User tim perutz - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T03:01:26Z http://mathoverflow.net/feeds/user/2356 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/125652/homology-in-the-a-infty-world/126791#126791 Answer by Tim Perutz for Homology in the $A_\infty$ World Tim Perutz 2013-04-07T17:15:06Z 2013-04-07T17:15:06Z <p>The following answer does not address question 1 in full, but it shows, I think, what one would need to think through.</p> <p>Take a dg algebra $(A,d)$ over a commutative ring $k$. Specify a splitting of the cocycles as cohomology plus coboundaries: $$\mathrm{ker} (d) = HA \oplus \mathrm{im}(d)$$ (such a splitting exists provided that $HA$ is projective), and let $i\colon HA\to A$ be the resulting inclusion. We can then construct canonically</p> <p>(i) An $A_\infty$ structure on $HA$ with differential $\mu^1=0$;</p> <p>(ii) an $A_\infty$ morphism $\mathcal{I} \colon HA \to A$ whose first term is the given inclusion $i$.</p> <p>So $HI \colon HA \to HA$ is the identity map. This is Kadeishvili's construction.</p> <p>These structures are defined by explicit recursive formulae. As such, they already have some desirable functoriality properties. For instance, if a group $G$ acts by automorphisms on $A$, and if the summand $i(HA)$ is $G$-invariant, then the $A_\infty$ data will be $G$-equivariant. </p> <p>Now suppose we specify in addition a splitting of $A$ as $\mathrm{ker} (d) \oplus A'$. We then have a projection $p\colon A \to HA$, and this extends canonically to </p> <p>(iii) an $A_\infty$ morphism $\mathcal{P}\colon A\to HA$, with</p> <p>(iv) a nullhomotopy of $\mathcal{P}\circ \mathcal{I}- \mathrm{id}_{HA}$.</p> <p>Moreover, there exists a nullhomotopy of <code>$\mathcal{I}\circ \mathcal{P}-\mathrm{id}_A$</code>, but I'm not sure how canonical this nullhomotopy is. A reference for these assertions is Paul Seidel's book <i>Fukaya categories and Picard-Lefschetz theory</i>, chapter 1. In general, $A_\infty$ quasi-isomorphisms induce quasi-equivalences of their module-categories, and this gives an affirmative answer to question 2.</p> <p>Now take a dg morphism $f\colon A \to B$, and suppose we're given splittings of $A$ and of $B$ as coboundaries plus cohomology plus complement and that $f$ respects these summands. Then we can construct an $A_\infty$ morphism <code>$$\mathcal{H}f = \mathcal{P}_B \circ (Hf) \circ \mathcal{I}_A \colon A\to B,$$</code> as indicated in the question. Under composition $g\circ f$ of splitting-respecting dg maps, there is a homotopy $\mathcal{H}g\circ \mathcal{H}f \simeq \mathcal{H}(g\circ f)$. The homotopy comes from the existence of a homotopy <code>$I_B \circ P_B \simeq \mathrm{id}_B$.</code> </p> <p>So we get a functor from the category of dga with splittings to the category of $A_\infty$-algebras and homotopy classes of morphisms. Presumably, if one can establish just how canonical the homotopy <code>$I_B \circ P_B \simeq \mathrm{id}_B$</code> is, one can sharpen the functoriality statement. </p> http://mathoverflow.net/questions/123619/explicit-model-of-bsu2-in-terms-of-singular-complex/123635#123635 Answer by Tim Perutz for Explicit model of BSU(2) in terms of singular complex Tim Perutz 2013-03-05T16:51:15Z 2013-03-05T16:51:15Z <p>The infinite quaternionic sphere $S(\mathbb{H}^\infty)=\bigcup_n{S(\mathbb{H}^n)}$ is contractible and carries a free action of the unit quaternions $SU(2)=Sp(1)$. The quotient $\mathbb{H}P^\infty$ is a model for $BSU(2)$, and $\mathbb{H}P^n$ is then an approximation suitable for computing homotopy groups or the cohomology ring in low degrees.</p> http://mathoverflow.net/questions/120819/open-problems-in-seiberg-witten-theory-on-4-manifolds/120827#120827 Answer by Tim Perutz for open problems in Seiberg-Witten Theory on 4-Manifolds Tim Perutz 2013-02-05T03:24:43Z 2013-02-05T03:24:43Z <p>One basic structural problem about the SW invariants is the question of <i>simple type</i>: suppose that $X$ is a simply connected 4-manifold with $b^+>1$, and $\mathfrak{s}$ a $\mathrm{Spin}^c$-structure such that $SW_X(\mathfrak{s})\neq 0$. Must $\mathfrak{s}$ arise from an almost complex structure? This is true when $X$ is symplectic (Taubes in "$SW\Rightarrow Gr$") but open in general.</p> <p>The 11/8-conjecture (that for a closed Spin 4-manifold $X$ of signature $\sigma$, one has $b_2(X)\geq 11|\sigma|/8$) is open. SW theory has yielded strong results in this direction (Furuta's 10/8 theorem); proving the conjecture via SW theory is very hard but might be possible. </p> <p>Essentially all of the fundamental questions about the classification of smooth 4-manifolds, or about the existence and uniqueness of symplectic structures on them, are open. We do not know how much Seiberg-Witten theory sees. For instance:</p> <p>Suppose $X$ is a closed 4-manifold with an almost complex structure $J$. Let $w\in H^2(X;\mathbb{R})$ be a class with $w^2>0$. Is there a symplectic form $\omega$ with compatible almost complex structure homotopic to $J$ and symplectic class $w$? The "Taubes constraints" are the following necessary conditions, which constrain the SW invariants in terms of $w$ and $c=c_1(TX,J)$ (see e.g. <a href="http://www.ams.org/journals/bull/1996-33-01/S0273-0979-96-00625-8/" rel="nofollow">Donaldson's survey</a> on the SW equations): (i) $SW(\mathfrak{s}_{can})=\pm 1$ (the sign can be made precise) where <code>$\mathfrak{s}_{can}$</code> is the $\mathrm{Spin}^c$-structure arising from $J$; (ii) $-c\cdot w\geq 0$; and (iii) if $SW(\mathfrak{s})\neq 0$ then $|c_1(\mathfrak{s})\cdot [\omega]| \leq -c \cdot [\omega]$, with equality iff $\mathfrak{s}$ is isomorphic to $\mathfrak{s}_{can}$ or its conjugate. The question is: if $X$ is simply connected, are these sufficient conditions? (Example: Fintushel-Stern knot surgery on an elliptically fibered K3 surface along a knot with monic Alexander polynomial.)</p> http://mathoverflow.net/questions/118479/orientations-for-pseudoholomorphic-curves-with-totally-real-boundary-condition/118483#118483 Answer by Tim Perutz for Orientations for pseudoholomorphic curves with totally real boundary condition Tim Perutz 2013-01-09T22:26:14Z 2013-01-18T23:13:22Z <p>1) The problem of orienting moduli spaces of pseudo-holomorphic discs with totally real boundary conditions is really a problem in index theory. It was solved Vin de Silva in his (unpublished) D. Phil. thesis, using Atiyah's Real K-theory, and independently by FOOO. There's an excellent account in Seidel's book (section 11, especially Lemma 11.7).</p> <p>A totally real Cauchy-Riemann problem is by definition a loop in the totally real Grassmannian $\mathrm{Gr}(V)$ of a complex vector space $V$ with a given real structure. The resulting space $L\mathrm{Gr}(V)$ parametrizes a family of Fredholm operators (the Cauchy-Riemann operator for functions on the closed disc, valued in $V$, with boundary conditions specified by the loop). Hence there is a determinant index bundle $$\underline{det} \to L \mathrm{Gr}(V),$$ and the basic orientation problem is to describe $w_1(\underline{det}) \in H^1(L \mathrm{Gr}(V);\mathbb{Z}/2)$. For the component $L_k\mathrm{Gr}(V)$ of Maslov index $k$ loops, $H^1(L_k \mathrm{Gr}(V);\mathbb{Z}/2)$ is 2 dimensional, as one calculates using a homotopy equivalence $\mathrm{Gr}(\mathbb{C}^n) \simeq U(n)/O(n)$, so there are just four possible answers to the orientation question for each $k$.</p> <p>The answer is simpler to state assuming $k$ is even. It is then as follows (I learned this from de Silva's thesis): Take a loop $\gamma\colon S^1\to L_k\mathrm{Gr}(V)$. Then $\langle w_1(\underline{det}), \gamma\rangle = \langle w_2, T_\gamma \rangle$, where $w_2$ is the second SW class of the universal totally real bundle on $\mathrm{Gr}(V)$, and $T_\gamma\colon S^1\times S^1\to \mathrm{Gr}(V)$ is the torus of boundary values swept out by $\gamma$.</p> <p>So, in a space of pseudo-holomorphic discs in a symplectic manifold $X$ attached to an orientable Lagrangian $\Lambda$, $w_1$ of the determinant bundle evaluates on a loop $\gamma$ by evaluating the torus of boundary values $T_\gamma$ on $w_2(T\Lambda)$. Essentially for this reason, it's natural to trivialize the determinant bundle by trivializing $w_2(\Lambda)$, i.e. specifying a Pin structure. More generally, one can trivialize $w_2(\Lambda)$ relative to a fixed background class $b\in H^2(\Lambda;\mathbb{Z}/2)$ which restricts to $w_2(\Lambda)$, i.e. specify a relative (or "twisted") Pin structure. That suffices essentially because the torus of boundary values is the boundary of a 3-chain in $X$, and so vanishes if $w_2$ is the restriction of $b$. Choosing Pin structures relative to a fixed background class $b$ gives a uniform way to orient moduli spaces of pseudo-holomorphic discs for relatively Pin Lagrangians.</p> <p>2) For Cauchy-Riemann operators on other curves, one can degenerate to a nodal union of discs and closed Riemann surfaces, combining the orientations for the space of discs with the complex orientation of the determinant line bundle over the moduli of closed curves. Thus no additional obstructions to orientation appear.</p> <p>To be precise, what goes into this is a gluing theorem for the index bundle, which is part of the linear analysis that underpins Floer theory and Gromov-Witten theory. It implies that the determinant index line over a connected sum is canonically isomorphic to the tensor product of the determinant index lines on the summands. Again, see Seidel's book, section 11 for the argument. For the underlying analysis, I'd recommend Donaldson's Floer homology book, chapter 3.</p> <p>3) I don't know, but there are further concrete calculations for real loci in the work of Welschinger and also Solomon.</p> http://mathoverflow.net/questions/116537/an-extended-question-of-gromov-every-generalized-open-almost-complex-manifold/116932#116932 Answer by Tim Perutz for an extended question of Gromov: Every **generalized open almost complex manifold** admits a **generalized symplectic structure**? Tim Perutz 2012-12-20T22:51:00Z 2012-12-20T22:51:00Z <p>In his thesis</p> <p><a href="http://arxiv.org/abs/math/0401221" rel="nofollow">http://arxiv.org/abs/math/0401221</a></p> <p>Marco Gualtieri explains that a generalized almost complex structure on an $n$-manifold $M$ is a reduction of the structure group of $TM \oplus T^\ast M$, which has its canonical hyperbolic quadratic form, from $O(n,n)$ to $U(n,n)$. He points out (p. 48) that since $U(n,n)$ retracts to its maximal compact subgroup $U(n)\times U(n)$, such a reduction implies a reduction of structure for $TM$ to $U(n)$, hence an almost complex structure. By Gromov's symplectic h-principle, an open manifold with a generalized almost complex structure therefore admits a symplectic form, which is an example of a generalized complex structure.</p> <p>I have nothing to say, however, about the more substantial question of whether the inclusion of the generalized complex structures into the generalized almost complex structures is a highly connected map. </p> http://mathoverflow.net/questions/115687/vortex-equations-on-cylinder/115724#115724 Answer by Tim Perutz for Vortex equations on cylinder Tim Perutz 2012-12-07T16:21:46Z 2012-12-07T16:21:46Z <p>For finite-energy vortices on a finite-type Riemannian surface with cylindrical ends, there is still a non-negative integer parameter, the vortex number $N$, and the moduli space is still canonically diffeomorphic to the $N$th symmetric product by the map that takes a gauge-equivalence class of vortices $[A,\phi]$ to $\phi^{-1}(0)$. One can prove that such vortices extend over the puncture, whereupon the usual methods apply.</p> <p>Some references:</p> <p>1) The case of the complex plane was treated in the book "Vortices and monopoles" by Jaffe-Taubes. </p> <p>2) The case of a cylinder is explicitly treated, by a different method, in a paper by Frauenfelder:</p> <p><a href="http://arxiv.org/abs/math/0507285" rel="nofollow">http://arxiv.org/abs/math/0507285</a></p> <p>3) One can regard the vortex equations as dimensional reductions of the Seiberg-Witten equations. There is a comprehensive treatment of those equations in the presence of cylindrical ends in Kronheimer and Mrowka's book "Monopoles and 3-manifolds". They also discuss Atiyah-Patodi-Singer boundary conditions in the case where there is a boundary. </p> http://mathoverflow.net/questions/111305/spin-c-structures-with-near-symplectic-forms/111382#111382 Answer by Tim Perutz for Spin-c Structures with Near-Symplectic Forms Tim Perutz 2012-11-03T14:56:03Z 2012-11-03T14:56:03Z <p>When $(X,\omega)$ is a near-symplectic oriented 4-manifold there is <b>always a canonical identification</b> between $\mathrm{Spin}^c(X)$ and the classes in $H_2(X,Z;\mathbb{Z})$ that bound $[Z]$, where $Z=\omega^{-1}(0)$. I call this identification the "Taubes map" $\tau$ in my "Lagrangian matching invariants" papers, since it comes from Taubes's work in near-symplectic geometry. A near-symplectic form $\omega$ can be defined intrinsically as a closed 2-form with $(\omega\wedge \omega) (x)>0$ except at the set $Z$ of points where $\omega(x)=0$; and at those points $\nabla \omega$ has rank 3, hence its image spans a maximal positive-definite subspace of $\Lambda^2 T^*_x X$.</p> <p>On $X-Z$ one has the canonical $\mathrm{Spin}^c$-structure $\mathfrak{s}_{\mathrm{can}}$ arising from an almost complex structure compatible with $\omega$. An arbitrary $\mathrm{Spin}^c$-structure $\mathfrak{s}$ on $X$ restricts to $X-Z$ as <code>$L\otimes \mathfrak{s}_{\mathrm{can}}$</code> for a unique line bundle $L$. Define $\tau(\mathfrak{s})\in H_2(X,Z;\mathbb{Z})$ to be the Lefschetz dual to $c_1(L)$. Since restriction $H^2(X;\mathbb{Z})\to H^2(X-Z;\mathbb{Z})$ is injective, the map $\tau\colon \mathrm{Spin}^c(X) \to H_2(X,Z;\mathbb{Z})$ is also injective; and its image is a coset of $H_2(X;\mathbb{Z})$. The structure of $\omega$ near a point of $Z$ is standard, so the image of $\tau$ will be the classes that bound $n[Z]$ for some standard $n$. </p> <p>That the answer is given by $n=1$ can be visualized by looking at an example on $S^1\times B^3$, with $\omega = dt\wedge \alpha+ \ast \alpha$, where $\alpha$ is a harmonic 1-form on the ball $B$ with one critical point; then gradient flow-lines for $\alpha$ into the critical point, crossed with $S^1$, represent classes in the image of $\tau$, and these precisely bound $Z$.</p> http://mathoverflow.net/questions/106553/why-is-the-base-of-slag-fibration-of-cy3-expected-to-be-s3/106569#106569 Answer by Tim Perutz for Why is the base of SLAG fibration of CY3 expected to be $S^3$? Tim Perutz 2012-09-07T04:56:07Z 2012-09-07T15:15:20Z <p>If your CY manifold is simply connected, the base of the torus-fibration will have to be simply connected too, since a homotopically non-trivial loop downstairs would lift to a loop upstairs which does not bound a disc. In 3 dimensions, that's the end of the story by the Poincar&eacute; conjecture.</p> <p>I'll try to explain via homological mirror symmetry (HMS) why, even in higher dimensions, the base of the SYZ fibration should be a rational homology-sphere. This only applies to "strict" CY manifolds.</p> <p>Say we have a special Lagrangian torus-fibration $\check{X}\to B$, and we would like to understand the Fukaya category as the derived category of a mirror $X$, defined over some field $K$ of characteristic zero (depending on the formulation of the Fukaya category, $K$ might be the field of rational or complex Novikov series; optimists think that $\mathbb{C}$ could also be a possibility for $K$). </p> <p>A basic aspect of HMS is the prediction that the mirror to a smooth torus-fiber $F_b$ will be a skyscraper sheaf $\mathcal{O}_{X,x}$ on $X$. That prediction gives rise to another: that the mirror $L$ to the structure sheaf $\mathcal{O}_X$ should be a Lagrangian section of $\check{X}\to B$. The reason is that <code>$\mathrm{Ext}^\ast(\mathcal{O}_X,\mathcal{O}_{X,x})=H^\ast(\mathcal{O}_{X,x})=K$</code>, so by HMS one should have $HF(L,F_b)=K$ for each fibre $F_b$. Taking Euler characteristics of the latter isomorphism, we get $[L]\cdot [F_b]=1$. So $L$ is at least a homology-section, and we guess that it should be a true section. In particular, $H^\ast(L;K)=H^\ast(B;K)$.</p> <p>By a "strict" CY $n$-manifold I mean that as well as trivial canonical bundle, one has $H^i(\mathcal{O}_X)=0$ for <code>$0&lt;i&lt;n$</code>. (In the setting of complex manifolds, this means that the holonomy is exactly $SU(n)$.) By Serre duality, $H^n(\mathcal{O}_X)=K$. Hence $\mathrm{Ext}^*(\mathcal{O}_X,\mathcal{O}_X)= H^\ast(\mathcal{O}_X)$ is isomorphic as a graded $K$-algebra to $H^\ast(S^n;K)$. On the other hand, $\mathrm{Ext}^*(\mathcal{O}_X,\mathcal{O}_X)\cong HF(L,L)$ by HMS. One makes the reasonable guess that $HF(L,L)\cong H^\ast(L;K)$, and infers that <code>$H^*(B;K) = H^*(L;K)\cong H^*(S^n;K)$</code>.</p> <p><i>Edit:</i> Ah, I think we don't need to guess at the last stage! The DGA of cochains computing <code>$\mathrm{Ext}^*(\mathcal{O}_X,\mathcal{O}_X)$</code> is <i>formal</i> - over $\mathbb{C}$, we get that from Deligne-Griffiths-Morgan-Sullivan plus Hodge by using a Dolbeault model. By HMS, $CF(L,L)$ is then formal as an $A_\infty$-algebra. The Oh spectral sequence $H^\ast(L;K) \Rightarrow HF(L,L)$ must then surely degenerate at $E_1$, so $H^\ast(L)\cong HF(L,L)$.</p> http://mathoverflow.net/questions/106212/elementary-computation-of-direct-image-sheaves/106244#106244 Answer by Tim Perutz for Elementary computation of direct image sheaves. Tim Perutz 2012-09-03T13:45:15Z 2012-09-03T21:23:08Z <p><i>Edited following algori's comment.</i></p> <p>In the context of the paper I think one can make these deductions on rather formal grounds - deep understanding of higher direct images is not needed.</p> <p>Gross assumes that the base $B$ is a simply connected closed 3-manifold (i.e., it's $S^3$). Let $i\colon B_0\to B$ be the inclusion of the (open and dense) regular locus, and $f_0\colon f^{-1}(B_0)\to B_0$ the restricted fibration. Gross also tells us that $f$ is "simple", meaning that $i_*(R^jf_{0\ast} \mathbb{R})\cong R^j f_\ast \mathbb{R}$ for each $j$. In degree $j=0$, simplicity is automatic: the fibers of $f_0$ are connected, and since disconnectedness of a fiber is an open property on $B$, the fibers of $f$ are also connected. So $f_{0\ast}\mathbb{R}=\mathbb{R}$ and $f_{\ast}\mathbb{R}=\mathbb{R}$.</p> <p>The fibers of $f_0$ are also oriented. Moreover, after choosing orientations for $X$ and $B$ we have a consistent notion of a "positive" fiber-orientation: one which, after wedging with the pullback of a positive orientation form on $B$, agrees with the orientation of $X$. Hence $R^3f_{0\ast}\mathbb{R}=\mathbb{R}$. By simplicity, $R^3 f_{\ast}\mathbb{R}\cong \mathbb{R}$. We can therefore complete the $0$th and $3$rd rows of $E_2$ - each is the de Rham cohomology of $S^3$: $$\mathbb{R}\quad 0\quad 0 \quad \mathbb{R}.$$ </p> <p>The spectral sequence converges to $H^\ast(X;\mathbb{R})$, and we know that $E_2^{2,0}=0$. We must have $E_2^{0,1}=0$ because otherwise it would survive to $E_\infty$ and we would have $H^1(X;\mathbb{R})\neq 0$. Similarly, since $E_2^{1,3}=0$ and $H^5(X;\mathbb{R})=0$, we must have $E_2^{3,2}=0$.</p> http://mathoverflow.net/questions/100263/why-is-heegaard-floer-homology-defined-in-terms-of-symg-sigma-g-instead-of-pic/102089#102089 Answer by Tim Perutz for Why is Heegaard Floer Homology defined in terms of Sym$^g\Sigma_g$ instead of Pic$^g\Sigma_g$? Tim Perutz 2012-07-13T01:47:48Z 2012-07-13T01:47:48Z <p>There is a tacit assumption behind this question, which I don't think is justified: that the Abel-Jacobi images of the Heegaard tori <code>$\mathbb{T}_{\alpha}$</code> and <code>$\mathbb{T}_{\beta}$</code> are Lagrangian with respect to some reasonable symplectic form on the Jacobian torus. </p> <p>One can make the Heegaard tori Lagrangian by using a Kaehler form on the symmetric product that is product-like outside some neighbourhood of the diagonal. And one can probably find symplectic forms for which Abel-Jacobi is a symplectomorphism outside a neighbourhood of the theta-divisor (this is certainly true in the genus 2 case). Doing both of these things at once would suffice to make the images Lagrangian, but this might be tricky to achieve - and it's perhaps not very natural?</p> http://mathoverflow.net/questions/28519/references-for-modern-proof-of-newlander-nirenberg-theorem/28522#28522 Answer by Tim Perutz for References for "modern" proof of Newlander-Nirenberg Theorem Tim Perutz 2010-06-17T15:27:54Z 2012-05-23T23:49:51Z <p>It's covered in Demailly's too-little-known book, <a href="http://mathonline.andreaferretti.it/books/view/19/Complex-analytic-and-algebraic-geometry" rel="nofollow">Complex analytic and differential geometry,</a> though the proof given there is apparently modelled on the references you cited.</p> <p><b>Edit</b>: I just noticed that the MathOnline link currently seems to be non-functional, so here's a link to <a href="http://www-fourier.ujf-grenoble.fr/~demailly/books.html" rel="nofollow">Demailly's webpage</a>. </p> http://mathoverflow.net/questions/96255/formality-of-ext-algebras-and-direct-sums/96258#96258 Answer by Tim Perutz for Formality of Ext algebras and direct sums Tim Perutz 2012-05-07T19:48:39Z 2012-05-21T23:32:56Z <p>My understanding is that formality of the DGA $Ext^\bullet(X\oplus Y,X\oplus Y)$ implies 1), but also formality of $Ext^\bullet(X,Y)$ and $Ext^\bullet(Y,X)$ as bimodules over $Ext^\bullet(X,X)$ and $Ext^\bullet(Y,Y)$, and that this is a much stronger condition.</p> <p>For instance, let $(E,p)$ be an elliptic curve over a field, work in the abelian category of coherent sheaves, let $X=\mathcal{O}$ and let $Y=\mathcal{O}_p$ be the skyscraper at $p$. </p> <p>Then $Ext^\bullet(X,X)$ and $Ext^\bullet(Y,Y)$ are both (intrinsically) formal, but $Ext^\bullet(X\oplus Y,X\oplus Y)$ knows the affine coordinate ring of $E\setminus{ p}$ for the cubic embedding into $\mathbb{P}^2$. That's because one can iteratively build $\mathcal{O}(np)$ for $n>0$ as a twisted complex in $X$ and $Y$ (namely, $\mathcal{O}((n+1)p)$ is the twist of $\mathcal{O}(np)$ along the spherical object $Y$). Over an algebraically closed field, this gives a $j$-line of quasi-isomorphism classes of $A_\infty$-algebras $Ext^\bullet(X\oplus Y,X\oplus Y)$.</p> <p>As requested a bit more detail on why 2) implies 1), probably by too clunky an argument. Let $A=Ext^\bullet(X\oplus Y, X\oplus Y)$. We can regard this as an ordinary graded $K$-algebra, in which case non-formality of the $A_\infty$-structure is detected by the primary deformation class in $HH^\bullet_K(A,A)$. That is: after transferring the DG structure to a minimal $A_\infty$-structure on $A$ using homological perturbation theory, the composition $\mu^3$ defines a Hochschild cocycle. If it is a coboundary then we can kill $\mu^3$ by a gauge transformation which leaves $\mu^1$ and $\mu^2$ untouched, whereupon $\mu^4$ is a cocycle; and so on. If the structure is not formal, one will eventually obtain a non-trivial Hochschild class, called the primary deformation class.</p> <p>We can alternatively regard $A$ as a 2-object graded-linear category, i.e., an algebra over $R=K\oplus K$, in which case non-formality is detected by a primary class in $HH^\bullet_R(A,A)$, defined similarly. But one checks using the bar resolution that $HH^\bullet_R(A,A)\cong HH^\bullet_K(A,A)$ as $K$-modules. Hence, if the algebra is formal, then so is the category; the restriction of the categorical primary deformation class to endomorphisms of $X$ is then trivial. </p> <p>The references I tend to use for this sort of thing are the first chapter of Seidel's book "Fukaya categories and Picard-Lefschetz theory", and also his paper "Homological mirror symmetry for the quartic surface", but there are certainly other possibilities.</p> http://mathoverflow.net/questions/92032/sft-compactness/92039#92039 Answer by Tim Perutz for SFT compactness Tim Perutz 2012-03-23T21:00:13Z 2012-03-23T21:00:13Z <p>For non-specialist readers:</p> <p>SFT = symplectic field theory </p> <p>BEHWZ = Bourgeois-Eliashberg-Hofer-Wysocki-Zehnder, the authors of the paper which establishes the basic compactness theorem for pseudo-holomorphic curves in symplectic manifolds with convex or concave ends.</p> <p>There is an attractive alternative approach to SFT compactness due to Cieliebak and Mohnke:</p> <p><b>Compactness for punctured holomorphic curves</b>. J. Symplectic Geom. 3 (2005), no. 4, 589-654. </p> <p>Their method is founded not on Deligne-Mumford-type degenerations of the source curves, but rather on the degenerations of their images in a symplectic manifold with a lengthening cylindrical neck. These degenerations are controlled by the Morse theory of the function on the source curve given by the holomorphic map followed projected to the cylindrical coordinate.</p> http://mathoverflow.net/questions/88729/mirror-to-the-dualizing-sheaf/88892#88892 Answer by Tim Perutz for Mirror to the dualizing sheaf Tim Perutz 2012-02-19T05:02:38Z 2012-02-19T05:02:38Z <p>I'll comment on the related question "what is the Serre functor for the Fukaya category?" </p> <p><b>Calabi-Yau setting</b></p> <p>The Serre functor $S$, by definition, satisfies $\mathsf{Hom}(X,SY) \cong \mathsf{Hom}(Y,X)^\vee$; since it's characterized categorically, it's preserved by the derived equivalences which arise in mirror symmetry. For the derived category $D^b\mathsf{Coh}(X)$ of a non-singular projective $n$-variety, $S= \cdot \otimes \omega_X [n]$, where $\omega_X$ is the dualizing sheaf. So, when $X$ is Calabi-Yau, it's simply a shift. </p> <p>For a Fukaya category of compact Lagrangians (of dimension $n$) in a symplectic manifold, $S$ is just a shift by $n$, because of the Floer-theoretic Poincare duality $HF^\ast(X,Y) \cong HF^{n-\ast}(Y,X)^\vee$. At the fully precise $A_\infty$-level, the claim that the Serre functor is a shift is partly conjectural. </p> <p>The mirror to $\mathcal{O}$ is a section $\sigma$ of the SYZ fibration, so the mirror to the canonical sheaf is $\sigma[n]$.</p> <p><b>LG models</b></p> <p>To get a more interesting answer, consider Fukaya categories of Landau-Ginzburg models, a.k.a. Fukaya-Seidel categories. These arise as mirrors to Fano manifolds. Out of caution, I'll assume that the L-G model is a symplectic Lefschetz fibration $E\to \mathbb{C}$. The objects of the category are Lagrangian submanifolds which map to eventually-horizontal paths in $\mathbb{C}$ (for instance, Lefschetz thimbles). Kontsevich proposed that the Serre functor should then be the "wrapping" or "monodromy" functor. This has been proved (at least at the level of objects, probably more), by Seidel (cf. his <a href="http://arxiv.org/abs/math/0609037" rel="nofollow">Symplectic homology as Hochschild homology</a> and <a href="http://arxiv.org/abs/math/0007115" rel="nofollow">Vanishing cycles and mutation</a>). </p> <p>The wrapping functor is defined as follows. Take a circle of large radius $R$ in $\mathbb{C}$, and consider the Dehn twist $\delta$ along this circle. So $\delta(z)=e^{i\rho(|z|)}z$, where the angle $\rho(|z|)$ runs from $0$ when $|z| &lt; R-1$ to $2\pi$ when $|z| > R+1$. There's a symplectomorphism $\Phi$ of $E$, covering $\delta$, given by symplectic parallel transport of the fibration over the arc from $z$ to $\delta(z)$. The wrapping functor takes a Lagrangian $L$ to $\Phi(L)$. It takes a standard Lefschetz thimble (fibering over a ray) to a "once-wrapped thimble", i.e. a thimble for a path that wraps once around the circle.</p> <p>In the case of the LG mirror to $\mathbb{CP}^2$, the mirror to $\mathcal{O}$ (which is one of the Beilinson generators of $D^b \mathsf{Coh}(\mathbb{CP}^2)$) is a thimble, so the mirror to the canonical sheaf is a once-wrapped thimble.</p> <p>The proof that the wrapping functor is the Serre functor invokes a general characterization of the Serre functor in triangulated categories with full exceptional collections in terms of the algebraic process of "mutation". The thimbles associated with a collection of vanishing paths form a full exceptional collection, and mutation corresponds to Hurwitz moves on vanishing paths.</p> http://mathoverflow.net/questions/87973/contactomorphism-of-s2n1-for-n1/88019#88019 Answer by Tim Perutz for contactomorphism of $S^{2n+1}$ for n>1 Tim Perutz 2012-02-09T17:50:58Z 2012-02-09T17:50:58Z <p>I know of no technique capable of bounding above the homotopy groups of a symplectomorphism group in dimension $\geq 6$, nor of a contactomorphism group in dimension $\geq 5$. </p> <p>There are, however, techniques for obtaining non-trivial elements in $\pi_i(\mathsf{Cont}(M))$. These were first explored, in the symplectic context, by Seidel, who showed that $\pi_1$ of the Hamiltonian automorphism group of a symplectic manifold has a natural representation on quantum cohomology. Analogues and extensions for contact manifolds, using linearized contact homology, have been developed by <a href="http://arxiv.org/abs/math/0407531" rel="nofollow">Bourgeois</a>. He finds a $\mathbb{Z}^3$ inside $\pi_1 \mathsf{Cont}(T^5)$, for instance. </p> <p>I think this is a promising research topic. For instance, one knows that $\pi_0\mathsf{Diff}(S^5)$ is trivial; what about $\pi_0\mathsf{Cont}(S^5)$? (Disclaimer: It's possible that this is already known.)</p> http://mathoverflow.net/questions/78275/automorphisms-of-riemann-surface-and-mapping-class/78286#78286 Answer by Tim Perutz for Automorphisms of Riemann surface and mapping class Tim Perutz 2011-10-16T20:39:00Z 2011-10-16T20:39:00Z <p>If a surface-diffeomorphism $h$ acts trivially on rational cohomology, the Lefschetz number of $h$ is equal to the Euler characteristic of the surface $\Sigma$. By the Lefschetz fixed-point formula, this number equals the intersection number, in $\Sigma \times \Sigma$, of the graph of the automorphism with the diagonal. </p> <p>In the case that $h$ is a non-trivial holomorphic automorphism, the intersections are isolated and the intersection multiplicities positive. This can occur only when the Euler characteristic is non-negative.</p> http://mathoverflow.net/questions/75381/relative-version-of-symplectic-thom-conjecture/75683#75683 Answer by Tim Perutz for Relative version of Symplectic Thom conjecture. Tim Perutz 2011-09-17T16:39:23Z 2011-09-17T16:39:23Z <p>This is a natural question, and I'm a bit startled to realise that, in this generality, I can't locate a reference for it.</p> <p>To frame it precisely, let's suppose that $X$ is a compact symplectic 4-manifold with convex contact-type boundary $Y$, and ask whether a compact symplectic surface $\Sigma$ in $X$, transverse to $Y$ and bounding a link $L\subset Y$ transverse to the contact structure, minimises minus the Euler characteristic among surfaces bounding $L$ and homologous to $\Sigma$ relative to $L$.</p> <p>There's lots in the literature about Bennequin-type inequalities for <i>Legendrian</i> links, notably Mrowka-Rollin's adjunction inequality: <a href="http://arxiv.org/abs/math/0410559" rel="nofollow">http://arxiv.org/abs/math/0410559</a>. But when considering boundaries of symplectic surfaces it seems more natural to take $L$ <i>transverse</i> to the contact structure.</p> <p><b>A sufficient condition.</b> Suppose that we can cap $X$ to a closed symplectic manifold $Z$, and cap $\Sigma$ inside $Z$ to a closed symplectic surface $S$. It then follows from the symplectic Thom conjecture in $Z$ that $\Sigma$ is genus-minimizing in the sense I indicated. </p> <p>A famous example is Kronheimer-Mrowka's proof (see <a href="http://www.math.harvard.edu/~kronheim/thom1.pdf" rel="nofollow">http://www.math.harvard.edu/~kronheim/thom1.pdf</a>) of the Milnor conjecture about the slice genus of algebraic links, in which one completes the (blown up) 4-ball to the (blown up) projective plane and applies the Thom conjecture there. [Experts will spot an anachronism in this summary.]</p> <p>We do know that any $X$ can be closed up symplectically; see, for instance, Eliashberg's article <a href="http://arxiv.org/pdf/math/0311459" rel="nofollow">http://arxiv.org/pdf/math/0311459</a>. It seems plausible that every pair $(X,\Sigma)$, where the boundary of $\Sigma$ is a transverse link, can be closed to a pair $(Z,S)$. Perhaps Eliashberg's argument can be refined to accomplish this.</p> http://mathoverflow.net/questions/73830/a-question-about-mathscinet-etiquette/73853#73853 Answer by Tim Perutz for A question about MathSciNet etiquette Tim Perutz 2011-08-27T16:21:22Z 2011-08-27T16:21:22Z <p>Like quid, I'm a regular reviewer, with about 60 reviews to date. The large majority of those reviews have been untouched since they first appeared, but in a few cases I have later asked Math Reviews to implement light edits to correct errors. The edited reviews appeared in due course. (That is, they appeared in MathSciNet; I don't know the policy about the print version.) On one occasion, I did this because the author contacted me to point out a mistake in my review, which I was glad to correct. </p> <p>In writing a review, I'm aiming for concision. But I try to touch on 1) the context of the work, 2) the main results, and 3) what sorts of ideas go into the proofs. I read only in as much detail as I need to do that. Occasionally, when I feel that I have a distinctive perspective on material in the paper I'm reviewing, or its relation to other papers, I comment on that perspective. I try to do so sensitively and sparingly: I am conscious that these comments should add to the information I convey about the article, not replace it. But I would argue that as a reviewer it is my privilege to explain the context of the work as I see it.</p> <p>If the problem could be resolved by straightforward corrections to the review, I would suggest that you contact the reviewer directly. If the review is so misleading as to make you think that the reviewer shouldn't be writing about papers in that area, you could instead make that case to Math Reviews - and perhaps enlist your expert colleague to do the same.</p> http://mathoverflow.net/questions/73673/rounding-the-corners-to-get-contact-boundary/73678#73678 Answer by Tim Perutz for "Rounding the corners" to get contact boundary Tim Perutz 2011-08-25T17:39:05Z 2011-08-25T17:39:05Z <p>In that generality, the answer is no: a symplectic form $\omega$ on $X$ which has contact-type boundary is exact on $\partial X$. Yet $\omega_1 \oplus \omega_2$ need not be exact on $M_1\times \partial M_2$, nor on $\partial M_1 \times M_2$.</p> <p>It is possible, however, if $M_1$ and $M_2$ are Liouville domains, i.e., if the symplectic form $\omega_i$ is given as $d\theta_i$ for 1-forms $\theta_i$ whose dual vector field $\lambda_i$ points strictly outwards along the boundary. In fact, if you round corners sensibly, $\theta_1 \oplus \theta_2$ will have those same properties on the product.</p> <p>Here's a relevant article by Alex Oancea: <a href="http://arxiv.org/abs/math/0403376" rel="nofollow">http://arxiv.org/abs/math/0403376</a></p> http://mathoverflow.net/questions/71794/where-are-and-infty-in-bordered-heegaard-floer-theory/71819#71819 Answer by Tim Perutz for Where are $+$, $-$ and $\infty$ in bordered Heegaard-Floer theory? Tim Perutz 2011-08-01T18:42:52Z 2011-08-01T18:48:44Z <p>A biased answer, based on Auroux's work <a href="http://arxiv.org/abs/1003.2962" rel="nofollow">http://arxiv.org/abs/1003.2962</a>. </p> <p>Auroux makes a connection between bordered Floer theory and an alternative approach, due to Lekili and myself, which is (still) under development, but which should include the $\pm$ and $\infty$ versions. We do have a preliminary paper out: <a href="http://arxiv.org/abs/1102.3160" rel="nofollow">http://arxiv.org/abs/1102.3160</a>.</p> <p><b>A general set-up:</b> Say you have a compact symplectic manifold $(X,\omega_X)$; and a codim 2 symplectic submanifold $D$, whose complement $M$ is exact: ${\omega_X}|_M=d\theta$, say. </p> <p>Key example: $X=Sym^g(F)$, where $F$ is a compact surface of genus $g$, and $\omega_X$ a suitable Kaehler form; $M=Sym^g(F-z)$, where $z\in F$. </p> <p><b>Forms of Floer cohomology</b>: There are various forms of Floer cohomology one can consider.</p> <p>(i) As in $\widehat{HF}$ Heegaard theory, one can consider $HF^\ast_M(L_0,L_1)$, the Floer cohomology in $M$ of a pair of (exact) compact Lagrangian submanifolds of $M$. When $L_0$ and $L_1$ are spin, this can be defined as a $\mathbb{Z}$-module.</p> <p>(ii) As in $HF^-$ Heegaard theory, one can consider the filtered Floer cohomology $HF^\ast_{X,D}(L_0,L_1)$ of a pair of compact Lagrangians $L_i\subset M$ as before. The coefficients are in $\mathbb{Z}[[U]]$. The differential counts holomorphic bigons in $X$, weighted by $U^n$ where $n$ is intersection number with $D$. </p> <p>(iii) One can consider non-compact Lagrangians $L_i\subset M$ which go to infinity nicely (following the Liouville flow). These have <i>wrapped</i> Floer cohomology $HW^\ast(L_0,L_1)$, as well as "partially wrapped" variants. Wrapping concerns how one chooses to perturb $L_0$ at infinity. This version takes place in $M$, and (AFAIK) can't naturally be extended to something that takes place in $X$.</p> <p><b>Invariants for 3-manifolds with boundary.</b> A basic idea is that a 3-manifold $Y$ bounding $F$ should define a (generalized) Lagrangian submanifold $L_Y$ where $X=Sym^{g(F)}F$, as in the "key example" above. The collection of filtered Floer modules $HF^*_{X,D}(\Lambda, L_Y)$ as $\Lambda$ ranges over Lagrangian submanifolds of $M$ (more precisely, the module, over the compact filtered Fukaya category of $(X,D)$, defined by $L_Y$) should be an invariant of $Y$. </p> <p>If one is interested only in the simpler groups $HF^*_M(\Lambda,L_Y)$, one can (in principle) determine these by looking at the finite collection of (partially wrapped) groups $HW^*(W_i,L_Y)$, where $W_i$ ranges over the thimbles for a certain Lefschetz fibration $M\to \mathbb{C}$. That is, one thinks of $L_Y$ as defining a module over the algebra $A_{LOT}$ formed by the sum of groups $HW^*(W_i,W_j)$. This follows from a deep theorem of Seidel about generating Fukaya categories by thimbles, adapted by Auroux. </p> <p>The algebra $A_{LOT}$ is (part of) what Lipshitz-Ozsvath-Thurston assign to a parametrized surface, and the module is what they call $\widehat{CFA}(Y)$. They arrived at it by a quite different route. They don't bother with constructing $L_Y$ itself, only the module it defines. Because they use the groups of type (iii) to form their algebra, their approach only works in $M$, not $X$. For that reason, they only capture the hat-theory. </p> <p>The great advantage of LOT's approach is its finiteness and computability. Lekili and I do construct $L_Y$. We can guess at finite collections of "test Lagrangians" sufficient to compute the module $HF^*_{X,D}(\cdot, L_Y)$, but have not yet proved that they are sufficient.</p> http://mathoverflow.net/questions/70429/is-canonical-class-a-topological-invariant/70445#70445 Answer by Tim Perutz for Is canonical class a topological invariant? Tim Perutz 2011-07-15T15:44:15Z 2011-07-16T01:20:47Z <p>This answer is about the case of complex surfaces $X$ and their diffeomorphisms (all my diffeos are assumed to be orientation-preserving!). </p> <p><b>(1) Examples of self-diffeomorphisms that reverse the sign of the canonical class.</b> </p> <p>Take $X=\mathbb{C}P^1\times \mathbb{C}P^1$. Let $\tau$ be reflection in the equator of $S^2=\mathbb{C}P^1$. Then $\tau \times \tau$ preserves orientation and acts as $-I$ on $H^2(X)$. It therefore sends $K_X$ to $-K_X$.</p> <p>One can also realise the automorphism $-I$ of $H^2(X)$ by a diffeomorphism when $X$ is the blow-up of the projective plane at $k$ points, $k = 2,3,\dots,9$. This follows from a result of C.T.C. Wall from</p> <p><i>Diffeomorphisms of 4-manifolds</i>, J. London Math. Soc. 39 (1964) 131–140, MR0163323</p> <p>Wall says that if $N$ is a simply connected, closed oriented 4-manifold with $b_2(N)&lt;9$, and $X$ is the connected sum of $N$ with $S^2 \times S^2$, then all automorphisms of the intersection form of $X$ are realised by diffeos. To apply this, recall that the 1-point blow-up of $\mathbb{C}P^1\times \mathbb{C}P^1$ is the 2-point blow up of the projective plane. (Wall's strategy, by the way, is to factor the automorphism into reflections along hyperplanes, and to realise those.)</p> <p><b>(2) Results from Seiberg-Witten theory.</b> </p> <p>These results tie complex geometry amazingly closely to differential topology. They say that the unsigned pair $\pm K_X$ is invariant under diffeomorphisms (Witten <a href="http://arxiv.org/abs/hep-th/9411102" rel="nofollow">http://arxiv.org/abs/hep-th/9411102</a> and others); so too is the Kodaira dimension; so too are the plurigenera (Friedman-Morgan <a href="http://arxiv.org/abs/alg-geom/9502026" rel="nofollow">http://arxiv.org/abs/alg-geom/9502026</a>). </p> <p>In Kodaira dimension $&lt;2$, one can take this further and prove that oriented-diffeomorphic surfaces are actually deformation-equivalent (to be safe, let me specify the simply connected case). But that's <i>not</i> the explanation in general: there are pairs of simply connected general-type surfaces that are diffeomorphic (by diffeos preserving the canonical class), which are not deformation-equivalent (Catanese-Wajnryb <a href="http://arxiv.org/abs/math/0405299" rel="nofollow">http://arxiv.org/abs/math/0405299</a>).</p> <p><b>(3) How it happens.</b></p> <p>The Seiberg-Witten invariant (for an oriented 4-manifold with $b^+(X)>1$) is a map $$SW: Spin^c(X)\to\mathbb{Z}$$ defined on the $H^2(X)$-torsor of $Spin^c$-structures. The overall sign is equivalent to a "homology orientation". It's natural under diffeomorphisms. It's also invariant under "conjugation" $\mathfrak{s}\mapsto \bar{\mathfrak{s}}$ of $Spin^c$-structures.</p> <p>For algebraic surfaces, there's a canonical spin-c structure $\mathfrak{s}$, so $Spin^c(X)$ is identified with $H^2(X)$. Witten (http://arxiv.org/abs/hep-th/9411102) observed that the elliptic equations that define $SW$ simplify drastically in the algebraic case; in evaluating $SW$ on a cohomology class represented by a complex line bundle $L\to X$, you're led to consider a moduli space of pairs consisting of a holomorphic structure on the line bundle and a holomorphic section of it, with an obstruction bundle on the moduli space. Conjugation-invariance becomes Serre duality. </p> <p>For general type surfaces, $\pm SW(\mathfrak{s}) = \pm SW(\bar{\mathfrak{s}}) = \pm 1$; all other spin-c structures have vanishing invariant. Since $c_1(\mathfrak{s})=-c_1(\bar{\mathfrak{s}})=-K$, one deduces diffeomorphism-invariance of $\pm K$. For lower Kodaira dimension, a more complicated analysis is needed.</p> http://mathoverflow.net/questions/68803/how-to-relate-equivariant-symplectic-cohomology-contact-homology-cyclic-homolog/69490#69490 Answer by Tim Perutz for How to relate equivariant symplectic cohomology, Contact Homology, Cyclic Homology and String Topology? Tim Perutz 2011-07-04T19:51:35Z 2011-07-05T00:19:52Z <p><b>Some blah on symplectic homology vs. cohomology.</b> There's an invariant $SH(M)$ of Liouville domains $M$ which some people call symplectic homology and some symplectic cohomology. This is the direct limit of Hamiltonian Floer groups associated with functions of increasing eventual slope. The dual theory has two rather unpleasant features: it involves inverse limits, hence one must worry about $\varprojlim^1$-terms; and in general it's not countably generated. It's not often used.</p> <p>Why the confusion about terminology? Well, depending on your convention for the sign of the symplectic action functional, you may regard this as Morse homology or compactly supported Morse cohomology of this function. From the perspective of Lagrangian Floer cohomology, consistency demands that one calls it symplectic cohomology; I do. However, symplectic field theorists (including Bourgeois-Oancea, I think) prefer the contrary convention.</p> <p><b>Blah about grading.</b> The integer grading on $SH^\ast(M)$ is defined when $c_1=0$, and is canonical when $H^1(M)=0$. One has Viterbo's map $H^\ast(M)\to SH^\ast(M)$, and one convention makes this preserve degree (I'll take that option), while another makes it shift degree by the complex dimension $n$ of $M$.</p> <p><b>Seidel's map for cotangent bundles.</b> (Edited: my first version was not correct.) There are actually <i>two</i> versions of Seidel's open-closed string map, derived from the same moduli spaces: $$\kappa: SH^\ast(M) \to HH^\ast(F(M),F(M))$$ and $$\lambda : HH_\ast(F(M),F(M)) \to SH^{n+\ast}(M).$$ Here $F(M)$ is the Fukaya category of exact, compact Lagrangians. Moreover, there are extensions of these maps to the wrapped Fukaya category $W(M)$. </p> <p>The absence of a dualisation - hence the connection of symplectic cohomology to both Hochschild homology and cohomology - looks strange. Mohammed Abouzaid points out in his comment below that this is a manifestation of a self-duality property for the wrapped category. He shows in <a href="http://arxiv.org/abs/1003.4449" rel="nofollow">http://arxiv.org/abs/1003.4449</a> that for $M=T^\ast L$ and the wrapped category, the map $\lambda$ is an isomorphism. </p> <p><b>Cyclic version.</b> My expectations are slightly different from those stated in the question. I'd guess that $\lambda$ extends to a map from cyclic homology to circle-equivariant symplectic cohomology, $$HC_\ast(F(M)) \to SH^{n+\ast}_{S^1}(M)$$ and that this should be an isomorphism when $\lambda$ is. </p> <p>For cotangent bundles $T^\ast L$ of simply connected, spin manifolds, one has $SH^\ast_{S^1}(M) \cong H_{n-*}^{S^1}(\mathcal{L}L)$. </p> <p>Linearised contact (co)homology is, according to Bourgeois-Oancea (if I have it right), the mapping cone of the (cochain level) Viterbo map $H^\ast(M; H^\ast_{S^1}(pt.)) \to SH^\ast_{S^1}(M)$. For cotangent bundles as before, Viterbo's map should be identifiable with the map induced by the equivariant inclusion of constant loops: $H^\ast(T^\ast L)[u] = H^\ast (L)[u] = H_{n-\ast}(L)[u] \to H_{n-\ast}^{S^1}(\mathcal{L}L)$. </p> http://mathoverflow.net/questions/68842/can-cotangent-bundles-see-exotic-smooth-structures/69249#69249 Answer by Tim Perutz for Can cotangent bundles see exotic smooth structures? Tim Perutz 2011-07-01T10:56:52Z 2011-07-01T10:56:52Z <p>I wrote a little expository piece about this and related matters in the Newsletter of the European Mathematical Society:</p> <p><a href="http://www.ems-ph.org/journals/newsletter/pdf/2010-03-75.pdf" rel="nofollow">http://www.ems-ph.org/journals/newsletter/pdf/2010-03-75.pdf</a></p> <p>The classical topology of $X:=T^\ast L$ can be taken to include a little more than its diffeomorphism type: there's also an almost complex structure $J$ on $X$, canonical up to homotopy. The pair $(X,J)$ knows the Pontryagin classes of $L$, because <code>$$c_{2k}(TX,J)|_{L}=c_{2k}(TL\otimes\mathbb{C})=(-1)^k p_k(TL),$$</code> so $(-1)^k p_k(TL)$ pulls back to $c_{2k}(L)$ under projection $X\to L$. However, even with this embellishment, the smooth topology of $X$ doesn't determine $L$.</p> <p><b>Faithfulness conjecture:</b> the exact symplectomorphism type of the cotangent bundle $(X,\omega=d\lambda_L)$ of a compact manifold $L$ determines $L$. </p> <p>An exact symplectomorphism $T^\ast L \to T^\ast L'$ is a diffeomorphism $f$ such that $f^*\lambda_{L'}-\lambda_L= dh$ for $h$ a compactly supported function. The conjecture (but not the name) is standard.</p> <p>Attempts to use symplectic <i>invariants</i> of $X$ to distinguish smooth structures on $L$ have so far been a complete failure. For example, the symplectic cohomology ring $SH^*(X)$ is isomorphic to loopspace homology $H_{-*}(\mathcal{L}L; w)$ (the coefficients are the local system $w$ of $\mathbb{Z}$-modules determined by $w_2$), with the string product. This invariant is determined by the homotopy type of $L$.</p> <p><b>"Arnol'd's conjecture"</b> (scare quotes because Arnol'd really made a much more circumspect conjecture). Any exact Lagrangian embedding $\Lambda \to X$ (with $\Lambda$ compact) is exact-isotopic to the embedding of the zero-section. </p> <p>This would immediately imply the faithfulness conjecture. </p> <p>There has been progress towards Arnol'd's conjecture of three kinds: </p> <p>(1) It's true for $L=S^2$ (Hind, <a href="http://arxiv.org/abs/math/0311092" rel="nofollow">http://arxiv.org/abs/math/0311092</a>). </p> <p>(2) The work of several authors (Fukaya-Seidel-Smith <a href="http://arxiv.org/abs/0705.3450" rel="nofollow">http://arxiv.org/abs/0705.3450</a>, Nadler <a href="http://arxiv.org/abs/math/0612399" rel="nofollow">http://arxiv.org/abs/math/0612399</a>, Abouzaid <a href="http://arxiv.org/abs/1005.0358" rel="nofollow">http://arxiv.org/abs/1005.0358</a>, and Kragh's work in progress) cumulatively shows that the projection from an exact Lagrangian to the zero-section is a homotopy equivalence. This is good evidence for the truth of the conjecture, but for the application to faithfulness one might as well make homotopy-equivalence an assumption. </p> <p>(3) As Andy mentioned, Abouzaid <a href="http://arxiv.org/abs/0812.4781" rel="nofollow">http://arxiv.org/abs/0812.4781</a> has shown that a homotopy $(4n+1)$-sphere $S$, such that $T^*S$ contains an exact embedded Lagrangian $S^{4n+1}$, bounds a parallelizable manifold. This is proved by a stunning analysis of the geometry of a space of pseudo-holomorphic discs.</p> <p>The existence of exact Lagrangian <i>immersions</i> is governed by homotopy theory (there is an h-principle which finds such an immersion given suitable homotopical data). Just as the subtleties of 4-manifold topology can be located at the impossibility of removing double points of immersed surfaces (the failure of the Whitney trick), so the subtlety of the symplectic structure of cotangent bundles comes down to the question of removability of double points of Lagrangian immersions.</p> http://mathoverflow.net/questions/58589/calabi-yau-manifolds-and-polygonal-linkage-configuration-spaces-related/58612#58612 Answer by Tim Perutz for Calabi-Yau manifolds and polygonal linkage configuration spaces: related? Tim Perutz 2011-03-16T05:07:23Z 2011-03-16T17:49:53Z <p>An illustrative example: <i>the moduli space $M$ of regular pentagons with edges of unit length</i>. This embeds as an open, dense subset of a compact complex surface $\bar{M}$ with a canonical Kaehler form. This surface, a 4-fold blow-up of $\mathbb{CP}^2$, is not Calabi-Yau (trivial canonical bundle) but Fano (ample anticanonical bundle).</p> <p>The compactified regular pentagon space $\bar{M}$ is the space of 5-tuples of unit vectors in $\mathbb{R}^3$ with centre of mass zero, modulo the diagonal action of $SO(3)$. Since we remember the order of these vectors, typical points represent regular pentagons with a distinguished vertex ("start here") and adjacent edge ("go this way"). There are also points which represent an equilateral triangle together with a pair of antipodal points, and these non-pentagon points form ten 2-spheres in $\bar{M}$. </p> <p>$\bar{M}$ has a natural symplectic structure, for which the ten 2-spheres are Lagrangian. Take the unique area-form on $S^2$, invariant under $SO(3)$, of total area 1 and inducing the complex orientation of $S^2=\mathbb{CP}^1$. The moment map $S^2\to \mathfrak{so}(3)^\ast \cong \mathbb{R}^3$ for the $SO(3)$-action is just the inclusion of $S^2$ into $\mathbb{R}^3$. The product $(S^2)^5$ carries the product symplectic form, again $SO(3)$-invariant, with moment map $\mu(x_1,\dots,x_5)=x_1+\dots + x_5\in \mathbb{R}^3$. The symplectic quotient $\mu^{-1}(0)/SO(3)$ is just $\bar{M}$.</p> <p>The action of $SO(3)=PU(2)$ respects the complex structure of $(\mathbb{CP}^1)^5$, and $\bar{M}$ inherits a complex structure by Kaehler reduction. It turns out to be isomorphic as a Kaehler surface to a blow up of $\mathbb{CP}^2$ at four special points with a Fano Kaehler form (but I haven't thought through which points). See Seidel's <a href="http://arxiv.org/abs/math/0309012" rel="nofollow">Lectures on 4-dimensional Dehn twists</a>, ex. 1.10. There's a natural action of the icosahedral group, permuting the $x_i$.</p> <p>If one wanted pentagons defined by some other linear equation, say $a_1x_1+\dots + a_5x_5=0$, one would give the $S^2$-factors areas $a_i$. </p> <p>One can also interpret $\bar{M}$ as an algebro-geometric (GIT) quotient of $(\mathbb{CP}^1)^5$ by $PSL_2(\mathbb{C})$. The quotient happens to be the Deligne-Mumford (or Grothendieck-Knudsen) compactification <code>$\bar{M}_{0,5}(\mathbb{C})$</code> of configurations of five points on $\mathbb{CP}^1$. The real points <code>$\bar{M}_{0,5}(\mathbb{R})$</code>, the fixed points of an anti-holomorphic involution of $\bar{M}$, are also interesting: their connected components are polyhedral and are copies of the 2-dimensional <a href="http://en.wikipedia.org/wiki/Associahedron" rel="nofollow">Stasheff associahedron</a> (a.k.a. pentagon).</p> <p><b>References:</b> </p> <p>F. Kirwan, <a href="http://books.google.com/books?id=4wfZBnlSaJ0C&amp;printsec=frontcover&amp;dq=kirwan+cohomology+of+quotients&amp;source=bl&amp;ots=WhidvP7PUy&amp;sig=4VbvWVflmsmYZ0F7PrdpO4-G1hg&amp;hl=en&amp;ei=C_aATYuYBovmsQP_g_SSBg&amp;sa=X&amp;oi=book_result&amp;ct=result&amp;resnum=2&amp;ved=0CCAQ6AEwAQ#v=onepage&amp;q&amp;f=false" rel="nofollow">Cohomology of quotients in symplectic and algebraic geometry</a>.</p> <p>J.-C. Hausmann and A. Knutson, <a href="http://arxiv.org/pdf/dg-ga/9602012v1" rel="nofollow">Polygon spaces and Grassmannians</a>.</p> http://mathoverflow.net/questions/57894/symplectic-4-manifolds-with-free-circle-action/57968#57968 Answer by Tim Perutz for symplectic 4-manifolds with free circle action Tim Perutz 2011-03-09T16:13:09Z 2011-03-09T16:13:09Z <p>Here's an example, using a construction of Fernandez, Gray and Morgan (1991): </p> <p>Take a closed surface $S$ with area form $\omega$, let $\phi$ be an area-preserving diffeomorphism, and $p\colon S_\phi \to S^1$ its mapping torus. This carries a closed 2-form $\omega_\phi$ induced by $\omega$, and a closed 1-form $p^\ast dt$. Take a class $e\in H^2(S_\phi;\mathbb{Z})$ which restricts trivially to $H^2(S;\mathbb{Z})$. Then $e$ has a de Rham representative of form $p^\ast dt\wedge a$, where $a$ is a closed 1-form. Take $L\to S_\phi$ be a hermitian line bundle with a connection form $i\eta$ of curvature $(-2\pi i) p^\ast dt\wedge a$, and let $\pi\colon M\to S_\phi$ be the unit circle bundle in $L$. Then the 4-manifold $M$ carries the $S^1$-invariant symplectic form $\Omega:= \pi^* \omega_\phi + \pi^*p^\ast dt\wedge \eta$.</p> <p>Let's take $S$ to have genus $>1$ and $\phi$ to be a Dehn twist along a non-separating curve. The Wang exact sequence identifies $\ker (H^2(S_\phi)\to H^2(S))$ with the cokernel of $(1-\phi^\ast)$ acting on $H^1(S)$. In this case, the cokernel is $\mathbb{Z}$, and we take $e$ to be the generator. The fibration by $S^1$-orbits is non-trivial, but we need to check that the resulting $M$ is not homeomorphic to $S^1\times N^3$ in some weirder fashion.</p> <p>Well, $\pi_1(M)$ is a non-trivial central $\mathbb{Z}$-extension of $\pi_1(S_\phi)$, and the latter is a semidirect product of $\pi_1(S)$ with $\mathbb{Z}$, where $1\in \mathbb{Z}$ acts on $\pi_1(S)$ by $\phi^{-1}$. If I'm not mistaken, $\pi_1(S_\phi)$ has trivial centre. Hence the centre of $\pi_1(M)$ is $\mathbb{Z}$, the subgroup generated by the $S^1$-fibre $\gamma$. If $M = S^1\times N$ then $\pi_1(M)$ is a trivial central $\mathbb{Z}$-extension of $\pi_1(N)$. But the central $\mathbb{Z}$-subgroup defined by this product splitting must be generated by a multiple of $\gamma$, and so the extension it defines is not trivial after all.</p> http://mathoverflow.net/questions/16632/stiefel-whitney-classes-over-integers/16639#16639 Answer by Tim Perutz for Stiefel-Whitney Classes over Integers? Tim Perutz 2010-02-27T22:21:52Z 2011-03-01T16:49:13Z <p><i>I'm grateful to Allen Hatcher, who pointed out that this answer was incorrect. My apologies to readers and upvoters. I thought it more helpful to correct it than delete outright, but read critically.</i></p> <p>If $X$ and $Y$ are cell complexes, finite in each degree, and two maps $f_0$ and $f_1\colon X\to Y$ induce the same map on cohomology with coefficients in $\mathbb{Q}$ and in $\mathbb{Z}/(p^l)$ for all primes $p$ and natural numbers $l$, then they induce the same map on cohomology with $\mathbb{Z}$ coefficients. To see this, write $H^n(Y;\mathbb{Z})$ as a direct sum of $\mathbb{Z}^{r}$ and various primary summands $\mathbb{Z}/(p^k)$, and note that the summand $\mathbb{Z}/(p^k)$ restricts injectively to the mod $p^l$ cohomology when $l\geq k$. One can take only those $p^l$ such that there is $p^l$-torsion in $H^\ast(Y;\mathbb{Z})$. (I previously claimed that one could take $l=1$, which on reflection is pretty implausible, and is indeed wrong.)</p> <p>We can try to apply this to $Y=BG$, for $G$ a compact Lie group. For example, $H^{\ast}(BU(n))$ is torsion-free (and Chern classes generate the integer cohomology), and so rational characteristic classes suffice. In $H^{\ast}(BO(n))$ and $H^{\ast}(BSO(n))$ there's only 2-primary torsion. That leaves the possibility that the mod 4 cohomology contains sharper information than the mod 2 cohomology. It does not, because, as Allen Hatcher has pointed out <a href="http://mathoverflow.net/questions/56932/what-characteristic-class-information-comes-from-the-2-torsion-of-hbsonz" rel="nofollow">in this recent answer</a>, all the torsion is actually 2-torsion.</p> <p>It's sometimes worthwhile to consider the integral Stiefel-Whitney classes $W_{i+1}=\beta_2(w_i)\in H^{i+1}(X;\mathbb{Z})$, the Bockstein images of the usual ones. These classes are 2-torsion, and measure the obstruction to lifting $w_i$ to an integer class. For instance, an oriented vector bundle has a $\mathrm{Spin}^c$-structure iff $W_3=0$.</p> <p>[I'm sceptical of your example in $2\mathbb{CP}^2$. So far as I can see, $3a+3b$ squares to 18, not 6, and indeed, $p_1$ is not a square.]</p> http://mathoverflow.net/questions/56884/does-there-exist-a-family-of-curves-or-abelian-varieties-on-the-punctured-line/56917#56917 Answer by Tim Perutz for Does there exist a family of curves (or abelian varieties) on the punctured line with specified monodromy on H^1? Tim Perutz 2011-02-28T18:45:25Z 2011-02-28T18:45:25Z <p>A special case of this situation: suppose one gives a sequence of simple closed curves $(\gamma_1,\dots, \gamma_N)$ on an oriented $C^\infty$ surface, such that the product of Dehn twists $\tau_{\gamma_N}\circ \dots \circ \tau_{\gamma_1}$ is isotopic to the identity. One can then build a closed 4-manifold $X$ and a topological Lefschetz fibration $f\colon X\to \mathbb{CP}^1$ whose monodromy, for a chosen basis of vanishing paths, is given by the specified Dehn twists. Moreover, $X$ carries essentially canonical symplectic forms (Gompf). </p> <p>A variant of question 3 (which shortcuts issues concerning the Torelli group) asks: <i>when one can take $X$ to be a Kaehler surface and $f$ to be an algebraic map?</i></p> <p>Donaldson proved that <i>every</i> symplectic 4-manifold with rational symplectic class admits (after blowing up enough points) a topological Lefschetz fibration. From this point of view, the question is about the disparity between symplectic and Kaehler structures on 4-manifolds - and this disparity is very wide.</p> <p>But here's a positive result about the genus 2 case. The genus 2 mapping class group maps onto the braid group on 6 strings, and if the $\gamma_i$ are non-separating and the twists $\tau_{\gamma_i}$ act transitively on the strings, then <a href="http://arxiv.org/pdf/math/0305343v1" rel="nofollow">Siebert and Tian</a> show that $(X,f)$ can be made algebraic.</p> http://mathoverflow.net/questions/55496/length-of-floer-flow-lines/55561#55561 Answer by Tim Perutz for Length of Floer flow lines Tim Perutz 2011-02-15T22:55:40Z 2011-02-15T22:55:40Z <p>In your symplectically aspherical setting, bounds on length will indeed exist. </p> <p>Suppose one has a sequence of solutions $u_n$ to Floer's equation, of bounded energy, and a sequence of points $t_n\in S^1$ with lengths $l(u_n(\cdot,t_n))\to \infty$. Gromov-Floer compactness tells us that after passing to a subsequence, this sequence converges in the Gromov-Floer topology to a broken trajectory plus bubbles. Each cylinder in the domain of the broken trajectory is associated with a time-translation $\sigma_k \colon s\mapsto s+s_k$, and the maps $\sigma_k^\ast u_n$ converge to a limiting cylinder $v_k$ <i>plus bubbles.</i> Outside a long but finite cylinder $[-T,T] \times S^1$, $\sigma_k^\ast u_n$ is exponentially close to $v_k$ (meaning $\leq ce^{-as}$, where $a>0$ depends only on the asymptotic limits $x$ and $y$, and $c$ only on $u_n(s_k,\cdot)$). Similar decay applies to the first derivative of $u_n$. The divergence of lengths must therefore occur inside the finite cylinder. </p> <p>Inside the finite cylinder but away from finitely many "bad points", $\sigma_k^\ast u_n$ converges to $v_k$, and for any $r$, convergence in $C^r$ is uniform on compact subsets. At the bad points, bubbles form. In the symplectically aspherical case, bubbling is impossible: there are no bad points. Therefore one has uniform $C^1$ bounds on $\sigma_k^\ast u_n$ over the finite cylinder, contradicting the divergence of the lengths.</p> <p>Maybe it would be interesting to write down a bubbling sequence of holomorphic maps $\mathbb{CP}^1\to \mathbb{CP}^1$ and watch what happens to the lengths. </p> http://mathoverflow.net/questions/55215/what-is-the-hochschild-cohomology-of-the-dg-category-of-perfect-complexes-on-a-va What is the Hochschild cohomology of the dg category of perfect complexes on a variety? Tim Perutz 2011-02-12T16:41:49Z 2011-02-12T19:16:40Z <p>Let $X$ be a quasi-projective variety over a field $k$. Let $D_{qcoh}$ be a dg enhancement of the unbounded derived category of quasi-coherent sheaves over $X$, and $D_{perf}$ its full subcategory of perfect complexes. </p> <p>This question is about Hochschild cohomology in the dg category sense. The question in the title gives the gist, but a more precise question is the following. There is a restriction homomorphism $$r\colon HH^\ast(D_{qcoh},D_{qcoh}) \to HH^\ast(D_{perf},D_{perf}).$$ This is clear if one computes Hochschild cohomology using the bar complex. </p> <p><b>Question:</b> Is $r$ an isomorphism? </p> <p>The left-hand side makes me nervous because of the unbounded complexes, but I think this is true in the case of smooth $X$. I'm interested in singular varieties (or schemes). What about the affine case? To be concrete, what about the case where $X$ is a local complete intersection? </p> <p><b>Comments:</b> The category $D_{qcoh}$ looks much more ferocious than $D_{perf}$. The reason for bringing it into the picture is that its Hochschild cohomology seems to be better understood. One possible reference (which also discusses similar results for perfect complexes, but only in the smooth case) is <a href="http://arxiv.org/abs/math/0408337" rel="nofollow">Toen's article on derived Morita equivalence.</a> If I understood correctly (did I?), $HH^\ast(D_{qcoh},D_{qcoh})$ is isomorphic to <code>$$HH^\ast(X):= Ext^\ast_{\mathcal{O}_{X\times X}}(\delta_{\ast}\mathcal{O}_X,\delta_{\ast}\mathcal{O}_X)$$</code> where $\delta\colon X\to X\times X$ is the diagonal. And this is the thing I really want to compare to $HH^\ast(D_{perf},D_{perf})$, for the reason that I know how to compute it in examples. There is a local-to-global Ext spectral sequence converging to $HH^\ast(X)$, and for local complete intersections one can use the Hochschild-Kostant-Rosenberg isomorphism to understand the Ext-sheaf.</p> <p><b>Motivation:</b> I've been looking at manifestations of homological mirror symmetry in which one has an embedding of the Fukaya category of a symplectic manifold into $D_{per}$ for a mirror variety. I'd like to compute Hochschild cohomology of the Fukaya category via algebraic geometry.</p> http://mathoverflow.net/questions/54338/plurisubharmonic-exhaustion-functions-without-critical-points-at-infinity/54888#54888 Answer by Tim Perutz for Plurisubharmonic exhaustion functions without critical points at infinity Tim Perutz 2011-02-09T14:40:59Z 2011-02-09T14:40:59Z <p><i>Expanded version of my earlier comment, which was directed at the unedited version of the question.</i> </p> <p>Stein manifolds are those complex manifolds $X$ which have a strictly psh exhaustion function, i.e., a proper, bounded below $C^\infty$ function $\psi$ such that the closed $(1,1)$-form $\omega:=-dd^c \psi$ is positive with respect to the complex structure $J$. (My convention is that $d^c f = J \circ df$ where $J$ is the complex structure acting on cotangent vectors.)</p> <p>Those for which one can take $\psi$ to have compact critical set are called "finite-type". They include smooth affine algebraic varieties $X\subset \mathbb{C}^N$, which one can see by compactifying to a projective variety $\bar{X}=X\cup D$ and considering $\log \|s\|^2$ where $s$ is a section of a hermitian holorphic line bundle cutting out the divisor $D$.</p> <p>Any open Riemann surface is Stein, but those of infinite genus (i.e. with infinite rank $H_1$) do not have finite type. If there were a psh exhaustion with compact critical set one could perturb it near the critical set so as to make it a Morse function. The downward gradient flow exists and converges to critical points, and so Morse theory bounds the rank of $H_\ast$ from above by the number of critical points.</p> http://mathoverflow.net/questions/26776/kahler-structure-on-cotangent-bundle/26781#26781 Comment by Tim Perutz Tim Perutz 2013-03-07T22:27:48Z 2013-03-07T22:27:48Z Right, but Whitney proved that every smooth manifold is diffeomorphic to a real analytic manifold. Kai's answer is indeed a good one (I didn't know of the work he cited). http://mathoverflow.net/questions/123619/explicit-model-of-bsu2-in-terms-of-singular-complex Comment by Tim Perutz Tim Perutz 2013-03-06T19:43:54Z 2013-03-06T19:43:54Z @Xiao-Gang: I found it hard to follow your proposal for the definition of $\pi$-cohomology. But algebraic topologists do have an assortment of theories that in some sense refine ordinary cohomology - notably, (co)bordism theories - which are accessible from cellular descriptions of the spaces, though harder to compute than ordinary cohomology. You might think about whether those are relevant to the problems you are working on. http://mathoverflow.net/questions/123619/explicit-model-of-bsu2-in-terms-of-singular-complex/123635#123635 Comment by Tim Perutz Tim Perutz 2013-03-05T21:00:47Z 2013-03-05T21:00:47Z This description should indeed yield a CW complex, and from that one could I suppose work out a triangulation. But, like Andr&#233; Henriques, I find it hard to believe that triangulation is an effective way to extract information (compare the standard treatments of $BU(1)$ and $B(\mathbb{Z}/2)$). http://mathoverflow.net/questions/122119/what-prevents-a-manifold-to-be-symplectic/122327#122327 Comment by Tim Perutz Tim Perutz 2013-02-19T20:47:47Z 2013-02-19T20:47:47Z (This should have come first, but anyway: welcome to MO!) http://mathoverflow.net/questions/122119/what-prevents-a-manifold-to-be-symplectic/122327#122327 Comment by Tim Perutz Tim Perutz 2013-02-19T20:34:37Z 2013-02-19T20:34:37Z &quot;As far as I know this is the only known geometric obstruction&quot;: A little pedantically, there are further obstructions in 4 dimensions, beyond the Taubes constraints w.r.t. $[\omega]$ and the a.c. structure: namely, the Taubes constraints on a finite covering space w.r.t to the pullback data! http://mathoverflow.net/questions/121928/off-topic-question-lefschetz-fibration Comment by Tim Perutz Tim Perutz 2013-02-15T21:24:15Z 2013-02-15T21:24:15Z The second description is certainly intuitive and important, but it's also inadequate; it doesn't fully characterize a Lefschetz fibration. My calculus students are irritated by the official definition of continuity... http://mathoverflow.net/questions/120819/open-problems-in-seiberg-witten-theory-on-4-manifolds/120827#120827 Comment by Tim Perutz Tim Perutz 2013-02-15T02:36:52Z 2013-02-15T02:36:52Z I omitted to say that the Taubes constraints apply when $b^+&gt;1$, and the question I mentioned at the end concerns that case. http://mathoverflow.net/questions/121610/analogue-of-picard-lefschetz-formulas-for-more-than-one-node/121639#121639 Comment by Tim Perutz Tim Perutz 2013-02-12T20:17:12Z 2013-02-12T20:17:12Z Indeed: in a nearby fibre, the vanishing cycles will be disjoint. Far from the critical points, the whole family over the disc is trivialized by radial parallel transport into the central fibre; near the critical points, one has a standard quadratic model. Show in this way that the smooth monodromy is a simultaneous Dehn twist along the vanishing spheres $V_i$. Then prove Picard-Lefschetz by observing that for any $n$-cycle $x$, $(S-1)(x)$ is alway homologous to a cycle supported in neighborhoods $T^*V_i$ of the vanishing cycles, hence to a linear combination of the $\delta_i$. http://mathoverflow.net/questions/120442/are-rational-varieties-simply-connected/120444#120444 Comment by Tim Perutz Tim Perutz 2013-01-31T23:10:57Z 2013-01-31T23:10:57Z Mohammad: being algebraically simply connected means that there are no non-trivial connected etale covers. Such covers are finite, while in topology one makes no finiteness condition on the covers. See <a href="http://en.wikipedia.org/wiki/&#201;tale_fundamental_group" rel="nofollow">en.wikipedia.org/wiki/&hellip;</a> http://mathoverflow.net/questions/120442/are-rational-varieties-simply-connected/120444#120444 Comment by Tim Perutz Tim Perutz 2013-01-31T19:50:00Z 2013-01-31T19:50:00Z Vesselin, Jason: thanks! http://mathoverflow.net/questions/120442/are-rational-varieties-simply-connected/120444#120444 Comment by Tim Perutz Tim Perutz 2013-01-31T19:30:34Z 2013-01-31T19:30:34Z Do I understand right that your paragraph beginning &quot;Alternatively&quot; shows that over the complex numbers the algebraic fundamental group is trivial, but leaves open the possibility that the classical fundamental group is non-trivial and without finite-index subgroups? Is the classical fundamental group of a rationally connected, smooth, complex projective variety known by other means to be trivial? http://mathoverflow.net/questions/119586/what-foliations-are-symplectic-foliations Comment by Tim Perutz Tim Perutz 2013-01-22T18:34:26Z 2013-01-22T18:34:26Z One of the main goals of symplectic geometry is to answer this question in the special case where $M$ is the unique leaf... http://mathoverflow.net/questions/118479/orientations-for-pseudoholomorphic-curves-with-totally-real-boundary-condition/118483#118483 Comment by Tim Perutz Tim Perutz 2013-01-21T16:34:55Z 2013-01-21T16:34:55Z Also, for the statement at the end of 1) the relatively Pin Lagrangians should still be orientable (thanks to Penka Georgieva for noting this). http://mathoverflow.net/questions/118479/orientations-for-pseudoholomorphic-curves-with-totally-real-boundary-condition/118483#118483 Comment by Tim Perutz Tim Perutz 2013-01-21T16:33:12Z 2013-01-21T16:33:12Z (Perhaps I should add in light of Aleksey's answer that, as I think is usual for an MO answer, this is an informal exposition of the situation as I understand it, and is certainly not comprehensive, mathematically or bibliographically. I mentioned what I take to be the two earliest sources, but among the possible later references I am biased towards the source that I myself learned from.) http://mathoverflow.net/questions/118479/orientations-for-pseudoholomorphic-curves-with-totally-real-boundary-condition/118483#118483 Comment by Tim Perutz Tim Perutz 2013-01-11T23:25:44Z 2013-01-11T23:25:44Z Chris, I've now added some details.