Why do A_\infty functors form an A_\infty category? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T01:42:49Z http://mathoverflow.net/feeds/question/58169 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/58169/why-do-a-infty-functors-form-an-a-infty-category Why do A_\infty functors form an A_\infty category? Sam Lisi 2011-03-11T15:51:01Z 2011-03-30T01:59:07Z <p>I am in a reading group studying Seidel's book (<i>Fukaya Categories and Picard-Lefschetz Theory</i>). All of the participants have backgrounds in symplectic topology/pseudoholomorphic curve methods. We are stuck in trying to understand the chapter presenting the algebraic background for Fukaya Categories.</p> <p>Seidel makes the following claim: Non-unital $A_\infty$ functors $\mathcal F: \mathcal A \rightarrow \mathcal B$ are themselves the objects of a non-unital $A_\infty$ category. The morphisms $\mathrm{hom}( \mathcal F_0, \mathcal F_1)$ are something he calls (following Fukaya) pre-natural transformations. (The morphisms $T$ for which $\mu_1(T) = 0$ are the natural transformations.) Seidel then provides the formulae for the compositions $\mu_d$. (This is discussed in Section (1d) of the book [page 10].)</p> <p>In our working group, we tried to check that these formulae for the compositions satisfied the $A_\infty$ associativity equations, but were unable to do so beyond $\mu_1$. </p> <p>I have two questions (that may be the same question):</p> <blockquote> <p>Why do these composition maps satisfy the $A_\infty$ associativity equations? Is there a way of understanding this geometrically?</p> </blockquote> http://mathoverflow.net/questions/58169/why-do-a-infty-functors-form-an-a-infty-category/59516#59516 Answer by Jeremy Pecharich for Why do A_\infty functors form an A_\infty category? Jeremy Pecharich 2011-03-25T05:18:16Z 2011-03-25T05:18:16Z <p>The reference that I know, which goes through the algebraic details, is by V. Lyubashenko: <a href="http://arxiv.org/abs/math/0210047" rel="nofollow">http://arxiv.org/abs/math/0210047</a>.</p> <p>The geometric interpretation is very interesting but I don't think all the details have been written down yet. In the $A_{\infty}$-algebra case, i.e. an $A_{\infty}$-category with one object, we have the following interpretation: an $A_{\infty}$-structure on a graded vector space $A$ is equivalent to a coderivation $Q$ on the coalgebra $T^c(A[1]):=\bigoplus_{n\geq 0} A[1]^{\otimes n}$. In geometric language, $T^c(A[1])$ is called a non-commutative thin scheme and the coderivation $Q$ is a homological vector field. To see this equivalent to an $A_{\infty}$-algebra you just expand $Q=Q_1+Q_2+\cdots$ and use that $Q^2=0$. The functor from algebras to sets given by $$<br> A\mapsto Hom_{Coalg}(A^*\otimes T^c(B),T^c(C))$$ is representable by a non-commutative thin scheme which clearly has a coderivation. The corresponding non-commutative thin scheme is denoted by $Maps(B,C)$ and every point of $Map(B,C)$ corresponds to an $A_{\infty}$-morphism. One now needs to take some sort of completion along the subscheme of these points to get the $A_{\infty}$-category structure. This is about as far as I understand the geometric interpretation of $A_{\infty}$-categories. The details about geometric interpretations of $A_{\infty}$-algebras are in Kontsevich and Soibelman's book "Deformation Theory" but such a description for $A_{\infty}$-categories has not appeared (to the best of my knowledge), but it would be very nice to have. </p> http://mathoverflow.net/questions/58169/why-do-a-infty-functors-form-an-a-infty-category/60014#60014 Answer by Eric Zaslow for Why do A_\infty functors form an A_\infty category? Eric Zaslow 2011-03-29T19:24:38Z 2011-03-30T01:59:07Z <p>I can explain the pictures I usually draw to think of $A_\infty$ functors, but I don't know if they're standard. Anyway, I'll describe what is just a rubric for ingesting the long formulas, nothing more.</p> <p>Let's consider first the Yoneda embedding $Y$, which re-thinks an object $L$ in an $A_\infty$-category $A$ as an $A$-module, or functor from $A^{op}$ to chain complexes. So $Y_L(M) = hom_A(M,L).$</p> <p>I confess that when I confront these formulas/concepts, I always think in terms of the Fukaya category, which is very amenable to pictures and for which the $A_\infty$ structures are geometric.</p> <p>So I draw a curve on a piece of paper and label it $L$. (The curve is literally a Lagrangian submanifold of my ${\mathbb R}^2$ piece of paper.) When I want to think of $L$ in terms of its Yoneda image, I draw the SAME curve, but as a squiggly line.</p> <p>So what is the data that the squiggly line gives us? For each object $M$ (a regular curve on my paper), we have the intersection points, which form a graded vector space $hom_A^*(M,L).$ This vector space has the structure of a chain complex (Floer), with differential given by football-shaped bi-gons with one regular side and one squiggly side. For a pair of other objects, $M_1, M_2,$ we get a map $$\mu^2: hom_A(M_2,L)\otimes hom_A(M_1,M_2) \rightarrow hom_A(M_1,L),$$ and so on for all the structure of a module (section 1j, p. 19).</p> <p>For the Fukaya category, the equations 1.19 follow (for non-squiggly lines) from studying degenerations of 1-parameter families of holomorphic polygons. Now squigglifying those same pictures gives 1.19 for an arbitrary module, and the equations are similar for not just modules but arbitrary functor between two $A_\infty$-categories.</p> <p>What data do we have if we have <em>two</em> squiggly lines $L_1$ and $L_2$? They should intersect at a morphism between functors (and it should have a degree). This morphism of functores gives more data, using the Fukaya perspective. If we added one normal line $M$, we'd have the spaces $Y_{L_1}(M)$ and $Y_{_2}(M)$, and have a triangle which is a map between them. Higher polygons and the relations between them (by considering one-parameter families) should give you all the equations and give you a hint as to verify them. (But no promises!)</p> <p>Hope that lengthy and pretty vague description was worth our time. (Oh, geez, this was a March 11 question? Probably stale by now!)</p>