of f must be supported at P and O only. This implies that P is a torsion point, a contradiction.
|
{
"source": [
"https://mathoverflow.net/questions/7153",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1107/"
]
}
|
7,155 |
Some famous quotes often give interesting insights into the vision of mathematics that certain mathematicians have. Which ones are you particularly fond of? Standard community wiki rules apply: one quote per post.
|
I heard this one while taking a differential geometry class in Mexico City. I love it. "Groups, as men, will be known by their actions" . -Guillermo Moreno.
|
{
"source": [
"https://mathoverflow.net/questions/7155",
"https://mathoverflow.net",
"https://mathoverflow.net/users/362/"
]
}
|
7,283 |
In a workshop about the geometry of $\mathbb{F}_1$ I attended recently, it came up a question related to a mysterious but "not-so-secret-anymore" seminar about... an hypothetical Topological Langlands Correspondence! I had never heard about this program; I have found this page via Google: http://www.math.jhu.edu/~asalch/toplang/ I only know a bit about the Number-Theoretic Langlands Program, and I still have a hard time trying to understand what is happening in the Geometrical one, so I cannot even start to draw a global picture out of the information dispersed in that site. So, the questions are: What do you know (or what can you infere from the web) about the Topological Langlands Correspondence? Which is the global picture? What are its analogies with the (original) Langlands Program? Is it doable, or just a "little game" for now? What has been proved until now? What implications would it have? (Note: It is somewhat difficult to tag this one, feel free to retag it if you have a better understanding of the subject than I have!)
|
I would cautiously venture that there is not really something we could call a topological Langlands program to outsiders at this point. My objection is to the final word - we don't really know what we're doing. For example, I don't think we even have a conjecture at this point relating representations of something to something else, or have the right idea of L-functions that are supposed to play a role. The "topological Langlands program" banner is more of an idea that the scattered pieces of number-theoretic content we see in stable homotopy theory should be part of a general framework. There is the appearance of groups whose orders are denominators of Bernoulli numbers in stable homotopy groups of spheres, Andrew Salch's calculations at higher chromatic levels that seem to be related to special values of L-functions, the relationship between K(1)-local orientation theory and measures p-adically interpolating Bernoulli numbers (see Mark Behrens' website for some material on this), the surprisingly canonical appearance of realizations of Lubin-Tate formal group laws due to work of Goerss-Hopkins-Miller in homotopy theory, et cetera, et cetera. Perhaps at this point I'd say that we have a "topological Langlands program" program, whose goal is to figure out why on earth all this arithmetic data is entering homotopy theory and what form the overarching structure takes in a manner similar to the Langlands program itself.
|
{
"source": [
"https://mathoverflow.net/questions/7283",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1234/"
]
}
|
7,315 |
So I've been skimming Bowen's 1972 paper "Symbolic Dynamics for Hyperbolic Flows" hoping it would give me some insight into how to build a Markov family for the cat flow (i.e., the Anosov flow obtained by suspension of the cat map with unit height). For the sake of completeness, the cat flow $\phi$ is obtained as follows: i. Consider the cat map $A$ on the 2-torus and identify points $(Ax,z)$ and $(x,z+1)$ to obtain a 3-manifold $M$ ii. Equip $M$ with a suitable metric (e.g., $ds^2 = \lambda_+^{2z}dx_+^2 + \lambda_-^{2z}dx_-^2 + dz^2$, where $x_\pm$ are the expanding and contracting directions of $A$ and $\lambda_\pm$ are the corresponding eigenvalues.) iii. Consider the flow generated by the vector field $(0,1)$ on $M$--that's the cat flow. Unfortunately I'm getting stuck at the first part of Bowen's quasi-constructive proof, which requires finding a suitable set of disks and subsets transverse to the flow. Rather than rehash the particular criteria for a set of disks and subsets used in Bowen's construction, I will relay a simpler but very similar set of criteria, for a proper family (which if it meets some auxiliary criteria is also a Markov family): $\mathcal{T} =$ {$T_1,\dots,T_n$} is called a proper family (of size $\alpha$) iff there are differentiable closed disks $D_j$ transverse to the flow s.t. the $T_j$ are closed $M = \phi_{[-\alpha, 0]}\Gamma(\mathcal{T})$, where $\Gamma(\mathcal{T}) = \cup_j T_j$ $\dim D_j = \dim M - 1$ diam $D_j < \alpha$ $T_j \subset$ int $D_j$ and $T_j = \bar{T_j^*}$ where $T_j^*$ is the relative interior of $T_j$ in $D_j$ for $j \ne k$, at least one of the sets $D_j \cap \phi_{[0,\alpha]}D_k$, $D_k \cap \phi_{[0,\alpha]}D_j$ is empty. I've been stuck on even constructing such disks and subsets (let alone where the subsets are rectangles in the sense of hyperbolic dynamics). Bowen said this sort of thing is easy and proceeded under the assumption that the disks and subsets were already in hand. I haven't found it to be so. The thing that's killing me is 6, otherwise neighborhoods of the Adler-Weiss Markov partition for the cat map would fit the bill and the auxiliary requirements for the proper family to be a Markov family. I've really been stuck in the mud on this one, could use a push.
|
Take Adler-Weiss on $0\times\mathbb T^2$, $1/3\times\mathbb T^2$ and $2/3\times\mathbb T^2$. Take neighborhoods of this tripled Adler-Weiss. Then this collection would satisfy all the properties with $\alpha=1/3$. I am not sure why are you particularly interested in suspension flow, everything is determined by the base Anosov diffeo. Edit: Indeed, this has to be tinkered a bit. Say Adler Weiss has two rectangles with neighborhoods $D_1$ and $D_2$. Then take collection
$0\times D_1$,
$1/3\times D_1$,
$2/3\times D_1$,
$\varepsilon\times D_2$,
$1/3+\varepsilon\times D_2$,
$2/3+\varepsilon\times D_2$. To ensure the second property take $\alpha=1/3+\varepsilon$.
|
{
"source": [
"https://mathoverflow.net/questions/7315",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1847/"
]
}
|
7,318 |
$\newcommand{\bb}{\mathbb}\DeclareMathOperator{\gal}{Gal}$
Before stating my question I should remark that I know almost nothing about etale cohomology - all that I know, I've gleaned from hearing off hand remarks and reading encyclopedia type articles. So I'm looking for an answer that will have some meaning to an etale cohomology naif. I welcome corrections to any evident misconceptions below. Let $E/\bb Q$ be an elliptic curve the rational numbers $\bb Q$: then to $E/\bb Q$, for each prime $\ell$, we can associate a representation $\gal(\bar{\bb Q}/\bb Q) \to GL(2n, \bb Z_\ell)$ coming from the $\ell$-adic Tate module $T_\ell(E/\bb Q)$ of $E/\bb Q$ (that is, the inverse limit of the system of $\ell^k$ torsion points on $E$ as $k\to \infty$). People say that the etale cohomology group $H^1(E/\bb Q, \bb Z_\ell)$ is dual to $T_\ell(E/\bb Q)$ (presumably as a $\bb Z_\ell$ module) and the action of $\gal(\bar{\bb Q}/\bb Q)$ on $H^1(E/\bb Q, \bb Z_\ell)$ is is the same as the action induced by the action of $\gal(\bar{\bb Q}/\bb Q)$ induced on $T_\ell(E/\bb Q)$. Concerning this coincidence, I could imagine two possible situations: (a) When one takes the definition of etale cohomology and carefully unpackages it, one sees that the coincidence described is tautological, present by definition. (b) The definition of etale cohomology (in the case of an elliptic curve variety) and the action of $\gal(\bar{\bb Q}/\bb Q)$ that it carries is conceptually different from that of the dual of the $\ell$-adic Tate module and the action of $\gal(\bar{\bb Q}/\bb Q)$ that it carries. The coincidence is a theorem of some substance. Is the situation closer to (a) or to (b)? Aside from the action $\gal(\bar{\bb Q}/\bb Q)$ on $T_\ell(E/\bb Q)$, are there other instances where one has a similarly "concrete" description of representation of etale cohomology groups of varieties over number fields and the actions of the absolute Galois group on them? Though I haven't seen this stated explicitly, I imagine that one has the analogy [$\gal(\bar{\bb Q}/\bb Q)$ acts on $T_\ell(E/\bb Q)$: $\gal(\bar{\bb Q}/\bb Q)$ acts on $H^1(E/\bb Q; \bb Z_\ell)$]::[$\gal(\bar{\bb Q}/\bb Q)$ acts on $T_\ell(A/K)$: $\gal(\bar{\bb Q}/\bb Q)$ acts on $H^1(A/K; \bb Z_\ell)$] where $A$ is an abelian variety of dimension $n$ and $K$ is a number field: in asking the last question I am looking for something more substantively different and/or more general than this. I've also inferred that if one has a projective curve $C/\bb Q$, then $H^1(C/\bb Q; \bb Z_\ell)$ is the same as $H^1(J/\bb Q; \bb Z_\ell)$ where $J/\bb Q$ is the Jacobian variety of $C$ and which, by my above inference I assume to be dual to $T_\ell(J/\bb Q)$, with the Galois actions passing through functorially. If this is the case, I'm looking for something more general or substantially different from this as well. The underlying question that I have is: where (in concrete terms, not using a reference to etale cohomology as a black box) do Galois representations come from aside from torsion points on abelian varieties? [Edit (12/09/12): A sharper, closely related question is as follows. Let $V/\bb Q$ be a (smooth) projective algebraic variety defined over $\bb Q$, and though it may not be necessary let's take $V/\bb Q$ to have good reduction at $p = 5$. Then $V/\bb Q$ is supposed to have an attached 5-adic Galois representation to it (via etale cohomology) and therefore has an attached (mod 5) Galois representation. If $V$ is an elliptic curve, this Galois representation has a number field $K/\bb Q$ attached to it given by adjoining to $\bb Q$ the coordinates of the 5-torsion points of $V$ under the group law, and one can in fact write down a polynomial over $\bb Q$ with splitting field $K$. The field $K/\bb Q$ is Galois and the representation $\gal(\bar{\bb Q}/\bb Q)\to GL(2, \bb F_5)$ comes from a representation $\gal(K/\bb Q) \to GL(2, \bb F_5)$. (I'm aware of the possibility that knowing $K$ does not suffice to recover the representation.) Now, remove the restriction that $V/\bb Q$ is an elliptic curve, so that $V/\bb Q$ is again an arbitrary smooth projective algebraic variety defined over $\bb Q$. Does the (mod 5) Galois representation attached to $V/\bb Q$ have an associated number field $K/\bb Q$ analogous to the (mod 5) Galois representation attached to an elliptic curve does? If so, where does this number field come from? If $V/\bb Q$ is specified by explicit polynomial equations is it possible to write down a polynomial with splitting field $K/\bb Q$ explicitly? If so, is a detailed computation of this type worked out anywhere? I'm posting a bounty for a good answer to the questions succeeding the "Edit" heading.
|
IMO, the scenario is closer to your (a). I'll sketch an explanation of the duality between $H^1(E,\mathbf{Z}_l)$ and the dual to the Tate module. We have $H^1(E,\mathbf{Z}_l)=\text{Hom}(\pi_1(E),\mathbf{Z}_l)$, where that $\pi_1$ means etale fundamental group with base point the origin $O$ of $E$. Thus the isomorphism we really want is between $\pi_1(E)\otimes\mathbf{Z}_l$ and $T_\ell(E)$. What is $\pi_1(E)$? In the topology world, we'd consider the universal cover $f\colon E'\rightarrow E$ and take $\pi_1(E)$ to be its group of deck transformations. Then $\pi_1(E)$ has an obvious action on $f^{-1}(O)$. If $E$ is the complex manifold $\mathbf{C}/L$ for a lattice $L$, this is just the natural isomorphism $\pi_1(E)\cong L$. But in the algebraic geometry world, there is no universal cover in the category of varieties, so the notion of universal cover is replaced with the projective system $E_i\to E$ of etale covers of $E$. Then $\pi_1(E)$ is the projective limit of the automorphism groups of $E_i$ over $E$. One nice thing about $E$ being an elliptic curve is that any etale cover $E'\rightarrow E$ must also be an elliptic curve (once you choose an origin on it, anyway); if $E\rightarrow E'$ is the dual map then the composition $E\rightarrow E'\rightarrow E$ is multiplication by an integer. So it's sufficient to only consider those covers of $E$ which are just multiplication by an integer. Since it's $\pi_1(E)\otimes\mathbf{Z}_l$ we're interested in, it's enough to consider the isogenies of $E$ given by multiplication by $l^n$. What are the deck transformations of the maps $l^n\colon E\rightarrow E$? Up to an automorphism of $E$, they're simply translations by $l^n$-division points. And now we see the relationship to the Tate module: A compatible system of deck transformations of these covers is the exact same thing as a compatible system of $l^n$-division points. Thus we get the desired isomorphism. Naturally, it's Galois compatible! In the end, we see that torsion points were tucked away in the construction of the etale cohomology groups, so it wasn't exactly a coincidence. Hope this helps. Re the edit: I believe your best bet is to work locally. First of all, you didn't mention which Galois representation you wanted exactly; let's say you want the representation on $H^i$ of your variety for a given $i$. Let's assume this space has dimension $d$. Step 1. For each prime $p$ at which your variety $V$ has good reduction, you can compute the local zeta function of $V/\mathbf{F}_p$ by counting points on $V(\mathbf{F}_{p^n})$ for $n\geq 0$. In this way you can compute the action of the $p^n$th power Frobenius on $H^i(V\otimes\overline{\mathbf{F}}_p,\mathbf{F}_5)$ for various primes $p$. Step 2. Do this enough so that you can gather up information on the statistics of how often the Frobenius at $p$ lands in each conjugacy class in the group $\text{GL}_d(\mathbf{F}_5)$. In this way you could guess the conjugacy class of the image of Galois inside $\text{GL}_d(\mathbf{F}_5)$. Step 3. Now your job is to find a table of number fields $F$ whose splitting field has Galois group equal to the group you found in the previous step. I found a table here: http://hobbes.la.asu.edu/NFDB/ . You already know which primes ramify in $K$ -- these are at worst the primes of bad reduction of $V$ together with 5 -- and you can distinguish your $F$ from the other number fields by the splitting behavior your found in Step 1. Then $K$ is the splitting field of $F$. A caveat: Step 1 may well take you a very long time, because unless your variety has some special structure or symmetry to it, counting points on $V$ is Hard. Another caveat: Step 3 might be impossible if $d$ is large. If $d$ is 2 then perhaps you're ok, because there might be a degree 8 number field $F$ whose splitting field has Galois group $\text{GL}_2(\mathbf{F}_5)$. If $d$ is large you might be out of luck here. You are free not to accept this answer because of the above caveats but I really do think you've asked a hell of a tough question here!
|
{
"source": [
"https://mathoverflow.net/questions/7318",
"https://mathoverflow.net",
"https://mathoverflow.net/users/683/"
]
}
|
7,320 |
We know that presheaves of any category have enough projectives and that sheaves do not, why is this, and how does it effect our thinking? This question was asked(and I found it very helpful) but I was hoping to get a better understanding of why. I was thinking about the following construction(given during a course); given an affine cover, we normally study the quasi-coherent sheaves, but in fact we could study the presheaves in the following sense: Given an affine cover of X, $Ker_2\left(\pi\right)\rightrightarrows^{p_1}_{p_2} U\rightarrow X$ then we can define $X_1:=Cok\left(p_1,p_2\right)$, a presheaf, to obtain refinements in presheaves where we have enough projectives and the quasi-coherent sheaves coincide. Specifically, if $X_1\xrightarrow{\varphi}X$ for a scheme $X$, s.t. $\mathcal{S}\left(\varphi\right)\in Isom$ for $\mathcal{S}(-)$ is the sheaffication functor, then for all affine covers $U_i\xrightarrow{u_i}X$ there exists a refinement $V_{ij}\xrightarrow{u_{ij}}U_i$ which factors through $\varphi$. This hinges on the fact that $V_{ij}$ is representable and thus projective, a result of the fact that we are working with presheaves. In sheaves, we would lose these refinements. Additionally, these presheaves do not depend on the specific topology(at the cost of gluing). In this setting, we lose projectives because we are applying the localization functor which is not exact(only right exact). However, I don't really understand this reason, and would like a more general answer. A related appearance of this loss is in homological algebra. Sheaves do not have enough projectives, so we cannot always get projective resolutions. They do have injective resolutions, and this is related to the use of cohomology of sheaves rather than homology of sheaves. In paticular, in Rotman's Homological Algebra pg 314, he gives a footnote; In The Theory of Sheaves, Swan writes "...if the base space X is not discrete, I know
of no examples of projective sheaves except the zero sheaf." In Bredon, Sheaf Theory :
on locally connected Hausdorff spaces without isolated points, the only
projective sheaf is 0 addressing this situation. In essence, my question is for a
heuristic or geometric explanation of
why we lose projectives when we pass
from presheaves to sheaves. Thanks in advance!
|
This is pretty much Dinakar's answer from a different view point: He says that it is too easy for a sheaf morphism to be an epi, so, since there are so many epis, it is now a stronger requirement that for every epi we find a lift - so strong that is not satisfied most of the times. I just want to call attention to the fact that this problem has nothing to do with module sheaves but is about sheaves of sets - and as such has the following nice interpretation: The condition of being a projective module sheaf can be split in two conditions: That of existence of the lifting map as a morphism of sheaves of sets and that of it being a morphism of module sheaves . In the category of sets the first condition is always satisfied; we have the axiom of choice which says that every epimorphism has a section and composing the morphism from our would-be projective with this section produces a lift - set-theoretically. Then one has to establish that one such lift is a module homomorphism. But in a sheaf category step one can fail. Sheaves (of sets) are objects in the category of sheaves. This category is a topos and can be seen as an intuitionistic set-theoretic universe (in a precise sense: there is a sound and complete topos semantics for intuitionistic logic, see e.g. this book ). Now in an intuitionistic universe of sets, the axiom of choice is not valid in general; there might not be a "set-theoretic" section of the epimorphism!
|
{
"source": [
"https://mathoverflow.net/questions/7320",
"https://mathoverflow.net",
"https://mathoverflow.net/users/348/"
]
}
|
7,330 |
My vote would be Milnor's 7-page paper "On manifolds homeomorphic to the 7-sphere", in Vol. 64 of Annals of Math. For those who have not read it, he explicitly constructs smooth 7-manifolds which are homeomorphic but not diffeomorphic to the standard 7-sphere. What do you think? Note: If you have a contribution, then (by definition) it will be a paper worth reading so please do give a journal reference or hyperlink! Edit: To echo Richard's comment, the emphasis here is really on short papers. However I don't want to give an arbitrary numerical bound, so just use good judgement...
|
A natural choice is Riemann's "On the Number of Primes Less Than a Given Magnitude" at only 8 pages long... https://en.wikipedia.org/wiki/On_the_Number_of_Primes_Less_Than_a_Given_Magnitude
|
{
"source": [
"https://mathoverflow.net/questions/7330",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1464/"
]
}
|
7,389 |
This is community wiki. In each answer, please list one word at the top and below that list as many different meanings of that word in mathematics as you can think of, preferably with links or definitions. ("Adjective" and "adjective noun" count as the same adjective.) People should edit previous answers as appropriate. (This is mostly just for fun, but I'm also curious if there have been successful attempts to rename concepts that involve overused words.) Edit: I may have been slightly unclear about the intent of this question. When I say "overused" I don't mean "used too often," I mean "used in too many different ways." So I'll change the title of the question to reflect this. Different concepts named after the same mathematician, while potentially confusing, are understandable. I mostly had in mind adjectives that get recycled in different disciplines of mathematics. Different uses of the same noun tend to be less confusing, e.g. the example of "space" below. I think it's good to be intentionally vague about what we consider a "space."
|
Regular. To start off: The regular representation of a group $G$ over a field $k$ is the action on $k[G]$ given by group multiplication. A topology is regular if a closed set and a point not in that set can be separated by disjoint open sets. A point $\zeta_0$ on the boundary of a domain in $\mathbb C$ is called regular if there exists a subharmonic barrier function $b(z)$ defined within $D$ near $\zeta$. This may not be the standard definition but Gamelin's complex Analysis defines it as a subharmonic function $\omega(z)$ on $\{|z-\zeta_0|<\delta\}\cap D$ which is negative everywhere, tends to 0 at $\zeta_0$, but $\limsup(\omega(z))<0$ as $z$ tends to any other boundary point of $D$ within distance $\delta$ of $\zeta_0$. I've borrowed/paraphrased the following from the Wikipedia disambiguation page but removed a couple that either are not too relevant to pure math or qualify the "regularity" more. Feel free to put them in too. Regular cardinal , a cardinal number that is equal to its cofinality Regular category Regular element , and regular sequence and regular immersion. Regular code , an algebraic code with a uniform distribution of distances between codewords Regular graph , a graph such that all the degrees of the vertices are equal The regularity lemma , which has nothing to do with regular graphs Regular polygon , and regular polyhedron Regular prime : a prime $p$ that does not divide the class number of the $p$th cyclotomic field $\mathbb Q[\zeta_p]$. Regular surface in algebraic geometry Regularity of an elliptic operator JS Milne's comment: A regular map is a morphism of algebraic varieties. Regular value of a differentiable map Regular ring (Note: this definition can be made noncommutative. A right noetherian ring R is said to be right regular if every finitely generated right R -module has finite global dimension. See Lam's Lectures in Modules and Rings , Section 5G.) (von Neumann) Regular ring Regular language , a language that can be accepted by a finite state machine. Absolutely regular is a synonym for $\beta$-mixing in stochastic processes. Regular matroid , a matroid which is representable over every field. In this sense, all graphs are regular (their cycle matroids are regular), which has nothing to do with regular graphs.
|
{
"source": [
"https://mathoverflow.net/questions/7389",
"https://mathoverflow.net",
"https://mathoverflow.net/users/290/"
]
}
|
7,470 |
There is a nice formula for the area of a triangle on the 2-dimensional sphere;
If the triangle is the intersection of three half spheres, and has angles $\alpha$, $\beta$ and $\gamma$, and we normalize the area of the whole sphere to be $4\pi$ then the area of the triangle is
$$
\alpha + \beta + \gamma - \pi.
$$
The proof is a cute application of inclusion-exclusion of three sets, and involves the fact
that the area we want to calculate appears on both sides of the equation, but with opposite signs. However, when trying to copy the proof to the three dimensional sphere the parity goes the wrong way and you get 0=0. Is there a simple formula for the volume of the intersection of four half-spheres of $S^3$ in terms of the 6 angles between the four bounding hyperplanes?
|
On the volume of a hyperbolic and spherical tetrahedron , by Murakami and Yano. The volume is obtained as a linear combination of dilogarithms and squares of logarithms. The origin of their formula is really interesting: Asymptotics of quantum $6j$ symbols. (These asymptotics have also been studied by many other people: D. Thurston, Roberts , Woodward, Frohman, Kania-Bartoszynska, etc.) Note that the 3-dimensional formula has to be much more complicated. The 2-dimensional formula comes from Euler characteristic and Gauss-Bonnet, but the Euler characteristic of the 3-sphere, or any odd-dimensional manifold, vanishes. In fact every characteristic class of a 3-sphere vanishes, because the tangent bundle is trivial. There can't be a purely linear treatment of volumes in isotropic spaces in odd dimensions. In even dimensions, there is always a purely linear extension from lower dimensions using generalized Gauss-Bonnet.
|
{
"source": [
"https://mathoverflow.net/questions/7470",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2229/"
]
}
|
7,477 |
I can show that if $X$ is a scheme such that all local rings $\mathcal{O}_{X,x}$ are integral and such that the underlying topological space is connected and Noetherian, then $X$ is itself integral. This doesn't seem to work without the "Noetherian" condition. But can anyone think about a nice counterexample to illustrate this? So I am looking for a non-integral scheme - with connected underlying topological space - having integral local rings.
|
Let me try to give a counterexample. (I don't know whether it is 'nice'). First, let us rewrite your properties for an affine scheme $X=Spec(A)$. Connectedness for $A$ means $A$ has no nontrivial idempotents; Integrality for $A$ is the usual one ($A$ is a domain); Local integrality means that whenever $fg=0$ in $A$, every point of $X$ has a neighborhood
where either $f$ or $g$ vanishes. Let us construct a connected locally integral ring that is not integral. Roughly speaking, the construction is as follows: let $X_0$ be the cross (the union of coordinate axes) on the affine plane. Then let $X_1$ be the (reduced) full preimage of $X_0$ on the blow-up of the plane ($X_1$ has three rational components forming a chain). Then blow up the resulting surface at the two singularities of $X_1$, and let $X_2$ be the reduced preimage of $X_1$
(which has five rational components), etc. Take $X$ to be the inverse limit. The only problem with this construction is that blow-ups glue in a projective line, so $X_1$ is not affine. Let us correct this by gluing in an affine line instead (so our scheme will be an open subset in what was described above). Here's an algebraic description: For every $k\ge 0$, let $A_k$ be the following ring: its elements are collections of
polynomials $p_i\in{\mathbb C}[x]$ where $i=0,\dots,2^k$ such that $p_i(1)=p_{i+1}(0)$.
Set $X_k=Spec(A_k)$. $X$ is a union of $2^k+1$ affine lines that meet transversally in a chain. (It may be better to index polynomials by $i/2^k$, but the notation gets confusing.) Define a morphism $A_k\to A_{k+1}$ by
$$(p_0,\dots,p_{2^k})\mapsto(p_0,p_0(1),p_1,p_1(1),\dots,p_{2^k})$$
(every other polynomial is constant). This identifies $A_k$ with a subring of $A_{k+1}$.
Let $A$ be the direct limit of $A_k$ (basically, their union). Set $X=Spec(A)$. For every
$k$, we have a natural embedding $A_k\to A$, that is, a map $X\to X_k$. Each $A_k$ is connected but not integral; this implies that $A$ is connected but not integral. It remains to show that $A$ is locally integral. Take $f,g\in A$ with $fg=0$ and $x\in X$. Let us construct a neighborhood of $x$ on which one of $f$ and $g$ vanishes. Choose $k$ such that $f,g\in A_{k-1}$ (note the $k-1$ index).
Let $y$ be the image of $x$ on $X_k$. It suffices to prove that $y$ has a neighborhood on
which either $f$ or $g$ (viewed as functions on $X_k$) vanishes. If $y$ is a smooth point of $X_k$ (that is, it lies on only one of the $2^k+1$ lines), this is obvious. We can therefore assume that $y$ is one of the $2^k$ singular points, so two components of $X_k$ pass through $y$. However, on one of these two components (the one with odd index), both $f$ and $g$ are constant, since they are pullbacks of functions on $X_{k-1}$. Since $fg=0$ everywhere, either $f$ or $g$ (say, $f$) vanishes on the other component.
This implies that $f$ vanishes on both components, as required.
|
{
"source": [
"https://mathoverflow.net/questions/7477",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1107/"
]
}
|
7,584 |
I suppose this question can be interpreted in two ways. It is often the case that two or more equivalent (but not necessarily semantically equivalent) definitions of the same idea/object are used in practice. Are there examples of equivalent definitions where one is more natural or intuitive? (I'm meaning so greatly more intuitive so as to not be subjective.) Alternatively, what common examples are there in standard lecture courses where a particular symbolic definition obscures the concept being conveyed.
|
Many topics in linear algebra suffer from the issue in the
question. For example: In linear algebra, one often sees the determinant of a
matrix defined by some ungodly formula, often even with
special diagrams and mnemonics given for how to compute it
in the 3x3 case, say. det(A) = some horrible mess of a formula Even relatively sophisticated people will insist that
det(A) is the sum over permutations, etc. with a sign for
the parity, etc. Students trapped in this way of thinking
do not understand the determinant. The right definition is that det(A) is the volume of the
image of the unit cube after applying the transformation
determined by A. From this alone, everything follows. One
sees immediately the importance of det(A)=0, the reason why
elementary operations have the corresponding determinant,
why diagonal and triangular matrices have their
determinants. Even matrix multiplication, if defined by the usual
formula, seems arbitrary and even crazy, without some
background understanding of why the definition is that way. The larger point here is that although the question asked about having a single wrong definition, really the problem is that a limiting perspective can infect one's entire approach to a subject. Theorems,
questions, exercises, examples as well as definitions can be coming
from an incorrect view of a subject! Too often, (undergraduate) linear algebra is taught as a
subject about static objects---matrices sitting there,
having complicated formulas associated with them and
complex procedures carried out with the, often for no
immediately discernible reason. From this perspective, many
matrix rules seem completely arbitrary. The right way to teach and to understand linear algebra is as a fully dynamic
subject. The purpose is to understand transformations of
space. It is exciting! We want to stretch space, skew it,
reflect it, rotate it around. How can we represent these
transformations? If they are linear, then we are led to
consider the action on unit basis vectors, so we are led
naturally to matrices. Multiplying matrices should mean
composing the transformations, and from this one derives
the multiplication rules. All the usual topics in
elementary linear algebra have deep connection with
essentially geometric concepts connected with the
corresponding transformations.
|
{
"source": [
"https://mathoverflow.net/questions/7584",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1521/"
]
}
|
7,586 |
Here $\zeta(s)$ is the usual Riemann zeta function, defined as $\sum_{n=1}^\infty n^{-s}$ for $\Re(s)>1$. Let $A_n=${$s\;:\;\zeta(s)=n$}. The behaviour of $A_0$ is basically just the Riemann hypothesis; my question concerns $A_n$ for $n\neq0$. 1) Is determining this just as hard as the Riemann hypothesis? 2) If we know the behaviour of some $A_n$, does it help in deducing the behaviour of other $A_m$? 3) For which $n$ is $A_n$ non-empty? Question 3 has now been answered for all strictly positive $n$ - it is non-empty, and has points on the real line to the left of $s=1$. For $n=0$, it is known to be non-empty. Any idea for negative $n$? (the same answer won't work, since $\zeta(s)$ is strictly positive on the real line to left of $s=1$.
Big Picard gives it non-empty for all but at most one $n$. How can we remove the 'at most one'?
|
Regarding 3), this "Big Picard" stuff is serious overkill. Think like an undergraduate real analysis student: The p-series $\zeta(p)$ converges for real $p > 1$ , whereas $\zeta(1)$ = sum of the harmonic series = oo. An easy argument using (e.g.) the integral test shows that $$lim_{p \rightarrow \infty} \zeta(p) = 1$$ The function $\zeta(p)$ is continuous in p [the convergence is uniform on right half-planes, hence on compact subsets], so by the intermediate value theorem it takes on every positive integer value $n \ge 2$ at least once -- and, since it is a decreasing function of p, exactly once -- on the real line. Thus $A_n$ is nonempty for all $n > 1$ . EDIT: Let me show that zeta(s) takes on all real values infinitely many times on the negative real axis. For this, note that for all $n > 0$ , $$\zeta(-(2n-1)) = - \frac{B_{2n}}{(2n)}$$ , where $B_{2n}$ is the $(2n)$ th Bernoulli number. It is known that the $B_{2n}$ 's alternate in sign and grow rapidly in absolute value: $$|B_{2n}| \sim 4 \sqrt{\pi n} \left(\frac{n}{(\pi e)^{2n}}\right)$$ The claim follows from this and the Intermediate Value Theorem.
|
{
"source": [
"https://mathoverflow.net/questions/7586",
"https://mathoverflow.net",
"https://mathoverflow.net/users/385/"
]
}
|
7,603 |
From a post to The Jouanolou trick : Are all topologically trivial (contractible) complex algebraic varieties necessarily affine? Are there examples of those not birationally equivalent to an affine space? The examples that come to my mind are similar to a singular $\mathbb P^1$ without a point given by equation $x^2 = y^3$. This particular curve is clearly birationally equivalent to affine line. Perhaps the "affine" part follows from a comparison between Zariski cohomology and complex cohomology?
|
No. Counterexamples were first constructed by Winkelmann, as quotients of $\mathbb A^5$ by algebraic actions of $\mathbb G_{\text{a}}$ . I learned this from Hanspeter Kraft's very nice article available here: Challenging problems on affine $n$ -space . Recently Aravind Asok and Brent Doran have been studying these kinds of examples in the setting of $\mathbb A^1$ -homotopy theory, on the arxiv as On unipotent quotients and some A^1-contractible smooth schemes .
|
{
"source": [
"https://mathoverflow.net/questions/7603",
"https://mathoverflow.net",
"https://mathoverflow.net/users/65/"
]
}
|
7,626 |
This seems like an obvious fact, but I'm not sure what the necessary meaning of "nice" is to get a result like this. I'm wondering if there is a theorem of the form: For any <1> field extension $K/F$, a map from $\phi:K\rightarrow F$ that satisfies <2> is the field norm (or trace). where <1> could be something like finite, algebraic, etc., and <2> could be anything (obviously there would be different <2>'s for norm and trace).
|
The field norm and trace exist when $K$ is a finite algebraic extension of $F$. In this case, an element $\alpha \in K$ can be interpreted as an $F$-linear map on $K$ by multiplication. The field norm is just the determinant of $\alpha$ as a linear map, while the trace is the trace of $\alpha$ as a linear map. This yields an evident generalization: Norm and trace are part of a family of nice maps, namely the coefficients of the characteristic polynomial of $\alpha$. Since Zev asks for a uniqueness theorem in the comments, here is one that shows both the merits and limitations of the characteristic polynomial as an answer. For simplicity let $F$ have characteristic 0. Let $K$ be a field extension of degree $n$ which is generic in the sense that the Galois group is $S_n$. Then any Galois-invariant polynomial in $\alpha \in K$ and its Galois conjugates, is a symmetric polynomial. The theorem is that the algebra of symmetric polynomials is generated by elementary symmetric polynomials, which are exactly the coefficients of the characteristic polynomial of $\alpha$. (This is using the fact that the eigenvalues of $\alpha$ as a map are itself and its Galois conjugates.) In particular, the trace is the unique linear such map up to a scalar; and any multiplicative polynomial of this type is a power of the norm. You can also describe the norm as the last Galois-invariant polynomial (the one of degree $n$) that provides new information. But if the Galois group is smaller, then the ring of invariant polynomials in $\alpha$ and its Galois conjugates is larger, and any of these other invariant polynomials is also "nice". These extras are somewhat hidden by the fact that, for any Galois group, the trace is still the only linear example and the norm is still the only multiplicative example. Well, the original question was open-ended. I think that this answer does fit one interpretation of the question, but maybe it is too standard and maybe there are also other interesting answers.
|
{
"source": [
"https://mathoverflow.net/questions/7626",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1916/"
]
}
|
7,656 |
Defining $$\xi(s) := \pi^{-s/2}\ \Gamma\left(\frac{s}{2}\right)\ \zeta(s)$$ yields $\xi(s) = \xi(1 - s)$ (where $\zeta$ is the Riemann Zeta function). Is there any conceptual explanation - or intuition, even if it cannot be made into a proof - for this? Why of all functions does one have to put the Gamma-function there? Whoever did this first probably had some reason to try out the Gamma-function. What was it? (Best case scenario) Is there some uniform way of producing a factor out of a norm on the rationals which yields the other factors for the p-adic norms and the Gamma factor for the absolute value?
|
To the best of my understanding, the answer is yes, and this uniform way consists of some integration over the local field. This is explained in John Tate's dissertation. One starts with a certain smooth rapidly decreasing function, for which one takes the characteristic function of the p-adic integers in the nonarchimedean case and the function $e^{-|x|^2}$ for an archimedean field. This is being multiplied with $|x|^s$ (approximately) and integrated over the Haar measure of the additive group of the field. This produces the $\Gamma$-factor for an archimedean field and $(1-p^{-s})^{-1}$ for a p-adic field.
|
{
"source": [
"https://mathoverflow.net/questions/7656",
"https://mathoverflow.net",
"https://mathoverflow.net/users/733/"
]
}
|
7,664 |
I am interested in the theory of reductive groups which is useful in the theory of automorphic forms. But the trouble boring me so long time is that I don't know the appropriate material for beginners or outsiders who wants to pave in this field and learn more about automorphic forms.
So my question is that what kind of introduction materials, books or papers, fits me as an introduction and for further reading? Thanks in advance!
|
Sit at a table with the books of Borel, Humphreys, and Springer. Bounce around between them: if a proof in one makes no sense, it may be clearer in the other. For example, Springer's book develops everything needed about root systems from scratch, and has lots of nice exercises relate to that stuff. On the other hand, Borel is better about systematically allowing general ground fields from early on (so one doesn't have to redo the proofs all over again upon discovering that it is a good idea to allow ground fields like $\mathbf{R}$, $\mathbf{Q}$, $\mathbf{F}_ p$, and $\mathbf{F} _p(t)$). Pay attention to the power of inductive arguments with centralizers and normalizers (especially of tori). Unfortunately, none makes good use of schemes, which clarifies and simplifies many things related to tangent space calculations, quotients, and positive characteristic. (For example, the definition of central isogeny in Borel's book looks a bit funny, and if done via schemes becomes more natural, though ultimately equivalent to what Borel does.) So if some proofs feel unnecessarily complicated, it may be due to lack of adequate technique in algebraic geometry. (Everyone has to choose their own poison!) Waterhouse's book has nothing serious to say about reductive groups, but the theory of finite group schemes that he discusses (including Cartier duality and structure in the infinitesimal case) is very helpful for a deeper understanding isogenies between reductive groups in positive characteristic. The exposes in SGA3 on quotients and Grothendieck topologies (etale, fppf, etc.) are helpful a lot too (some of which is also developed in the book "Neron Models"). Galois cohomology is also useful when working with rational points of quotients.
|
{
"source": [
"https://mathoverflow.net/questions/7664",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1930/"
]
}
|
7,689 |
I was in a lecture not long ago given by C. Teleman and at some point he said "Well, since Riemann-Roch is an index problem we can do..." Then right after that he argued in favour of such a sentence. Could anyone tell me what did he mean exactly?. That is to say, in this case what is elliptic operator like, what is the heuristic idea which such a result relies on? ...and a little bit of more details about it. As usual references will be appreciated. ADD: Thanks for the comments below, but I think they do not answer the question of title : Why is RR an Index problem?. Up to this point, what I can see is that two numbers happened to be the same.
|
Here is a sketch of the argument as I learned it in a complex analysis class: For a Riemann surface $X$ and a holomorphic line bundle $L$, we want
$$\text{dim}H^0(X,L)-\text{dim}H^0(X,L\otimes\Lambda^{0,1})=c_1(L)+\frac{1}{2}\chi(X)$$
You have an operator $\overline{\partial}$ (differentiation with respect to $d\overline{z}$ taking $\Gamma(X,L)$ to $\Gamma(X,L\otimes\Lambda^{0,1})$. Then $H^0(X,L)$ is the kernel of $\overline{\partial}$ and $H^0(X,L\otimes\Lambda^{0,1})$ is the kernel of its adjoint, $\overline{\partial}^+$. Now define $\Delta^+=\overline{\partial}\overline{\partial}^+$ and $\Delta^-=\overline{\partial}^+\overline{\partial}$. Their spectra are the same, except for the kernels, and we get
$$\text{Tr}(e^{-t\Delta^+})-\text{Tr}(e^{-t\Delta^-})=\text{dim}(\text{ker}\Delta^+)-\text{dim}(\text{ker}\Delta^-)$$
We also have that the kernel of $\overline{\partial}$ is the kernel of $\Delta^-$, and the kernel of $\overline{\partial}^+$ is the kernel of $\Delta^+$, so it's enough to get your hands on the left-hand side. Then you write those traces as integrals of heat kernels, take the limit as $t\rightarrow 0^+$, and show that the integrals go to $c_1(L)+\frac{1}{2}\chi(X)$. And that's possible because we can interpret Chern classes and Euler characteristics of Riemann surfaces as integrals of curvatures of line bundles. Of course, then there's more work to turn $c_1(L)+\frac{1}{2}\chi(X)$ into it's more familiar form.
|
{
"source": [
"https://mathoverflow.net/questions/7689",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1547/"
]
}
|
7,715 |
I am starting on a Phd program and am supposed to read Colliot Thelene and Sansuc's article
on R-equivalence for tori. I find it very difficult and although I have some knowledge over schemes , I am completely baffled by this scalar restriction business of having a field extension $K/k$ , a torus over $K$ and "restricting" it to $k$.
I would be very gratefull for a reference or even better by some explanation . I found nothing in my standard books (Hartshorne, Qing Liu, Mumford etc) so I hope this question is appropriate for the site. Thank you.
|
As said, the sought after concept is also known as Weil restriction. In a word, it is the algebraic analogue of the process of viewing an $n$ -dimensional complex variety as a $(2n)$ -dimensional real variety. The setup is as follows: let $L/K$ be a finite degree field extension and let $X$ be a scheme over $L$ . Then the Weil restriction $W_{L/K} X$ is the $K$ -scheme representing the following functor on the category of K-algebras: $A\mapsto X(A \otimes_K L)$ . In particular, one has $W_{L/K} X(K) = X(L)$ . By abstract nonsense (Yoneda...), if such a scheme exists it is uniquely determined by the above functor. For existence, some hypotheses are necessary, but I believe that it exists whenever $X$ is reduced of finite type. Now for a more concrete description. Suppose $X = \mathrm{Spec} L[y_1,...,y_n]/J$ is an affine scheme. Let $d = [L:K]$ and $a_1,...,a_d$ be a $K$ -basis of $L$ . Then we make the following "substitution": $$y_i = a_1 x_{i1} + ... + a_d x_{id},$$ thus replacing each $y_i$ by a linear expression in d new variables $x_{ij}$ . Moreover,
suppose $J = \langle g_1,...,g_m \rangle$ ; then we substitute each of the above equations into $g_k(y_1,...,y_n)$ getting a polynomial in the $x$ -variables, however still with $L$ -coefficients. But now using our fixed basis of $L/K$ , we can regard a single polynomial with $L$ -coefficients as a vector of $d$ polynomials with $K$ coefficients. Thus we end up with $md$ generating polynomials in the $x$ -variables, say generating an ideal $I$ in $K[x_{ij}]$ , and we put $\mathrm Res_{L/K} X = \mathrm{Spec} K[x_{ij}]/I$ . A great example to look at is the case $X = G_m$ (multiplicative group) over $L = \mathbb{C}$ (complex numbers) and $K = \mathbb{R}$ . Then $X$ is the spectrum of $$\mathbb{C}[y_1,y_2]/(y_1 y_2 - 1);$$ put $y_i = x_{i1} + \sqrt{-1} x_{i2}$ and do the algebra. You can really see that the
corresponding real affine variety is $\mathbb{R}\left[x,y\right]\left[(x^2+y^2)^{-1}\right]$ , as it should be: see e.g.
p. 2 of http://alpha.math.uga.edu/~pete/SC5-AlgebraicGroups.pdf for the calculations. Note the important general property that for a variety $X/L$ , the dimension of the Weil restriction from $L$ down to $K$ is $[L:K]$ times the dimension of $X/L$ . This is good to keep in mind so as not to confuse it with another possible interpretation of "restriction of scalars", namely composition of the map $X \to \mathrm{Spec} L$ with the map $\mathrm{Spec} L \to Spec K$ to
give a map $X \to \mathrm{Spec} K$ . This is a much weirder functor, which preserves the dimension but screws up things like geometric integrality. (When I first heard about "restriction of
scalars", I guessed it was this latter thing and got very confused.)
|
{
"source": [
"https://mathoverflow.net/questions/7715",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2292/"
]
}
|
7,732 |
Let $M$ be a closed Riemannian manifold.
Assume $\tilde M$ is a connected Riemannian $m$-fold cover of $M$.
Is it true that
$$\mathop{diam}\tilde M\le m\cdot \mathop{diam} M\ ?\ \ \ \ \ \ \ (*)$$ Comments: This is a modification of a problem of A. Nabutovsky. Here is yet related question about universal covers. You can reformulate it for compact length metric space --- no difference. The answer is YES if the cover is regular (but that is not as easy as one might think). The estimate $\mathop{diam}\tilde M\le 2{\cdot}(m-1){\cdot} \mathop{diam} M$ for $m>1$ is trivial. We have equality in $(*)$ for covers of $S^1$ and for some covers of figure-eight.
|
I think I can prove that $diam(\tilde M)\le m\cdot diam(M)$ for any covering. Let $\tilde p,\tilde q\in\tilde M$ and $\tilde\gamma$ be a shortest path from $\tilde p$ to $\tilde q$. Denote by $p,q,\gamma$ their projections to $M$. I want to prove that $L(\gamma)\le m\cdot diam(M)$. Suppose the contrary. Split $\gamma$ into $m$ arcs $a_1,\dots,a_n$ of equal length: $\gamma=a_1a_2\dots a_m$, $L(a_i)=L(\gamma)/m>diam(M)$. Let $b_i$ be a shortest path in $M$ connecting the endpoints of $a_i$. Note that $L(b_i)\le diam(M)< L(a_i)$. I want to replace some of the components $a_i$ of the path $\gamma$ by their "shortcuts" $b_i$ so that the lift of the resulting path starting at $\tilde p$ still ends at $\tilde q$. This will show that $\tilde\gamma$ is not a shortest path from $\tilde p$ to $\tilde q$, a contradiction. To switch from $a_i$ to $b_i$, you left-multiply $\gamma$ by a loop $l_i:=a_1a_2\dots a_{i-1}b_i(a_1a_2\dots a_i)^{-1}$. More precisely, if you replace the arcs $a_{i_1},a_{i_2},\dots,a_{i_k}$, where $i_1< i_2<\dots< i_k$, by their shortcuts, the resulting path is homotopic to the product $l_{i_1}l_{i_2}\dots l_{i_k}\gamma$.
So it suffices to find a product $l_{i_1}l_{i_2}\dots l_{i_k}$ whose lift starting from $\tilde p$ closes up in $\tilde M$. Let $H$ denote the subgroup of $\pi_1(M,p)$ consisting of loops whose lifts starting at $\tilde p$ close up. The index of this subgroup is $m$ since its right cosets are in 1-to-1 correspondence with the pre-images of $p$. While left cosets may be different from right cosets, the number of left cosets is the same $m$. Now consider the following $m+1$ elements of $\pi_1(M,p)$: $s_0=e$, $s_1=l_1$, $s_2=l_1l_2$, $s_3=l_1l_2l_3$, ..., $s_m=l_1l_2\dots l_m$. Two of them, say $s_i$ and $s_j$ where $i< j$, are in the same left coset. Then $s_i^{-1}s_j=l_{i+1}l_{i+2}\dots l_j\in H$ and we are done.
|
{
"source": [
"https://mathoverflow.net/questions/7732",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1441/"
]
}
|
7,782 |
Let $H$ be a group. Can we find an automorphism $\phi :H\rightarrow H$ which is not an inner automorphism, so that given any inclusion of groups $i:H\rightarrow G$ there is an automorphism $\Phi: G\rightarrow G$ that extends $\phi$ , i.e. $\Phi\circ i=i\circ \phi$ ?
|
The answer is that the inner automorphisms are indeed characterized by the property of the existence of extensions to larger groups containing the original group. I learned as much from this blog entry (in Russian). The reference is Schupp, Paul E. , A characterization of inner automorphisms , Proc. Am. Math. Soc. 101, 226-228 (1987). ZBL0627.20018 ..
|
{
"source": [
"https://mathoverflow.net/questions/7782",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2224/"
]
}
|
7,836 |
I have this question coming from an earlier Qiaochu's post . Some answers there, especially David Lehavi's one, were drawing the analogy bundles and varieties versus modules and rings. So I am just wondering, is there any big reason why the study of bundles would give information about varieties? (I suppose that for this matter, I should actually replace varieties by manifolds?) I have heard about some invariants, like the Picard group for complex manifolds. But given my inexperience in these concepts, I don't really know why they should be important. So for those who are thinking about "cooking up invariants", some more detailed explanations on why they are useful (and hopefully some elementary examples!) would be appreciated. Thanks!
|
Well, in algebraic geometry, here's a couple of reasons: 1) Subvarieties: Take a vector bundle, look at a section, where is it zero? Lots of subvarieties show up this way (not all, see this question ) but generally, we can get lots of information out of vector bundles regarding subvarieties. 2) Invariants of spaces: The Picard group of Line bundles and more generally the Grothendieck group/ring is a useful invariant for differentiating spaces and analyzing the geometry indirectly. On smooth spaces, in fact, complexes of vector bundles can be used to replace coherent sheaves entirely (I believe by the Syzygy Theorem). 3) Maps into Projective Space: This one is line bundle specific. Let $V\to\mathbb{P}^n$ be any imbedding, say, then the pullback of $\mathcal{O}(1)$ is a line bundle on $V$. The nice thing is, the global sections of this line bundle determine and are determined by the map (we can get degenerate mappings by taking subspaces, but lets ignore that, and base loci for the moment). It turns out that we can define a line bundle to be ample, a condition just on the bundle, and that suffices to say that a power of it gives a morphism to $\mathbb{P}^n$, so understanding maps into projective space is the same thing as studying ample line bundles on a variety. Hope that helps, there's a lot more, but those are the first three things that came to mind.
|
{
"source": [
"https://mathoverflow.net/questions/7836",
"https://mathoverflow.net",
"https://mathoverflow.net/users/-1/"
]
}
|
7,917 |
Suppose I tried to take Hartshorne chapter II and re-do all of it with non-commutative rings rather than commutative rings. Is this possible? Which parts work in the non-commutative setting and which parts don't? Edit: I also welcome any comments/references regarding any reasonable notions whatsoever of "non-commutative algebraic geometry".
|
I think it is helpful to remember that there are basic differences between the commutative and non-commutative settings, which can't be eliminated just by technical devices. At a basic level, commuting operators on a finite-dimensional vector space can be simultaneously diagonalized [added: technically, I should say upper-triangularized,
but not let me not worry about this distinction here], but this is not true of non-commuting operators. This already
suggests that one can't in any naive way define the spectrum of a non-commutative ring.
(Remember that all rings are morally rings of operators, and that the spectrum of a commutative ring has the same meaning as the [added: simultaneous] spectrum of a collection of commuting operators.) At a higher level, suppose that $M$ and $N$ are finitely generated modules over a
commutative ring $A$ such that $M\otimes_A N = 0$, then $Tor_i^A(M,N) = 0$ for
all $i$. If $A$ is non-commutative, this is no longer true in general. This reflects the fact
that $M$ and $N$ no longer have well-defined supports on some concrete spectrum of $A$.
This is why localization is not possible (at least in any naive sense) in general in the non-commutative setting. It is the same phenomenon as the uncertainty principle in quantum mechanics, and manifests itself in the same way: objects cannot be localized at points in the non-commutative setting. These are genuine complexities that have to be confronted in any study of non-commutative geometry. They are the same ones faced by beginning students when they first discover that in general matrices don't commute. I would say that they are real, fascinating, and difficult, and people have put, and are currently putting, a lot of effort into understanding them. But it is a far cry from just generalizing the statements in Hartshorne.
|
{
"source": [
"https://mathoverflow.net/questions/7917",
"https://mathoverflow.net",
"https://mathoverflow.net/users/83/"
]
}
|
7,921 |
Smoothing theory fails for topological 4-manifolds, in that a smooth structure on a topological 4-manifold $M$ is not equivalent to a vector bundle structure on the tangent microbundle of $M$. Is there an explicit compact counterexample, i.e., are there two compact smooth 4-manifolds which are homeomorphic, have isomorphic tangent bundles, but are not diffeomorphic? (The uncountably many smooth structures on $\mathbb{R}^4$ should give a noncompact counterexample, since $Top(4)/O(4)$ does not have uncountably many components.) Addendum to question, added 12/11/09: I'm also interested in the other type of counterexample, of a nonsmoothable topological 4-manifold whose tangent microbundle does admit a vector bundle structure. Does someone know such an example? Tim Perutz's answer to my first question, below, says that homeomorphic smooth 4-manifolds have isomorphic tangent bundles. If it's not true that all topological 4-manifolds have vector bundle refinements of their tangent microbundle, what is the obstruction in the homotopy of $Top(4)/O(4)$?
|
For a pair of smooth, simply connected, compact, oriented 4-manifolds $X$ and $Y$, Any isomorphism of the intersection lattices $H^2(X)\to H^2(Y)$ comes from an oriented homotopy equivalence $Y\to X$ (Milnor, 1958). Any oriented homotopy equivalence is a tangential homotopy equivalence (Milnor, Hirzebruch-Hopf 1958). Any oriented homotopy equivalence comes from an h-cobordism (Wall 1964). Any oriented homotopy equivalence comes from a homeomorphism (Freedman). It need not be the case that $X$ and $Y$ are diffeomorphic (Donaldson). Many examples are now known: e.g., Fintushel-Stern knot surgery on a K3 surface gives a family of exotic K3's parametrized by the Alexander polynomials of knots. Here's a sketch of why homotopy equivalences preserve tangent bundles: $X$ and $Y$ have three characteristic classes: $w_2$, $p_1$ and $e$. However, $e[X]$ is the Euler characteristic, and $p_1[X]$ three times the signature. By the Wu formula, $w_2$ is the mod 2 reduction of the coset of $2H^2(X)$ in $H^2(X)$ given by the characteristic vectors, hence is determined by the lattice. In trying to construct an isomorphism of tangent bundles over a given homotopy equivalence, the obstructions one encounters are in $H^2(X;\pi_1 SO(4))=H^2(X;Z/2)$ and in $H^4(X;\pi_3 SO(4))=Z\oplus Z$, and these can be matched up with the three characteristic classes.
|
{
"source": [
"https://mathoverflow.net/questions/7921",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2327/"
]
}
|
8,039 |
This question comes out of the answers to Ho Chung Siu's question about vector bundles. Based on my reading, it seems that the definition of the term "section" went through several phases of generality, starting with vector bundles and ending with any right inverse. So admittedly I'm a little confused about which level of generality is the most useful. Some specific questions: Why can we think of sections of a bundle on a space as generalized functions on the space? (I'm being intentionally vague about the kind of bundle and the kind of space.) What's the relationship between sections of a bundle and sections of a sheaf? How should I think about right inverses in general? I essentially only have intuition for the set-theoretic right inverse. Pointers to resources instead of answers would also be great.
|
To your first question, "function on a space" $X$ usually means a morphism from $X$ to one of several "ground spaces" of choice, for example the reals if you work with smooth manifolds, Spec(A) if you work with schemes over a ring, etc. (This is a fairly selective use of the word "function" which used to confuse me.) A section $\gamma$ of a (some-kind-of) bundle $E\to X$ is thought of as a "generalized function" on $X$ by thinking of it as a funcion with "varying codomain", i.e. at each point $x\in X$, it takes value in the fibre $E_x\to x$. If one is talking about locally free / locally trivial bundles, meaning $E$ is locally (over open sets $U\subset X$) isomorphic to some product $U\times T$, then we can locally identify the fibres with $T$. Thus locally a section just looks like a function with codomain $T$, which is often required to be nice. To your second question, I generally take the "right-inverse" or "pre-inverse" definition from category theory, because it relates back to others in the following precise way: Say $\pi: Y\to X$ is a space over $X$ (intentionaly vague). The word "over" is used to activate the tradition of suppressing reference to the map $\pi$ and refering instead to the domain $Y$. For $U\subseteq X$ open, the notation $\Gamma(U,Y)$ denotes sections of the map $\pi$ over $U$ , i.e. maps $U\to Y$ such that the composition $U \to Y\to X$ is the identity (thus necessarily landing back in $U$). It's not hard to see that $\Gamma(-,Y)$ actually forms a sheaf of sets on $X$. Conversely, given any sheaf of sets $F$ on a space $X$, one can form its espace étalé , a topological space over $X$, say $\pi: \acute{E}t(F) \to X$. Then for an open $U\subseteq X$, the elements of $F(U)$ correspond precisely to sections of the map $\pi$, which by the above notation is written $\Gamma(U,\acute{E}t(F)$. That is to say, $F(-)\simeq\Gamma(-,\acute{E}t(F))$ as sheaves on $X$. This explains why people often refer to sheaf elements as "sections" of the sheaf. Moreover, what we now denote by $\acute{E}t(F)$ actually used to be the definition of a sheaf, so people tend to identify the two and write $\Gamma(-,F)$ a instead of $\Gamma(-,\acute{E}t(F))$. This explains the otherwise bizarre tradition of writing $\Gamma(U,F)$ instead of the the more compact notation $F(U)$. $\Big($ Unfortunate linguistic warning: Many people incorrectly use the term "étale space". However, the French word "étalé" means "spread out", whereas "étale" (without the second accent) means "calm", and they were not intended to be used interchangeably in mathematics. This is unfortunate, because the espace étalé has very little to with with étale cohomology. More unfortunate is the annoying coincidence that when dealing with schemes the projection map from the espace étalé happens to be an étale morphism, because it is locally on its domain an isomorphism of schemes, a much stronger condition.$\Big)$ To your third question, I think the observation that $\Gamma(-,Y)$ forms a sheaf on $X$ gives a nice context in which to think of sections $X$ to $Y$: they "live in" the sheaf $\Gamma(-,Y)$ as its globally defined elements.
|
{
"source": [
"https://mathoverflow.net/questions/8039",
"https://mathoverflow.net",
"https://mathoverflow.net/users/290/"
]
}
|
8,042 |
The title pretty much says it all: suppose $R$ is a commutative integral domain such that every countably generated ideal is principal. Must $R$ be a principal ideal domain? More generally: for which pairs of cardinals $\alpha < \beta$ is it the case that: for any commutative domain, if every ideal with a generating set of cardinality at most $\alpha$ is principal, then any ideal with a generating set of cardinality at most $\beta$ is principal? Examples: Yes if $2 \leq \alpha < \beta < \aleph_0$; no if $\beta = \aleph_0$ and $\alpha < \beta$:
take any non-Noetherian Bezout domain (e.g. a non-discrete valuation domain). My guess is that valuation domains in general might be useful to answer the question, although I promise I have not yet worked out an answer on my own.
|
No such ring exists. Suppose otherwise. Let $I$ be a non-principal ideal, generated by a collection of elements $f_\alpha$ indexed by the set of ordinals $\alpha<\gamma$ for some $\gamma$. Consider the set $S$ of ordinals $\beta$ with the property that the ideal generated by $f_\alpha$ with $\alpha<\beta$ is not equal to the ideal generated by $f_\alpha$ with $\alpha\leq \beta$. $I$ is generated by the $f_\beta$ with $\beta \in S$, so if $S$ is finite, then $I$ is finitely generated and thus is principal. On the other hand, if $S$ is infinite, then take a countable subset $T=
\{\beta_1<\beta_2<\dots\}$ of $S$. If the ideal generated by the corresponding set of $f_\beta$'s were principal, its generator would have to be in some $\langle f_{\beta_k}
\mid k\leq i \rangle$ for some $i$ (since any element of $\langle f_{\beta}\mid \beta \in T\rangle$ is a finite combination of $f_\beta$'s and therefore lies in some such ideal).
Now no $\beta_j$ with $j>i$ could be in $T$. ** The same argument shows that all rings for which any countably generated ideal is finitely generated, have all their ideals finitely generated. ** Corrected thanks to David's questions.
|
{
"source": [
"https://mathoverflow.net/questions/8042",
"https://mathoverflow.net",
"https://mathoverflow.net/users/1149/"
]
}
|
8,052 |
I sort of understand the definition of a spectral sequence and am aware that it is an indispensable tool in modern algebraic geometry and topology. But why is this the case, and what can one do with it? In other words, if one were to try to do everything without spectral sequences and only using more elementary arguments, why would it make things more difficult?
|
Let's say you have a resolution $0\to A\to J^0\to J^1\to\dots$ (of a module, a sheaf, etc.) If $J^n$ are acyclic (meaning, have trivial higher cohomology, resp. derived functors $R^nF$), you can use this resolution to compute the cohomologies of $A$ (resp. $R^nF(A)$). If $J^n$ are not acyclic, you get a spectral sequence instead, and that's the best you can do. Let us say you have two functors $F:\mathcal A\to\mathcal B$ and $G:\mathcal B\to \mathcal C$. Let us say you know the derived functors for $F$ and $G$ and would like to compute them for the composition $GF$. Answer: Grothendieck's spectral sequence. 1 and 2 account for the vast majority of applications of spectral sequences, and provide plenty of motivation -- I am sure you will agree. The reason for the spectral sequences in both cases is the same. Intuitively, $A$ in case 1 (resp. $F(A)$ in case 2) is made of parts which are not themselves elementary. Instead, they are made (via an appropriate filtration) from some other elementary, "acyclic" objects. So there is a 2-step process here. You can do the first step and the second step separately but they are not exactly independent of each other. Instead, they are entangled somehow. The spectral sequence gives you a way to deal with this situation.
|
{
"source": [
"https://mathoverflow.net/questions/8052",
"https://mathoverflow.net",
"https://mathoverflow.net/users/344/"
]
}
|
8,056 |
It seems that knowing French is useful if you're an algebraic geometer. More generally, I've sometimes wished I could read German and Russian so I could read papers by great German and Russian mathematicians, but I don't know how useful this would actually be. What non-English languages are good for a generic mathematician to know? Are there specific languages associated to disciplines other than algebraic geometry? (This question is a little English-centric, but I figure it's okay because this website is run in English.)
|
C and LaTeX. Speak them like it's the mother tongue.
|
{
"source": [
"https://mathoverflow.net/questions/8056",
"https://mathoverflow.net",
"https://mathoverflow.net/users/290/"
]
}
|
8,091 |
You and I decide to play a game: To start off with, I provide you with a frictionless, perfectly spherical sphere, along with a frictionless, unstretchable, infinitely thin magical rope. This rope has the magical property that if you ever touch its ends to each other, they will stick together and never come apart for all eternity. You only get one such rope, but you are allowed to specify its length. Next, I close my eyes and plug my ears as you do something to the rope and the sphere. When you are done with whatever you have decided to do, you give me back the sphere and rope. Then I try my best to remove the rope from the sphere (i.e., make the smallest distance from a point on the rope to a point on the sphere at least 1 meter). Of course, since the rope is not stretchable, the total length of the rope cannot increase while I am trying to remove it from the sphere. If I succeed in removing the rope from the sphere, I win. Otherwise, you win. Who has the winning strategy? EDIT: To clarify, Zeb is looking for an answer with a finite length, piecewise smooth rope, and the sphere should be rigid.
|
Without loss of generality we can assume that rope is everywhere tangent to the sphere.
Then some infinitesimal Möbius tranform of its surface will shorten the wrapping length while preserving the crossing pattern (and therefore will loosen the rope without causing it to pass through itself).
Once it is proved, moving in this direction will eventually allow the sphere to escape. Proof. let $u$ be conformal factor.
Since Möbius tranform preservs total area $\oint u^2=1$ .
Thus, $\oint u<1$ .
It follows that for a sutable rotation of $S^2$ , we get length decreasing family of Möbius tranforms. Comments The same proof works for link made out of 3 circles. It is easy capture sphere in a link from 4 circles. (4 strings go around 4 faces of regular tetrahedron on $S^2$ and link at the vertexes as in the answer of Anton Geraschenko, the first picture ) BTW, Can one capture a convex body in a knot?
|
{
"source": [
"https://mathoverflow.net/questions/8091",
"https://mathoverflow.net",
"https://mathoverflow.net/users/2363/"
]
}
|
8,145 |
A stick knot is a just a piecewise linear knot. We could define "stick isotopy" as isotopy that preserves the length of each linear piece. Are there stick knots which are topologically trival, but not trivial via a stick isotopy?
|
Yes, there are. See "Locked and Unlocked Polygonal Chains in 3D", T. Biedl, E. Demaine, M. Demaine, S. Lazard, A. Lubiw, J. O'Rourke, M. Overmars, S. Robbins, I. Streinu, G. Toussaint, S. Whitesides, arXiv:cs.CG/9910009 , figure 6.
|
{
"source": [
"https://mathoverflow.net/questions/8145",
"https://mathoverflow.net",
"https://mathoverflow.net/users/3/"
]
}
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.