Is a groupoid determined by its Hopfish algebra? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T08:54:51Z http://mathoverflow.net/feeds/question/31742 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/31742/is-a-groupoid-determined-by-its-hopfish-algebra Is a groupoid determined by its Hopfish algebra? Theo Johnson-Freyd 2010-07-13T18:27:21Z 2010-07-14T10:10:41Z <p>This is a follow up to my question <a href="http://mathoverflow.net/questions/31248/" rel="nofollow">What is the precise relationship between groupoid language and noncommutative algebra language?</a>. I will briefly review some definitions; for details, a good place to look is <a href="http://arxiv.org/abs/math/0701499v1" rel="nofollow">Christian Blohmann, Alan Weinstein. Group-like objects in Poisson geometry and algebra. 2007. arXiv:math/0701499v1</a>. And actually, there are two versions of my question, one for groupoids and the other for categories. So that I can avoid all analysis, I will restrict my attention to finite things; if you know the answer in, say, topological spaces, or smooth manifolds, or..., then I'm also interested.</p> <p>A <strong>category</strong> is a span of sets <code>$C = \{ C_0 \overset l \leftarrow C_1 \overset r \rightarrow C_0\}$</code> which is an algebra object in the category of <code>$C_0,C_0$</code> spans. I.e. there are maps of spans <code>$i: \{C_0 = C_0 = C_0\} \to C$</code> and <code>$m: C \underset{C_0}\times C \to C$</code> making the usual diagrams commute. A category is a <strong>groupoid</strong> if additionally there is an involution <code>${^{-1}} : \{ C_0 \overset l \leftarrow C_1 \overset r \rightarrow C_0\} \to \{ C_0 \overset r \leftarrow C_1 \overset l \rightarrow C_0\}$</code> satisfying some condition. A category $C$ is <strong>finite</strong> if both <code>$C_0$</code> and <code>$C_1$</code> are finite.</p> <p>A finite-dimensional algebra $A$ (over a fixed field $\mathbb K$) is <strong>sesqui</strong> if it is equipped with a bimodule <code>${_A \Delta _{A\otimes A}}$</code> and an "associativity isomorphism" <code>$$\varphi: {_A \Delta _{A\otimes A}} \underset{A\otimes A}\otimes \bigl( {_A A _A} \underset{\mathbb K}\otimes {_A \Delta _{A\otimes A}}\bigr) \overset\sim\to {_A \Delta _{A\otimes A}} \underset{A\otimes A}\otimes \bigl( {_A \Delta _{A\otimes A}} \underset{\mathbb K}\otimes {_A A _A} \bigr)$$</code> of <code>$A, A^{\otimes 3}$</code> bimodules, which satisfies a pentagon. There should also be a "counit" bimodule <code>$_A \epsilon _{\mathbb K}$</code>, some triangle isomorphisms, and some more equations. A sesquialgebra is <strong>hopfish</strong> if a hard-to-write-down condition is satisfied; see <a href="http://arxiv.org/abs/math/0510421v2" rel="nofollow">Xiang Tang, Alan Weinstein, Chenchang Zhu. Hopfish algebras. 2006. arXiv:math/0510421v2</a>. Let <code>${_A {\Delta^{\rm flip}} _{A\otimes A}}$</code> denote the bimodule $\Delta$ with the two right $A$-actions flipped. A sesquialgebra is <strong>symmetric</strong> if it comes equipped with a bimodule isomorphism <code>$\psi: {_A \Delta _{A\otimes A}} \overset\sim\to {_A {\Delta^{\rm flip}} _{A\otimes A}}$</code> so that $\varphi,\psi$ satisfy two hexagons. A sesquialgebra is <strong>finite</strong> if $A,\Delta, \dots$ are finite-dimensional over $\mathbb K$.</p> <p>Let <code>$C = \{ C_0 \overset l \leftarrow C_1 \overset r \rightarrow C_0\}$</code> be a finite category. Then it gives rise to a finite symmetric sesquialgebra as follows. The algebra <code>$A$</code> is given by the vector space <code>$\mathbb K C_1$</code> with the convolution product (given on the basis by $a\otimes b \mapsto ab$ if <code>$(a,b)$</code> is a composable pair of morphisms, and $a\otimes b \mapsto 0$ otherwise). The bimodule $\Delta$ is given as the vector space with basis all pairs <code>$(a,b) \in C_1 \times C_1$</code> with $l(a) = l(b)$. I will let you work out the rest: the actions, the associator $\varphi$ and symmetrizer $\psi$, etc. If $C$ is actually a groupoid, then <code>$\mathbb K C_1$</code> is hopfish. This construction extends to a 2-functor, and so sends equivalences of categories to Morita equivalences of sesquialgebras.</p> <blockquote> <p><em>Question:</em> It is well known that a groupoid $C$ cannot be recovered from the algebra <code>$\mathbb K C_1$</code>; compare for example the group with two elements, thought of as a groupoid with one object, and the set with two elements, thought of as a groupoid with only identity morphisms. But the examples I know can be distinguished by remembering the hopfish structure.</p> <ul> <li>Can a finite category be recovered from its symmetric sesquialgebra?</li> <li>If not, can a finite groupoid be recovered from its symmetric hopfish structure?</li> <li>Can an equivalence class of finite categories be recovered from the Morita-equivalence class of finite symmetric sesquialgebras?</li> <li>If not, do we at least have the corresponding statement for groupoids/hopfish algebras?</li> </ul> </blockquote> <p>In a footnote, <a href="http://arxiv.org/abs/math/0701499v1" rel="nofollow">Blohmann and Weinstein</a> suggest that they do not know the answers to the above questions. But that was three years ago; perhaps there has been more recent work?</p> http://mathoverflow.net/questions/31742/is-a-groupoid-determined-by-its-hopfish-algebra/31754#31754 Answer by David Ben-Zvi for Is a groupoid determined by its Hopfish algebra? David Ben-Zvi 2010-07-13T19:31:11Z 2010-07-13T19:31:11Z <p>I believe this is the subject of Tannakian reconstruction (as in <a href="http://mathoverflow.net/questions/3446/tannakian-formalism" rel="nofollow">this question</a>)? i.e. if I understand correctly the Hopfish algebra attached to a groupoid is built so that its category of modules as a tensor category is the category of vector bundles on the groupoid, in the discrete case, or of (quasi)coherent sheaves in the algebraic case, or some topological substitute? If so one can try to reconstruct the groupoid as the groupoid of tensor functors from this tensor category to vector spaces (more generally you can construct a functor of points out of the groupoid of tensor functors to R-modules for varying rings R, or to vector bundles of the appropriate type on general "test" spaces). For discrete groupoids, or for quasicompact stacks with affine diagonal, this reconstruction works to reconstruct the space from the tensor category, i.e. from the Hopfish algebra.</p> http://mathoverflow.net/questions/31742/is-a-groupoid-determined-by-its-hopfish-algebra/31825#31825 Answer by Bugs Bunny for Is a groupoid determined by its Hopfish algebra? Bugs Bunny 2010-07-14T10:10:41Z 2010-07-14T10:10:41Z <p>Let $A=KC_0$ be the direct sum of fields indexed by the base of the groupoid. Then the algebra $H=KC_1$ is an $A|K$-Hopf algebra in terminilogy going back to Sweedler. The groupoid can be recovered from it as the set of grouplike elements, in the same way as a group can be recovered from its Hopf group algebra.</p>