User paul johnson - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T11:28:23Z http://mathoverflow.net/feeds/user/1102 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/97058/do-fixed-point-sets-in-equivariant-crepant-resolutions-have-the-same-cohomology Do fixed point sets in equivariant crepant resolutions have the same cohomology? How about for the specific case of Nakajima quiver varieties? Paul Johnson 2012-05-15T22:11:17Z 2012-06-15T21:17:02Z <p>A crepant resolution $f:Y\to X$ is a resolution of singularities with $f^*(K_X)=K_Y$. Crepant resolutions do not always exist, and when they exist they may not be unique. However, different crepant resolutions $Y_1$ and $Y_2$ share many properties. In particular, Kontsevich introduced motivic integration to prove that the $Y_i$ have the same Betti numbers. </p> <blockquote> <p>Suppose that $X$ has a $\mathbb{C}^*$ action, and that $f_i:Y_i\to X; i=1,2$ are equivariant crepant resolutions -- that is, the $Y_i$ have $\mathbb{C}^*$ actions and the $f_i$ are equivariant maps. Let $F_i\subset Y_i$ be the fixed point sets. Do the $F_i$ have the same Betti numbers? </p> </blockquote> <h2>How about for Nakajima quiver varieties?</h2> <p>This may be too much to expect in general, so let me mention that the case of particular interest to me is when $Y$ and $X$ are Nakajima quiver varieties (see <a href="http://arxiv.org/abs/0905.0686" rel="nofollow">Ginzburg</a>).</p> <p>More specifically, $X$ is the space $(\mathbb{C}^2/G)^n/S_n$, where a generator $g\in G=\mathbb{Z}/r\mathbb{Z}$ acts on $\mathbb{C}^2$ by $g(x,y)=(\omega x, \omega^{-1} y)$ for $\omega$ a primitive $r$th root of unity, and the $Y_i$ are certain other quiver varieties. Nakajima considers the $\mathbb{C}^*$ action given by $t(x,y)=(tx,y)$. I care about a more general action $t(x,y)=(t^a x,t^b y), a,b>0$. This $X$ is a Nakajima quiver variety, and other Nakajima quiver varieties provide natural equivariant crepant resolutions. </p> <p>Note that in this case it is known that the $Y_i$ are in fact diffeomorphic; however they are not equivariantly diffeomorphic for my torus action. Also note that using the action of the larger torus $(\mathbb{C}^*)^2$, we can see that $\chi(F_1)=\chi(Y_1)=\chi(F_2)$, so they at least have the same euler characteristic.</p> <p>For these Nakajima quiver varieties you can compute the $H^*(F_i)$ on a case by case basis using the obvious $(\mathbb{C}^*)^2$ action, and the question seems to have an affirmative answer. However, actually proving it in general using this method boils down to a difficult combinatorial question about partitions that is what I was originally considering. I only recently came up with the current formulation of the question, which seems like quite a natural question, and I was hoping that I would find the answer to my question already in the literature (for instance, in Kaledin's work on the symplectic McKay correspondence or symplectic resolutions more generally), but have had no luck so far, and so I turn to MO.</p> http://mathoverflow.net/questions/76044/intuition-behind-the-age-grading-in-quantum-cohomology-of-orbifolds/76055#76055 Answer by Paul Johnson for Intuition behind the age grading in quantum cohomology of orbifolds Paul Johnson 2011-09-21T12:46:54Z 2011-09-21T12:46:54Z <p>It is my understanding that the original motivation did come from GW theory: Chen and Ruan came across the orbifold cohomology ring, and its product, as a byproduct and first step of defining the orbifold quantum cohomology. The strongest motivation I have for the grading also comes from this perspective, and maybe we have to start with the question of why Chen and Ruan were trying to define orbifold GW theory to begin with.</p> <p>The fairy tale story I have in my head is that string theorists found that they could do string theory on mildly singular spaces. Often, rather than dealing with singularities, we instead get rid if it, by either a) resolving the singularity, or b) deforming the equations to something smooth. If our singular space is actually a smooth as an orbifold, orbifold cohomology gives us a third option, c).</p> <p>Now, the physicists also had intuition that, in nice cases and if we look at it the right way, all three approaches should essentially give us the same thing. So, for instance, passing from a) to b) directly can go by the name "conifold transition".</p> <p>To get back to the age grading: a) and c) are related by what has come to be known as the Crepant Resolution conjecture: if $f:Y\to X$ is a crepant resolution of an orbifold $X$ (i.e. $f$ is a resolution and $f^*(K_X)=K_Y$), then $X$ and $Y$ should have "the same" quantum cohomology. In particular, the base graded vector spaces of $X$ and $Y$ should be the same. </p> <p>A strong motivation for the age grading of quantumn cohomology is that it makes this true: if $X$ is an orbifold that has a crepant resolution, then the graded vectorspace $H^{orb}(X)$ is isomorphic to the graded vector space $H^*(Y)$, for any crepant resolution $Y$. This statement about the graded vector space was proved using motivic integration by <a href="http://arxiv.org/abs/math/0110228" rel="nofollow">Yasuda</a>, and independently by Lupercio and Poddar, and provides a lot of evidence that the age shifting is the "right" shifting. I should note that the stronger statement about the orbifold quantumn cohomlogy rings being "the same" is still open, and in the general case requires some work to even state correctly, as in this <a href="http://mathoverflow.net/questions/47322/quantum-cohomology-rings-as-invariants/47352#47352" rel="nofollow">answer</a> by Jim Bryan.</p> <p>I feel like a really good answer to your very last question should be possible and not that hard, but I don't have it at my fingertips now. So I'll just give a short "yes": the obstruction bundles used to define the orbifold cup product are exactly cohomology spaces of bundles on orbifolds, and so the RR formula is going to pop up there. If you're okay with the age grading showing up in the RR formula, then from the definition of the quantum product it follows that the age grading of $H^{orb}$ has to be what it is. I feel like I'm glib and just sort of restating your question -- I'm trying to say I think you're right.</p> http://mathoverflow.net/questions/66929/number-of-weighted-trivalent-trees/67034#67034 Answer by Paul Johnson for number of weighted trivalent trees Paul Johnson 2011-06-06T12:35:01Z 2011-06-06T12:35:01Z <p>This is just building off the previous discussion -- I will just give some indication of why Dan's $f$ and $g$ are Legendre transforms of each other. This is essentially a fleshed out version of some of what Frédéric was saying that I wanted to work out in detail -- it would look too small and ugly as a comment. </p> <p>We're going to use some basic facts about the Lambert W-function.<a href="http://en.wikipedia.org/wiki/Lambert_W_function" rel="nofollow">Lambert W-function</a> and the <a href="http://en.wikipedia.org/wiki/Legendre_transformation" rel="nofollow">Legendre transformation</a>, but nothing past what you can find in those Wikipedia links.</p> <p>First, showing that $f$ and $g$ are Legendre transforms of each other is essentially showing that their derivatives are compositional inverses of each other. So we will show this for $f^\prime$ and $g^\prime$, and not use anything else about the Legendre transformation.</p> <p>The Lambert $W$ function $W(x)$ is defined by the functional equation $x=W e^W$. You recognize it's coming into play when you start seeing $n^n$ terms, as they show up in the taylor expansion of it and simple expressions related to it. In particular, wikipedia tells us that:</p> <p>$$\left(\frac{W}{x}\right)^r = e^{-rW}=\sum_{n\geq 0} r(n+r)^{(n-1)} \frac{(-x)^n}{n!}$$</p> <p>The first equation is the functional equation, and in general to get the taylor expression you want to use Lagrange inversion and the functional equation.</p> <p>In any case, if we set $r=-1$, take $0^0=1$ and hopefully not make any sign errors, we should obtain</p> <p>$$g^\prime(x)=1-e^{W(-x)}.$$ Simple calculus gives $$f^\prime=-(1-x)\ln(1-x).$$ From these expressions it is immediate that $f^\prime(g^\prime(x))=x$ by using the functional equation for $W$.</p> http://mathoverflow.net/questions/63318/orbifold-covering/63325#63325 Answer by Paul Johnson for orbifold covering Paul Johnson 2011-04-28T19:05:13Z 2011-04-28T19:05:13Z <p>Kevin's comment gives easy counterexamples -- take an orbifold that's topologically a sphere, but has lots of Z_2 points -- we could give it any negative orbifold characteristic that we want, and yet this would obvious never cover a two holed torus with no orbifold structure, say.</p> <p>You might rule that specific example out with also asking something about the euler characteristic of the coarse moduli spaces, but I don't think it would help much.</p> <p>A little more broadly, you have the usual condition of covering spaces, that the fundamental group of the cover is a subgroup of the fundamental group downstairs. It would be interesting to know whether this was sufficient: if $\pi_1(Y)$ is a subgroup of $\pi_1(X)$, does $Y$ cover $X$? </p> <p>I haven't thought about this hard, though. I suspect this is probably known, somewhere in the literature on Fuchsian groups. Again, not being careful, but you might hope this reformulated question would be related to something along the lines of whether if G was abstractly a subgroup of H as groups, and now we consider them as subgroups of the automorphisms of the hyperbolic plane, can we conjugate G into H? </p> http://mathoverflow.net/questions/63197/known-mirror-calabi-yau-pairs/63200#63200 Answer by Paul Johnson for Known Mirror Calabi-Yau pairs Paul Johnson 2011-04-27T17:21:01Z 2011-04-27T17:21:01Z <p>I'm not an expert at this at all, but a couple of observations:</p> <p>1) Rather than just hypersurfaces I believe Batyrev and Borisov have a more general description for complete intersections in toric fano thing manifolds.</p> <p>2) This <a href="http://arxiv.org/abs/1011.5006" rel="nofollow">recent paper</a> of Alan Stapledon works on orbifolding the Batyrev-Borisov correspondence, and begins with a series of references to known and conjectured mirror pairs, that I think is probably close to the state of the art. Examining those references will probably help, and a key phrase involved in one is the "pfaffian-Grassmannian correspondence" </p> http://mathoverflow.net/questions/63095/how-does-one-understand-grr-grothendieck-riemann-roch/63157#63157 Answer by Paul Johnson for how does one understand GRR? (Grothendieck Riemann Roch) Paul Johnson 2011-04-27T12:25:14Z 2011-04-27T12:25:14Z <h2>Introduction: GRR gives relations in the tautological ring</h2> <p>I can't speak directly to the potential applications you had in mind as far as Kodaira dimension, but I can say something about Mumford's application of GRR to the moduli space of curves. It seems that it's quite close to what you imagine, and in fact it's very important in the study of (a certain part of) the cohomology ring of the moduli spaces $\mathcal{M_g}$, and its relatives such as Deligne-Mumford space $\overline{\mathcal{M}}_{g,n}$ (now the curve has $n$ marked points, and we compactify by adding certain nodal curves should the points try to collide or the complex structure of the curve degenerate), and even further into Gromov-Witten theory. I'm going to give an overview of this story, building out of your question (I hope).</p> <p>The part of the cohomology ring of $\mathcal{M_g}$ I'm talking about is called the <em>tautological ring</em>, and a gentle survey-introduction is Ravi Vakil's <a href="http://math.stanford.edu/~vakil/files/cimemay26.pdf" rel="nofollow">The moduli space of curves and Gromov-Witten theory</a>. Those notes do not explicitly mention GRR, but they do quote a result that comes directly from Mumford's GRR calculation, and what I'm going to try to do is explain this a little bit. </p> <p>I'd also like to mention that, as far as I understand it, this direction of application is essentially what Mumford had in mind. The paper he does this calculation is, after all, entitled <a href="http://books.google.com/books?hl=en&amp;lr=&amp;id=df_X80mKSi0C&amp;oi=fnd&amp;pg=PA235&amp;dq=mumford+twoard+an+enumerative&amp;ots=IKtif2znLe&amp;sig=CD36-cUyAAx9Aj48G4QEdi5PDVY#v=onepage&amp;q&amp;f=false" rel="nofollow">"Toward an Enumerative Geometry of the Moduli Space of Curves"</a>.</p> <h2>Warm-up: Grassmannian</h2> <p>You can see in the first paragraph of Mumford's paper that he is explicitly modeling what he's doing after the cohomology of the Grassmannian, so I'm going to spend a paragraph on them, to motivate what's coming in the moduli space of curves. </p> <p>On the one hand, we have the schubert cycles, given by the loci of planes that intersect the a fixed flag with given dimensions. On the other hand, since each point represents a vector space, these vector spaces fit together to give a tautological vector bundle, and we can take cycles representing the chern classes of this bundle, and get different classes -- it's not necessarily clear at all that these different tautological cycles should be related, but they are. </p> <p>Mumford, and many after him, are trying to find similar relations between different tautological classes in $\mathcal{M}_g$. Mumford ends his first paragraph with "Moreover, it appears that many geometrically natural classes are expressible in terms of a small number of basic classes" -- this is akin to that description of the Grassmannian, and it's what GRR will give us.</p> <h2>We start with your basic idea</h2> <p>Rather than get into all the technical details of it, I just want to point out that he proceeds essentially exactly as you imagined here: </p> <blockquote> <p>A good relative riemann roch theorem would then relate the universal canonical sheaf on the total space, to the canonical sheaves on the curve fibers, and the cohomology of the push down of the universal sheaf to the base space, the moduli space of curves.</p> </blockquote> <p>Rephrased slightly differently: let $\pi:\mathcal{M_{g,1}}\to\mathcal{M_g}$ be the map from a moduli space of curves with one marked point to the moduli space of curves with no marked points -- it turns out this is exactly the universal family. Then we have the universal canonnical sheaf $\omega_\pi$ on $\mathcal{M_{g,1}}$ that you were discussing. We can use this to get cohomology classes in $H^*(\mathcal{M_g})$ in two different ways: first take its chern character, and then push down to $\mathcal{M_g}$, or first push down to $\mathcal{M_g}$, and then take the chern character. </p> <p>These two alternatives give rise to a priori very different looking cohomology classes on $\mathcal{M_g}$, but GRR says that, after fiddling with Todd classes, they are the same.</p> <h2>Chern then pushforward</h2> <p>Let's see what happens when we take the first path. Since $\omega_\pi$ is one dimensional, taking the chern character is simply exponentiating $c_1(\omega_\pi)$. The class $c_1(\omega_\pi)$ is called the psi class $\psi$. Note that usually this is defined as the first chern class of the tangent bundle to the curve at the marked point, but through the identification of the universal curve with $\mathcal{M_{g,1}}$, these are equivalent. But I should warn you that this is no longer quite true if we start adding more marked points or nodes to our curves. So taking the Chern character of $\omega_\pi$ gives powers of $\psi$ on $\mathcal{M_{g,1}}$, and now if we push these forward we get the Morita-Mumford-Miller kappa classes $\kappa_i=\pi_*(\psi^{i+1})\in H^{2i}(\mathcal{M_g})$ -- indeed, this is the definition of $\kappa_i$.</p> <h2>Pushforward then Chern</h2> <p>Now, what happens if we go in the other direction? To pushforward $\omega_\pi$, we take the cohomology of $\mathcal{M_g}$ -- since $h^0(C, \omega_C)=g$, independent of the curve $C$, we have $\pi_*(\omega_\pi)=\mathbb{E}$, where $\mathbb{E}$ is a dimension $g$ vector bundle on $\mathcal{M_g}$ known as the hodge bundle. More simply, the fiber of $\mathbb{E}$ over a curve $C$ are the sections of the canonical bundle of $C$. The chern <em>classes</em> of the $\mathbb{E}$ are known as the $\lambda$ classes: $\lambda_i=c_i(\mathbb{E})$. Taking the Chern character of the $\mathbb{E}$ then would then give us a bit of a mess of polynomials in the $\lambda$ classes.</p> <h2>Comparing them</h2> <p>So taking the two different paths from $K(\mathcal{M_{g,1}})$ to $H^*(\mathcal{M_g})$ gives two different looking types of tautological classes, the $\kappa$ classes and the $\lambda$ classes. Since we set this up with GRR in mind, we should now see a relation between them. </p> <p>In this case it turns out that this relationship cleans up rather nicely if we package it in a generating function, and a better answer might explain how, but I'll just note that since we were working with the relative cotangent bundle to begin with, the relative Todd class can be manipulated into just giving us more $\kappa$ classes, and then with more mucking around with characteristic classes, it turns out we can express this relationship very beautifully in terms of generating functions:</p> <p>$$\sum_{i=0}^\infty \lambda_i t^i=\exp\left(\sum_{j=i}^\infty \frac{B_{2j}\kappa_{2j-1}}{2j(2j-1)}t^{2j-1}\right).$$</p> <p>Here $B_{2j}$ are the Bernoulli numbers, coming from the Todd class. This is the formula in Ravi's notes I alluded to earlier: he cites Faber for this particular expression.</p> <h2>Extensions</h2> <p>I've done this just for $\mathcal{M_g}$ for simplicity, but I want to indicate here that you can get a lot more gas out of the same basic idea.</p> <p>First, you can add marked points and boundary points and it essentially goes through the same. The universal curve is still just adding another marked, and then forgetting it. What gets a little complicated is that the relative dualizing sheaf $\omega_\pi$ stops being equal to just the cotangent line at the extra point when our curve becomes singular, but we can understand how they differ, and so we get some additional contributions involving the boundary strata -- I think Mumford already started to deal with this, and Faber and Pandharipande certainly dealt with it.</p> <p>Also, you can extend this to Gromov-Witten theory, and consider moduli spaces of stable maps, and consider a curve with marked points together with a map $f:C\to X$, and play the same game there. Or better, we can first pull back a bundle from $E\to X$, and play the game described above with $f^*(E)$. In case $E$ is a line bundle, $s:X\to E$ a line bundle, and $Y=s^{-1}(0)$ the vanishing set, this can give relationships between the Gromov-Witten invariants of $X$ and $Y$ in terms of the chern classes of $E$. And in case $E$ is a negative, this can express the Gromov-Witten invariants of the total space of $E$ in terms of the Gromov-Witten invariants of $X$ and the chern classes of $E$. This is worked out in Tom Coates's thesis, and in his joint Annals paper with advisor, Givental: <a href="http://annals.math.princeton.edu/2007/165-1/p02" rel="nofollow">Quantum Riemannâ€“Roch, Lefschetz and Serre</a>, and is very important to GW theory, as its the method by which we can understand $X$ that is a complete intersection in a toric variety $Y$ -- this method relates the GW theory of $X$ to that of $Y$, and since $Y$ is toric we can localize with respect to the torus action to compute its Gromov-Witten invariants. </p> http://mathoverflow.net/questions/62764/a-natural-refinement-of-the-a-n-arrangement-is-to-consider-all-2n-1-hyperpla A natural refinement of the $A_n$ arrangement is to consider all $2^n-1$ hyperplanes given by the sums of the coordinate functions. Have you seen this arrangement? Is it completely intractable? Paul Johnson 2011-04-23T18:36:26Z 2011-04-23T21:43:19Z <h2>The short version</h2> <p>Here is an extremely natural hyperplane arrangement in $\mathbb{R}^n$, which I will call $R_n$ for <em>resonance arrangement</em>. </p> <p>Let $x_i$ be the standard coordinates on $\mathbb{R}^n$. For each nonempty $I\subseteq [n]={1,\dots,n}$, define the hyperplane $H_I$ to be the hyperplane given by $$\sum_{i\in I} x_i=0.$$ The resonance arrangement is given by all $2^n-1$ hyperplanes $H_I$. The arrangement $R_n$ is natural enough that it arises in many contexts -- to first order, my question is simply: have you come across it yourself? </p> <p>This feels rather vague to be a good question, and after giving some background on where I've seen this I'll try to be a bit more specific about what I'm looking for, but my point is this arrangement has a rather simple and natural definition, and so crops up in multiple places, and I'd be curious to hear about more of them even if you can't specifically connect it to what follows.</p> <h2>What I knew until this week</h2> <p>I came across this arrangement in my study of double Hurwitz numbers -- they are piecewise polynomial, and the chambers of the resonance arrangement are the chambers of polynomiality. I don't want to go into this much more, as it's unimportant to most of what follows Though I will say that conjecturally double Hurwitz numbers could be related to compactified Picard varieties in a way which would connect this arrangement up with birational transformations of those. Also, the name "resonance arrangement" was essentially introduce in this context, by Shadrin, Shapiro and Vainshtein.</p> <p>It apparently comes up in physics -- I only know this because the number of regions of $R_n$, starting at $n=2$, is 2, 6, 32, 370, 11292, 1066044, 347326352 ... Sloane sequence <a href="http://oeis.org/A034997" rel="nofollow">A034997</a>, which you will see was entered as "Number of Generalized Retarded Functions in Quantum Field Theory" by a physicist.</p> <p>You might expect by that rate of growth that this hyperplane arrangement is completely intractable, and more specifically, what I would love from an answer is some kind qualitative statement about how ugly the $R_n$ get. Which brings us to:</p> <h2>Connection to the GGMS decomposition</h2> <p>I got to thinking about this again now and decided to post on MO of it because of <a href="http://mathoverflow.net/questions/62052/finding-the-vertices-of-a-polyhedral-complex-coming-from-a-git-wall-and-chamber-d" rel="nofollow">Noah's question</a> about the vertices of a variation of GIT problem, where this arrangement is lurking around -- see there for more detail. Allen's brief comment there prompted me to skim some of his and related papers to that general area, and I found the introduction to <a href="http://arxiv.org/abs/0903.3694" rel="nofollow">Positroid varieties I: juggling and geometry</a> most enlightening, together with the discussion at Noah's question could give another suggestion why $R_n$ is perhaps intractable. Briefly:</p> <p>The arrangement $R_n$ is a natural extension of the $A_n$ arrangement. One common description of the $A_n$ arrangement is as the $\binom{n+1}{2}$ hyperplanes $y_i-y_j=0, i,j\in [n]$ and $y_i=0, i\in [n]$. However, one can consider the triangular change of variables $$y_k\mapsto \sum_{j\leq k} y_j$$. </p> <p>This changes the hyperplanes to $$\sum_{i\leq k \leq j} y_k=0.$$ </p> <p>These hyperplanes are no longer invariant under permutation of the coordinates, and if we proceed to add the entire $S_n$ orbit of them, we get the resonance arrangement $R_n$.</p> <p>From the discussion on Noah's question and the introduction to the Positroid paper, we see that this manipulation is a shadow of the GGMS decomposition of the Grassmannian, and describing this decomposition in general seems to be intractable in that it requires identifying whether matroids are representable or not. So, what I'd really like to know how is much of the "GGMS abyss", as they refer to it, is reflected in the resonance arrangement? Is it hopeless to describe and count its chambers? </p> http://mathoverflow.net/questions/62052/finding-the-vertices-of-a-polyhedral-complex-coming-from-a-git-wall-and-chamber-d/62763#62763 Answer by Paul Johnson for Finding the vertices of a polyhedral complex coming from a GIT wall and chamber decomposition Paul Johnson 2011-04-23T18:35:11Z 2011-04-23T18:35:11Z <p>This is not a real answer. However, I'd seen something closely related to this before, but they were tangential to what I cared about and I didn't sit and think about it. This question made me stop and think about it again (for way to long now), and I'm posting some of what I've learned as it may help, and that so I can get back to what I <em>should</em> have been doing the last few days -- though I really enjoyed thinking about this and would be happy to think more about it later. </p> <p>So, all I really have is a slight reformulation that "reduces" your question to something I had seen before, that this has prompted me to ask MO about myself. You can see a tiny bit more structure after this reduction. Reduce is in quotes because I'm not sure it's any easier, or that the reduction is particularly effective computationally, but I did find it pleasing conceptually. </p> <p>The way my previous experience has connections with your question is if we forget some of the structure you have, and simply consider the following central hyperplane arrangement in $\mathbb{R}^n$. Let the $x_i$ be the coordinate functions on $\mathbb{R}^n$. The hyperplane arrangement has one hyperplane for each of the $2^n-1$ nonempty subset $I\subset [n]$, with the corresponding equation given by: $$\sum_{i\in I} x_i=0.$$</p> <p>Another description of the hypersimplex is to slice the unit $n$-cube by the $n-1$ hyperplanes $\sum x_i=k, 1\leq k \leq n$. Then the $n$-pieces that result are hypersimplices. So, rather than consider your problem one at a time, I'm going to stack up all the hypersimplices for fixed $n$ and varying $k$, and get a decomposition of the unit cube, and discuss the vertices of this decomposition. In a way this will give slightly more structure than you had looking at one hypersimplex at a time.</p> <p>One could connect your problem to the hyperplane arrangement I've described above by also adding translations of the hyperplanes above and restricting to the unit cube, but here's a more pleasing way: let's take the quotient of $\mathbb{R}^n$ by the usual lattice $\mathbb{Z}^n$, and consider the same set of hyperplanes. The hyperplanes will now wrap around and give the translations of the hyperplanes as you ask above, giving a polyhedral decomposition of the $n$-torus -- you're asking for the vertices of this decomposition.</p> <p>I should say that this trick of viewing this on the torus is in my head in part from skimming Alexeev's paper "Compactified Jacobians and the Torelli map", and that I first saw this hyperplane arrangement in my work on double Hurwitz numbers, which are piecewise polynomial, with walls given by this hyperplane arrangement, and which are rather conjecturally related to wall crossing behavior for some kind of universal compactified Jacobian.</p> <p>One way to describe a vertex would be a set $J$ of $n$ hyperplanes in that arrangement intersecting transversally. Note that the same set of $n$ hyperplanes, as they wrap around, will in general intersect multiple times on the torus, so that the set $J$ will in general label some set $G_J$ of vertices of the decomposition. I have called this set $G_J$, because in fact they form a finite subgroup of the torus: the definition of $G_J$ essentially makes it the kernel of a surjective homomorphism from $T^n$ to $T^n$. And it is easy to figure out what this group $G_J$ is from the set $J$ of hyperplanes -- it's the usual way to get a finite abelian group from a cone from toric geometry.</p> <p>So, though I found all of that pleasing, and at least some more can be done with it, I'm not sure it's actually helpful -- the hyperplane arrangement is huge and a mess, and I'm not saying we can just take the cones that correspond to chambers of the hyperplane arrangement -- I'd be surprised of that arrangement is simplicial, and even if it I see no reason that we could just consider the cones of regions of the arrangement anyway, and not arbitrary sets of transverse hyperplanes.</p> <p>I will mention something else I learned, that you may already know and that Allen was pointing at and that I found out only in trying to understanding what Allen was saying. There's a rather natural decomposition of the hypersimplex into simplices, with vertices only the vertices of the hypersimplices, but it is not $S_n$ invariant, and your decomposition is what you get by acting by the $S_n$ action. I learned this just recently from the first few pages of Lam and Postnikov, <a href="http://arxiv.org/abs/math/0501246" rel="nofollow">Alcoved Polytopes I</a> -- and rather than write it out I'll just send you there, though I'll note that I found it more convenient to imagine this happening on the torus rather than on the unit cube.</p> <p>This naturally relates to Allen's comment connecting this to the moment map from the Grassmannian -- you're probably more aware of this than I am, but I found the introduction to his paper with Lam and Speyer on Positroid Varieties pertinent, as it talks about what this story means upstairs in the Grassmannian corresponds to, and that it's a huge mess, although perhaps it's somehow cleaner downstairs.</p> http://mathoverflow.net/questions/62088/products-of-conjugacy-classes-in-s-n/62106#62106 Answer by Paul Johnson for Products of Conjugacy Classes in S_n Paul Johnson 2011-04-18T11:02:41Z 2011-04-18T11:02:41Z <p>Short answer: Yes, on Hurwitz spaces.</p> <p>Let's set these numbers up as the structure constants of $Z_d=Z(\mathbb{C}[S_d])$, the center of the group ring of the symmetric group $S_d$. The ring $Z_d$ has basis $K_{\mu}$, where $\mu$ is a partition of $d$, and $K_\mu$ represents the sum of all permutations of cycle type $\mu$. Then multiplication gives </p> <p>$$K_\mu K_\nu=\sum_{\lambda} C^\lambda_{\mu,\nu} K_\lambda$$ for some numbers $C^\lambda_{\mu,\nu}$, which are what you're interested in. I've seen these called <em>connection coefficients</em>, in work of Goulden and Jackson, for instance, their paper <a href="http://www.jstor.org/stable/2161793?seq=1" rel="nofollow">Transitive Factorisations into Transpositions and Holomorphic Mappings on the Sphere</a> , which starts to get you a simple connection to geometry: by looking at ramified covers of the sphere. I'll talk about this a bit first, giving a rough sketch and some pointers, and then I'll address some of Mariano's comments.</p> <p>This easiest connection to geometry is what you asked about in your <a href="http://mathoverflow.net/questions/9415/hurwitz-encoding" rel="nofollow">previous question</a> as "Hurwitz encoding", and David's answer there was good so I'll take that as background. You can start turning this into a problem about intersection theory by looking at Hurwitz Spaces. </p> <p>You can make various flavours of these, but lets call the most basic one $H_{g,d}$, the moduli space of all holomorphic maps $\pi:\Sigma\to \mathbb{P}^1$ of degree $d$ from a smooth genis $g$ Riemann surface $\Sigma$ to the Riemann sphere $\mathbb{P}^1$. Generically, such maps will all have simple ramification, and by the Riemann-Hurwitz formula there will be $r=2g-2+2d$ such points of ramification, and so we see that $H_{g,d}$ will have complex dimension $r$. We will be able to view your numbers as suitable intersections on the Hurwitz space $H_{g,d}$. </p> <p>There is a map from $H_{g,d}$ to $\mathbb{P}^r=(\mathbb{P}^1)^r/S_r$ that forgets $\Sigma$ and just remembers the $r$ branch points, (the critical values of $\pi$), counted with multiplicity. This is sometimes called the branch map, and I believe it is essentially what is known as the Lyashko-Looijenga map, and so I'll call this map LL. </p> <p>The degree of the map LL is what is known as a Hurwitz number, and translating everything into monodromy we see that it counts the number of tuples of $r$ transpositions $t_i$ in $S_d$ with the product of the $t_i$ being the identity, divided by $d!$ coming from automorphisms of the cover, or choosing a labeling of the $d$ sheets of the cover, depending on your viewpoint. </p> <p>To understand your connection coefficients geometrically, for a partition $\mu$ and a point $p\in \mathbb{P}^1$ we could define a cohomology class $\alpha(\mu, p)$ to consist of those maps in $H_{g,d}$ where $\pi$ has ramification profile $\mu$ over $p$. Then, if we've set up $g$ correctly with respect to $\mu, \nu$ and $\lambda$, the numbers $C^{\lambda}_{\mu,\nu}$ should be, again, up to some factor of automorphisms, the number of points in the triple intersection $\alpha(\mu,p_1)\cap \alpha(\nu,p_2)\cap \alpha(\lambda, p_3)$. </p> <p>I'm not addressing some stuff (for isntance, connected versus discnonected covers) or necessarily giving you the most useful view in practice, but this is the simplest way to something like what you want, I think -- the Hurwitz space $H_{g,d}$ is not compact, and we'd want to compactify it (admissible covers is the first way, but this winds up not being normal, and you can use some orbifold Gromov-Witten theory and compactify with see <a href="http://arxiv.org/abs/math/0106211" rel="nofollow">twisted stable maps</a> to get the normalization). But hopefully that's some idea of how this would go. To see this viewpoint used in practice, there are for instance, papers for instance of Lando and Zvonkine on the Arxiv -- I'm not sure where exactly you'd want to start. Through something known as the ELSV formula this story gets connected to intersection numbers on the moduli space of curves, which might be what you had in mind...</p> <p>To connect into what Mariano was saying in comments, you'd want to get into the of a permutation $\sigma$ -- the minimal number of transpositions $\sigma$ factors into. Let's call this the weight of sigma -- for a permutation of cycle type $\mu$, it is equal to $|\mu|-\ell(\mu)$, where $\ell(\mu)$ is the number of parts of $\mu$. The center of the group ring $Z_d$ is filtered by the weight, and the "top" coefficients are ones where the weight adds -- where $d-\ell(\lambda)=d-\ell(\mu)+d-\ell(\nu)$. </p> <p>In our geometric viewpoint, the weight is the amount of ramification above a certain point, and the top coefficients correspond to covers where all components are genus zero, the coefficients where the weight is off by two means we have a genus one cover, and similarly -- this filtration is geometrically filtering by the genus of our cover. The "top" coefficients are particularly nice in that they are independent of $d$ and so when you take the associated graded for each $s_d$ this plays well with the natural inclusions between the $S_d$ and you get some universal ring out of all the $S_d$ the Farahat-Higman ring.</p> <p>Mariano's mention of the hilbert scheme of points in the plane is a bit of a different longer story here -- the brief outline as I like to think about it is that we can view $Z_d$ as the Chen=-Ruan orbifold cohomology of the $\mathcal{B}S_d=point/S_d$. The stack $\mathbb{C}^{d}/S_d$ will have the same vector space of cohomology, but the grading will be different -- this is what algebraic geometers call "age" that Mariano referred to. This age induces exactly the filtration above. The filtration is just doubled for $\mathbb{C}^{2d}$, and the Hilbert Scheme of points is a crepant resolution of this space, and so you get the relation on homology above. This is a long story, and seems slightly off from what you want.</p> http://mathoverflow.net/questions/39475/gromov-witten-and-integrability-2/39724#39724 Answer by Paul Johnson for Gromov-Witten and integrability 2. Paul Johnson 2010-09-23T10:39:20Z 2010-09-23T10:39:20Z <p>Here's a sketch of my understanding of where the difficulty lies with higher genus curves. It got kind of long and vague, at parts, but hopefully it explains a few problems.</p> <p>In Gromov-Witten theory, I'm aware of two or three general approaches to integrability currently. Certainly there's overlap among these approaches:</p> <ol> <li><p>The quantumn cohomology of $X$ turns the cohomology of $X$ into a Frobenius manifold, and then Dubrovin has connections to various integrable hierarchies.</p></li> <li><p>In some cases, the Gromov-Witten theory can be nicely expressed in terms of various Fock-spaces and the infinite wedges, and then you can get connections to the integrable hierarchies related to various infinite dimensional lie groups from the Kyoto school.</p></li> <li>Matrix Models, and in particular the work of Eynard and Orantin. This seems to have mostly implemented in terms of the Topological vertex, which is rephrasing things in terms of combinatorics.</li> </ol> <p>For higher genus curves, approach 1 is completely a no-go: there are no positive degree maps from a sphere to a higher genus curve, so the quantum cohomology is just the usual cohomology.</p> <p>The GW-theory of higher genus curves is computed by Okounkov and Pandharipande very much in the general method of 2. The GW/Hurwitz correspondence shows that what they call the "stationary sector" (descendents of point classes only -- not the identity or odd cohomology classes. Is this the same as the stable sector?) is equivalent to Hurwitz theory, which is completely computable in terms of the symmetric group. There are nice ways of computing these characters, and this is where the connections to the infinite dimensional lie algebras and come up. However, this computation becomes much more complicated as we increase the genus. I'd like to explain how it's an entirely different beast for genus 2 or bigger.</p> <p>The GW/Hurwitz correspondence turns insertions of point classes into ramification data, and hence as far as counting goes, into multiplying elements in the symmetric group. The ramification that shows up is nice and is easy to write in terms of the infinite wedge (free fermion) and I think can be written in terms of matrix model type stuff as well. One particular nice bit is that any given insertion only produces permutations with bounded supports: even if we let the degree get big (so considering symmetric groups $S_d$ for $d$ large, we're only going to have a few nontrivial cycles in these permutations. Most points in $S_d$ will be fixed. </p> <p>For genus 0, we're only multiplying these special elements in the symmetric group, and this is why everything is so beautiful here.</p> <p>For higher genus, one way to keep everything in terms of just the symmetric group is degenerate the curve by pinching off $g$ cycles, so that we again have a genus zero curve, but now with $g$ pairs of points identified. At each of these identified pairs of points, we're going to need to insert inverse permtuations, and we're going to have to sum over all permutations in $S_d$ this way for each pair of points.</p> <p>This is where things get ugly -- in higher genus, we have to consider these arbitrary permutations.</p> <p>In genus one, things aren't all that bad: we only have two arbitrary permutations. So we can start with one of them, multiply in turn by the nice permutations that we know how to do, and then at end, instead of multiplying two arbitrary permutations, we only have to check that we have the same permutation, as the whole product has to be the identity. Essentially, we're taking the trace of some nice operator on the infinite wedge, and this why the quasi-modular forms show up here -- work of Bloch-Okounkov shows these are quasimodular forms.</p> <p>Once we get to genus two though, we loose this as well. We really have to be able to multiply two arbitrary permutations of $S_d$. I haven't read Mironov-Morozov's stuff on this closely at all, but seemed to recall them having some type of non-integrablity results for the general multiplication of three permutations, but couldn't find exactly this statement. The start of section three of <a href="http://front.math.ucdavis.edu/0904.4227" rel="nofollow">this paper</a> might touch on this, though.</p> <p>The representation theory viewpoint is probably better for integrable systems, but I think it's similarly more difficult. </p> http://mathoverflow.net/questions/38293/virasoro-constraints-for-the-generating-function-of-hurwitz-numbers/39026#39026 Answer by Paul Johnson for Virasoro constraints for the generating function of Hurwitz numbers. Paul Johnson 2010-09-16T21:53:12Z 2010-09-16T21:53:12Z <p>This is something I've long been meaning to think about seriously, so maybe I can get back to you with something better later. In the meantime... </p> <p>I wasn't aware of anything explicitly in the literature about this until googling just a bit ago, when I found the paper "Virasoro constraints for Kontsevich-Hurwitz partition function" by Mironov and Morozov. </p> <p>I haven't fully digested their paper yet, but they seem to be using the viewpoint of the Kazarian paper on Hodge integrals and KP hierarchy I mentioned in my answer to your other question. </p> <p>Briefly: the ELSV formula relates single Hurwitz numbers to Hodge integrals, essentially the GW theory of a point. Kazarian shows how this transformation can be done explicitly by a certain operator M&amp;M call $\hat{U}$ (the quantization of a quadratic function). </p> <p>M and M's point seems to be that since the generating functions are related by this operator, and one of them satisfies Virasoro, we can conjugate the Virasoro operators by $\hat{U}$ and get Virasoro operators for the other generating function. The particular form they take seems to be a bit of a mess, and I worry about some details, but I've only skimmed that paper very quickly.</p> <p>But philosophically, this seems to be going about things backwards: the Hurwitz side is really simpler, and the above setup is often used to show that the GW of a point satisfies Virasoro. I feel we should be able to construct Virasoro operators for single Hurwitz numbers more easily with some kind of direct approach.</p> <p>Hurwitz theory is all about statements about the symmetric group, and there are constructions (I read about it in a Frenkel-Wang paper) that build Virasoro actions out of the symmetric group. </p> <p>I'm not fully motivating this, but single and Hurwitz theory is very conveniently done on a certain Fock space. Basically, you have the operator that multiplies by a transposition, (called $M_0$ by Kazarian, $\mathcal{F}_2$ by Okounkov-Pandharipande, physicists have some other notation for it...), and you have the operators $\alpha_n, n\in\mathbb{Z}$ that add or remove cycles of length $n$ from a conjugacy class, and that form a heisnberg algebra. </p> <p>These operators are exactly what you need to do Hurwitz theory. But they're also what Frenkel and Wang use to construct a Virasoro algebra -- essentially, the commutators $L_n=[\alpha_n, M_0]$. So we might hope that some similar construction would give us easier to understand Virasoro constraints for single Hurwitz numbers. But I haven't spent the necessary time trying to nail it down.</p> <p>As far as double Hurwitz numbers go, I'm a little less hopeful for the above vague ideas. All I know is that Goulden, Jackson and Vakil have a few lines about trying and failing to construct Virasoro operators in their "Towards the Geometry of Double Hurwitz Numbers" paper.</p> http://mathoverflow.net/questions/38294/gromov-witten-and-integrability/39023#39023 Answer by Paul Johnson for Gromov-Witten and integrability. Paul Johnson 2010-09-16T21:25:10Z 2010-09-16T21:25:10Z <p>Short answer: essentially, the point and $P^1$ are the only spaces where the GW generating function is a tau-function. However, you mentioned two variations on this spaces: equivariant orbifold versions. And there are other variations that go a bit further -- twisted and relative invariants, and, a little wilder, Landau-Ginzburg theory. But in all cases, everything seems to be a reduction of either the KP hierarchy or the 2-Today hierarchy.</p> <h2>Twisted Invariants</h2> <p>Let $L_i$ be some line bundles on $X$, and let $B$ be the total space of $\oplus L_i$. The GW theory of $B$ doesn't make sense since it's not compact, and so neither will $\overline{\mathcal{M}}_{g,n}(B, \beta)$, but we can take a $\mathbb{C}^*$ on $B$ that fixes $X$ and just acts on the fibers, and then use localization on $\overline{\mathcal{M}}_{g,n}(B, \beta)$. </p> <p>The fixed point locus will be $\overline{\mathcal{M}}_{g,n}(X, \beta)$, but the invariants will be different because of extra terms coming from the euler class of the normal bundle. We call and so it will make sense to integrate, and we define these to be $GW(B)$, and call them twisted invariants of $X$. </p> <p>A Grothendieck-Riemann-Roch calculation begun by Mumford and extended and expanded by Faber-Pandharipande and Coates-Givental will reduce $GW(B)$ to $GW(X)$, but in a messy way that would for instance change the invariants, and certainly change around the integrable hierarchy.</p> <p>So, for instance, in the case of a point, we'll just get integrals over $\overline{\mathcal{M}}_{g,n}$, the normal bundle will essentially be copies of the Hodge bundle, and so we'll get $\lambda$ classes appearing in our integrals along with the usual $\psi$ classes: these are called Hodge integrals.</p> <p>In the case that there is just one line bundle (so $B=\mathbb{C}$, the integral will be linear in the $\lambda$ classes, and will be the integral appearing in the ELSV formula. <a href="http://arxiv.org/abs/0809.3263" rel="nofollow">Kazarian</a> shows that the resulting GW theory satisfies the KP hierarchy.</p> <p>The case with three line bundles ($B=\mathbb{C}^3$) is the case covered by the topological vertex. As far as I know, there is no known integrable hierarchy satisfying the whole thing. But <a href="http://arxiv.org/abs/math/0310408" rel="nofollow">Zhou</a> has shown that certain specializations lead to KP and 2-Toda type equations. </p> <p>Similarly, we can twist the GW theory of $\mathbb{P}^1$ and hope to get integrable hierarchies here. <a href="http://arxiv.org/abs/1002.0582" rel="nofollow">Brini</a> has made some progress here for the direct sum of two line bundles, so that $B$ is a three-fold. In particular, for the resolved conifold (the total space of $\mathcal{O}(-1)\oplus\mathcal{O}(-1)$, he gets connections with the Ablowitz-Ladik hierarchy, apparently some reduction of 2-Toda.</p> <h2>Relative Invariants</h2> <p>Another variation we can play is to use relative invariants, working relative to a divisor.</p> <p>In dimension zero this doesn't work, but we can work relative to 0 and $\infty$ on $\mathbb{P}^1$. Okounkov and Pandharipande have shown that this flavor also satisfies some 2-Toda type hierarchy.</p> <p>In the paper of Zhou mentioned above, he also sets up certain relative generating functions for toric varieties that satisfy KP and 2-Toda hierarchies.</p> <h2>Landau-Ginzburg</h2> <p>In another direction, if we move slightly away from GW theory we can get more interesting examples. Fan, Jarvis and Ruan have recently finished rigorously constructing a Landau-Ginzburg A-model -- essentially, a Gromov-Witten theory for hypersurface singularities. </p> <p>These theories have a central charge that acts as a dimension. The central charge 0 case has an ADE classification. Witten conjectured that the theory of these should satisfy the corresponding Kac-Wakimoto / Drinfeld-Sokolov reduction of the KP hierarchy. Note that for $A_1$, this is the Gromov-Witten theory of a point, and the usual KdV hierarchy.</p> <p>In the $A_r$ case, the analytic machinery of FJR is not needed to define the theory, and it goes under the name $r$-spin curves -- essentially, we're doing integrals over the moduli space of orbifold curves with an chosen $r$-th root of the canonical bundle. Witten's conjecture was proven here by Faber, Shadrin and Zvonkine. FJRW have recently proven the D and E cases.</p> http://mathoverflow.net/questions/38671/how-to-recover-partition-from-its-multiset-of-hook-lengths/38808#38808 Answer by Paul Johnson for How to recover partition from its multiset of hook lengths? Paul Johnson 2010-09-15T11:22:36Z 2010-09-15T11:22:36Z <p>I wanted to at least give a systematic way to show your given multiset is not a set of hook lengths after my flubbed comment. So: taking the $n$ quotients of a partition gives us a constraint on its possible hook lengths.</p> <p>In particular, take the set of all hook lengths that are divisible by $n$, and then divide each of them by $n$. This new set of numbers will be the set of hook lengths of the $n$ different partitions that are its $n$-quotients. </p> <p>If your multiset came from a partition, then together its two 2-quotients would have hook lengths 1,2,2,4,5, which can't be the hook lengths of two partitions.</p> <p>Alternatively, since 5 of the hook lengths were divisible by 2, together the two 2-quotients will be a partition of 5, and when translated back to the original partition will account for 10 of the 12. Therefore, the 2-core must have had size 2. But the 2-cores are exactly the staircase positions, and so 2 isn't a 2-core.</p> <p>I wish I had a good source for explaining cores and quotients. I think of them by translating through the Maya diagrams , as in Figure 5 on Page 49 of <a href="http://arxiv.org/abs/1008.4205" rel="nofollow">this paper.</a> Maybe <a href="http://books.google.co.uk/books?id=e9YJVd2cwBQC&amp;pg=PA120&amp;lpg=PA120&amp;dq=maya+diagram&amp;source=bl&amp;ots=Ke5ii6iXU1&amp;sig=0t399UFhxUhbtQioesU2ed-LLZE&amp;hl=en&amp;ei=mKSQTL_uPKaH4gatlMTfDQ&amp;sa=X&amp;oi=book_result&amp;ct=result&amp;resnum=7&amp;ved=0CCcQ6AEwBg#v=onepage&amp;q=maya%2520diagram&amp;f=false" rel="nofollow">The description here</a> helps, too.</p> <p>As far as an algorithm to construct the possible partitions with a given hook length set thinking in terms of Maya diagrams and possibly cores and quotients could possibly provide a useful way to control the searching and branching. </p> <p>I can imagine an algorithm that starts by immediately taking just the even hook lengths and dividing them all by 2, to find the hook-lengths of the 2-quotients. Then we can find the 2-core if it exists. There would now be a lot of branching: we have to distribute the hook-lengths of the 2-quotients over all possible splittings. But once we've chosen a splitting, we can recursively call our program again on the 2-quotients, which will be much smaller partitions. </p> <p>When it reaches the end, you then have to glue the original partition back together from the core and quotient and check whether the parition has the right hook lengths -- this seems quite expensive.</p> <p>I make no claims that this algorithm is any good -- I don't know whether the division helps compared to the branching. But as someone who hasn't programmed much the recursion seems relatively easy to code.</p> http://mathoverflow.net/questions/32409/examples-of-folk-theorems/33640#33640 Answer by Paul Johnson for Examples of "folk theorems" Paul Johnson 2010-07-28T10:12:31Z 2010-07-28T10:12:31Z <p>The example I first learned was the following: a 2-D TQFT is equivalent to a Frobenius algebra.</p> <p>This is discussed and stated as a folk theorem by <a href="http://xxx.lanl.gov/abs/hep-th/9401023" rel="nofollow">Voronov</a>; later, a careful proof was <a href="http://home.gwu.edu/~labrams/docs/tqft.ps" rel="nofollow">written up</a> and published by Lowell Abrams. See also the <a href="http://mat.uab.es/~kock/TQFT.html" rel="nofollow">book</a> by Joachim Kock. </p> http://mathoverflow.net/questions/26788/shuffling-decks-of-cards-where-not-all-cards-are-distinguishable/26808#26808 Answer by Paul Johnson for Shuffling decks of cards where not all cards are distinguishable Paul Johnson 2010-06-02T10:32:33Z 2010-06-02T10:32:33Z <p>A place to start is the recent preprint:</p> <p><a href="http://front.math.ucdavis.edu/0905.4698" rel="nofollow">Riffle shuffles of a deck with repeated cards</a> Sami Assaf, Persi Diaconis, K. Soundararajan.</p> <p>They also have another preprint about reducing this to a "rule of thumb", and cite some earlier work of Conger and Viswanath, available <a href="http://www.math.lsa.umich.edu/~divakar/" rel="nofollow">here</a>.</p> <p>Pulling carelessly from tables in the different papers without checking exactly what I'm saying, it seems that in the total variation distance, (the measure which yields seven shuffles for a standard deck), four shuffles are already sufficient for blackjack, while for the separation distance (the measure yielding eleven shuffles for a standard deck), it still takes nine shuffles. </p> http://mathoverflow.net/questions/8158/a-question-about-gromov-witten-invariant/8180#8180 Answer by Paul Johnson for a question about Gromov-Witten invariant Paul Johnson 2009-12-08T11:55:02Z 2009-12-08T12:04:57Z <p>To further qualify Charles's yes: that these moduli spaces are orbifolds instead of manifolds does result in rational numbers, but this is quite natural and not much of a problem. The orbifolds here are just resulting because we're counting things that have automorphisms, here, for instance, the map from P^1 to P^1 given by the polynomial z^d has Z_d as its automorphisms: we can multiply a point in P^1 by a dth root of unity and not change where it maps to). Whenever you count things with automorphisms it's quite natural to count each thing weighted by 1/(the size of its automorphism groups), or to rigidify the things we're counting by adding some kind of extra structure so they no longer have automorphisms. </p> <p>As an example: Cayley's formula that there are n^(n-2) trees on n <em>labeled</em> vertices - the labeling of the vertices guarantees that the objects we're counting do not have automorphisms, and we get an integer - we've rigidified the problem. If we wanted to count the number of trees on n <em>unlabeled</em> vertices, the problem is much more difficult. However, if we weight each such tree by the inverse of its automorphism group, then the problem has a nice answer again: it's simply n^(n-2)/n!. My point is: the rationality is not the ugly part of what's going on.</p> <p>The ugly part is that these moduli spaces of maps are not even orbifolds: they have much worse singularities, and can have different components of different dimension. From deformation theory, we expect these moduli spaces to have a certain dimension. To get a finite number, we put conditions on the map that cut this dimension down until its zero. Geometrically, you should think of each of these conditions as a cycle on the moduli space, and we want to intersect them. Doing this intersection naively doesn't work when the space is singular, and furthermore the moduli space might be smooth but have a dimension different than what we were expecting. But a lot of hard work shows that these spaces have a "virtual fundamental class" of the dimension that we expect, and using this we can proceed as above to get a number. But in doing this, we've lost the sense in that we're counting something.</p> <p>But it strikes me that perhaps that's not necessarily what the questioner was after; most typically this is done for smooth, projective varieties of C, but somehow the part that really matters is the symplectic structure: Gromov-Witten invariants can be defined for any symplectic manifold - they will all have almost complex structures J that "play nicely" with the symplectic form omega, and we're "counting" these maps. Or: all this works for orbifolds (which are really smooth objects), but not singular spaces.</p> <p>The over $\mathbb{C}$ bit is pretty necessary, I think - people have looked a little at doing in positive characteristic, but one big problem is that the orbifold stuff, which I was just telling you isn't really a problem, can be a big problem in positive characteristic if the order of your automorphisms aren't coprime with the characteristic. </p> http://mathoverflow.net/questions/6891/gerbes-for-a-cyclic-group-or-maybe-g-m-too/6906#6906 Answer by Paul Johnson for Gerbes for a cyclic group. (or maybe G_m too) Paul Johnson 2009-11-26T18:13:30Z 2009-11-26T18:13:30Z <p>This probably deserves to be worked out in more detail, and with more sophistication, but quickly and easily...</p> <p>The gerbes you get from your construction are all banded - so certainly any nonbanded gerbe would work, but this is kind of silly so I'll construct a banded example for you.</p> <p>When thinking about gerbes I find it useful to think about the zero dimensional case: then we're really doing group theory, which feels more familiar. </p> <p>In other words, let's think about BG: a line bundle over BG is just a one dimensional representation of G; a (banded) H-gerbe over BG is a (central) extension of G by H. </p> <p>Let's take G=Z/2Z X Z/2Z, and H=Z/2Z. Which extensions do we get from your construction? Well, a representation will either be trivial, or have one element that acts by multiplication by (-1). In the group extension that we get, this representation should have a nontrivial square root. In the trivial representation case we can take just take the trivial extension G X H, and the representation where (0,0, 1) act by multiplication by -1. as our nontrivial square root. For the nontrivial representations, the extension will be (Z/4Z) X (Z/2Z), and the square root will be the representation where (1,0) acts by multiplication by i.</p> <p>In particular, we notice that all the extensions we got in this manner were abelian groups. However, G has a central extension by H that isn't abelian: D_4, the dihedral group of order 8 - rotation by 180 degrees is central. So BD_4 is a banded H gerbe over BG that doesn't come from your construction.</p> <p>One can then easily modify this example to get something a little more geometric: take your favorite space X with a free G action. Give it a D_4 action by having rotation by first mapping to D_4 to G and then acting on X, so rotation by 180 acts trivially. Then the global quotient [X/D_4] should be a banded Z/2Z gerbe over [X/(Z/2Z X Z/2Z)] that doesn't come from your construction. </p> http://mathoverflow.net/questions/5936/whats-so-great-about-blackboards/5966#5966 Answer by Paul Johnson for What's so great about blackboards? Paul Johnson 2009-11-18T13:27:29Z 2009-11-18T13:27:29Z <p>As a speaker, I have to say that a lot of my preference for blackboards over whiteboards (and, to a lesser degree, computers) is aesthetic. I have always sort of rolled my eyes at writers waxing on about the physical feeling of a pen in their hand, or the force required to press the keys of a typewriter as opposed to those on a computer, but perhaps I should take it easy on them, because I have similar feelings about chalk.</p> http://mathoverflow.net/questions/2461/line-bundles-on-moduli-spaces/2470#2470 Answer by Paul Johnson for Line bundles on moduli spaces Paul Johnson 2009-10-25T13:14:09Z 2009-10-25T13:14:09Z <p>Or why would you want line bundles on anything? If you wanted to show it was projective, for instance, you would want an ample line bundle. In general, I think a lot of the motivation for all the technical studying of divisors on M_{g,n} is to try to understand its birational geometry.</p> http://mathoverflow.net/questions/97058/do-fixed-point-sets-in-equivariant-crepant-resolutions-have-the-same-cohomology Comment by Paul Johnson Paul Johnson 2012-06-15T11:11:39Z 2012-06-15T11:11:39Z Hiraku, I could have been clearer. We certainly have $\chi(F_1)=\chi(X)=\chi(F_2)$ by localization. But it appears that more is true -- namely, the betti numbers of the $F_i$ are equal. I've edited the question to make this a bit clearer. http://mathoverflow.net/questions/97058/do-fixed-point-sets-in-equivariant-crepant-resolutions-have-the-same-cohomology/97186#97186 Comment by Paul Johnson Paul Johnson 2012-05-17T17:46:25Z 2012-05-17T17:46:25Z Thanks for the reference -- unfortunately, it appears that with my group action they aren't in fact equivariantly diffeomorphic. I'm really looking at just the ordinary hilbert scheme of points on C^2. Z_r acting antidiagonal on C^2 induces an action on the hilbert scheme, and the components of the fixed point set are the quiver varieties. The C^* action I have is the induced action from a C^$action on C^2. With the whole (C^*)^2, you know the fixed point sets agree -- they are finite sets. But for different resolutions the normal bundles of the fixed point sets are different. http://mathoverflow.net/questions/76044/intuition-behind-the-age-grading-in-quantum-cohomology-of-orbifolds/76055#76055 Comment by Paul Johnson Paul Johnson 2011-09-22T09:50:23Z 2011-09-22T09:50:23Z You are correct. The point is that$X$will not have a crepant resolution unless$H^{orb}(X)$is actually integrally graded. In fact, for$X$to have a crepant resolution, all the degree shifting numbers should be even. To see that this is reasonable, note that over a point$x$in$X$, the isotropy group$G_x$will act on$K_X$. To be crepant, we want$f^*(K_X)=K_Y$-- since$K_Y$has no orbifold structure, it seems that we should have that$K_X$is the trivial representation of$G_x$, which means that$G_x$acts on$T_xX$with determinant 1. http://mathoverflow.net/questions/63626/sticking-subgroups-in-different-directions-symmetric-group Comment by Paul Johnson Paul Johnson 2011-05-06T18:42:44Z 2011-05-06T18:42:44Z Jordan -- the standard use of &quot;support&quot; for an element$g$of$S_n$that I've seen is just the elements of$[n]$that$g$moves -- the support is the complement of the fixed point set. Presumably that's what they mean... http://mathoverflow.net/questions/3591/mathematicians-who-were-late-learners-list/3595#3595 Comment by Paul Johnson Paul Johnson 2011-05-06T11:45:16Z 2011-05-06T11:45:16Z @dan: Smale's PhD is from Michigan -- perhaps you were thinking of the letter from Ray Wilder that appears at the bottom of the page here, and mostly on the top of the next page: <a href="http://books.google.co.uk/books?id=MJ22fDnalXEC&amp;lpg=PA37&amp;ots=PbP5jaTaQM&amp;dq=stephen%20smale%20wife%20letter&amp;pg=PA37#v=onepage&amp;q&amp;f=false" rel="nofollow">books.google.co.uk/&hellip;</a> The book is &quot;Stephen Smale: the mathematician who broke the dimension barrier&quot; vy Steve Batterson http://mathoverflow.net/questions/63318/orbifold-covering/63325#63325 Comment by Paul Johnson Paul Johnson 2011-04-28T21:16:04Z 2011-04-28T21:16:04Z Ah, this is much easier than I could see at the beginning. I didn't process that essentially I was asking if the fundamental group determined the orbifold surface, and it gets much cleaner when you think of it that way. Thanks! http://mathoverflow.net/questions/63197/known-mirror-calabi-yau-pairs/63200#63200 Comment by Paul Johnson Paul Johnson 2011-04-28T09:38:56Z 2011-04-28T09:38:56Z Again, not an expert on this -- but the introduction to Alan's paper claims they do, and cites their Inventiones paper, on the Arxiv here: <a href="http://arxiv.org/abs/alg-geom/9509009" rel="nofollow">arxiv.org/abs/alg-geom/9509009</a> The abstract is: We prove in full generality the mirror duality conjecture for string-theoretic Hodge numbers of Calabi-Yau complete intersections in Gorenstein toric Fano varieties. So, from everything I can see, they do. http://mathoverflow.net/questions/62764/a-natural-refinement-of-the-a-n-arrangement-is-to-consider-all-2n-1-hyperpla Comment by Paul Johnson Paul Johnson 2011-04-27T19:41:21Z 2011-04-27T19:41:21Z Just commenting to thank Richard -- this is one kind of thing I wanted to hear: that people had put some real thought into it and not gotten anything interesting. I wasn't up-voting that you didn't get anywhere. http://mathoverflow.net/questions/62764/a-natural-refinement-of-the-a-n-arrangement-is-to-consider-all-2n-1-hyperpla Comment by Paul Johnson Paul Johnson 2011-04-23T21:56:27Z 2011-04-23T21:56:27Z @Theo: Also, you are half responsible for the long title: somewhere you mentioned that a title can be about a &quot;tweet and a half&quot; long, and that was an odd enough description that it stuck in my head. Good work. http://mathoverflow.net/questions/62764/a-natural-refinement-of-the-a-n-arrangement-is-to-consider-all-2n-1-hyperpla Comment by Paul Johnson Paul Johnson 2011-04-23T21:48:32Z 2011-04-23T21:48:32Z Think I fixed the TeX. And I was a bit torn about it not being specific enough myself, but it was never going to get asked if I kept worrying about it. Next time will be better. http://mathoverflow.net/questions/62052/finding-the-vertices-of-a-polyhedral-complex-coming-from-a-git-wall-and-chamber-d/62763#62763 Comment by Paul Johnson Paul Johnson 2011-04-23T18:36:57Z 2011-04-23T18:36:57Z Here's the question I asked about this hyperplane arrangement: <a href="http://mathoverflow.net/questions/62764/a-natural-refinement-of-the-a-n-arrangement-is-to-consider-all-2n-1-hyperpla" rel="nofollow" title="a natural refinement of the a n arrangement is to consider all 2n 1 hyperpla">mathoverflow.net/questions/62764/&hellip;</a> http://mathoverflow.net/questions/62088/products-of-conjugacy-classes-in-s-n/62266#62266 Comment by Paul Johnson Paul Johnson 2011-04-19T16:33:22Z 2011-04-19T16:33:22Z Taking$\mu=\nu$and then summing over all$\mu$has connections to covers of a genus 1 curve, rather than a sphere -- essentially, you're degenerating a loop in the torus to a point, or cutting it along the loop, to get a sphere with two punctures. You dont' know what the original monodromy was around the loop you cut (summing over all of them), but to glue together you need them to be the same ($\mu=\nu$). Dijkgraaf connects with modular forms: <a href="http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.8.4194" rel="nofollow">citeseerx.ist.psu.edu/viewdoc/&hellip;</a> Mike Roth has a gentle intro <a href="http://www.mast.queensu.ca/~mikeroth/notes/covers.pdf" rel="nofollow">mast.queensu.ca/~mikeroth/notes/covers.pdf</a> http://mathoverflow.net/questions/62088/products-of-conjugacy-classes-in-s-n Comment by Paul Johnson Paul Johnson 2011-04-19T09:21:21Z 2011-04-19T09:21:21Z That was really vague -- I can fill it out a bit more if you're a bit more precise on what you're looking for/curious about. http://mathoverflow.net/questions/62088/products-of-conjugacy-classes-in-s-n Comment by Paul Johnson Paul Johnson 2011-04-19T09:20:23Z 2011-04-19T09:20:23Z John -- what do you mean, &quot;get&quot;? My answer below is mostly about the geometry, and though I'm not sure how useful it is, it does set up a geometric meaning for arbitrary$\mu,\nu,\lambda$-- change the genus,$g$. As for algorithmically computing them, I'm am not sure what the state of the art is. I'd want to use character theory to do this, as multiplication is semisimple in that basis. My gut feel for this is that things are nice when you force$\ell(\mu)$to be close to zero (most characters vanish) or$|\mu|\$ (the permutation is mostly the identity), but messier in between. http://mathoverflow.net/questions/62088/products-of-conjugacy-classes-in-s-n Comment by Paul Johnson Paul Johnson 2011-04-18T11:09:06Z 2011-04-18T11:09:06Z Amritanshu -- I address this a bit at the end of my answer, but he's talking in part about: M. Lehn and C. Soerger. Symmetric groups and the cup product on the cohomology of Hilbert schemes.