User ah - MathOverflowmost recent 30 from http://mathoverflow.net2013-05-18T13:43:14Zhttp://mathoverflow.net/feeds/user/332http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/87402/nice-proofs-of-the-poincarebirkhoffwitt-theoremNice proofs of the Poincaré–Birkhoff–Witt theoremAH2012-02-03T05:41:51Z2013-02-06T16:19:02Z
<p>Let $\mathfrak{g}$ be a finite-dimensional Lie algebra with an ordered basis $x_1 < x_2 < ... < x_n$.</p>
<p>We define the universal enveloping algebra $U(\mathfrak{g})$ of $\mathfrak{g}$ to be the free noncommutative algebra $k\langle x_1,...,x_n\rangle$ modulo the relations $(x_ix_j - x_jx_i = [x_i,x_j])$.</p>
<p>The Poincaré–Birkhoff–Witt theorem states that $U(\mathfrak{g})$ has a basis consisting of lexicographically ordered monomials i.e. monomials of the form $x_1^{e_1}x_2^{e_2}...x_n^{e_n}$. Checking that this basis spans $U(\mathfrak{g})$ is trivial, so the work lies in showing that these monomials are linearly independent.</p>
<p>One standard proof of PBW is to construct a $\mathfrak{g}$-action on the commutative polynomial ring $k[y_1,...,y_n]$ by setting $x_1^{e_1}x_2^{e_2}...x_n^{e_n}\cdot 1 = y_1^{e_1}y_2^{e_2}...y_n^{e_n}$ and verify algebraically that this gives rise to a well-defined representation of $\mathfrak{g}$. Details can be found in Dixmier's book on enveloping algebras.</p>
<p>What other proofs of PBW are there out there?</p>
<p>Are there nice reformulations of the above proof from a different perspective, such as one that emphasizes the universal property of $U(\mathfrak{g})$?</p>
<p>However, I would be especially interested in learning about proofs which are not just repackaged versions of the same algebraic manipulations used in the above proof (for example, geometric proofs which appeal to some property of $U(\mathfrak{g})$ as differential operators, etc.). If we allow ourselves more tools than just plain algebra, what other proofs of PBW can we get?</p>
http://mathoverflow.net/questions/115617/notes-for-botts-1963-lectures-on-morse-theoryNotes for Bott's 1963 lectures on Morse theoryAH2012-12-06T15:23:49Z2012-12-13T12:27:44Z
<p>Would anybody happen to know where I could obtain a scanned version of</p>
<p>Lectures on Morse theory - [revised and expanded version of notes of lectures delivered at Professor R. Bott's topology seminar at Harvard in February and March of 1963], taken by Richard S Palais?</p>
<p>As far as I am aware, these notes were never published.</p>
http://mathoverflow.net/questions/54932/how-were-moduli-spaces-defined-before-functorsHow were moduli spaces defined before functors?AH2011-02-09T21:38:00Z2012-11-17T22:46:46Z
<p>People today in algebraic geometry will typically define a moduli space to be the space which represents the functor of families of whatever object they are interested in studying.</p>
<p>However, I am fairly certain that moduli spaces in algebraic geometry have been around for much longer than functors have. So my questions are:</p>
<p>How did algebraic geometers define moduli spaces before functors? Was there a precise definition they employed? Or was it more informal i.e. "Difficult to define, but we'll know one when we see one"? If there wasn't a precise definition, then how could they employ moduli spaces in actual proofs?</p>
http://mathoverflow.net/questions/1006/motivation-interpretation-for-quillens-q-constructionMotivation/interpretation for Quillen's Q-construction?AH2009-10-18T05:48:07Z2011-12-11T15:20:53Z
<p>This question has been on my mind for a while. As I understand it, the Q-construction was the first definition for higher algebraic K-theory. Some details can be found here.</p>
<p><a href="http://en.wikipedia.org/wiki/Algebraic_K-theory" rel="nofollow">http://en.wikipedia.org/wiki/Algebraic_K-theory</a></p>
<p>I have always wondered what train of thought led Quillen to come up with this definition. Does anyone know an interpretation of the Q-construction that makes it seem natural?</p>
http://mathoverflow.net/questions/74279/example-of-a-measure-preserving-system-with-dense-orbits-that-is-not-ergodicExample of a measure-preserving system with dense orbits that is not ergodicAH2011-09-01T18:35:25Z2011-09-01T19:26:13Z
<p>Let $X$ be a Borel probability space (i.e. equipped with a measure $\mu$ on the Borel $\sigma$-algebra such that $\mu(X) = 1$) with a measure-preserving transformation $T$ such that every point has a dense orbit. Does it follow that the measure-preserving system $(X,\mu,T)$ is ergodic?</p>
<p>I have heard that the answer is no, but I haven't been able to think of any examples.</p>
http://mathoverflow.net/questions/988/for-which-hypersurfaces-in-projective-space-does-the-complement-admit-an-algebraiFor which hypersurfaces in projective space does the complement admit an algebraic group structure?AH2009-10-18T02:56:32Z2011-06-14T00:34:46Z
<p>For example, if $H$ is a hyperplane, then $\mathbb{P}^n - H = \mathbb{A}^n$, which is a vector space.</p>
<p>If $n = m^2 - 1$, then we can regard $\mathbb{A}^{n+1}$ as the space of $m \times m$ matrices and take the hypersurface $H$ in $\mathbb{P}^n$ corresponding to the singular matrices. The complement $\mathbb{P}^n - H$ is $\mathbf{PGL}_n$.</p>
<p>If we restrict ourselves to irreducible $H$, are there any more examples besides the two above?</p>
<p>If we allow reducible hypersurfaces, then we can get a few more. We can realize the multiplicative group $\mathbb{G}_m$ as $\mathbb{P}^1$ minus two points, and removing the union of two distinct lines from $\mathbb{P}^2$ will give us $\mathbb{G}_m \times \mathbb{A}^1$. What can we say about the situation here?</p>
<p>The complement of a hypersurface is affine, so only linear algebraic groups will arise.</p>
<p>I haven't put much thought into the base field, so we can just start with $\mathbb{C}$.</p>
http://mathoverflow.net/questions/5892/what-is-convolution-intuitively/18954#18954Answer by AH for What is convolution intuitively?AH2010-03-21T21:04:04Z2011-06-13T19:15:55Z<p>I want to expand on a special case of Terry's answer which I think is particularly intuitive.</p>
<p>Suppose there is a function $f$ that you want to understand, but perhaps it is not smooth. Convolution gives you a way to construct new, possibly nicer functions which approximate $f$.</p>
<p>If you let $g$ be a bump function centered at the origin, then the convolution $f*g$ is a new function whose value at $x$ is given by averaging the values of $f$ around $x$. What do we mean exactly by "averaging"? Well, you use $g$ as your measure; translate it over so that it is centered at $x$, and then the integral $$f * g(x) = \int_{\mathbb{R}^n} f(y)g(y - x) dy$$ in the convolution corresponds to the $g$-weighted average of the values of $f$ around $x$ (i.e. in the small ball where $g$ doesn't vanish).</p>
<p>The convolution $f*g$ in this case has the advantage that it is much smoother than $f$. Intuitively, this should be not surprising since the value of $f*g(x)$ was gotten by averaging nearby $f$-values of $x$. Furthermore, you can approximate $f$ by smooth(er) things by considering a sequence of convolutions $f*(g_n)$ where $g_n$ is a sequence of bump functions which are more and more concentrated at the origin.</p>
<p>If you think of the second function $g$ in the convolution $f*g$ as a measure, then you can think of convolutions as $g$-weighted averages of $f$.</p>
http://mathoverflow.net/questions/64602/techniques-for-computing-cup-products-in-singular-cohomologyTechniques for computing cup products in singular cohomologyAH2011-05-11T12:45:40Z2011-05-14T19:28:40Z
<p>Suppose that we are given a CW complex X in terms of the cells and the gluing maps. My understanding is that computing the cup product of the singular cohomology ring from this information is a non-trivial task. I know of two basic strategies that one might take:</p>
<p>1) If the X is homotopy equivalent to a closed oriented manifold, then we can translate from cup product into intersection product and the problem becomes easier to visualize.</p>
<p>2) If X is not too complicated, then we can try to find a simple presentation of X as a finite simplicial complex and compute the cup product explicitly for all the cochains.</p>
<p>My question is: what are other techniques/tricks that can be used to find the cup product?</p>
<p>Surely there must be some general approaches beyond the naive ones I mentioned. Feel free to strengthen the hypotheses or consider specific situations, as I don't expect there to be one trick which works for everything.</p>
http://mathoverflow.net/questions/62159/relationship-between-hilbert-schemes-and-deformation-spaces/62165#62165Answer by AH for Relationship between Hilbert schemes and deformation spacesAH2011-04-18T20:06:25Z2011-04-18T20:28:08Z<p>When you say "deformation functor", you have to be careful to specify exactly which functor you are thinking about. There are two relevant deformation functors at play here.</p>
<p>One is the functor $D$ of abstract deformations of $Y$. If $A$ is an Artin local $k$-algebra, then $D(A)$ is the set of isomorphism classes of flat families $Y_A$ over $A$ whose central fiber is isomorphic to $Y$.</p>
<p>The other functor $D'$ is the functor of deformations of $Y$ inside $\mathbb{P}^n_k$. Here, $D'(A)$ is the set of isomorphism classes of flat families $Y_A$ together with an embedding $Y_A \rightarrow \mathbb{P}^n_k \times Spec A$ such that the central fiber is $Y \rightarrow \mathbb{P}^n_k$ (not isomorphic to $Y$, but exactly $Y$, as we are talking about subschemes of $\mathbb{P}^n_k$).</p>
<p>The functor $D$ does not have any direct relationship to the Hilbert scheme, but the functor D' certainly does.</p>
<p>Given any scheme $X$ and a point $x \in X$, the local ring $\mathcal{O}_x$ defines a functor $F: (Art)_k \rightarrow (Sets)$ by setting $F(A) = Hom(Spec A, Spec \mathcal{O}_x)$. We say that $F$ is <em>pro-represented</em> by $\mathcal{O}_x$.</p>
<p>If you take the local ring $\mathcal{O}_{[Y]}$ of the Hilbert scheme (which you probably know to be representable by an honest scheme by a theorem of Grothendieck) at the point $[Y]$, then the functor it pro-represents is none other than $D'$.</p>
<p>Being pro-representable is stronger than having a versal deformation space; it means that the functor has a <em>universal deformation space</em>, which is the formal completion of the point whose local ring is doing the pro-representing.</p>
<p>So the (uni)versal deformation space of $D'$ is the formal completion of the point $[Y]$ in the Hilbert scheme. As for the versal deformation space of $D$, I am not really sure what it looks like, but there is a morphism $D' \rightarrow D$ of deformation functors given by forgetting the embedding of $Y_A \rightarrow \mathbb{P}^n_k \times Spec A$.</p>
<p>In general, I think of a versal deformation space as an infinitesimal object; it is only keeping track of what happens when you deform $Y$ a little bit. The Hilbert scheme, on the other hand, is global; it is keeping track of all subschemes with the same Hilbert polynomial as $Y$, including ones which might be far away from $Y$ (if $Y$ was a point, for instance). Thus, you would expect the versal deformation space of the appropriate functor to map into the Hilbert scheme, rather than the other way around.</p>
http://mathoverflow.net/questions/56025/doing-explicit-computations-with-coordinate-ringsDoing explicit computations with coordinate ringsAH2011-02-20T01:18:38Z2011-02-20T22:39:11Z
<p>Suppose that we are given an integral $k$-algebra $A$ of finite type explicitly, by which I mean that we are given the generators of the defining ideal $J$ where $A = k[x_1,...,x_n]/J$. What kinds of tools are out there to compute the integral closure of $A$? I would like the answer as explicitly as possible i.e. generators of the defining ideal.</p>
<p>While suggestions of computer programs are welcome, I want to be able to do these on my own, so I am looking for results which let me prove the answer. I'm asking from the perspective of someone who knows very little computational commutative algebra.</p>
<p>As a related question, if I have two such rings $A$, $B$ given explicitly as above, together with an explicit homomorphism between them, how can I go about determining the kernel and cokernel explicitly? Also, how about if we localize everything at a maximal ideal?</p>
<p>Broadly speaking, I would like to know about what kinds of computational methods are available for rings which arise from studying complex algebraic varieties. Have people out there settled these kinds of computations completely, or is this a hard question in general?</p>
http://mathoverflow.net/questions/56104/early-two-author-math-papers/56108#56108Answer by AH for Early Two-Author Math PapersAH2011-02-20T21:15:13Z2011-02-20T21:22:10Z<p>The famous paper of Dedekind and Weber:</p>
<p>R. Dedekind, H. Weber: Theorie der algebraischen Funktionen einer Veraendlichen, J.
Reine Angew. Math 92 (1882) 181-290.</p>
<p>is the first place where the points of a Riemann surface are described in terms of ideals of the ring of functions. To put this into context, Dedekind had only invented the notion of ideal a few years earlier. They also give an algebraic proof of the Riemann-Roch theorem.</p>
<p>I think the analogy between function fields and number fields started here.</p>
http://mathoverflow.net/questions/55725/interesting-examples-of-flasque-sheavesInteresting examples of flasque sheaves?AH2011-02-17T12:13:03Z2011-02-18T17:50:10Z
<p>Does anyone know any interesting examples of flasque sheaves? Ideally, I would like to see one that both arises naturally and is geometric in some sense. On the other hand, I know so few examples other than direct products of stalks that I would be happy to see anything new.</p>
http://mathoverflow.net/questions/53988/what-is-the-motivation-for-a-vertex-algebraWhat is the motivation for a vertex algebra?AH2011-02-01T14:52:34Z2011-02-02T14:46:13Z
<p>The mathematical definition of a vertex algebra can be found here:</p>
<p><a href="http://en.wikipedia.org/wiki/Vertex_operator_algebra" rel="nofollow">http://en.wikipedia.org/wiki/Vertex_operator_algebra</a></p>
<p>Historically, this object arose as an axiomatization of "vertex operators" in "conformal field theory" from physics; I don't know what these phrases mean.</p>
<p>To date, I haven't been able to gather together any kind of intuition for a vertex algebra, or even a precise justification as to why anyone should care about them <em>a priori</em> (i.e. not "they come from physics" nor "you can prove moonshine with them").</p>
<p>As far as I am aware, theoretical physics is about finding mathematical models to explain observed physical phenomena. My questions therefore are:</p>
<p>What is the basic physical phenomenon/problem/question that vertex operators model?</p>
<p>What is the subsequent story about vertex operators and conformal field theory, and how can we see that this leads naturally to the axioms of a vertex algebra?</p>
<p>Are there accessible physical examples ("consider two particles colliding in an infinite vacuum...", etc.) that illustrate the key ideas?</p>
<p>Also, are there alternative, purely mathematical interpretations of vertex algebras which make them easier to think about intuitively?</p>
<p>Perhaps people who played a role in their discovery could say a bit about the thinking process that led them to define these objects?</p>
http://mathoverflow.net/questions/53511/geometric-meaning-of-small-extensions/53633#53633Answer by AH for Geometric meaning of small extensions ?AH2011-01-28T16:45:08Z2011-01-28T17:06:15Z<p>If you think of elements in a local Artin $k$-algebra $R$ as, say functions on the origin of $k^n$ which remember some (finite amount of) higher order information in the various $n$ directions, then a small extension $R'$ of $R$ is just another such ring with functions that remember "at most one order higher".</p>
<p>For example, let $R = k[x,y]/((x,y)^2)$ and consider the small extension $R' = k[x,y]/((x,y)^3)$. The elements of $R$ are functions which remember up to 1st order in the directions $x,y$. The elements of $R'$ are what we get if we take functions from $R$ and stick on some 2nd order terms in the $x,y$.</p>
<p>You can also just extend only in one direction. For example, $R'' = k[x,y]/(x^3,xy,y^2)$ is also a small extension of $R$, but the only 2nd order term we added was $x^2$, not $xy$ nor $y^2$.</p>
<p>However, if you take $A = k[x]/(x^2)$ and $A'' = k[x]/(x^4)$, then this is not a small extension because we went two orders up, from 1st order to 3rd order.</p>
<p>Generally, I interpret a small extension as one that only thickens our fat point by an order of at most 1 in each direction.</p>
<p>Things are more complicated if you add a new direction. If you take $R = k[x,y]/((x,y)^2)$ as before, and consider $R[z]/(z^2)$, then this is not a small extension because of the cross terms $xz$ and $yz$. However, if you get rid of them, then $R[z]/(z^2,xz,yz)$ is a small extension of $R$.</p>
<p>You could say that adding a new direction involves making two small extensions. The first one adds the new variable with no cross terms, and the second adds the cross terms. For instance, let $R = k[x]/(x^2)$, $R' = R[y]/(xy,y^2)$, and $R'' = R[y]/(y^2)$.</p>
http://mathoverflow.net/questions/41867/is-a-submodule-of-the-sheaf-of-sections-of-a-smooth-vector-bundle-necessarily-fin/41884#41884Answer by AH for Is a submodule of the sheaf of sections of a smooth vector bundle necessarily finitely generated?AH2010-10-12T08:50:04Z2010-10-12T08:50:04Z<p>One simple counterexample is to let X be the real line, let V be the trivial line bundle, and consider the submodule of \Gamma(V) of smooth sections with compact support. The same story works for any vector bundle on any non-compact manifold.</p>
http://mathoverflow.net/questions/616/what-is-an-example-of-a-presheaf-p-where-p-is-not-a-sheaf-only-a-separated-preWhat is an example of a presheaf P where P^+ is not a sheaf, only a separated presheaf?AH2009-10-15T17:12:18Z2010-07-15T21:55:48Z
<p>There is a standard way to construct the sheafification of a presheaf on a Grothendieck topology which involves matching families. Details may be found here:</p>
<p><a href="http://ncatlab.org/nlab/show/matching+family" rel="nofollow">http://ncatlab.org/nlab/show/matching+family</a></p>
<p>In short, there is a functor + sending presheaves to separated presheaves and then separated presheaves to sheaves. So P^++ is always a sheaf.</p>
<p>Gelfand/Manin's Methods of Homological Algebra has a wrong proof that P^+ is a sheaf, and I have seen in several places a proof that P^++ is a sheaf. However, it seems that for any presheaf P I run into, P^+ is already a sheaf.</p>
<p>Does anyone know an example of a presheaf P where P^+ is not a sheaf i.e. where you actually need to apply the functor + twice to get a sheaf?</p>
http://mathoverflow.net/questions/19418/can-i-become-a-good-mathematician/19439#19439Answer by AH for Can I become a good mathematician?AH2010-03-26T15:59:08Z2010-03-26T15:59:08Z<p><a href="http://en.wikipedia.org/wiki/Raoul_Bott" rel="nofollow">This guy</a> also did electrical engineering (all the way to a PhD) before he decided to go into math. The general consensus is that he ended up being a pretty good mathematician.</p>
http://mathoverflow.net/questions/6874/what-if-anything-happened-to-intersection-homology/6876#6876Answer by AH for What (if anything) happened to Intersection Homology?AH2009-11-26T06:34:41Z2009-11-26T06:34:41Z<p>Intersection homology quickly found applications in representation theory, starting with the Kazhdan-Lusztig conjectures. Today, the theory of perverse sheaves is an important tool in geometric representation theory.</p>
http://mathoverflow.net/questions/2124/good-introductory-references-on-algebraic-stacks/2189#2189Answer by AH for Good introductory references on algebraic stacks?AH2009-10-23T21:09:05Z2009-10-23T21:09:05Z<p>I am not sure if the book I am about to suggest is the half-finished text you are hinting at, but there is a book in progress by Kai Behrend, Brian Conrad, Dan Edidin, William Fulton, Barbara Fantechi, Lothar Göttsche and Andrew Kresch. You can find a link to it here:</p>
<p><a href="http://www.math.uzh.ch/index.php?pr%5Fvo%5Fdet&key1=1287&key2=580&no%5Fcache=1" rel="nofollow">Book</a></p>
<p>It is the most complete reference on algebraic stacks in English that I am aware of. It also has the advantage of being addressed to the beginner.</p>
<p>I think that beyond the basic things, anything deeper you learn about stacks typically involves specific stacks, with certain applications or questions in mind.</p>
http://mathoverflow.net/questions/358/examples-of-great-mathematical-writing/608#608Answer by AH for Examples of great mathematical writingAH2009-10-15T16:22:52Z2009-10-15T16:22:52Z<p>Real Analysis by Elias Stein and Rami Shakarchi</p>
<p>I absolutely hated analysis until I read the Stein/Shakarchi analysis series (Fourier, Complex, Real Analysis). Now I find the subject to be very beautiful and full of deep ideas, and it is these books that really convinced me.</p>
http://mathoverflow.net/questions/372/derived-categories-and-homotopy-categories/604#604Answer by AH for Derived categories and homotopy categoriesAH2009-10-15T15:51:53Z2009-10-15T16:01:10Z<p>I think you don't want any bounded condition. I don't see how the category of chain complexes with bounded cohomology could be a model category. It doesn't have all small colimits; just take longer and longer chain complexes with trivial differentials, and you get something with unbounded cohomology.</p>
http://mathoverflow.net/questions/115617/notes-for-botts-1963-lectures-on-morse-theoryComment by AHAH2012-12-07T01:05:15Z2012-12-07T01:05:15ZCarl, I don't think these are the same notes. Bott also has notes giving an exposition of Morse theory, which is what I am after.http://mathoverflow.net/questions/372/derived-categories-and-homotopy-categories/604#604Comment by AHAH2011-09-09T13:46:05Z2011-09-09T13:46:05ZRegardless, that is what I am using.http://mathoverflow.net/questions/372/derived-categories-and-homotopy-categories/604#604Comment by AHAH2011-09-09T08:22:56Z2011-09-09T08:22:56ZThe definition I am working with is the one here:
<a href="http://en.wikipedia.org/wiki/Model_category" rel="nofollow">en.wikipedia.org/wiki/Model_category</a>http://mathoverflow.net/questions/74843/how-to-interpret-the-euler-characteristic-of-complex-algebraic-varieties/74856#74856Comment by AHAH2011-09-08T16:47:05Z2011-09-08T16:47:05Z"any projective variety can be defined by determinental quadratic equations in some large projective space." Would you know a place where this is proven?http://mathoverflow.net/questions/64602/techniques-for-computing-cup-products-in-singular-cohomology/65003#65003Comment by AHAH2011-05-15T12:06:46Z2011-05-15T12:06:46ZWould you happen to know a place where I can see this bootstrapping argument worked out carefully for a serious example?http://mathoverflow.net/questions/64602/techniques-for-computing-cup-products-in-singular-cohomologyComment by AHAH2011-05-11T17:04:49Z2011-05-11T17:04:49ZIn my experience, any manifold that I understand well enough to be able to compute the de Rham algebra of is probably a manifold that I understand well enough to work out the intersection product on. Would you happen to know of an example which doesn't fit this description?http://mathoverflow.net/questions/62159/relationship-between-hilbert-schemes-and-deformation-spaces/62165#62165Comment by AHAH2011-04-21T04:35:19Z2011-04-21T04:35:19Zmdeland - I was not referring specifically to D or D' when I said "versal deformation space". Rather, I meant to point out that versal deformation spaces tend to look like formal completions of schemes at a point, so they are really small, local objects such as Spf O_x. One would not expect a big scheme like the Hilbert scheme to map into them in an interesting way.http://mathoverflow.net/questions/62159/relationship-between-hilbert-schemes-and-deformation-spaces/62165#62165Comment by AHAH2011-04-21T04:30:38Z2011-04-21T04:30:38ZSL - I don't know what your precise definition of a deformation functor is, but the functor F: (Art)_k \to (sets) given by F(A) = Hom(Spec A, Spec O_x) is considered the most canonical example of a deformation functor. If your definition does not admit this example, then it is at odds with the designs of deformation theory.
I think the only universal requirement for a deformation functor is that F(k) = {pt}. Some people go further and require that some of the Schlessinger conditions be fulfilled. Here, F is pro-representable by definition, so it satisfies all of the Schlessinger conditions.http://mathoverflow.net/questions/56104/early-two-author-math-papers/56109#56109Comment by AHAH2011-02-20T23:09:15Z2011-02-20T23:09:15ZYes, of course. I had not been aware that Brill-Noether theory was that old.http://mathoverflow.net/questions/56104/early-two-author-math-papers/56109#56109Comment by AHAH2011-02-20T21:51:53Z2011-02-20T21:51:53ZI never realized that the Noether of "Brill-Noether" referred to Max Noether, not Emmy Noether.http://mathoverflow.net/questions/55725/interesting-examples-of-flasque-sheavesComment by AHAH2011-02-19T11:18:38Z2011-02-19T11:18:38ZI'm really just looking for expand my basis of examples. That said, I am looking for ones that are neither constants or products or sums of stalks.http://mathoverflow.net/questions/55725/interesting-examples-of-flasque-sheavesComment by AHAH2011-02-18T21:23:28Z2011-02-18T21:23:28ZI seem to have done a really bad job with phrasing and titling this question. I don't need to be convinced that the flasque sheaves I know of are interesting; I already believe it. I'm just looking for other sources of flasque sheaves so I can get some sense of how common/uncommon such objects are. You could say I am more pondering the limits of this definition than trying to do any actual algebraic geometry, for instance.
Apologies to all around. If I get another complaint, I will rewrite (and retitle) the question to mean what I originally meant.http://mathoverflow.net/questions/55725/interesting-examples-of-flasque-sheaves/55762#55762Comment by AHAH2011-02-18T12:13:09Z2011-02-18T12:13:09ZThis sounds interesting. Would you mind saying more?http://mathoverflow.net/questions/55725/interesting-examples-of-flasque-sheavesComment by AHAH2011-02-17T17:01:34Z2011-02-17T17:01:34ZI probably should have been less vague in my statement. Other than direct products of stalks or constant sheaves on irreducible spaces, I don't know of any other examples of flasque sheaves. So I was hoping for something other than these.http://mathoverflow.net/questions/53988/what-is-the-motivation-for-a-vertex-algebraComment by AHAH2011-02-01T16:19:47Z2011-02-01T16:19:47ZBut is there a "fundamental example" that captures what a QFT is supposed is to do (or be)? I would be happy just to hear a story about a particular physical system, in a way that illustrates the important features of QFT.