User domenico fiorenza - MathOverflowmost recent 30 from http://mathoverflow.net2013-05-23T06:15:23Zhttp://mathoverflow.net/feeds/user/8320http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/128325/invariant-definition-of-graded-poisson-bracket/128349#128349Answer by domenico fiorenza for Invariant definition of graded Poisson bracketdomenico fiorenza2013-04-22T12:51:02Z2013-04-22T12:51:02Z<p>If one adopts the convention that the Hamiltonian vector field $X_h$ is defined by the equation $\iota_{X_h}\omega=dh$ (as both Cattaneo-Fiorenza-Longoni and Cattaneo-Schaetz do), and that $\iota_X(df)=X(f)$, then yes, there is a missing $(-1)^{|f|+1}$ factor in front of the formula in Cattaneo-Fiorenza-Longoni. But I can imagine other graded symplectic geometry sign conventions in which <code>$\{f,g\}=\iota_{X_f}\iota_{X_g}\omega$</code> is the correct formula. For instance, in ordinary symplectic geometry, it is customary to define Hamiltonian vector fields by the equation $\iota_{X_h}\omega+dh=0$ rather than by $\iota_{X_h}\omega=dh$, and with this convetion one has <code>$\iota_{X_f}\iota_{X_g}\omega=-X_f(g)$</code> with no need of additional sign corrections. </p>
http://mathoverflow.net/questions/48477/never-appeared-forthcoming-papersNever appeared forthcoming papersdomenico fiorenza2010-12-06T19:58:30Z2013-03-08T17:37:28Z
<p>This has been inspired by this MO question: <a href="http://mathoverflow.net/questions/48174/harmonic-maps-into-compact-lie-groups" rel="nofollow">http://mathoverflow.net/questions/48174/harmonic-maps-into-compact-lie-groups</a></p>
<p>Just for joking: which is your favourite never appeared forthcoming paper?</p>
<p>(do not hesitate to close this question if unappropriate)</p>
http://mathoverflow.net/questions/114090/what-is-the-dgla-controlling-the-deformation-theory-of-a-complex-submanifold/122618#122618Answer by domenico fiorenza for What is the DGLA controlling the deformation theory of a complex submanifold?domenico fiorenza2013-02-22T09:43:10Z2013-02-22T09:43:10Z<p>I'd too suggest the paper by Donatella Iacono as a basic reference. I'll just add
here a few lines to complement Urs Schreiber's answer by explaining in which
sense the $\infty$-groupoids point of view clarifies what happens here (if my
memory is not playing a bad trick on my, in the paper with Elena Martinengo we do
not address the problem of deformations of submanifalds, although the needed
technology is there). So these few lines are to be read as a "How to read
Donatella Iacono's results by an $\infty$-groupoids point of view".</p>
<p>What one does is moving from Set-valued deformation functors to
$\infty$-groupoids valued deformation functors (these are called "formal moduli
problems" in Lurie's DAG X). One recovers the classical deformation functor from
the $\infty$-groupoid valued one simply by taking $\pi_0$. However (and this is
the reason to prefer the $\infty$-groupoid valued version), things are behaved
much more naturally in the $\infty$-groupoids setting since here one can make homotopy invariant constructions. </p>
<p>More precisely, if we denote by $Def_{\mathfrak{g}}$ the $\infty$-groupoids-valued deformation functor associated with a differential graded Lie algebra $\mathfrak{g}$, then the association $\mathfrak{g}\mapsto Def_{\mathfrak{g}}$ establishes an equivalence of $(\infty,1)$ categories between differential graded Lie algebras and formal moduli problems (see Lurie's DAG X or Pridham's "Unifying derived deformation theories"). This means in particular that if the deformation problem we are interested in arises as a (homotopy) limit of simpler deformation problems for which we know differential graded Lie algebras $\mathfrak{g}_i$ governing them, then the problem we are interested in will be governed by the (homotopy) limit of the $\mathfrak{g}_i$'s.</p>
<p>For instance, consider the problem of $Def_{Z\hookrightarrow X}$ of infinitesimal deformations of a complex submanifold $Z$ inside a complex manifold $X$. Such a deformation is equivalently the datum of a deformation of the pair $(Z,X)$ together with the datum of a trivialization of the deformation of $X$. So if we denote by $Def_{(Z,X)}$ and by $Def_X$ the deformation functors describing infinitesimal deformations of the pair and of $X$ respectively, we see that the problem $Def_{Z\hookrightarrow X}$ we are interested in is the (homotopy) fiber of the forgetful morphism $Def_{(Z,X)}\to Def_X$. For both $Def_X$ and $Def_{(Z,X)}$ it is simple to describe dglas governing them. They are the Dolbeault complex on $X$ with coefficients in the tangent sheaf of $X$ and the sub-dgla of this given by the kernel of the natural morphism to the Dolbeault complex on $Z$ with coefficients in the normal sheaf of $Z$ (i.e., by differential forms on $X$ with coefficients in the tangent sheaf of $X$ which, restricted to $Z$ are tangent to $Z$, thus inducing a deformation of $Z$). The dgla governing $Def_{Z\hookrightarrow X}$ will therefore be the homotopy fiber of this inclusion.</p>
<p>Since the model category structure on differential graded Lie algebras is induced by that on chain complexes, as a chain complex the homotopy fiber of the inclusion of the kernel above is quasi-isomorphic to the Dolbeault complex on $Z$ with coefficients in the normal sheaf of $Z$ shifted by one degree. And by the homotopical transfer of $L_\infty$-structures this means that there is a natural $L_\infty$-algebra structure on the shifted Dolbeault complex of $Z$ with values in $N_{X/Z}$, extending the chain complex structure, such that the associated deformation functor is $Def_{Z\hookrightarrow X}$. This answers Question 2 above.</p>
<p>In particular one recovers the well known fact that the tangent space to $Def_{Z\hookrightarrow X}$ is $H^0(Z, N_{X/Z})$: this is a degree 1 cohomology group for the shifted Dolbeault complex. Similarly one sees that $H^1(Z, N_{X/Z})$ is an obstruction space for $Def_{Z\hookrightarrow X}$.</p>
http://mathoverflow.net/questions/114585/an-explicit-homotopy-equivalence-between-the-de-rham-complex-and-the-cech-de-rhamAn explicit homotopy equivalence between the de Rham complex and the Cech-de Rham total complexdomenico fiorenza2012-11-26T21:28:09Z2012-11-27T13:53:50Z
<p>I'm currently in need an explicit formula in classical cohomology which I'm pretty sure is well known, but which I've been unable to find in the references I am aware of. </p>
<p>Let $X$ be a smooth manifold and let $\mathcal{U}={U_\alpha}$ be a fixed open cover of $X$ such that all the finite intersections $U_{\alpha_1}\cap\cdots U_{\alpha_n}$ are contractible. Consider the following two cochain complexes:</p>
<p>the de Rham complex $\Omega^\bullet(X)$ of $X$ </p>
<p>the total complex of the Cech-de Rham bicomplex $\Omega^\bullet(\mathcal{U}_\bullet)$.</p>
<p>The restriction of a global form on $X$ to the open sets $U_\alpha$ gives a linear map</p>
<p>$j: \Omega^\bullet(X) \to Tot^\bullet(\Omega^\bullet(\mathcal{U}_\bullet))$</p>
<p>which, if I'm not wrong here, is a injective quasi-isomorphism of cochain complexes. I've been able to prove this (if I've not made mistakes), by brute force: i.e. by showing that $j$
is bijective in cohomology. But I'd like to have a fancier proof by writing an explicit "globalization" morphism</p>
<p>$\pi : Tot^\bullet(\Omega^\bullet(\mathcal{U}_\bullet)) \to \Omega^\bullet(X)$</p>
<p>such that</p>
<p>$\pi j= id_{\Omega^\bullet(X)}$</p>
<p><code>$j \pi = id_{Tot^\bullet(\Omega^\bullet(\mathcal{U}_\bullet))} + [d_{tot},K]$</code></p>
<p>with $K$ some explicit morphism of graded vector spaces <code>$Tot^\bullet(\Omega^\bullet(\mathcal{U}_\bullet)) \to Tot^\bullet(\Omega^\bullet(\mathcal{U}_\bullet))[-1]$</code>.</p>
<p>I guess one should be able to build $K$ by using a partition of the unit subordinate to the
cover $\mathcal{U}$, but somehow I got lost in the computation. Since I feel this should be a well known fact, I'm asking here for direct references before attempting back to write $K$ myself. </p>
http://mathoverflow.net/questions/114585/an-explicit-homotopy-equivalence-between-the-de-rham-complex-and-the-cech-de-rham/114659#114659Answer by domenico fiorenza for An explicit homotopy equivalence between the de Rham complex and the Cech-de Rham total complexdomenico fiorenza2012-11-27T13:53:50Z2012-11-27T13:53:50Z<p>Thanks to an email by Chris Rogers, I now know that my question above is precisely the subject of Proposition 9.5 in Bott-Tu, Differential Forms in Algebraic Topology., where an explicit formula for the homotopy operator in terms of a partition of unit subordinate to the given open cover is given.</p>
<p>They also write "The not very intuitive formulas below were obtained, after repeated tries, by a careful bookkeeping of the inductive steps in the proof of Proposition 8.8." (where Proposition 8.8 is the statement that $j$ is a quasiisomorphism), which comforts me a lot in view of my failed attemps. Yet, since Bott-Tu is a basic reference on the subject I should have checked it, too, before asking on MO. Sorry for having not done.</p>
http://mathoverflow.net/questions/110435/pull-back-and-push-forward-of-higher-local-systemsPull-back and push-forward of higher local systemsdomenico fiorenza2012-10-23T14:43:17Z2012-10-25T21:26:39Z
<p>This is a follow up to the following two MO questions: <a href="http://mathoverflow.net/questions/106896/higher-holonomies-for-higher-local-systems" rel="nofollow">q1</a>,<a href="http://mathoverflow.net/questions/50497/fully-dualizable-objects-in-classical-field-theories" rel="nofollow">q2</a></p>
<p>What I'm interesting in understanding is the universal property (if any) of the morphism $Sum_n:Fam_n(\mathcal{C})\to \mathcal{C}$ considered by Freed-Hopkins-Lurie-Teleman in <a href="http://arxiv.org/abs/0905.0731" rel="nofollow">Topological Quantum Field Theories from Compact Lie Groups</a>. Namely, the existence and the more or less explicit form of such a monoidal functor is in itself such an amazing fact that one one could easily forget that if just a monoidal functor $Fam_n(\mathcal{C})\to \mathcal{C}$ is what one seeks, then this problem would be trivially solved by the trivial (or zero) functor. This would be clearly a trivial solution and one could say that what is remarkable in FHLT cosnstruction is that it is not trivial, but this would be quite a weak statement. Rather I suspect (and the question is precisely "How much are these suspects correct and which are references for them?") that $Sum_n$ secretely satisfies an adjointness property. More precisely, one has the forgetful functor $Fam_n(\mathcal{C})\to Fam_n$ and so a fully extended tqft with target $Fam_n(\mathcal{C})$ determines a "classical background" $X$, as the image of the point via the composition $Bord_n \to Fam_n(\mathcal{C})\to Fam_n$ (here I'm omitting eventual framings and orientations fore ease of writing/reading). A lifting from $Bord_n \to Fam_n$ to $Bord_n \to Fam_n(\mathcal{C})$ is the datum of a $\mathcal{C}$-local system on $X$ )or at least, it should be, if I'm not mistaken here), and $\mathcal{C}$ itself can be identified with $\mathcal{C}$-local system on the point. </p>
<p>So from this point of view the $Sum_n$ construction seems to be a push-forward to the point of a $\mathcal{C}$-local system on $X$. This suggests to consider the more general situation of a morphism $f:X\to Y$ of classical backgrounds, and to consider the pull-back of $\mathcal{C}$-local systems (this should be the easy direction) and wondering whether this has an adjoint. Then $Sum_n$ in FHLT should map the $\mathcal{C}$-local system $A:X\to \mathcal{C}$ of the pushforward of $A$ along the terminal morphism $X\to *$, and all the other values of $Sum_n$ should be uniquely determined by the cobordism hypothesis.</p>
<p>Is this correct? References?</p>
http://mathoverflow.net/questions/106896/higher-holonomies-for-higher-local-systemsHigher holonomies for higher local systemsdomenico fiorenza2012-09-11T10:01:34Z2012-09-12T10:30:18Z
<p>In Jacob Lurie's classification of tqfts, one finds a version of the cobordism hypothesis for $(X,\zeta)$-structure, where an $(X,\zeta)$ structure on a manifold $M$ is the datum of a continuous map $f:M\to X$ and of an isomorphism $\alpha$ of $\mathbb{R}$-vector bundles $T_M\oplus \mathbb{R}^{n-\dim(M)}\simeq f^*\zeta$. Here $\zeta$ is a fixed rank $n$ vector bundle on $X$ and $\dim(M)\leq n$. This version of the cobordism hypothesis states that for any symmetric monoidal $(\infty,n)$-category $\mathcal{C}$ with duals there is an equivalence of $(\infty,0)$ categories
$Fun^\otimes(Bord_n^{(X,\zeta)},\mathcal{C}) \simeq Hom_{O(n)}(\tilde{X},\mathcal{C}^{\sim})$, where $\tilde{X}$ is the total space of the principal $O(n)$-bundle on $X$ associated with the vector bundle $\zeta$ and $\mathcal{C}^{\sim}$ is the $(\infty,0)$ category obtained from $\mathcal{C}$ by discarding all the noninvertible morphisms.</p>
<p>In particular, this says (if I'm not wrong here) that given an $O(n)$-equivariant $\mathcal{C}$-valued local system $E$ on $\tilde{X}$ (i.e., an element $E$ in $Hom_{O(n)}(\tilde{X},\mathcal{C}^{\sim})$), to any manifold $M$ of dimension less or equal to $M$ equipped with an $(X,\zeta)$-structure is canonically associated an element $Z(E;M;f,\alpha)$ in $\Omega^{\dim M}\mathcal{C}$.</p>
<p>Now, fix an $n$-dimensional closed manifold $M$ and let $(X,\zeta)=(M,T_M)$. Then $M$ has a canonical $(X,\zeta)$-structure: the one given by the identity morphism! So the above general result on the cobordism hypothesis would say that to any $O(n)$-equivariant $\mathcal{C}$-valued local system $E$ on $\tilde{M}$ is canonically associated an element in $\Omega^n\mathcal{C}$, which I guess one should think of as the $n$-dimensional holonomy of the given local system. In other words, one should have a canonical morphism</p>
<p>$hol_M: Hom_{O(n)}(\tilde{M},\mathcal{C}^{\sim}) \to \Omega^n \mathcal{C}$.</p>
<p>More in general, if $\dim M\leq n$, one can take $(X,\zeta)=(M,T_M\oplus\mathbb{R}^{n-\dim M})$ and again the identity morphism provides a canonical $(X,\zeta)$-structure on $M$. Thus one should have a canonical morphism </p>
<p>$hol_M: Hom_{O(n)}(\tilde{M},\mathcal{C}^{\sim}) \to \Omega^{n-\dim M} \mathcal{C}$</p>
<p>for any closed manifold $M$ of dimension less or equal to $n$.</p>
<p>questions: </p>
<p>i) is this argument correct?</p>
<p>ii) can one give a direct description of the morphism $hol_M: Hom_{O(n)}(\tilde{M},\mathcal{C}^{\sim}) \to \Omega^n \mathcal{C}$ (i.e., without invoking the cobordism hypothesis)? note that some form of the cobordism hypothesis, namely the framed version, seems to be necessary even to state that there is a $O(n)$-action on $\mathcal{C}^\sim$. </p>
<p>iii) references?</p>
http://mathoverflow.net/questions/98291/are-fully-extended-tqfts-generalized-cohomology-theoriesAre fully extended TQFTs generalized cohomology theories?domenico fiorenza2012-05-29T16:42:06Z2012-06-18T12:14:04Z
<p>Forgive the naiveness of this question. Whatever an $n$-vector space exactly is, one expects that the basic example of fully extended $n$-dimensional tqft is a symmetric monoidal functor $Cob_n\to n$-Vect. Now, whatever an $n$-vector space exactly is, one expects $(n-1)$-Vect to be the based loop space of $n$-Vect. This suggests that the $n$-categories of $n$-vector spaces organize themselves in an hypothetical spectrum <b>Vect</b> and that the tqft invariants one computes are actually cohomology classes for the corresponding generalized cohomology. For instance, the fact that a fully extended tqft is completely determined by its value on a point would be in this perspective an analogue of Mayer-Vietoris. Also, the combinatorial constructions of the Dijkgraaf-Witten model would be an analogue of operations in simplicial cohomology. So it seems there is some general abstract nonsense supporting the above point of view.</p>
<p>Question: are there references addressing/formalizing/developing this point of view? </p>
http://mathoverflow.net/questions/92066/tqfts-with-target-category-of-higher-type-than-the-sourceTQFTs with target category of higher type than the sourcedomenico fiorenza2012-03-24T08:16:43Z2012-05-16T10:43:31Z
<p>In the classical version of the Cobordism Hypothesis, such as, e.g., in Jacob Lurie's <a href="http://arxiv.org/abs/0905.0465" rel="nofollow">On the Classification of Topological Field Theories</a>, one considers the $\infty$-category of symmetric monoidal functors from the $(\infty,n)$-symmetric monoidal category of (framed) $n$-cobordism to some $(\infty,n)$-symmetric monoidal target category $\mathcal{C}$. One of the first things one proves is that this category of functors is actually an $(\infty,0)$-category. For instance, if $n=1$ one can take $\mathcal{C}=Vect_k$ and one sees that the datum of a 1-dimensional tqft is the choice of a finite dimensional (i.e. fully dualizable) vector space $V$, so that the datum of a natural transformation between two such tqfts is a morphism of finite dimensional vector spaces $f:V\to W$. These are the data attached to the (oriented) point. Next one looks at what happens for the data attached to 1-dimensional manifolds and sees that $f$ is constrained to be an isomorphism (the quickest test here is to notice that $f$ has to induce an isomorphism between the dimension of $V$ and the dimension of $W$, in the category whose objects are elements of $k$ and whose morphism are only the identities). As this simple example shows, what is crucial here is that 1-dimensional manifolds has come into play. This means that if we had restricted our attention to 0-dimensional manifolds instead, i.e., we would have considered a symmetric monoidal functor from the symmetric monoidal $(\infty,0)$-category of 0-dimensional cobordims to the symmetric monoidal $(\infty,1)$-category of vector spaces over $k$, we would have had complete freedom in the choice of $f$.</p>
<p>This suggests that in general $n$-dimensional cobordism "eats" $n$ non-invertible levels in the target, so that the $\infty$-category of of symmetric monoidal functors from the $(\infty,n)$-symmetric monoidal category of (framed) $n$-cobordism to some $(\infty,n+k)$-symmetric monoidal target category $\mathcal{C}$ is actually an $(\infty,k)$-category.</p>
<p>Apart from the interest in this result in itself, I'm interested into it for the following possible application to the characters of finite groups: a representation $(V,\rho)$ of a finite group $G$ can be seen as a $k$-$k[G]$-bimodule, and so as a morphism from $k$ to $k[G]$ in the $(\infty,2)$-category of algebras, bimodules, bimodule morphisms. If this naturally induces a morphism of 1-dimensional tqfts from the tqft $Z_k$ defined by assigning to the point the algebra $k$ and the tqft $Z_{k[G]}$ defined by assigning to the point the algebra $k[G]$, then we would also have a natural morphism $\varphi_\rho:Z_k(S^1)\to Z_{k[G]}(S^1)$, i.e., from $k$ to the class functions of $G$. This should be nothing but the trace of $\rho$. </p>
<p>Is this correct? Is this point of view already expanded in some reference?</p>
http://mathoverflow.net/questions/94426/geometric-natural-transformations-between-fourier-mukai-transformsGeometric natural transformations between Fourier-Mukai transformsdomenico fiorenza2012-04-18T16:48:08Z2012-04-18T19:27:00Z
<p>Given two schemes $X$ and $Y$ one can consider additive functors (eventually with some nice additional property) between the categories of <code>$\mathcal{O}_X$</code>-modules and of <code>$\mathcal{O}_Y$</code> modules. Among these one has those "of a geometric nature", in particular Furier-Mukai-type functors, i.e., those of the form <code>$\Phi_{Z,\mathcal{P}}:=g_*(\mathcal{P}\otimes f^*-)$</code>, where $(f,g):Z\to X\times Y$ and $\mathcal{P}$ is some "twisting" sheaf on $Z$. With suitable finiteness assumptions (e.g., if $Z$ is Noetherian), $\Phi_{Z,\mathcal{P}}$ is a functor $QC(X)\to QC(Y)$. Also, one can consider a derived variant of this, and look at $\Phi_{Z,\mathcal{P}}$ as to a functor $D^b(Coh(X))\to D^b(Coh(Y))$ (all this is very well known and studied, I'm writing it only to provide a setting for my question).</p>
<p>Given two Furier-Mukai-type functors $\Phi_{Z,\mathcal{P}}$ and $\Phi_{W,\mathcal{Q}}$ between categories of $\mathcal{O}_X$-modules and of $\mathcal{O}_Y$-modules, are there natural "geometric" transformations between these functors? which are the main examples of these? (here "geoemtric" is somehow vague: it pretends to mean "something with the same flavor of Fourier-Mukai transform, e.g., maybe something involving a morphism $T\to Z\times W$ and some twisting kernel over $T$)</p>
http://mathoverflow.net/questions/94308/pushforwards-of-stacks-of-algebrasPushforwards of stacks of algebras?domenico fiorenza2012-04-17T17:22:29Z2012-04-17T23:20:34Z
<p>This is a refined/sheafified version of this <a href="http://mathoverflow.net/questions/67121/sections-of-2-vector-bundles" rel="nofollow">previos question</a> of mine.</p>
<p>Let $(X,\mathcal{O}_X)$ be a ringed space or more in general a ringed stack, where the structure sheaf $\mathcal{O}_X$ is a sheaf of $\mathbb{K}$-algebras for some field $\mathbb{K}$. Then a global section of $\mathcal{O}_X$ can be thought of as a "scalar field" on $X$. In the "categorical progression" of higher $\mathbb{K}$-vector spaces, the field $\mathbb{K}$ is the 0th level. 1st level are classical $\mathbb{K}$-vector spaces, and so the "1st level version" of a section of $\mathcal{O}_X$ is (at least roughtly) a field of $\mathbb{K}$-vector spaces on $X$. A natural formalization of this rough idea is that of going from the sheaf $\mathcal{O}_X$ to the stack $\mathcal{O}_X$-Mod of sheaves of $\mathcal{O}_X$-modules on $X$ (maybe with suitable assumptions, e.g., one could be considering only quasi-coherent sheaves of $\mathcal{O}_X$-modules). A remarkable property of $\mathcal{O}_X$-modules is that they can be pushed forward along morphisms of ringed spaces. Now, the structure sheaf $\mathcal{O}_X$ we started with is in particular an $\mathcal{O}_X$-module, and pushing it forward along the terminal morphism $\pi:X\to {pt}$ we precisely get sections of $\mathcal{O}_X$ mentioned above. </p>
<p>Now we can make a further step, and go from the stack of $\mathcal{O}_X$-modules to the 2-stack of $\mathcal{O}_X$-algebras (with $\mathcal{O}_X$-$\mathcal{O}_X$-bimodules as 1-morphisms and morphisms of bimodules as 2-morphisms). My question is: can $\mathcal{O}_X$-algebras be pushed forward along morphisms of ringed spaces? (under which hypothesis?). In particular, considering $\mathcal{O}_X$ as an $\mathcal{O}_X$-algebra, what is the $\mathbb{K}$-algebra <code>$\pi_*\mathcal{O}_X$</code>? (the prototypical conjectural example of this in my mind is the following: if $G$ is a finite group, then <code>$\pi_*\mathcal{O}_{pt//G}$</code> is Morita equivalent to $\mathbb{K}[G]$, the group algebra of $G$).</p>
<p>In the above, an $\mathcal{O}_X$-algebra is to be thought as of a placeholder for its category of modules, so in a non-affine situation it is actually deceitful to reason in terms of algebras and a better description would be thinking of the "second step" as $\mathcal{O}_X$-linear categories, with additive functors and natural transformations of these. In this more general setting, the pushforward to a point of $\mathcal{O}_X$ would be the "pushforward of the 2-category of $\mathcal{O}_X$-linear categories" and this should be a $\mathbb{K}$-linear category, but not necessarily the category of representations of an algebra. In particular, by abstract nonsense I would expect this pushforward to be the category of $\mathcal{O}_X$-modules.</p>
<p>Note that going one step backwards instead of one step forward, the existence of a pushforward is nontrivial: the pushforward of a section of $\mathcal{O}_X$ along $X\to {pt}$ is an integration of fuctions on $X$, so it requires additional data to be defined (a "measure").</p>
http://mathoverflow.net/questions/92356/a-infty-basic-reference/92396#92396Answer by domenico fiorenza for $A_\infty$ basic referencedomenico fiorenza2012-03-27T17:44:01Z2012-03-27T17:44:01Z<p>To me the best path to $A_\infty$-categories is via topology. One first gets familiar with the notion of $A_\infty$-space as a natural generalization of that of topological monoid (the classical reference by Jim Stasheff is probably still where one should have a look for a complete account on this). Next one considers topological categories as a natural generalization of topological monoids (the only difference being that the product is not defined on $M\times M$ but on a fibered product $M_1\times_{M_0}M_1$, the latter reducing to the first for the space of objects $M_0$ consisting of a single point). Then one mixes these two generalizations of topological monoid and gets the definition of "$A_\infty$ topological category" in the most natural possible way (in my opinion). Once one is familiar with this, one sees that nothing changes if instead of being in the topological setting one works in any setting where "homotopies" are meaningful. For instance one can work with dg-categories, and this gives the version of $A_\infty$-category one usually meets.</p>
http://mathoverflow.net/questions/91327/spaces-with-no-topological-monoid-structure-which-are-homotopy-equivalent-to-topoSpaces with no topological monoid structure which are homotopy equivalent to topological monoidsdomenico fiorenza2012-03-15T19:38:05Z2012-03-21T10:43:28Z
<p>In motivating $A_\infty$-spaces to my students I'm going to insist on the homotopy invariance of the notion, saying that "being $A_\infty$ is the homotopy invariant version of being a topological monoid" and to stress this I'd like to say that if $X$ is a topological monoid and $Y$ is a space homotopy equivalent to $X$ then $Y$ will carry an $A_\infty$-structure making it equivalent to $X$ as an $A_\infty$-space, but in general not a topological monoid structure with this property. But at this point I see to my shame that I miss an explicit example of this! </p>
<p>Clearly the most dramatic example would be that of a space $Y$ which is homotopy equivalent to a topological monoid $X$, but such $Y$ carries no topological monoid structure at all, not to have to go into the equivalence issue. For a while I thought the closed interval could be an example of this (double shame: there are at least two very simple and well known topological monoid structures on $[0,1]$!), so I'm completely without examples, and I do not either know if such a space $Y$ does actually exist at all.</p>
<p>Any suggestion?</p>
<p>edit: despite I originally formulated my question in the most dramatic possible form, an example where, given a homotopy equivalence $f:Y\to X$ there is no monoid structure on $Y$ such that $\pi_0(f)$ is an isomorphism of monoids $\pi_0(Y)\to \pi_0(X)$ is even better for what I need to explain, namely that going from monoids to $A_\infty$-spaces not only $Y$ is naturally endowed with an $A_\infty$-space structure, but $f$ is promoted to an equivalence of $A_\infty$-spaces. So I will now leave the original question open as a general topology question which may have its interest in its own (despite it is admittedly an odd question), while for myself I'll be perfectly satisfied with the very nice answer by Tyler below.</p>
http://mathoverflow.net/questions/68863/rigorous-recursive-definition-of-m-algebrasRigorous recursive definition of $m$-algebrasdomenico fiorenza2011-06-26T17:08:23Z2012-03-13T18:59:38Z
<p>Naively, $m$-algebras over a field $\mathbb{k}$ are easily defined recursively: the category of 0-algebras is the symmetric monoidal category of $\mathbb{k}$-vector spaces, and for $m>0$ an $m$-algebra is an algebra object in the symmetric monoidal category of $(m-1)$-algebras.</p>
<p>This definition is based on the fact that algebra objects in a symmetric monoidal category should form themselves a symmetric monoidal category. Actually this is too a naive statement, and trying to make it rigorous one drops into higher categories: a semi-rigorous statement would be: "algebra objects in a symmetric monoidal $(\infty,n)$-category form a symmetric monoidal $(\infty,n+1)$-category" (this kind of statement can be found in Section 7 of Freed-Hopkins-Lurie-Teleman's <a href="http://arxiv.org/abs/0905.0731" rel="nofollow">Topological Quantum Field Theories from Compact Lie Groups</a>: "Recall first that vector spaces and linear maps form a symmetric tensor category, and that
algebra objects in a symmetric tensor category form in turn a symmetric tensor category (of one
level higher, but who’s counting?)."</p>
<p>My problems here is:</p>
<p>Recall from where? I'm not aware of a rigorous reference for this statement, and actually even not for the notion of symmetric monoidal $(\infty,n)$-category. For the latter, I'm familiar with Lurie's <a href="http://arxiv.org/abs/0905.0465" rel="nofollow">On the Classification of Topological Field Theories</a>, but there one first finds the question "(c) What is a symmetric monoidal structure on an $(\infty, n)$-category, and what does it mean for a functor to be symmetric monoidal?" but then one finds "In the interest of space, we will gloss over (b) and (c) (for an
extensive discussion of (c) in the case n = 1 we refer the reader to [<a href="http://arxiv.org/abs/math/0703204" rel="nofollow">16</a>])."</p>
<p>I'm ok with the intuitive and sketchy definitions above, but I'd like to know whether a rigorous treatment is available somewhere (and if not available yet, if there is interest in having such a rigorous treatment: I'm pondering this as a possible PhD thesis subject)</p>
http://mathoverflow.net/questions/69175/principal-g-bundles-as-fully-extended-tqfts-and-n-representationsPrincipal $G$-bundles as fully extended TQFTs, and $n$-representationsdomenico fiorenza2011-06-30T11:09:58Z2012-03-13T18:58:42Z
<p>This is a follow up to this MO question: <a href="http://mathoverflow.net/questions/50497/fully-dualizable-objects-in-classical-field-theories" rel="nofollow">http://mathoverflow.net/questions/50497/fully-dualizable-objects-in-classical-field-theories</a></p>
<p>Assuming the notation there (which in turn come from <a href="http://arxiv.org/abs/0905.0731" rel="nofollow">Topological Quantum Field Theories from Compact Lie Groups</a>), let $G$ be a finite group. The one-object groupoid $*//G$ is then an object of the symmetric monoidal category $Fam_n$ for any fixed $n$. Then by the cobordism hypotesis, there is at most one (up to isomorphism) symmetric monoidal functor $F: Bord_n^{SO}\to Fam_n$ with <code>$F(pt^+)=*//G$</code>, and there exists such a functor precisely when $*//G$ is a fully dualizable object in $Fam_n$. </p>
<p>So since the functor $Bun_G$ assigning to a manifold $X$ the groupid of principal $G$-bundles on $X$ is clearly a symmetric monoidal functor $F: Bord_n^{SO}\to Fam_n$ with <code>$F(pt^+)=*//G$</code> this should mean that:</p>
<p>i) for any finite group $G$, the groupoid <code>$*//G$</code> is a fully dualizable object in $Fam_n$;</p>
<p>ii) $Bun_G$ is the unique <code>$Fam_n$</code>-valued fully extended TQFT determined by $G$ (i.e., with <code>$F(pt^+)=*//G$</code>).</p>
<p>If so, a classical (fully extended) topological field theory from a finite group $G$ in the sense of Freed-Hopkins-Lurie-Teleman would reduce to the datum of a (fully dualizable) $n$-representation $G\to \mathcal{C}$, for $\mathcal{C}$ a symmetric monoidal $n$-category. </p>
<p>My question is: are there other classical examples of these (fully dualizable) $n$-representations of finite groups than those considered in Freed-Hopkins-Lurie-Teleman's paper? which are the TQFTs associated with these examples?</p>
http://mathoverflow.net/questions/38968/betti-numbers-of-moduli-spaces-of-smooth-riemann-surfacesBetti numbers of moduli spaces of smooth Riemann surfacesdomenico fiorenza2010-09-16T12:48:10Z2012-02-15T12:59:19Z
<p>Where can I find a list of the known Betti numbers of the moduli spaces $\mathcal{M}_{g,n}$ of genus $g$ Riemann surfaces with $n$ marked points? I need it to cross check results by an implemented algorithm which should be producing them using Kontevich's graph complex. </p>
<p>I am interested in the "open" moduli space consisting of smooth connected surfaces, not in its Deligne-Mumford compactification <code>$\overline{\mathcal{M}}_{g,n}$</code>. Also, I'm interested in the single Betti numbers and not in the Euler characteristic, which I know from e.g. Harar-Zagier and Bini-Gaiffi-Polito, and which I used to have a first check of the results of the algorithm. </p>
<p>Thanks.</p>
<p>Edit: Riccardo Murri's paper with the algorithm and its implementation has now appeared on arXiv: <a href="http://arxiv.org/abs/1202.1820" rel="nofollow">http://arxiv.org/abs/1202.1820</a></p>
http://mathoverflow.net/questions/84218/lie-algebra-valued-1-forms-and-pointed-maps-to-homogeneous-spacesLie algebra valued 1-forms and pointed maps to homogeneous spacesdomenico fiorenza2011-12-24T10:05:08Z2011-12-26T12:52:47Z
<p>Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$, and let $(M,p_0)$ be a simply connected pointed smooth manifold. A $\mathfrak{g}$-valued 1-form $\omega$ on $M$ can be seen as a connection form on the trivial principal $G$-bundle $G\times M\to M$. Assume that this connection is flat. Then, parallel transport along a path $\gamma$ in $M$ from $p_0$ to $p$ determines an element in $G$, which actually depends only on the endpoint $p$ since we are assuming that the connection is flat and that $M$ is simply connected. Thus we get a pointed map $\Phi_\omega:(M,p_0)\to (G,e)$, where $e$ is the identity element of $G$. Now, the Lie group $G$ carries a natural flat $\mathfrak{g}$-connection on the trivial princiapl $G$-bundle $G\times G\to G$, namely the one given by the Maurer-Cartan 1-form $\xi_G$. Using $\Phi_\omega$ to pull-back $\xi_G$ on $M$ we get a $\mathfrak{g}$-valued 1-form $\omega$ on $M$ which, no surprise, is $\omega$ itself. So one has a natural bijection between <code>$\{\omega\in \Omega^1(M,\mathfrak{g}) | d\omega+\frac{1}{2}[\omega,\omega]=0\}$</code> and pointed maps from $(M,p_0)$ to $(G,e)$.</p>
<p>(all this is well known; I'm recalling it only for set up)</p>
<p>If now $\omega$ is not flat but has holonomy group at $p_0$ given by the Lie subgroup $H$ of $G$, then we can verbatim repeat the above construction to get a pointed map $\Phi_\omega:(M,p_0)\to (G/H,[e])$. I therefore suspect by analogy that there should be a natural bijection </p>
<p><code>$
\{\omega\in \Omega^1(M,\mathfrak{g}) | \text{some condition}\} \leftrightarrow C^\infty((M,p_0),(G/H,[e])),
$</code></p>
<p>but I've been so far unable to see whether this is actually true, nor to make explicit what "some condition" should be (it should be something related to the Ambrose-Singer holonomy theorem and to Narasimhan-Ramanan results on universal connections, but I've not been able to see this neatly, yet). I think that despite my unability to locate a precise statement for the above, this should be well known, so I hope you will be able to address me to a reference.</p>
http://mathoverflow.net/questions/83147/is-the-first-differential-pontryagin-class-a-morphism-of-stacksIs the first differential Pontryagin class a morphism of stacks?domenico fiorenza2011-12-10T21:36:05Z2011-12-14T12:24:09Z
<p>In <a href="http://projecteuclid.org/DPubS?verb=Display&version=1.0&service=UI&handle=euclid.cmp/1104286562&page=record" rel="nofollow">Cech Cocycles for Characteristic Classes</a>, Jean-Luc Brylinski and Dennis McLaughlin provide explicit formulas for Cech cocycles for characteristic classes of real and complex vector bundles, and show how to refine these to Cech-Deligne cocycles for differential characteristic classes in differential cohomology. </p>
<p>When the compact Lie group $G$ involved is enough connected, e.g., if one is interested in the second Chern class of a principal $SU(n)$-bundle, the Brylinski-McLaughlin formula drastically simplifies, and it can be shown that in this case it actually gives a morphism of stacks from the classifying stack of principal $G$-bundles with connections to the (higher) stack of principal $U(1)$-$n$-gerbes with connections (here $n+2$ is the degree of the characteristic class involved).</p>
<p>This stacky interpretation is emphatised, e.g., in the follow-up <a href="http://arxiv.org/abs/1011.4735" rel="nofollow">Cech cocycles for differential characteristic classes - An $\infty$-Lie theoretic construction</a>, by Urs Schreiber, Jim Stasheff and myself, where it is obtained (as the title suggests) via integration of $L_\infty$-algebras to higher Lie groups. For this approach, the connectivity of $G$ is essential. For instance one can see that the first fractional differential Pontryagin class $\frac{1}{2}\hat{p}_1$ is a morphism of stacks from the stack of princiapl Spin bundles with connection to the 3-stack of $U(1)$-2-gerbes with connection, and this precisely reproduces Brylinski-McLaughlin construction, the but one cannot see the Brylinski-McLaughlin cocycle for the first differential Pontryagin class $\hat{p}_1$ for princiapal $SO$-bundles with connections via Lie integration: $SO$ is not enough connected to allow this. This pheneomenon is in a sense not surprising: for instance the "identity" morphism from the Lie algebra of $O(2)$ to the Lie algebra of $SO(2)$ cannot be integrated to a morphism of Lie groups from $O(2)$ to $SO(2)$ due to "lack of connectivity" reasons. </p>
<p>Yet the fact that a particular technique fails does not mean that a statement is false. So here is my question: is there a natural interpretation of Brylinski-McLaughlin cocycle for the first differential Pontryagin class $\hat{p}_1$ as a morphism of stacks (from $SO$-bundles with connections to $U(1)$-2-gerbes with connections)? or is there a natural interpretation of Brylinski-McLaughlin cocycle for the second differential Chern class $\hat{c}_2$ as a morphism of stacks from $U$-bundles (not $SU$!) with connections to $U(1)$-2-gerbes with connections)? </p>
<p>My feeling is that once there are no topological obstruction (i.e., once the characteristic classes are defined at the non-very-conected level, as in these cases), the morphism of stacks (which surely exists at the highly connected level) descends from the higher connected cover of $G$ involved to the original $G$. So, for instance since $\frac{1}{2}p_1$ is not an integral class for $SO$-bundles one can not make $\frac{1}{2}\hat{p}_1$ descend from principal Spin-bundles with connections to princiapal $SO$-bundles with connections; but since $p_1$ is an integral class for $SO$-bundles, then it should be possible that $\hat{p}_1$ descends. But all my attemps towards a rigorous proof of this have failed so far, so I've begun thinking that I may be wrong, and that $\hat{p}_1$ can be given no natural interpretation as a morphism of stacks after all.</p>
<p>Any suggestion, reference or criticism concerning this problem is welcome. </p>
http://mathoverflow.net/questions/83175/periodic-matrices-in-sl3-z/83186#83186Answer by domenico fiorenza for Periodic matrices in SL(3,Z)domenico fiorenza2011-12-11T17:19:52Z2011-12-11T20:46:23Z<p>ok, let me expand Geoff's suggestion. Let $A\in SL(3,\mathbb{Z})$ be such that $A^n=Id$ for some positive integer $n$. Since the characteristic polynomial of $A$ is a cubic polynomial of the form $-t^3+\cdots +1$, it has a positive real root; and since all roots of $A$ are roots of the unit, 1 is an eigenvalue of $A$. Moreover, since $A^n=Id$, $A$ is semisimple over $\overline{\mathbb{Q}}$ and so its minimal polynomial over $\mathbb{Q}$ is of the form $(t-1)p(t)$ with $p(t)\in \mathbb{Z}[t]$ a cyclotomic polynomial of degree at most 2 with $p(1)\neq 0$. This leaves only the following possibilities: $1$, $t+1$, $t^2+t+1$, $t^2+1$ or $t^2-t+1$, corresponding to $A$ having period $1,2,3,4$ or $6$ respectively. The period 1 case is trivial. For the period 6 case one can work as follows: we have a splitting of $\mathbb{Q}^3$ as $V\oplus W$, where $V$ and $W$ are $A$-stable $\mathbb{Q}$-vector spaces, with $\dim_{\mathbb{Q}}(V)\geq 1$ and with $A$ acting as the identity on $V$. Moreover, since $p(1)=1$, we can find polynomials $a(t)$ and $b(t)$ in $\mathbb{Z}[t]$ such that $a(t)p(t)+b(t)(t-1)=1$. This implies that $\mathbb{Z}^3=(\mathbb{Z}^3\cap V)\oplus (\mathbb{Z}^3\cap W)$. The abelian subgroup $\mathbb{Z}^3\cap V$ of $\mathbb{Z}^3$ is free since subgroups of free abelian groups are free, and has rank $\dim_{\mathbb{Q}}V$, since $V$ is a $\mathbb{Q}$-subspace of $\mathbb{Q}^3$. So we can find a $\mathbb{Z}$-basis $B_V$ for $\mathbb{Z}^3\cap V$ consisting of $\dim_{\mathbb{Q}}V$ elements. The same for $\mathbb{Z}^3\cap W$. The two basis $B_V$ and $B_W$ together are a $\mathbb{Z}$-basis of $\mathbb{Z}^3$ and up to multiplying by $-1$ one of the vectors in this basis we may assume that the change of basis matrix $P$ from the standard basis of $\mathbb{Z}^3$ to the basis $B_V\cup B_W$ is an element of $SL(3;\mathbb{Z})$. By construction, $PAP^{-1}$ is a block-diagonal matrix in $SL(3;\mathbb{Z})$, with an upper $1\times 1$ block $(1)$ and a lower $2\times 2$ block in $SL(2;\mathbb{Z})$. </p>
<p>This method, however, clearly does not apply to the remaining three cases. </p>
http://mathoverflow.net/questions/74593/3-2-1-tqfts-and-verlinde-algebras(3,2,1)-TQFTs and Verlinde algebrasdomenico fiorenza2011-09-05T16:08:30Z2011-09-05T17:40:28Z
<p>Given a modular category $\mathcal{C}$ there are two natural ways to get a Frobenius algebra out of $\mathcal{C}$. One is to take the Verlinde algebra (or `fusion algebra') of $\mathcal{C}$. The other consist in considering the $(3,2,1)$-dimensional TQFT associated with $\mathcal{C}$, and to get out of it a $(2,1)$-dimensional TQFT by multiplication by $S^1$ (and a $(2,1)$-dimensional TQFT is the same thing as the datum of a Frobenius algebra). It is well known in fully extended TQFT folklore that these two constructions coincide. Is anyone aware of a reference I could cite as a source for this statement? (I know Dan Freed's <i>
The Verlinde algebra is twisted equivariant K-theory</i>, where this can be read between the lines)</p>
http://mathoverflow.net/questions/71805/algebra-deformations-and-maurer-cartan-elements/71831#71831Answer by domenico fiorenza for Algebra Deformations and Maurer-Cartan elementsdomenico fiorenza2011-08-01T21:18:07Z2011-08-01T21:25:54Z<p>A good way of rewriting the gauge action is $e^{f}(\mu+\mu')e^{-f}=\mu+e^{f}*\mu'$, which makes the equivalence manifest. There are several references for the computation above. The first coming to my mind is Marco Manetti's <a href="http://arxiv.org/abs/math/0507286" rel="nofollow">Lectures on deformations of complex manifolds</a> where, if I'm not wrong, it's spelled out in detail.</p>
http://mathoverflow.net/questions/69863/action-on-tensor-power-and-element-notation-in-monoidal-categories/69868#69868Answer by domenico fiorenza for Action on tensor power and "element notation" in monoidal categoriesdomenico fiorenza2011-07-09T11:01:43Z2011-07-09T14:19:06Z<p>There's a point of view on symmetric monoidal categories (dating back to Graeme Segal's "Categories and cohomology theories" and then taken up by Bertrand To\"en in "Dualit\'e de Tannaka superieure, I: Structures monoidales" to define symmetric monoidal $n$-categories) which stresses the symmetric group action from the very beginning. It goes as follows: let $\Gamma$ be the category of pointed finite sets with maps of pointed sets as morphisms. Denote by $[1]$ the set <code>$\{0,1\}$</code> pointed at $0$. Then for any finite set $X$ and any element $x$ in $X$ there is a unique map <code>$u_x:[1]\to X\coprod\{*\}$</code> in $\Gamma$ mapping $1$ to $x$ (where <code>$*$</code> is the distinguished point of <code>$X\coprod\{*\}$</code>). With these notations, a symmetric monoidal category is a functor $F:\Gamma^{op}\to Categories$ such that </p>
<p>i) <code>$F(\{*\})=*$</code> (the terminal category);</p>
<p>ii) the natural map $F(X\coprod{*})\to F([1])^{|X|}$ induced by the morphisms $u_x:[1]\to X$ and the universal property of the product is an equivalence.</p>
<p>The correspondence between this and the classical definition of symmetric monoidal category is (clearly) given by $F \to F[1]$.
The natural action of the symmetric group <code>$\mathfrak{S}_n$</code> is then induced by the isomorphism <code>$\mathfrak{S}_n\cong \mathrm{Aut}_\Gamma([n])$</code>, where <code>$[n]=\{0,1,\dots,n\}$</code> is pointed at $0$. </p>
http://mathoverflow.net/questions/69560/fundamental-groups-of-spaces-of-maps-to-eilenberg-maclane-spacesFundamental groups of spaces of maps to Eilenberg-MacLane spacesdomenico fiorenza2011-07-05T18:32:46Z2011-07-06T08:03:29Z
<p>Let $X$ be a finite CW-complex and $A$ an abelian group, and consider the space $Maps(X,K(A,n))$ of continuous maps from $X$ to $K(A,n)$ endowed with the compact-open topology, so that it represents the functor $Y\mapsto C(X\times Y,K(A,n))$. Let $Maps_0(X,K(A,n))$ be the path-connected component of $Maps(X,K(A,n))$ corresponding to the zero in $H^n(X,A)$, and let $*$ any point in $Maps_0(X,K(A,n))$. I'm interested in the fundamental group $\pi_1( Maps_0(X,K(A,n)),*)$.</p>
<p>i) is it abelian?</p>
<p>ii) does it act trivially on $\pi_i( Maps_0(X,K(A,n)),*)$ for $i\geq 1$?</p>
<p>If the answer to these two were "yes", then one would have that for $X$ a closed compact connected smooth oriented manifold of dimension $k\lt n$, the Postnikov tower of <code>$Maps(X,K(\mathbb{Z},n))$</code> begins with <code>$K(\mathbb{Z},n-k)$</code>, so that there would be a canonical $(n-k-2)$-gerbe on the (moduli) space of $(n-2)$-gerbes on $X$</p>
http://mathoverflow.net/questions/69352/topology-of-su3/69355#69355Answer by domenico fiorenza for Topology of SU(3)domenico fiorenza2011-07-02T19:17:44Z2011-07-02T22:23:46Z<p>Apart from jokes, an answer which may satisfy you is the following: $SU(3)$ is a $S^3$-bundle over $S^5$. To see this just consider the defining representation of $SU(3)$ on $\mathbb{C}^3$; this induces a transitive action of $SU(3)$ on the unit sphere of $\mathbb{C}^3$, which is $S^5$. Since the stabilizer of a point for this action is $SU(2)$ this exhibits $SU(3)$ as an $SU(2)$-bundle over $S^5$, and as you wrote $SU(2)$ is diffeomophic to $S^3$. Now, the next question is: which $SU(2)$-bundle over $S^5$ is $SU(3)$? to answer this, recall that isomorphic classes of principal $SU(2)$-bundles over (a not too wild) topological space $X$ are in bijection with the set $[X,BSU(2)]$ of homotopy classes of maps from $X$ to the classifying space of $SU(2)$. So in the case at hand you are interested in <code>$[S^5,BSU(2)]= \pi_5(BSU(2))= \pi_4(SU(2))= \pi_4(S^3)= \mathbb{Z}/2\mathbb{Z}$</code>. So there are only two $S^3$-bundles over $S^5$, the trivial one and the nontrivial one: $SU(3)$ is the nontrivial one (otherwise one would have $\pi_4(SU(3))=\mathbb{Z}/2\mathbb{Z}$, which is not the case: it is $\pi_4(SU(3))={0}$).</p>
http://mathoverflow.net/questions/67121/sections-of-2-vector-bundlesSections of 2-vector bundlesdomenico fiorenza2011-06-07T12:20:17Z2011-06-20T17:01:56Z
<p>If we work over a field $k$, and take the recursive definition of $n$-vector spaces (as, e.g. in Topological Quantum Field Theories from Compact Lie Groups, arXiv:0905.0731) then a $2$-vector space is a $k$-algebra $A$, to be thought as a placeholder for its categoty of modules; morphisms between $A$ and $B$ are $B$-$A$-bimodules, ad 2-morphisms are morphisms of bimodules.</p>
<p>Now consider a 2-vector bundle over some space $X$. How does one see that its global sections are a 2-vector space? </p>
<p>The answer somehow depends on the notion of section one adopt, but in any case the relation between the various definitions should be investigated. Basically there are two notions coming to my mind: </p>
<p>i) natural transformations from the trivial 2-bundle to the given bundle. This is a very neat object, but it is not clear (to me) that this is a 2-vector space: which is the underlying algebra?</p>
<p>ii) the limit in 2-Vect of the functor from the nerve of a good open cover of $X$ to 2Vect defining the 2-bundle. This is manifestly a 2-vector space, but it is not clear (to me) that this limit exists.</p>
<p>Clearly the dream statement here would be that i) has a natural structure of 2-vector space, and that this 2-vector space represents the limit ii), but I'm unable to prove this.</p>
<p>(or the dual version of the above, under suitably finiteness assumptions)</p>
http://mathoverflow.net/questions/67861/differential-forms-on-the-simplex-which-are-constant-towards-the-boundaryDifferential forms on the simplex which are "constant towards the boundary"domenico fiorenza2011-06-15T12:45:32Z2011-06-15T13:02:48Z
<p>Let $\Delta^k$ the standard $k$-dimensional simplex, $\Delta^k={(x^0,\dots,x^k)\in \mathbb{R}^{k+1}|\sum_{i=0}^kx^i=1}$, and let $\Omega^\bullet(\Delta^k)$ be the de Rham complex of smooth differential forms on $\Delta^k$ (as usual a differential form is smooth on $\Delta^k$ if it admits a smooth extension to an neighborhood of $\Delta^k$). It is well known, dating back at least to Poincare', that $\Omega^\bullet(\Delta^k)$ is an acyclic complex (in strictly positive degree).</p>
<p>Now consider the following graded subspace $\Omega^\bullet_{ctb}(\Delta^k)$ of $\Omega^\bullet(\Delta^k)$: for any point $p$ in $\Delta^k$, let $\Delta_p$ be the lowest dimensional simplicial subset of $\Delta^k$ the point $p$ belongs to and let $\pi_p$ the orthogonal projection on the linear subspace determined by the $\Delta_p$; and say that a differential form $\omega$ is "constant towards the boundary" if for any point $p$ in $\Delta^k$ there is a neighborhood $U$ of $p$ such that $\omega=\pi_p^*(\omega\vert_{\Delta_p})$ on $U$ (these differential forms are called "differential forms with sitting instants" in <a href="http://arxiv.org/abs/1011.4735" rel="nofollow">Cech cocycles for differential characteristic classes</a>).</p>
<p>Since the de Rham differential commutes with pullback (and restriction is a special case of pullback), it is immediate to see that $\Omega^\bullet_{ctb}(\Delta^k)$ is actually a subcomplex of $\Omega^\bullet(\Delta^k)$, and my question is: is $\Omega^\bullet_{ctb}(\Delta^k)\hookrightarrow \Omega^\bullet(\Delta^k)$ a quasiisomorphism? (equivalently, is $\Omega^\bullet_{ctb}(\Delta^k)$ acyclic in positive degree?). Or, in completely explicit terms, is it true that if $\omega$ is a closed differential form on $\Delta^k$ which is constant towards the boundary, one can find a primitive $\eta$ which is constant towards the boundary? </p>
<p>I've worked out the lowest $k$ cases by hand, and the answer is yes. For $k=1$ this is almost trivial and for $k=2$ it is just a bit more tricky, working inductively. Since no change in the homotopy typ happens as dimension of the simplex increases, I guess acyclicity of $\Omega^\bullet_{ctb}(\Delta^k)$ will be true for any $k$, but the inductive argument I have in mind gets more and more involved at every step, so that I rapidly loose control over it. On the other hand I am convinced that a clever application of Poincare' lemma (i.e. by choosing a contraction of the simplex to a point along a vector field which is orthogonal to the boundary and with a sink at the center of the simplex), but I'm not completely sure of this argument. </p>
<p>(do not hesitate to close this question if it is too localized)</p>
http://mathoverflow.net/questions/65352/model-category-structures-on-the-category-of-l-infty-algebrasModel category structures on the category of $L_\infty$-algebrasdomenico fiorenza2011-05-18T17:13:54Z2011-05-26T22:14:13Z
<p>Let $k$ be a characteristic zero field. Then it is known that the forgetful functor $dgla(k)\to chain(k)$ from differential graded Lie algebras (over $k$) to cochain complexes induces a model category structure on $dgla(k)$ with "the same" fibrations and weak equivalences as on $chain(k)$, i.e., fibrations are surjective dgla morphisms and weak-equivalences are quasi-isomorphisms. </p>
<p>Also on the category of $L_\infty$-algebras over $k$ there is a forgetful functor $L_\infty(k)\to chain(k)$, which picks the linear part of an $L_\infty$-morphism.
Then, on $L_\infty(k)$ we have two natural functors: the forgetful functor $L_\infty(k)\to chain(k)$ and the embedding $L_\infty(k)\hookrightarrow dgcu(k)$, where $dgcu(k)$ is the category of differential graded counitary cocommutative coalgebras over $k$. </p>
<p>This suggests we could have two natural model category structures on $L_\infty(k)$, and my question is: how are they related? do they coincide? in particular, is a morphism of $L_\infty$-algebras whose linear part is surjective a fibration in the $dgcu(k)$ model structure? is a quasi-isomorphism of $L_\infty$-algebras (i.e., a morphism of $L_\infty$-algebras whose linear part is a quasi-isomorphism) a weak-equivalence in the the $dgcu(k)$ model structure?</p>
http://mathoverflow.net/questions/65352/model-category-structures-on-the-category-of-l-infty-algebras/65508#65508Answer by domenico fiorenza for Model category structures on the category of $L_\infty$-algebrasdomenico fiorenza2011-05-20T08:24:53Z2011-05-20T08:24:53Z<p>I've been discussing this with Jonathan Pridham, who pointed my attention to his Unifying derived deformation theories, where a model category structure is described on a suitable subcategory $DG_{\mathbb{Z}}Sp(k)$ of $dgcu(k)$. (actually, the definition of $DG_{\mathbb{Z}}Sp$ is more general, but on a characteristic zero field $k$ it is naturally a subcategory od $dgcu(k)$). </p>
<p>The category $DG_{\mathbb{Z}}Sp(k)$ has a few remarkable properties: on the one hand it is Quillen equivalent to the larger category $dgcu(k)$ (endowed with the Hinich's model structure); on the other hand $L_\infty$-algebras over $k$ are precisely the fibrant objects in $DG_{\mathbb{Z}}Sp(k)$ and a morphism $\varphi$ between $L_\infty$-algebras is a fibration (resp. a weak equivalence) in $DG_{\mathbb{Z}}Sp(k)$ if its image via the "linearization" functor $L_\infty(k)\to chains(k)$ is a fibration (resp. a weak equivalence) in $chains(k)$, i.e. if the linearization of $\varphi$ is surjective (resp. a quasi-isomorphism).</p>
http://mathoverflow.net/questions/51290/calculating-normal-bundle/51300#51300Answer by domenico fiorenza for Calculating normal bundledomenico fiorenza2011-01-06T12:39:13Z2011-01-06T13:19:27Z<p>A detailed treatment can be found in </p>
<p>Sheldon Katz, On the finiteness of rational curves on quintic threefolds, Compositio Math 60, 151-162 (1986) available as archive.numdam.org/article/CM_1986__60_2_151_0.pdf</p>
http://mathoverflow.net/questions/50992/wall-crossing-in-physics-and-mathematics/51057#51057Answer by domenico fiorenza for Wall Crossing in Physics and Mathematicsdomenico fiorenza2011-01-03T20:15:03Z2011-01-04T08:40:44Z<p>In extremely vague, rough, oversimplified and imprecise terms, wall-crossing phenomena can be described as follows. </p>
<p>Given some linear category <code>$\mathcal{C}$</code> (vector bundles, coherent sheaves, etc.), from a physicist's point of view it is quite natural to consider the <em>space <code>$\mathcal{M}_{\mathcal{C}}$</code> of all objects of <code>$\mathcal{C}$</code></em>, and to take averages of naturally defined quantities on <code>$\mathcal{M}_{\mathcal{C}}$</code>, obtaining characteristic numbers <code>$I_\mathcal{C}$</code> (I've said I'm really oversimplifying things). </p>
<p>To a mathematician, the space <code>$\mathcal{M}_{\mathcal{C}}$</code> makes no sense in the realm of manifolds or schemes: it is a kind of derived object known as (higer) stack. However, things are quite under control if <code>$\mathcal{M}$</code> is given a stability condition, i.e., a notion of stable objects. Now, the idea of Tom Bridgeland (derived from Michael Douglas investigations on D-branes) is that there is not a distinguished stability condition, but a whole space $\Sigma$ of stability conditions (which turns out to be a topological manifold). For any <code>$P$</code> in $\Sigma$ one can then consider the moduli space <code>$\mathcal{M}_P$</code> of $P$-stable objects, and this leads to a formalization of the physicist's space <code>$\mathcal{M}_\mathcal{C}$</code> as a bundle of moduli spaces over the space $\Sigma$ of stability conditions: the fiber over $P$ is <code>$\mathcal{M}_P$</code>. </p>
<p>Now, the geometry of $\mathcal{M}_P$ can abruptly change as $P$ moves in $\Sigma$. This is easily explained: if $P$ and $Q$ are two distinct points in $\Sigma$, there could be some object $E$ in <code>$\mathcal{C}$</code> which is $P$-stable but not $Q$ stable. So in the moduli space <code>$\mathcal{M}_P$</code> there will be a point representing $E$, but in <code>$\mathcal{M}_Q$</code> there will be not such a point. Conversely there will be some object $F$ which is $Q$-stable but not $P$-stable. So, if we join $P$ and $Q$ by a path and follow $E$ along this path, then we see that at some point along the path $E$ must become unstable and so disappear from the moduli space of stable objects and be substituted by $F$. And this happens for each path between $P$ and $Q$: there's a <em>wall</em> separating the region where $E$ is stable and $F$ is unstable from the region where $E$ is unstable and $F$ is stable. </p>
<p>As we cross the wall the geometry of the moduli space abruptly changes. However this change in geometry is expected to be quite regular: more precisely it is expected to be a Mukai flop. This is too, quite easy to imagine from abstract nonsense (proving it formally is another kind of affair...). Namely, for a fixed wall, one expects that there is a whole linear space of objects which become unstable on the wall. And since we are considering objects up to isomorphisms, and we are considering linear categories, nonzero scalars will act as isomorphisms and so will have to be quotiented out. This leaves us with a projective space of objects that gets unstabilized by the wall, and so disappears as we cross the wall. But for what we just said above, there will be also a projective space of objects appearing as we cross the wall. So crossing the wall, there is a <code>$\mathbb{P}^n$</code> replaced by another <code>$\mathbb{P}^n$</code>, and this is, very roughtly and basically, the idea of a Mukai flop. The idea of wall crossing transition of moduli spaces by flops goes back at least to Thaddeus' influential work on the Verlinde formula, studying the dependence of geometric invariant theory quotients on the choice of stability (linearization of the action), and similar wall crossings were important e.g. in Donaldson theory.</p>
<p>Let us now come back again to the invariants $I_\mathcal{C}$. From the mathematician's perspective we have not a single invariant, but a quantitu $I_P$ for any stability condition $P$. What happens when we cross a wall? the spaces <code>$\mathcal{M}_P$</code> and <code>$\mathcal{M}_Q$</code> are geometrically different, so the quantities $I_P$ and $I_Q$ are a priori completely unrelated. On the other hand, the physicist's ill-defined quantity $I_\mathcal{C}$ was defined before stability conditions, so it <em>does not knows</em> about the wall! this means that the well-defined quantities $I_P$
and $I_Q$ must coincide (or better be canonicaly related): in colloquial terms, <em>the invariant $I$ crosses the wall</em>, and the relation between $I_P$ and$I_Q$ ia a <em>wall-crossing phenomenon</em>. </p>
<p>Actually, this is an huge oversimplification even of the physiscist's intuition. A better description of the physicist's point of view is the following (see the excellent answer by Andy Neitzke below for details). The relevant quantity one is interested in does not originally live on $\mathcal{M}$, but in a larger space of superconformal field theories. And it depends on a paramter $t$. Let us denote this quantity by the symbol $\Omega(t)$. As $t$ varies into the parameter space of SCFTs, the quantity $\Omega(t)$ varies continuosly. On the other hand, as said above, $\Omega(t)$ is the average of some quantity $J(t)$, and this average can be computed by localization formulae as suitable integrals on the space of critical points of $J(t)$ (the spaces of vacua of the theory). It turns out that at least certain values of $t$ can be identified with stability conditions on <code>$\mathcal{M}$</code>; if $t=t_P$, this identifies the space of critical points of $J(t_P)$ with the moduli space $\mathcal{M}_P$ and the localization formulae relate in a precise way $\Omega(t_P)$ with $I(P)$. Now, the geometry of the set of critical points of $J(t)$ may abruptly change as the parameter $t$ varies (this is nothing strange: this happens already for families of smooth functions $f_t:\mathbb{R}\to\mathbb{R}$, as every first year calculus student knows). So the formulae for $I_P$ and $I_Q$ will be quite different if there's a wall between $P$ and $Q$. On the other hand, they are related in a precise way to $\Omega(t_P)$ and $\Omega(t_Q)$, and these quantities change in a well-behaved way as $t$ goes from $t_P$ to $t_Q$. Reading this good behaviour at the level of $I_P$ and $I_Q$ gives the wall-crossing formulae. </p>
http://mathoverflow.net/questions/128325/invariant-definition-of-graded-poisson-bracket/128349#128349Comment by domenico fiorenzadomenico fiorenza2013-04-22T18:11:21Z2013-04-22T18:11:21ZIndeed that could be a reason, but I have to admit not be expert at all on motivations from classical mechanics. From a purely mathematical point of view, using the defining equation <code>$\iota_{X_h}\omega+dh=0$</code> one has <code>$X_{\{f,g\}}=[X_f,X_g]$</code> and so mapping a smooth function to its Hamiltonian vector field is a Lie algebra homomorphism from the Poisson algebra of the symplectic manifold to the Lie algebra of vector fields, rather than a Lie algebra antihomomorphism as is the case using <code>$\iota_{X_h}\omega=dh$</code>.http://mathoverflow.net/questions/114585/an-explicit-homotopy-equivalence-between-the-de-rham-complex-and-the-cech-de-rham/114625#114625Comment by domenico fiorenzadomenico fiorenza2012-11-27T06:45:24Z2012-11-27T06:45:24ZHi Sasha,
thanks! That's a very elegant argument, which is actually a very neat and clear way of reorganize my brute force computation on cohomology classes. Thanks a lot!
However what I'd really need here is an explicit homotopy $K$ (which I think one should be able to build by unwinding the proof that $j$ is a quasiisomorphisms, but which I also suspect is already written explicitly somewhere, that's why I'm asking for references here).http://mathoverflow.net/questions/106896/higher-holonomies-for-higher-local-systemsComment by domenico fiorenzadomenico fiorenza2012-09-11T15:41:39Z2012-09-11T15:41:39Zright! I should re-read more carefully before posting! thanks, damienhttp://mathoverflow.net/questions/106896/higher-holonomies-for-higher-local-systemsComment by domenico fiorenzadomenico fiorenza2012-09-11T13:47:16Z2012-09-11T13:47:16Zright, thanks! I've now added "with duals" to the description of the $(\infty,n)$-category $\mathcal{C}$http://mathoverflow.net/questions/98291/are-fully-extended-tqfts-generalized-cohomology-theories/98298#98298Comment by domenico fiorenzadomenico fiorenza2012-05-29T19:44:07Z2012-05-29T19:44:07ZThanks a lot, Damien!http://mathoverflow.net/questions/94742/examples-of-interesting-false-proofs/94743#94743Comment by domenico fiorenzadomenico fiorenza2012-04-21T17:15:08Z2012-04-21T17:15:08ZThis false proof is so good I've got used to proposing to my students $q(t)=tr(tI-A)$ as an antidote.http://mathoverflow.net/questions/94553/a-question-about-maximal-subgroups/94554#94554Comment by domenico fiorenzadomenico fiorenza2012-04-20T05:33:21Z2012-04-20T05:33:21ZI need apologize for having read the totally nonsense "less or equal to $n$" in place of "greater or equal to $n$". I'm now voting to remove the completely unuseful answer I wrote above. Thanks having made me notice my mistake. http://mathoverflow.net/questions/94426/geometric-natural-transformations-between-fourier-mukai-transforms/94450#94450Comment by domenico fiorenzadomenico fiorenza2012-04-18T20:07:49Z2012-04-18T20:07:49ZHi David, enlightening and informative as usual, thanks a lot! Just one more thing: do I also have a pushforward map between integral kernels induced by $W\to Z$? (eventually under suitable hypothesis?). This way one would associate a natural transformation to a correspondence $Z\leftarrow T \to W$, too.
(as you will have surely read between the lines, my latest MO questions are secretly related to an attempt of mine to get familiar with the higher tqft constructions by Francis, Nadler and yourself)
http://mathoverflow.net/questions/94308/pushforwards-of-stacks-of-algebras/94336#94336Comment by domenico fiorenzadomenico fiorenza2012-04-18T05:39:03Z2012-04-18T05:39:03ZThanks a lot, David!http://mathoverflow.net/questions/94308/pushforwards-of-stacks-of-algebrasComment by domenico fiorenzadomenico fiorenza2012-04-17T19:21:55Z2012-04-17T19:21:55ZYes, but in this context an algebra is to be thought as a placeholder for its category of modules. Now that I write it I see that in the non-affine case one should not expect an algebra just like a quasi-coherent $\mathcal{O}_X$-module on $X$ is not the sheaf of modules associated with an $\mathcal{O}_X(X)$-module. I'll now edit accordingly my question. Thanks again. http://mathoverflow.net/questions/94308/pushforwards-of-stacks-of-algebrasComment by domenico fiorenzadomenico fiorenza2012-04-17T18:02:50Z2012-04-17T18:02:50ZHi Martin, thanks. Here I'm thinking of $\mathcal{O}_X$-algebras as the 2-category having $\mathcal{O}_X$-$\mathcal{O}_X$-bimodules as 1-morphisms and bimodule morphisms as 2-morphisms, not of 𝕆X-algebras as a subcategory of $\mathcal{O}_X$-modules. But I may be misunderstanding your comment: let me know about this.http://mathoverflow.net/questions/92356/a-infty-basic-reference/92396#92396Comment by domenico fiorenzadomenico fiorenza2012-03-27T20:02:39Z2012-03-27T20:02:39ZThat's right, but it is precisely since the question does not specify whether the motivations are algebraic or a reference is sought for the general idea of $A_\infty$ category that I suggested to start from where everything begun: my fear is that with no background on $A_\infty$-spaces at all, the whole of $A_\infty$-categories may remain too abstract and out of focus, a bit like the definition of homotopy between morphisms of chain complexes without having seen a proof of Poincare' lemma before.http://mathoverflow.net/questions/92066/tqfts-with-target-category-of-higher-type-than-the-sourceComment by domenico fiorenzadomenico fiorenza2012-03-24T20:49:25Z2012-03-24T20:49:25ZHi Kevin, hi Noah, thanks a lot for the references. Noah, you're suggesting to think of the k-mod-k[G] bimodule as a codimension-one defect separating two tqfts, right? That's a nice point of view, thanks for the suggestion. In Jacob's paper this is mentioned in Example 4.3.23, but anyway the literature on tqfts with defects is quite ample, I'll have a look there.http://mathoverflow.net/questions/91327/spaces-with-no-topological-monoid-structure-which-are-homotopy-equivalent-to-topo/91814#91814Comment by domenico fiorenzadomenico fiorenza2012-03-23T09:39:21Z2012-03-23T09:39:21ZHi Gabriel,
very nice answer, thanks! Let me summarize and simplify a bit your argument:
Let M be a monoid; then: i) for each pair x,y of invertible elements there exist a homeomorphism of M in itself mapping x to y; ii) either M is a group or there exist an element which has neither a left inverse nor a right inverse; iii) if M is path connected but M-{e} is not connected then all elements of M except at most those in one connected component of M-{e} are invertible.
Thus a monoid structure on the infinite ternary graph would give a homeomorphism mapping an edge internal point onto a vertex. http://mathoverflow.net/questions/91327/spaces-with-no-topological-monoid-structure-which-are-homotopy-equivalent-to-topo/91353#91353Comment by domenico fiorenzadomenico fiorenza2012-03-16T08:37:59Z2012-03-16T08:37:59ZHi Neil, thanks for pointing my attention to that. But I'm not sure that completely solves the question, yet: it seems to me the answers there suggest that if such a space X does exist, then one of the simplest examples of such a space is likely to be a tree; but it seems that the question whether there actually exist trees with no admissible monoid structure is left open. Or am I missing something?