User justin curry - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T21:13:19Z http://mathoverflow.net/feeds/user/1622 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/120729/cosheafification/124929#124929 Answer by Justin Curry for Cosheafification Justin Curry 2013-03-19T05:07:59Z 2013-03-19T05:25:37Z <p>I have had some conversations with Jon Woolf (the author of the paper referenced by Davidac897). He has pointed out that the forgetful functor from $for:\mathrm{Vect}\to\mathrm{Set}$ preserves limits, but not colimits. Thus a cosheaf of vector spaces need not define a cosheaf of sets and in particular cosheafifying for pre-cosheaves valued in one data category $\mathcal{D}$, may "look different" depending on what $\mathcal{D}$ is.</p> <p>One can view a pre-cosheaf $\hat{F}:\mathrm{Open}(X)\to\mathrm{Vect}$ as a pre-sheaf $F:\mathrm{Open}(X)^{op}\to\mathrm{Vect}^{op}$ and try to use Grothendieck's sheafification prescription, but this will not work either. The requirements for Grothendieck's sheafification (outlined on page 24 of Schapira's <a href="http://people.math.jussieu.fr/~schapira/lectnotes/Shv.pdf" rel="nofollow"><strong>notes</strong></a>) is that the data category be one in which filtered colimits and finite limits commute, which for $\mathrm{Vect}^{op}$ boils down to the false statement that cofiltered limits and finite colimits commute in $\mathrm{Vect}$.</p> <p>The solution the following two papers get at is to work in the pro-object category, because there cofiltrant limits and finite colimits do commute, so the Grothendieck construction goes through.</p> <p><a href="http://arxiv.org/pdf/1105.3167.pdf" rel="nofollow">http://arxiv.org/pdf/1105.3167.pdf</a></p> <p><a href="http://kyokan.ms.u-tokyo.ac.jp/users/preprint/pdf/2001-33.pdf" rel="nofollow">http://kyokan.ms.u-tokyo.ac.jp/users/preprint/pdf/2001-33.pdf</a></p> <p>However, for some people (myself included), this is an unappealing solution. Pro-objects are diagrams in themselves, so a pre-cosheaf of pro-Vector spaces would assign to each open set a diagram of vector spaces.</p> <p>So here is something one can do: One can check abstractly whether cosheafification exists. This is equivalent to asking whether the inclusion functor from the category of cosheaves into the category of pre-cosheaves has a right adjoint. Freyd's general adjoint functor theorem says that, modulo set-theoretic issues, a functor has a right adjoint (is a left adjoint) if it preserves colimits. Since the category of cosheaves is clearly closed under colimits (one just defines open-by-open it to be the colimit, and since colimits commute, the cosheaf axiom holds for this colimit pre-cosheaf, i.e. the colimit is a cosheaf), then the inclusion functor does have a right adjoint.</p> <p>Of course, the devil is in the details, so I have written up the details and put them on my website <a href="http://www.math.upenn.edu/~jucurry/papers/abstract_cosheafification.pdf" rel="nofollow"><strong>here</strong></a>. I use an <a href="http://ncatlab.org/nlab/show/coreflective+subcategory" rel="nofollow"><strong>easier-to-check version</strong></a> of the adjoint functor theorem given by Vopenka's principle, but I think one could use the proof outlined to appeal to just Freyd's theorem. </p> <p>For more on cosheaves and their possible uses, the following might be of interest:</p> <p><a href="http://arxiv.org/abs/1303.3255" rel="nofollow">http://arxiv.org/abs/1303.3255</a></p> http://mathoverflow.net/questions/99969/cosheaf-homology-and-a-theorem-of-beilinson-in-a-paper-on-mixed-tate-motives/119381#119381 Answer by Justin Curry for Cosheaf homology and a theorem of Beilinson (in a paper on Mixed Tate Motives) Justin Curry 2013-01-20T00:40:07Z 2013-01-20T00:40:07Z <p>Cosheaves are indeed mysterious gadgets. On the one hand, cosheaves are everywhere, but on the other hand, someone used to thinking sheaf-theoretically may have some problems. I am very close to finishing an exposition on cosheaves, but need another week or so to put it on the arxiv. Bredon's book on sheaf theory has the most complete reference on cosheaves, so you might look there if you like.</p> <p>AS you may know, pre-cosheaves are just covariant functors $\hat{F}:\mathrm{Open}(X)\to\mathcal{D}$ where $\mathcal{D}$ is some "data category" like Vect, Ab, or what have you. Cosheaves send covers (closed under intersection) to colimits and different covers of the same open set get sent to isomorphic colimits. The Mayer-Vietoris axiom is a good way of thinking about cosheaves and since homology commutes with direct limits, one can see that $H_0(-,k)$ is always a cosheaf. In particular, $H_0(-,\mathcal{L})$ is a cosheaf whenever $\mathcal{L}$ is a local system.</p> <p>As you observed, since cosheaves are fundamentally colimit-y, they have left-derived functors rather than right-derived ones. Thus the answer to (1) is yes.</p> <p>In regards to (2), one must be careful. I believe the answer is yes, but allow me to pontificate on the problem.</p> <p><em>Filtered limits and finite colimits do not commute in most categories like Ab, Vect, or Set</em>. This has serious ramifications through the theory of cosheaves. </p> <p>For example, it is not necessarily true that a sequence of cosheaves is exact iff it is exact on costalks. Here costalks are defined using (filtered) inverse limits rather than direct ones. </p> <p>Another very serious consequence is that Grothendieck's sheafification procedure <em>cannot</em> be dualized to give cosheafification. Thus the usual phrase </p> <p>"let <em>blah</em> by the cosheaf associated to the pre-cosheaf <em>blah</em>" </p> <p>is not necessarily well-founded because it is unclear how to cosheafify! People have solved this problem in the past by working with pro-objects (which corrects for this "filtered limits not commuting with finite colimits" asymmetry) and then they use Grothendieck's construction. However, for abstract categorical reasons one can check that cosheafification does exist for data categories like Vect (i have worked out a proof and haven't found in the literature anyone who claims to have proved this), we just don't have an explicit construction. That said, the usual description of the left-derived functor of the push-forward should still hold. </p> <p>On the other hand, if one works in the constructible setting, one can get the statements you would like. In particular, it is true that cosheaves constructible with respect to a cell structure are derived equivalent to sheaves constructible with respect to the same cell structure. I discovered independently <a href="http://www.math.upenn.edu/~jucurry/papers/co_sheaf_dereq.pdf" rel="nofollow">my own proof</a>, only to find that at least two other people have proved this before. However, in my opinion, the equivalence is the "correct" form of Verdier duality. A larger and updated exposition should be available soon.</p> http://mathoverflow.net/questions/9799/how-much-work-does-it-take-to-be-a-successful-mathematician How Much Work Does it Take to be a Successful Mathematician? Justin Curry 2009-12-26T16:16:34Z 2012-06-14T20:21:19Z <p>Hi Everyone,</p> <p>Famous anecdotes of G.H. Hardy relay that his work habits consisted of working no more than four hours a day in the morning and then reserving the rest of the day for cricket and tennis. Apparently his best ideas came to him when he wasn't "doing work." Poincare also said that he solved problems after working on them intensely, getting stuck and then letting his subconscious digest the problem. This is communicated in another anecdote where right as he stepped on a bus he had a profound insight in hyperbolic geometry.</p> <p>I am less interested in hearing more of these anecdotes, but rather I am interested in what people consider an appropriate amount of time to spend on doing mathematics in a given day if one has career ambitions of eventually being a tenured mathematician at a university. </p> <p>I imagine everyone has different work habits, but I'd like to hear them and in particular I'd like to hear how the number of hours per day spent doing mathematics changes during different times in a person's career: undergrad, grad school, post doc and finally while climbing the faculty ladder. "Work" is meant to include working on problems, reading papers, math books, etcetera (I'll leave the question of whether or not answering questions on MO counts as work to you). Also, since teaching is considered an integral part of most mathematicians' careers, it might be good to track, but I am interested in primarily hours spent on learning the preliminaries for and directly doing research.</p> <p>I ask this question in part because I have many colleagues and friends in computer science and physics, where pulling late nights or all-nighters is commonplace among grad students and even faculty. I wonder if the nature of mathematics is such that putting in such long hours is neither necessary nor sufficient for being "successful" or getting a post-doc/faculty job at a good university. In particular, does Malcom Gladwell's 10,000 hour rule apply to mathematicians?</p> <p>Happy Holidays!</p> http://mathoverflow.net/questions/68647/understanding-the-wiki-page-on-verdier-duality/94478#94478 Answer by Justin Curry for Understanding (the wiki page on) Verdier duality Justin Curry 2012-04-19T00:39:00Z 2012-04-19T00:39:00Z <p>I just revamped what was written. Perhaps now it is more understandable:<a href="http://en.wikipedia.org/wiki/Verdier_duality" rel="nofollow">New Wiki Entry on Verdier duality.</a></p> http://mathoverflow.net/questions/85717/segals-original-definition-of-a-topological-category Segal's Original Definition of a Topological Category Justin Curry 2012-01-15T05:31:45Z 2012-01-15T21:10:51Z <p>Nowadays we can associate to a topological space $X$ a category called the fundamental (or Poincare) $\infty$-groupoid given by taking $Sing(X)$.</p> <p>There are many different categories that one can associate to a space $X$. For example, one could build the small category whose object set is the set of points with only the identity morphisms from a point to itself. It is claimed that the classifying space of this category returns the space: $BX=X$</p> <p>The inspiration for these examples comes from three primary sources: Graeme Segal's famous 1968 paper <em>Classifying Spaces and Spectral Sequences</em>, Raoul Bott's Mexico notes (taken by Lawrence Conlon) <em>Lectures on characteristic classes and foliations</em>, and a 1995 pre-print called <em>Morse Theory and Classifying Spaces</em> by Ralph Cohen, G. Segal and John Jones. </p> <p>In each of these papers there is a notion of a topological category. It is not just a category enriched in <strong>Top</strong>, since the set of objects can have non-discrete topology. Here is the definition that I can gleam from these articles:</p> <p>A <strong>topological category</strong> consists of a pair of spaces $(Obj,Mor)$ with four continuous structure maps:</p> <ul> <li>$i:Obj\to Mor$, which sends an object to the identity morphism</li> <li>$s:Mor\to Obj$, which gives the source of an arrow</li> <li>$t:Mor\to Obj$, which gives the target of an arrow</li> <li>$\circ:Mor\times_{t,s}Mor\to Mor$, which is composition.</li> </ul> <p>Were $i$ is a section of both $s$ and $t$, and all the axioms of a small category hold.</p> <p><strong>Is the appropriate modern terminology to describe this a <a href="http://ncatlab.org/nlab/show/Segal+space" rel="nofollow">Segal Space</a>? What would Lurie call it?</strong> Based on reading <a href="http://mathoverflow.net/questions/29728/a-model-category-of-segal-spaces" rel="nofollow">Chris Schommer-Pries MO post</a> and elsewhere this seems to be true. Would the modern definition of the above be a Segal Space where the Segal maps are identities? Also, why do we demand that the topology on objects be discrete for Segal Categories? <strong>Is there something wrong with allowing the object sets to have topologies?</strong></p> http://mathoverflow.net/questions/78850/derived-equivalence-of-sheaves-and-homotopy Derived Equivalence of Sheaves and Homotopy Justin Curry 2011-10-22T21:26:59Z 2011-10-22T23:41:43Z <p>This question loosely elaborates on an <a href="http://mathoverflow.net/questions/66098/sheaves-with-isomorphic-cohomology-but-not-quasi-isomorphic" rel="nofollow">earlier question</a>. It is pretty silly, but I'd like to hear some authoritative answers.</p> <p>Recall that if $f:S^{\bullet}\to T^{\bullet}$ is a quasi-isomorphism of sheaves over $X$, which is, say, a manifold, then for every open set $U\subset X$, we have an induced isomorphism $R\Gamma(U,S^{\bullet})\to R\Gamma(U,T^{\bullet})$, so $H^i(U,S^{\bullet})\cong H^i(U,T^{\bullet})$ and in particular $H^i(X,S^{\bullet})\cong H^i(X,T^{\bullet})$.</p> <p>To what extent is the converse true? At the coarsest level, when does a canonical isomorphism $R\Gamma(X,S^{\bullet})\to R\Gamma(X,T^{\bullet})$ reflect an underlying derived equivalence?</p> <p>For a counterexample to the coarsest case, I believe the following serves: Consider a space $X$. Consider the constant sheaf on $k_X$. Let $f:X\to x_0$ be the retraction to a point $x_0\in X$. By standard theorems, we know that $H^i(X,Rf_*k_X)\cong H^i(X,k_X)$, but evaluating $Rf_*k_X(U)$ on any open subset $U$ missing $x_0$ assigns zero, as the fiber is empty. So in general these sheaves are not derived equivalent. What if $X$ deformation retracts to $x_0$? Is $k_X$ and $Rf_*k_X$ derived equivalent then? What if the homotopy doesn't have some Vietoris-Begle type behavior? See Kashiwara Schapira 2.7.8.</p> http://mathoverflow.net/questions/12469/group-structure-on-cpinfinty Group Structure on CP^infinty Justin Curry 2010-01-21T00:27:37Z 2011-06-16T09:20:33Z <p>I was inspired by the following algebraic topology orals question:</p> <p>"Is $S^1$ the loop space of another space?"</p> <p>This is easy to see if you recognize that $S^1$ is a $K(\mathbb{Z},1)$, and the loop space of any $K(G,n)$ is a $K(G,n-1)$. </p> <p>I then also remembered that the loop space functor is a functor from pointed topological spaces and continuous maps to the category of H-spaces and continuous homomorphisms. H-spaces being topological spaces that satisfy the axioms of a group up to homotopy (see Spanier, Chapter 1, Section 5).</p> <p>I have three questions:</p> <ol> <li>Is there a useful criterion for when an H-space is actually a topological group?</li> <li>Seeing that $S^1$,$S^3$, and $S^7$ are the only spheres that support group structures, it doesn't seem coincidental that $S^1$ is a loop space, because it is in fact an H-space. Since $CP^{\infty}$ is the loop space of $K(Z,3)$ it too is an H-space, but is it known if it is a topological group? </li> <li>Even if not, is there a way (other than concatenation of loops) to "see" this structure on $CP^{\infty}$?</li> </ol> <p>Thanks!</p> http://mathoverflow.net/questions/16632/stiefel-whitney-classes-over-integers Stiefel-Whitney Classes over Integers? Justin Curry 2010-02-27T20:18:53Z 2011-05-28T21:30:55Z <p>An interesting thing happened the other day. I was computing the Stiefel-Whitney numbers for $\mathbb{C}P^2$ <strong>connect sum</strong> $\mathbb{C}P^2$ to show that it was a boundary of another manifold. Of course, one can calculate the signature, check that it is non-zero and conclude that it can't be the boundary of an <em>oriented</em> manifold. I decided it might be interesting to calculate the first and only Pontrjagin number to check that it doesn't vanish. I believe Hirzebruch's Signature Theorem can be used to show that it is 6, but I was interested in relating the Stiefel-Whitney classes to the Pontrjagin classes.</p> <p>I believe one relation is </p> <blockquote> <p>$p_i (\mathrm{mod} 2) \equiv w_{2i}^2$ (pg. 181 Milnor-Stasheff)</p> </blockquote> <p>So I went ahead and did a silly thing. I took my first Chern classes of the original connect sum pieces say 3a and 3b, used the fact that the inclusion should restrict my 2nd second "Stiefel-Whitney Class" (scare quotes because we haven't reduced mod 2) on each piece to these two to get $w_2(connect sum)=(3\bar{a},3\bar{b})$. I can use the intersection form to square this and get $3\bar{a}^2+3\bar{b}^2=6c$ since the top dimensional elements in a connect sum are identified. Evaluating this against the fundamental class gives us exactly the first Pontrjagin number! <strong>This is false. Of course this is wrong because it should be 9+9=18 as pointed out below. This does away with my supposed miracle example. My Apologies!</strong></p> <p>This brings me to a broader question, namely of defining Stiefel-Whitney Classes over the integers. This was hinted at in <a href="http://mathoverflow.net/questions/13813/construction-of-the-stiefel-whitney-and-chern-classes" rel="nofollow">Ilya Grigoriev's response to Solbap's question</a> when he says </p> <blockquote> <p>On thing that confuses me: why are the pullbacks of the integer cohomology of the real Grassmanian never called characteristic classes?</p> </blockquote> <p>Of course the natural reason to restrict to $\mathbb{Z}/2$ coefficients is to get around orientability concerns. But it seems like if we restrict our orientation to orientable bundles we could use a construction analogous to those of the Chern classes where Milnor-Stasheff inductively declare the top class to be the Euler class, then look at the orthogonal complement bundle to the total space minus its zero section and continue. I suppose the induction might break down because the complex structure is being used, but I don't see where explicitly. If someone could tell me where the complex structure is being used directly, I'd appreciate it. Note the Euler class on odd dimensional fibers will be 2-torsion so this might produce interesting behavior in this proposed S-W class extension.</p> <p>Another way of extending Stiefel-Whitney classes would be to use Steenrod squares. Bredon does use Steenrod powers with coefficient groups other than $\mathbb{Z}/2$ (generally $\mathbb{Z}/p$ $p\neq 2$), but this creates awkward constraints on the cohomology groups. Is this an obstruction to extending it to $\mathbb{Z}$ coefficients? It would be interesting to see what these two proposed extensions of S-W classes do and how they are related.</p> http://mathoverflow.net/questions/66098/sheaves-with-isomorphic-cohomology-but-not-quasi-isomorphic Sheaves with isomorphic cohomology, but not quasi-isomorphic Justin Curry 2011-05-26T19:43:37Z 2011-05-27T22:26:45Z <p>Suppose I have two (constructible) sheaves of vector spaces $F$ and $G$ over the same base space that have isomorphic cohomology (degree by degree), but no sheaf map inducing this isomorphism (i.e. they are <em>not</em> quasi-isomorphic). </p> <p>Now imagine I apply any of Grothendieck's 6 operations (or other functors) to $F$ and $G$, will the resulting sheaves, say $\Psi(F)$ and $\Psi(G)$, have isomorphic cohomology as well? </p> <p>Thanks!</p> <p><em>EDIT:</em> I suppose the answer in general is no. Consider an injective (acyclic) sheaf and any functor that doesn't preserve injectives. A concrete example would be nice though.</p> <p><em>EDIT 2:</em> Thanks to Algori's comments I need to substantially limit the sheaves of interest. Assume additionally that the sheaves have identical support and furthermore that when we take the function that assigns to a point the euler characteristic of the stalk over that point (thereby producing a constructible function) to these two sheaves, these functions are identical. If we are using real sub-analytic partitions then a theorem of Kashiwara's says that the Grothendieck group of the bounded derived category of real constructible sheaves is isomorphic to the group of constructible functions. If I understand things correctly this would mean that these two sheaves are equivalent in the Grothendieck group (which seems weird since this function ignores gluing data). I am dealing with particular examples of two constructible (cellular) sheaves over a simplicial complex, which are not isomorphic yet produce identical constructible functions.</p> http://mathoverflow.net/questions/44271/geometry-and-integrability-in-other-bundles Geometry and Integrability in Other Bundles Justin Curry 2010-10-30T19:52:32Z 2011-03-24T01:06:23Z <p><strong>Background:</strong> Suppose $E=TM$ is the tangent bundle to some differentiable manifold $M^n$. If we specify some subbundle $D\subset TM$ (distribution of $k$-planes) then there are two natural situations that arise. We may have that at some point $p$ there is an immersed submanifold $N^k\subset M^n$ passing through $p$ such that for all $q\in N$, $T_q N=D_q$. If this is true then we say $D$ is a <strong>integrable</strong> at $p$. Otherwise it is non-intregrable.</p> <p>Frobenius Integrability states that every involutive distribution is completely integrable, i.e. $M$ is foliated by integral manifolds. One way of phrasing involutivity is to require that any lie bracket of vector fields lying in $D$ stay in $D$. There is another version, which is the one that I prefer, using differential forms. Note that any $k$-distribution is cut out locally by $n-k$ independent 1-forms $\theta_1,\ldots,\theta_{n-k}$. Call $\Theta$ the ideal generated by these, then we say $\Theta$ is involutive if $d\Theta\subset \Theta$,i.e. it is a differential ideal.</p> <p><strong>Situation:</strong> This last version generalizes to arbitrary vector bundles (with connection) easily. Suppose $E$ is a rank $N>n$ vector bundle over $M^n$. Then I can specify any sub-vector-bundle $D\subset E$ by $$D_x:=\lbrace v\in E_x | \theta_1(v)=\ldots=\theta_{n-k}(v)=0\rbrace$$</p> <p>for some collection of 1-forms, i.e. sections of the dual bundle $E^*$. I can extend the above definition of involutive distribution to this subbundle in the obvious way. Let me take this as a definition of integrable subbundle.</p> <p>Question: <strong>If the geometric concept associated to an integrable subbundle of $TM$ is a foliation, what is the geometric concept associated to an integrable subbundle of $E$?</strong></p> <p>I am only beginning to dig into exterior differential systems, so any well articulated answer/exposition is appreciated.</p> <p><strong>EDIT:</strong> The differential one gets from a connection actually just lands in vector valued forms, i.e. the Twisted de Rham complex. So the complex one would like to get (which looks like a Koszul complex possibly) is not obtained in a canonical way with a connection. Other differential operators would be needed. </p> http://mathoverflow.net/questions/53000/cohomology-of-structure-sheaves-algebraic-constructible-and-more Cohomology of Structure Sheaves: Algebraic, Constructible and more Justin Curry 2011-01-23T22:20:32Z 2011-01-24T20:16:55Z <p>I am not an algebraic geometer, but I am a topologist who uses sheaves. I have studied some algebraic geometry and am interested in what happens as I reduce the amount of rigidity in the structure sheaves on a space. Specifically I want to know what the cohomology of the following structure sheaves tell you. <strong>Please do things over characteristic zero.</strong></p> <p>If $X$ is a topological space then the natural structure sheaf of continuous functions has no interesting cohomology because of the existence of <strong>partitions of unity.</strong> Consequently, $C^k$, smooth and topological manifolds have structure sheaves with no interesting cohomology.</p> <p>To contrast, schemes ($X,\mathcal{O}_X$) with their structure sheaves of regular functions have lots of interesting information. In particular, there is non-trivial higher cohomology. However, I am still unsure what these groups tell you (aside from what all sheaf cohomology means - obstructions to extending sections). For example Hartshorne exercise 4.3 tells you that $H^1(U,\mathcal{O}_U)$ is infinite dimensional (spanned by $x^iy^j|i,j&lt;0$) where $U=\mathbb{A}^2_k-(0,0)$. For $X$ a curve then the dimension of $H^1(X,\mathcal{O}_X)$ tells you the genus. For affine pieces this cohomology is trivial, so the cohomology of the structure sheaf detects "non-triviality" of a space. <strong>Are there any other characterizations of the higher cohomology groups of the structure sheaf?</strong></p> <p>I am actually interested in the definable/o-minimal/constructible setting. So I want to consider a constructible space $X$ along with it's structure sheaf of ($\mathbb{R}$ or $\mathbb{Z}$-valued) constructible functions as a ringed space. Since one implementation of definable spaces is the semi-algebraic (or semi-analytic) setting, I would like to know that the cohomology of the structure sheaf here tells you. <strong>So if someone could address any of the following:</strong></p> <ul> <li><p>a <strong>real</strong> analytic space with the sheaf of analytic functions (no partitions of unity, so potential higher cohomology?) Question: Since regular functions in AG are defined as locally being the quotient of polynomials, would regular for analytic be locally the ratio of analytic? (EDIT: I am interested primarily in the real case since GAGA shows in some cases complex analytic spaces are "as rigid as" complex algebraic ones.)</p></li> <li><p>a semi-analytic space with the above structure sheaf(ves)</p></li> <li><p>a semi-algebraic space with its structure sheaf (Which is? Do Nash functions come into play here?)</p></li> <li><p>for a cell complex, is there a natural way of considering it as a ringed space? If so what would that cohomology tell you?</p></li> </ul> <p>I apologize for the wide spread of questions. Partial answers will be voted up.</p> http://mathoverflow.net/questions/20764/representations-of-pi-1-g-bundles-classifying-spaces Representations of \pi_1, G-bundles, Classifying Spaces Justin Curry 2010-04-08T17:31:08Z 2010-10-29T03:38:28Z <p>This question is inspired by a statement of Atiyah's in "Geometry and Physics of Knots" on page 24 (chapter 3 - Non-abelian moduli spaces). </p> <p>Here he says that for a Riemann surface <code>$\Sigma$</code> the first cohomology <code>$H^1(\Sigma,U(1))$</code>, where <code>$U(1)$</code> is just complex numbers of norm 1, parametrizes homomorphisms <code>$$\pi_{1}(\Sigma)\to U(1).$$</code> </p> <p>This is fine by me, after all by Brown Representability we know <code>$$H^1(\Sigma,G)\cong [\Sigma,BG]=[B\pi_{1}\Sigma,BG]$$</code> since we know that Riemann surfaces are <code>$K(\pi_{1},1)$</code>s. We then use the handy fact from Hatcher Prop. 1B.9 (pg 90) that shows that... </p> <blockquote> <p>For $X$ a connected CW complex and $Y$ a $K(G,1)$ every homomorphism $\pi_1(X,x_0)\to\pi_1(Y,y_0)$ is induced by a map $(X,x_0)\to(Y,y_0)$ that is unique up to homotopy fixing $x_0$.</p> </blockquote> <p>So group homomorphisms <code>$\pi_{1}(\Sigma)\to U(1)$</code> correspond on the nose with first cohomology of the <code>$\Sigma$</code> with coefficients in $U(1)$.</p> <p>EDIT: To be accurate the Hatcher result shows we have an injection of group homomorphisms into homotopy classes of maps. Does anyone know if every homotopy class of maps is realized by a group homomorphism?</p> <p>What bothers me is what comes next. Now replace <code>$U(1)$</code> with <code>$G$</code> - any compact simply connected Lie group, take <code>$G=SU(n)$</code> for example. Now Atiyah claims that <code>$H^1(\Sigma,G)$</code> parametrizes <strong>conjugacy classes</strong> of homomorphisms <code>$\pi_{1}(\Sigma)\to G$</code>.</p> <p>Now if the fundamental group or the Lie group were abelian this would be the same statement, but higher genus Riemann surfaces (genus greater than 1) have non-abelian fundamental groups and Atiyah is looking specifically at non-abelian <code>$G$</code>. Also, it seems that the statement of Brown Representability requires abelian coefficient groups, so I am stuck. </p> <p>Does anyone know a clean way of proving Atiyah's claim?</p> <p>EDIT2: I renamed the question to draw in the "right" people. I think the answer has to do with the fact that principal G-bundles over a Riemann surface are determined by maps of $\pi_1(\Sigma)\to G$ (can someone explain why?). This is related to local systems and/or flat connections, which I don't understand well. Thanks!</p> http://mathoverflow.net/questions/36576/simplicial-covering-map Simplicial Covering Map Justin Curry 2010-08-24T18:10:53Z 2010-08-24T18:41:48Z <p>In Rezk's paper <a href="http://arxiv.org/abs/math/9811037" rel="nofollow">"A model for the homotopy theory of homotopy theory"</a> numerous references to simplicial covering maps are made. It's first appearance being at the bottom of page 8. Unfortunately no definition is provided in the paper and I was wondering if there is a purely combinatorial definition for this concept or whether we have to pass to the geometric realization.</p> <p>Maps of simplicial sets already match cells of the same dimension (roughly speaking), but it is the evenly covered concept that requires some work (I imagine).</p> <p>Any help would be appreciated.</p> http://mathoverflow.net/questions/28158/a-learning-roadmap-request-from-high-school-to-mid-undergraduate-studies/32352#32352 Answer by Justin Curry for A Learning Roadmap request: From high-school to mid-undergraduate studies Justin Curry 2010-07-18T12:00:30Z 2010-07-18T12:00:30Z <p>If you are looking for a break from Calculus and don't want to dive into Category Theory right away, my favorite book is Alan Beardon's <a href="http://www.amazon.com/Algebra-Geometry-Alan-F-Beardon/dp/0521890497" rel="nofollow">Algebra and Geometry</a>. What follows is an enhanced Amazon review I wrote as an undergrad. The Cambridge Schedules provide a wonderful guide to further study so do have a look.</p> <p>Beardon's book has become a bible of sorts to first-year <a href="http://www.maths.cam.ac.uk/undergrad/course/" rel="nofollow">students studying mathematics at Cambridge University</a> (please refer to the <a href="http://www.maths.cam.ac.uk/undergrad/course/schedules.pdf" rel="nofollow">Schedules</a> for a beautiful play-by-play of topics and books by perhaps the best foundational math curriculum the world-over). Its quality as a text cannot be doubted, although its usefulness for further years of algebra is limited. This is precisely the book to study from if you are doing vector calculus and differential equations, but still aren't sure about doing mathematics seriously. If you have not taken a course in linear algebra or abstract algebra, buy the paperback copy (~$50) of this book and start reading right away. Beardon starts with (what I believe is the best way) the study of permutations (think about shuffling a deck of cards) to develop an intuition of the basic notions of a group. From here the fundamentals for further study in mathematics is laid. I won't repeat the table of contents here, as you can look for yourself, but believe me when I say that mastering the concepts in this book will serve you very well. </p> <p>I truly wish I had a course which devoted itself to the complete digestion of this book. I used it for self-study and found that it served me very well. It does not fall easily into the structure of most American math sequences, as these departments are often forced to "modularize" mathematics into semester-bite-sized pieces. I believe that this often has a negative impact on the appreciation of mathematics as a whole, especially at the nexus between doing basic calculus and appreciating proof, rigor and beauty in mathematical structures. </p> <p>The book may not go into the same depth as, say, Artin's "Algebra", but rather the foundational concepts for the study of algebra and geometry are emphasized in a variety of settings. This is very important as the study of "abstract algebra" is precisely that if you do not have a wide-selection of examples and contexts to draw from. This book has plenty of exercises of varying difficulty, and everything in this book is accessible to the beginning student of mathematics. </p> <p><strong>Bottom line:</strong> If you are someone interested in learning linear algebra, geometry, group theory, Mobius transformations, complex variables all in a rigorous yet introductory level, this is the book for you. Developing a robust mental model for mathematics requires building several thin layers at a time. This means not going too deep too quickly, but rather snorkeling around the entire reef, before you gear up for further exploration.</p> http://mathoverflow.net/questions/31732/folk-functorial-figuring Folk Functorial Figuring Justin Curry 2010-07-13T16:32:52Z 2010-07-14T06:20:52Z <p>In the CRM Proceedings &amp; Lecture Notes Volume 50 "A Celebration of the Mathematical Legacy of Raoul Bott" Herbert Shulman writes (p. 48):</p> <blockquote> <p>"[Bott] taught many of us to think functorially, like thinking of a group as a category with one object and a morphism for each element, <strong>a manifold as a category of pairs (open set, point in the open set), and a bundle as an equivalence class of functors.</strong> When someone asked him who invented functors, he said 'functors are prehistoric!'. He talked about 'folk' theorems... theorems everyone knew, but were never written down."</p> </blockquote> <p>As I've highlighted in bold the latter two everyday examples of categories seem less well-known to me.</p> <p>The manifold-as-category seems clear enough, the open set is meant to be a coordinate neighborhood of the point. A morphism between two objects is just the transition maps between coordinate neighborhoods. What seems more natural to me would be to associate a category to a given atlas on a manifold. <strong>This makes me wonder what the classifying space of these categories looks like. How do they behave as you pass to the maximal atlas? Does anyone know?</strong> Since the transition maps (the morphisms) are homeomorphisms/ diffeomorphisms/ biholomorphisms (depending on what type manifold we have) the morphisms in this category are all invertible and so our category is a groupoid. For a connected manifold the classifying space should be a$BG$or$K(G,1)$. What is$G$?</p> <p>Finally, <strong>Can someone explain why a bundle is an equivalence class of functors?</strong> The functor part seems sort of clear, because the projection map is an open map. More explanation would be nice.</p> http://mathoverflow.net/questions/29964/which-spheres-are-complex-manifolds/29977#29977 Answer by Justin Curry for Which Spheres are Complex Manifolds? Justin Curry 2010-06-29T22:52:47Z 2010-07-01T14:20:38Z <p>It's a nice exercise in characteristic classes to show that S^4k for all k are NOT complex manifolds.</p> <p><strong>EDIT:</strong> I will answer Charlie's comment here and provide a sketch of the proof.</p> <p>Let$\omega=TS^{4k}$be the tangent space to the$4k$-sphere. If$S^{4k}$was actually a complex manifold then$\omega$would be a complex vector bundle. In this case the complexification of the underlying real vector bundle$\omega_{\mathbb{R}}$would be canonically isomorphic to the Whitney sum$\omega\oplus \bar{\omega}$(Milnor&amp;Stasheff page 176). Now by corollary 15.5 in Milnor&amp;Stasheff $$p_k(\omega_{\mathbb{R}})=c_k^2(\omega)-2c_{k-1}c_{k+1}(\omega)+\cdots\mp 2c_{2k}(\omega)$$</p> <p>This then shows that the top Pontrjagin number $$&lt; p_k,[S^{4k}]>=&lt;\mp 2c_{2k},[S^{4k}]>=\mp 4$$ but we also know that spheres are boundries of an oriented manifold and thus have higher Pontrjagin number 0. Contradiction.</p> <p>On another note, according to C.C. Hsiung's book <a href="http://books.google.com/books?id=-yz03cskUOgC&amp;printsec=frontcover#v=onepage&amp;q=complex%2520structures%2520on%2520spheres&amp;f=false" rel="nofollow"><em>Almost Complex and Complex Structures</em></a> on page 233 he says "In fact, the absence of an almost complex structure on$S^{4k}$for$k\geq 1$and$S^{2n}$for$n\geq 4$was proved by Wu and jointly Borel and Serre respectively."</p> http://mathoverflow.net/questions/29499/yet-another-roadmap-style-request-a-second-bite-of-the-cherry/29503#29503 Answer by Justin Curry for Yet another 'roadmap' style request- a second bite of the cherry Justin Curry 2010-06-25T12:09:42Z 2010-06-25T12:09:42Z <p>I sympathize with your case. A 2.1 is really not bad. You shouldn't denigrate yourself and view your peripatetic interests as requiring redemption.</p> <p>Taking on a big unsolved problem without guidance or the background of a PhD student seems doomed to fail. Locking yourself in a library with all the world's books is unlikely to produce anything of merit. I have never heard of a case of a student producing something of "intriguing/charming/advanced" and using that to gain graduate admission. The romantic, amateur heroic view of math is largely bunk as pointed out by <a href="http://terrytao.wordpress.com/career-advice/does-one-have-to-be-a-genius-to-do-maths/" rel="nofollow">Terry Tao.</a></p> <p>There is still hope. I know that undergraduate research is less common in the UK, but I would expect that if you email lots of professors in areas of interest to you and basically offer yourself as cheap or free labor (undergrad student level), there is a good chance that you'll be taken on as an unofficial research student by someone. I know many cases of people in math and science using this sort of informal contact to start research projects that eventually develop into PhD positions. Making yourself known to a tenured professor who can write you a strong recommendation is probably enough to get you a PhD position somewhere (in the US, UK or Europe). It is unlikely that claiming to solve a big problem or do research on your own is going to be trusted by graduate committees. You need recommendations from people trusted in the academic community.</p> <p>There are MO users who have taken a decade or more off from education and successfully started PhD positions at Princeton and other top research institues. Good luck.</p> http://mathoverflow.net/questions/28695/what-should-we-teach-to-liberal-arts-students-who-will-take-only-one-math-course/28848#28848 Answer by Justin Curry for What should we teach to liberal arts students who will take only one math course? Justin Curry 2010-06-20T14:41:26Z 2010-06-20T14:41:26Z <p>In surveying the other responses to date it seems like many people have assumed that without assuming calculus the most we can hope for teaching undergraduate students is probability, statistics, fractions/percentages, brain teasers and puzzles.</p> <p><strong>Aren't we shooting too low?</strong></p> <p>For an extreme example of how far we might actually push such a course consider a quote of Arnol'd's in <em>On teaching mathematics</em>:</p> <blockquote> <p>By the way, in the 1960s I taught group theory to Moscow schoolchildren. Avoiding all the axiomatics and staying as close as possible to physics, in half a year I got to the Abel theorem on the unsolvability of a general equation of degree five in radicals (having on the way taught the pupils complex numbers, Riemann surfaces, fundamental groups and monodromy groups of algebraic functions). This course was later published by one of the audience, V. Alekseev, as the book The Abel theorem in problems.</p> </blockquote> <p>Set aside the hairy issue that this high school, which is already far more specialized than US high schools, was one of the premier math/physics high schools in Russia. Rather, pay attention to the fact that the first 220 pages of Alekseev's book is self-contained and no calculus is necessary.</p> <p>Also, consider the idea of a year-long course following Penrose's <em>Road to Reality</em>. Showing people mathematics' role in unraveling the secrets of the universe has always seemed far cooler to me than other tactics for inspiration.</p> <p>Let me be perfectly clear that I think actually requiring a course such as Arnol'd's across the board is overly optimistic. However, I think that by selecting the topics of such a course to reflect what mathematicians generally value could go a long way towards providing both <strong>a cultural appreciation of modern mathematics</strong> (as Lockhart's Lament would like) and <strong>an opportunity to think rigorously</strong> about initially simple objects (groups) and then more complex but visual objects (Riemann surfaces).</p> <p>If people are pessimistic that "liberal arts majors" don't have the goods to think about group theory, then I would much rather have a history of mathematics course that follows something like Stillwell's <em>Mathematics and Its History</em>, leaving students with the impression that mathematics has a rich philosophical undercurrent, than bore them with a meaningless pursuit of graph theory, probability and a smorgasbord of seemingly unrelated topics.</p> <p>The prevalent idea among many people seems to be that we have to make sure that basic numeracy is in place and that this is the math department's job. I don't think that innumeracy is a problem with the school system as it stands. People just forget their middle school and high school math because to them mathematics is an uninspired dead subject that is just plain boring. Let's change that. </p> http://mathoverflow.net/questions/22619/perverse-sheaves-macpherson-lecture-notes Perverse Sheaves - MacPherson Lecture Notes Justin Curry 2010-04-26T16:54:44Z 2010-04-27T11:40:05Z <p>I keep running across papers that refer to a set of lecture notes by Robert MacPherson at MIT during the fall of 1993 on Perverse Sheaves. There might also be a set of notes from lectures in Utrecht in 1994 taken by Goresky. For references to these elusive, unpublished notes see the work of Maxim Vybornov and A. Polishchuk.</p> <p>Does anyone have a copy or know how I could obtain a copy?</p> <p>Thanks!</p> http://mathoverflow.net/questions/21025/proofs-that-require-the-existence-of-large-finite-numbers/21075#21075 Answer by Justin Curry for Proofs that require the existence of large finite numbers Justin Curry 2010-04-12T06:54:01Z 2010-04-12T06:54:01Z <p>This isn't addressed to logicians, but it may be of interest. I happen to know of an example in PDE that was necessary in proving the well-posedness of radial solutions of the Nonlinear Schrodinger Equation:</p> <p><code>$$i u_{t}+\Delta u=|u|^{4}u$$</code></p> <p>for which J. Bourgain was awarded his Fields Medal for treating. (J. Bourgain, Global well-posedness of defocusing 3D critical NLS in the radial case, JAMS 12 (1999), 145-171).</p> <p>In one of the many many critical steps required in this proof, a bound on energy is required. A team (J. COLLIANDER, M. KEEL, G. STAFFILANI, H. TAKAOKA, and T. TAO) have now treated the non-radial case and make explicit the large ordinals used for bounding the energy. I quote from page 36 of their paper "Global well-posedness and scattering for the energy-critical nonlinear Schrödinger equation in R^3":</p> <blockquote> <p>"If one then runs the induction of energy argument in a direct way (rather than arguing by contradiction as we do here), this leads to very rapidly growing (but still finite) bound for M(E) for each E, which can only be expressed in terms of multiply iterated towers of exponentials (the Ackermann hierarchy). More precisely, if we use X ↑ Y to denote exponentiation X^Y, X↑↑Y :=X↑(X↑...↑X) to denote the tower formed by exponentiating Y copies of X, X↑↑↑Y :=X↑↑(X↑↑...↑↑X) to denote the double tower formed by tower-exponentiating Y copies of X, and so forth, then we have computed our final bound for M(E) for large E to essentially be M(E) ≤ C ↑↑↑↑↑↑↑↑ (CE^C). This rather Bunyanesque bound is mainly due to the large number of times we invoke the induction hypothesis Lemma 4.1, and is presumably not best possible."</p> </blockquote> <p><a href="http://arxiv.org/abs/math/0402129" rel="nofollow">http://arxiv.org/abs/math/0402129</a></p> http://mathoverflow.net/questions/17778/books-you-would-like-to-see-translated-into-english/17900#17900 Answer by Justin Curry for Books you would like to see translated into English. Justin Curry 2010-03-11T19:27:54Z 2010-03-11T19:27:54Z <p><em>Analysis Situs</em> by Poincare. </p> <p>This is the foundation of algebraic topology and illustrates its historical connection with dynamics.</p> <p>According to <a href="http://en.wikipedia.org/wiki/Henri_Poincar%C3%A9#Poincar.C3.A9.27s_writings_in_English_translation" rel="nofollow">Wikipedia</a> it has been translated, but I can't find a copy in English.</p> http://mathoverflow.net/questions/3134/whats-your-favorite-equation-formula-identity-or-inequality/17150#17150 Answer by Justin Curry for What's your favorite equation, formula, identity or inequality? Justin Curry 2010-03-05T01:34:11Z 2010-03-05T01:34:11Z <p>I learned Quantum Mechanics and Linear Algebra in tandem, so Schrodinger's linear time-independent equation has always had a special place in my heart. It shows that eigenvalues and eigenvectors are fundamental to our description of atomic physics. Also treating observables as operators was a great conceptual revolution.</p> <blockquote> <p>$H\psi=E\psi$</p> </blockquote> http://mathoverflow.net/questions/2144/a-single-paper-everyone-should-read/15347#15347 Answer by Justin Curry for A single paper everyone should read? Justin Curry 2010-02-15T17:22:25Z 2010-02-15T17:22:25Z <p>I had recommended to me from several prominent faculty the paper:</p> <blockquote> <p><strong>The Yang-Mills Equations over Riemann Surfaces</strong> Author(s): M. F. Atiyah and R. Bott Source: Philosophical Transactions of the Royal Society of London. Series A, Mathematical and Physical Sciences, Vol. 308, No. 1505 (Mar. 17, 1983), pp. 523-615 Published by: The Royal Society Stable URL: <a href="http://www.jstor.org/stable/37156" rel="nofollow">http://www.jstor.org/stable/37156</a></p> </blockquote> <p>One professor called it "the basis for truly 21st century mathematics." It is also reportedly accessible by beginning graduate students with some exposure to differential geometry and suitable for independent study or as a reading course. It is a 93 page paper and develops a lot of fundamental constructions and ideas from scratch. Here is <a href="http://www.ams.org/mathscinet-getitem?mr=702806" rel="nofollow">Martin Guest's review on MathSciNet.</a></p> http://mathoverflow.net/questions/14903/every-manifold-cobordant-to-a-simply-connected-manifold Every Manifold Cobordant to a Simply Connected Manifold Justin Curry 2010-02-10T14:42:29Z 2010-02-10T18:40:30Z <p>I am wondering if it is true that <strong>every compact, connected, oriented manifold is cobordant to a simply connected manifold.</strong></p> <p>I believe that some sort of surgery will do the trick. Roughly speaking, I want to add handles so that I can kill representative loops. However, I don't know if my surgery process builds a cobordism and it is hard to for me to see what the new boundary is. Another possibility is to build a Morse function that constructs the cobordism for free.</p> <p>It is hard for me to get an intuition for what is going on, because all the compact, oriented 2-manifolds are boundaries and consequently cobordant to the empty set. CP^2 seems like the "easiest" test case, but it is already simply-connected for cellular reasons. Lens Spaces might be a nice candidate, but the 3 dimensional ones can be realized as boundaries of some disc bundle on S^2.</p> <p>If possible, I'd prefer a constructive procedure, but any answer that helps elucidate the material is welcome.</p> http://mathoverflow.net/questions/14703/pde-on-manifolds/14716#14716 Answer by Justin Curry for PDE on manifolds Justin Curry 2010-02-09T00:11:17Z 2010-02-09T00:11:17Z <p>There are lots of possible answers to your question, but maybe here are some ideas. They aren't papers, but good projects.</p> <ul> <li><p>Method of Characteristics in First Order Nonlinear PDE can be interpreted very cleanly using contact topology and symplectic forms. This frees one up from coordinates, but you can then use the geometry to write down the full-blown Hamilton-Jacobi equations. See Vladimir Arnold's "Lectures on Partial Differential Equations" Chapter 2. In general a lot of dynamical systems problems can be recast completely in differential form theoretic notation. For a physics perspective Jose and Saletan's "Classical Dynamics: A Contemporary Approach" has some of this.</p></li> <li><p>Depending on how much you've done, one can prove the Hodge Decomposition Theorem using basic Sobolev space theory, Lax-Milgram and Fredholm Alternative. This isn't coordinate-independent per se, but just uses general functional-analytic machinery. We did this in a PDE class recently, and I only have my notes as a reference, but Griffiths and Harris's "Principles of Algebraic Geometry" seems to do the proof starting on page 84.</p></li> <li><p>You could also look at Nash's original paper on his embedding theorem, it basically reduces to a fix-point problem. However, this is necessarily coordinate-driven.</p></li> </ul> <p>Good luck. </p> http://mathoverflow.net/questions/13209/can-minimal-surfaces-be-characterized-by-some-universal-property/13218#13218 Answer by Justin Curry for Can minimal surfaces be characterized by some universal property? Justin Curry 2010-01-28T01:52:18Z 2010-01-28T02:14:12Z <p>I'm not sure if this answer provides you with the universal property that you desire, but there is such a category that unifies these concepts that you are after.</p> <p>Cohen, Jones and Segal introduced a concept known as the "Flow Category" in the paper <em>Morse Theory and Classifying Spaces</em>, which associates to any manifold with a Morse Function a category whose objects are the critical points of the Morse function and whose morphisms are the gradient trajectories of some gradient-like vector field. Here is the reference:</p> <p><a href="http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.38.5003" rel="nofollow">http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.38.5003</a></p> <p>You can get the paper on Ralph Cohen's page if you don't have university access:</p> <p><a href="http://math.stanford.edu/~ralph/papers.html" rel="nofollow">http://math.stanford.edu/~ralph/papers.html</a></p> <p>Recall that Morse Theory was invented by Marston Morse to study geodesics on manifolds. Geodesics correspond precisely to critical points of the Energy functional. I imagine that any variational problem fits into this framework. </p> <p>As a word of caution, understanding the space of gradient trajectories lies at the heart of Floer Theory, so if you want to understand Morse Theory on infinite dimensional spaces, prepared to get your hands dirty with some serious analysis. Comment if you want more references. Also, most of the above article is concerned with proving a very elegant result about the classifying space of this category for certain situations. It is very slick!</p> http://mathoverflow.net/questions/12684/switching-research-fields/12731#12731 Answer by Justin Curry for Switching Research Fields Justin Curry 2010-01-23T06:20:03Z 2010-01-23T06:25:46Z <p>I think our questioner is aware of the difficulties of switching fields and if not he or she soon will be, so let me be naive and try to be constructive. </p> <p>For Quantum Computation, Isaac Chuang and Michael Nielsen's "Quantum Computation and Quantum Information" has become a standard introduction to the subject, suitable for a graduate student in either mathematics, physics or computer science.</p> <p>Since I have no idea what background you have in PDEs (you could be a specialist in D-modules for all I know and find these suggestions childish), here are some texts I've become acquainted with:</p> <p>-V.I. Arnold's "Lectures on Partial Differential Equations" gives a beautifully geometric and intuitive understanding of PDEs, introducing and weaving together contact and symplectic geometry. The table of contents looks quite basic, but it contains the depth you should expect from Arnold.</p> <p>-Lawrence C. Evans "Partial Differential Equations" is nice and contains the basic notions from Functional Analysis, Sobolev Spaces, Weak Theory and Regularity Theory. It does a good job of being self-contained and trying to give physical interpretations of various PDEs.</p> <p>-Gilbarg and Trudinger have the classic "Elliptic PDEs of Second Order", which is dense, but a classic nonetheless.</p> <p>As a mathematician you don't need to learn how physicists think in the next year. Physicists have different ways of looking at problems and are constrained to their own paradigms just as mathematicians are. It is often quicker to pick up advanced physics if you know advanced mathematics, with many excellent bridge texts by world-class mathematicians. Examples that come to mind are Bott's "Morse Theory Indomitable" which includes an exposition of some of Witten's ideas for a mathematician. Atiyah's "Geometry and Physics of Knots" is also an excellent example of this. Feynman's Lectures are great, but won't advance you to research. It's more like a Caltech undergraduate degree bound in 3 volumes.</p> <p>Finally, as a note of inspiration, I have heard of at least two new faculty who self-studied PDEs in their post-doctoral years. One was supplanting a thesis in deformation theory and integrable systems, the other in knot theory and Floer homology. It is definitely a hard path to follow, but is sometimes necessary for growth. Also, bear in mind that Ed Witten was a history major as an undergrad, dropped out of economics grad school before applying for Princeton applied math and then switching to physics. Raoul Bott switched from electrical engineering to mathematics after his PhD (a much harder path, one might argue). Finally, my personal hero, Douglas Hofstadter, after quitting his Berkeley math PhD and finishing a 7+ year physics PhD in Oregon, then lived at home for a few years re-tooling himself as an AI researcher. Now he has a Pulitzer and tenure at a university -- not too shabby.</p> <p>Good luck!</p> http://mathoverflow.net/questions/2917/where-does-a-math-person-go-to-learn-quantum-mechanics/11604#11604 Answer by Justin Curry for Where does a math person go to learn quantum mechanics? Justin Curry 2010-01-13T01:46:15Z 2010-01-13T01:46:15Z <p>I agree with Scott firmly here. The Lagrangian and Hamiltonian formulation of mechanics is a beautiful subject with immediate doorways to symplectic geometry, but unnecessary for the appreciation of the basic postulates of QM ("The Universe is a vector space..."). Honestly, if Qiaochu wants to appreciate this clean formulation and not worry about "perturbative expansions of blah blah blah," this is sufficient.</p> <p>I am also a huge fan of Griffith's "Introduction to Quantum Mechanics" as a first book. It can be read without serious knowledge of electromagnetism and classical mechanics. I know that the MIT physics undergraduate curriculum does <em>not</em> require Lagrangian/Hamiltonian Classical Mechanics before their 3 term QM sequence. This means for Qiaochu that 8.01 and 8.02 is sufficient, 18.03 is more important, and I have heard of many a MIT student taking 8.05 without 8.04, but don't expect a physics professor to say that to you. (My apologies for the MIT-speak)</p> <p>The problem with learning "grown-up" mechanics (at MIT 8.07 or Sussman's 6.946J, which I recommend highly) before QM is that this path leads more naturally to understanding more differential-geometric concepts and consequently takes serious time. This is fine for a physicist and is probably a wise move before taking GR, but for a growing mathematician, I would advise understanding smooth manifold theory <em>before</em> trying to learn the more sophisticated and elegant approach to mechanics. This inevitably entails a much more firm grounding in ODE theory and I recommend V.I. Arnold's ODEs book for that.</p> <p>Once you have these two perspectives in hand you can then sit back and wonder how the QM Universe-as-a-Hilbert-space viewpoint and the GR Universe-as-a-C^2-manifold description can ever be reconciled.</p> http://mathoverflow.net/questions/11375/cw-structures-and-morse-functions-a-reference-request/11432#11432 Answer by Justin Curry for CW-structures and Morse functions: a reference request Justin Curry 2010-01-11T14:00:29Z 2010-01-11T14:00:29Z <p>Seeing that algori is asking for a reference, I'd like to offer Liviu Nicolaescu's "Invitation to Morse Theory" as a superb modern treatment of the subject. I am fairly certain the result you are looking for is in there.</p> http://mathoverflow.net/questions/10282/alternative-undergraduate-analysis-texts/11013#11013 Answer by Justin Curry for Alternative Undergraduate Analysis Texts Justin Curry 2010-01-07T05:58:34Z 2010-01-07T05:58:34Z <p>I'm surprised no one has said Marsden and Hoffman's "Elementary Classical Analysis", but perhaps it is too elementary or classical. I didn't learn from it as an undergrad, but I did find myself turning to it as I worked problems from "Berkeley's Problems in Mathematics" by de Souza and Silva. M&amp;H fleshes out a lot more detail, which Rudin spares for the sake of elegance or relegates to the exercises. I wish I had it or Korner's book "A First Second or Second First Course in Analysis" alongside Rudin when I first studied analysis. In particular, I prefer Marsden and Hoffman's treatment of Arzela-Ascoli over Rudin's.</p> http://mathoverflow.net/questions/120729/cosheafification/124929#124929 Comment by Justin Curry Justin Curry 2013-03-20T21:13:06Z 2013-03-20T21:13:06Z @arsmath: Yes, you are right. It has a pretty strong consistency requirement. If I understand correctly, the strength of Vopenka's principle lies between Reinhardt cardinals and unmeasurable cardinals, but I don't have any committed opinions as to how controversial this is. I tend not to worry too much about these things, but perhaps I should. http://mathoverflow.net/questions/120729/cosheafification/124973#124973 Comment by Justin Curry Justin Curry 2013-03-20T21:07:59Z 2013-03-20T21:07:59Z Nice! Does this proof assume Vopenka's principle? http://mathoverflow.net/questions/120729/cosheafification/124929#124929 Comment by Justin Curry Justin Curry 2013-03-20T20:48:09Z 2013-03-20T20:48:09Z @Ryan: No, a functor$F:C\to D$always can be used formally to define a functor$F^{op}:C^{op}\to D^{op}$. The assignment of objects remains the same, so$F^{op}(x)=F(x)$, but now a morphism$f:x\to y$in$C$becomes a morphism$f^{op}:y\to x$. The functor$F^{op}$sends$f^{op}$to$F(f)^{op}$.$F(f):F(x)\to F(y)$defines a morphism$F(f)^{op}:F(y)\to F(x)$in$D^{op}$, which is equal to$F^{op}(f^{op}):F^{op}(y)\to F^{op}(x)\$. I don't know what you mean by &quot;destroys covers.&quot; http://mathoverflow.net/questions/2150/exactness-of-filtered-colimits Comment by Justin Curry Justin Curry 2012-11-14T15:34:34Z 2012-11-14T15:34:34Z In case you are still curious about your &quot;vague principle,&quot; you might be thinking of the Freyd-Mitchell Embedding Theorem: <a href="http://en.wikipedia.org/wiki/Mitchell" rel="nofollow">en.wikipedia.org/wiki/Mitchell</a>'s_embedding_theorem http://mathoverflow.net/questions/87291/inverse-limit-of-spectral-sequences/87313#87313 Comment by Justin Curry Justin Curry 2012-08-19T21:27:15Z 2012-08-19T21:27:15Z &quot;inverse limit of directed systems of locally finite dimensional graded vector spaces is an exact functor&quot; - really? You don't need to assume a Mittag-Leffler condition? http://mathoverflow.net/questions/25054/different-ways-of-thinking-about-the-derivative Comment by Justin Curry Justin Curry 2012-01-28T04:02:23Z 2012-01-28T04:02:23Z Has anyone asked Thurston if this list exists? I'd love to see the original building up to #37. http://mathoverflow.net/questions/85717/segals-original-definition-of-a-topological-category Comment by Justin Curry Justin Curry 2012-01-15T15:04:50Z 2012-01-15T15:04:50Z @Tyler, But couldn't we define X_2 to be the fibered pullback over s and t? Couldn't we do something similar for the higher X_n so that it is a pullback (on the nose)? http://mathoverflow.net/questions/85717/segals-original-definition-of-a-topological-category/85718#85718 Comment by Justin Curry Justin Curry 2012-01-15T14:53:01Z 2012-01-15T14:53:01Z @Tim, Thanks for pointing out the Vaughan&lt;-&gt;John mistake. http://mathoverflow.net/questions/78850/derived-equivalence-of-sheaves-and-homotopy Comment by Justin Curry Justin Curry 2011-10-23T15:28:55Z 2011-10-23T15:28:55Z I realize that, but for someone who is more familiar with algebraic topology, where quasi-isomorphism refers to inducing isomorphisms on cohomology groups of a space, i.e. sheaf cohomology of the constant sheaf, then one is tempted to think that quasi-iso means a map inducing isomorphisms on sheaf cohomology and not cohomology sheaves (although the presheaf description of the latter is close to the former). I have to admit that although the definition was unambiguous, my semantic web conflated my intuition of the two. Thanks to Sam's answer, we now know when the two agree. http://mathoverflow.net/questions/78850/derived-equivalence-of-sheaves-and-homotopy/78854#78854 Comment by Justin Curry Justin Curry 2011-10-23T03:15:07Z 2011-10-23T03:15:07Z This is an excellent answer. Thank you. http://mathoverflow.net/questions/66098/sheaves-with-isomorphic-cohomology-but-not-quasi-isomorphic/66231#66231 Comment by Justin Curry Justin Curry 2011-05-27T22:43:06Z 2011-05-27T22:43:06Z So these two local systems have isomorphic cohomology, but are not q.i. Can they ever be distinguished by applying any standard endofunctors and then taking cohomology? http://mathoverflow.net/questions/66098/sheaves-with-isomorphic-cohomology-but-not-quasi-isomorphic Comment by Justin Curry Justin Curry 2011-05-26T21:59:36Z 2011-05-26T21:59:36Z @algori, What is the functor that you are applying? Yes, maps in the derived category, but other examples are good. http://mathoverflow.net/questions/44271/geometry-and-integrability-in-other-bundles/58916#58916 Comment by Justin Curry Justin Curry 2011-03-23T15:15:33Z 2011-03-23T15:15:33Z I ended up finding out a little about Lie Algebroids, but thanks for posting. Also I am familiar with some of the Marsden work on Dirac structures, but never thought of it in an integrable context. Thanks! http://mathoverflow.net/questions/53000/cohomology-of-structure-sheaves-algebraic-constructible-and-more/53021#53021 Comment by Justin Curry Justin Curry 2011-01-24T21:19:41Z 2011-01-24T21:19:41Z thank you for your insights and for being general. do you have a paper/notes that elaborates on the proof of the above statements? http://mathoverflow.net/questions/53000/cohomology-of-structure-sheaves-algebraic-constructible-and-more/53033#53033 Comment by Justin Curry Justin Curry 2011-01-24T18:24:21Z 2011-01-24T18:24:21Z I was just trying to emphasize to others that in my view GAGA says that complex analytic functions are as rigid as polynomials. I wanted to step away from this rigidity and consider real analytic functions (among other things). Also, I didn't know that we use &quot;meromorphic&quot; in the real case, thus the confusion.