Categorical primitive forms of Calabi–Yau $A_\infty$-categories with semi-simple cohomology

Abstract

We study categorical primitive forms for Calabi–Yau $A_\infty$ categories with semi-simple Hochschild cohomology. We classify these primitive forms in terms of certain grading operators on the Hochschild homology. We use this result to prove that, if the Fukaya category ${{\textsf {Fuk}}}(M)$ of a symplectic manifold M has semi-simple Hochschild cohomology, then its genus zero Gromov–Witten invariants may be recovered from the $A_\infty$-category ${{\textsf {Fuk}}}(M)$ together with the closed-open map. An immediate corollary of this is that in the semi-simple case, homological mirror symmetry implies enumerative mirror symmetry.

Appendices

Appendix A. Hochschild invariants of Calabi–Yau $A_\infty$-algebras

In this appendix we prove Theorem 2.4. We start with the following proposition.

Proposition A.1

Given a Hochschild cochain class $\varphi \in HH^{\bullet }\left( A \right)$ and chains $\alpha , \beta \in HH_{\bullet }\left( A\right)$ we have

$$\begin{aligned} \langle \varphi \cap \alpha , \beta \rangle _{\mathsf{Muk} } =(-1)^{|\varphi ||\alpha |}\langle \alpha , \varphi \cap \beta \rangle _{\mathsf{Muk} } \end{aligned}$$

In other words, capping with a fixed class is a self-adjoint map.

Proof

The proof is identical to that of Lemma 5.39 in [34]. Given a closed cochain $\varphi$ we define the maps $H_1, H_2, H_3: CC_\bullet (A)^{\otimes 2}\rightarrow CC_\bullet (A)$, as follows, for $\alpha =\alpha _0|\alpha _1|\ldots |\alpha _r$ and $\beta =\beta _0|\beta _1|\ldots |\beta _s$, we set

$$\begin{aligned}&H_1(\alpha , \beta )\\&\quad = \sum {\textsf {tr}}\left[ c \rightarrow (-1)^{\dagger _1}{\mathfrak {m}}_* \left( \alpha _j,\ldots \varphi _*(\alpha _k,\ldots ),\ldots ,\alpha _0,\ldots , {\mathfrak {m}}_*(\alpha _i,\ldots ,c,\beta _n,\ldots ,\beta _0,\ldots ),\ldots \right) \right] \end{aligned}$$

where $\dagger _1= |c||\beta | +|\varphi |'+ |\varphi |(|\alpha _j|'+\cdots +|\alpha _{k-1}|') + |\alpha _0|'+\cdots +|\alpha _{i-1}|'+|\alpha _k|'+\cdots +|\alpha _{r}|' + @$.

$$\begin{aligned}&H_2(\alpha , \beta )\\&\quad = \sum {\textsf {tr}}\left[ c \rightarrow (-1)^{\dagger _2} {\mathfrak {m}}_*\left( \alpha _j,\ldots ,\alpha _0,\ldots ,{\mathfrak {m}}_* \left( \alpha _i,\ldots ,c,\beta _n,\ldots ,\varphi _* (\beta _p,\ldots ), \ldots ,\beta _0,\ldots \right) ,\ldots \right) \right] \end{aligned}$$

where $\dagger _2 = |c|(|\beta |+|\varphi |') +|\varphi ||\alpha |+|\varphi |'(|\beta _n|'+\cdots +|\beta _{p-1}|') +|\alpha _i|'+\cdots +|\alpha _{j-1}|'+ @$.

$$\begin{aligned}&H_3(\alpha , \beta )\\&\quad = \sum {\textsf {tr}}\left[ c \rightarrow (-1)^{\dagger _3} \varphi _*\left( \alpha _j,\ldots ,{\mathfrak {m}}_*\left( \alpha _k,\ldots , \alpha _0,\ldots ,{\mathfrak {m}}_*(\alpha _i,\ldots ,c,\ldots ,\beta _0,\ldots ),\ldots \right) , \beta _n,\ldots \right) \right] \end{aligned}$$

where $\dagger _3= 1+|c||\beta | + |\alpha _0|'+\cdots +|\alpha _{i-1}|'+ |\alpha _k|'+\cdots +|\alpha _{r}|'+@$.

Finally we define $H:=H_1+H_2+H_3$. Then the result follows from the following statement: for any chains $\alpha$ and $\beta$,

$$\begin{aligned} \langle \varphi \cap \alpha , \beta \rangle _{{\textsf {Muk}}} -(-1)^{|\varphi ||\alpha |}\langle \alpha , \varphi \cap \beta \rangle _{{\textsf {Muk}}} + H(b(\alpha )\otimes \beta + (-1)^{|\alpha |}\alpha \otimes b(\beta ))=0 \end{aligned}$$

This is a direct, albeit long, computation (that we omit) using only the $A_\infty$ relations, the closedness of $\varphi$ and the fact that ${\textsf {tr}}(A\circ B)={\textsf {tr}}(B\circ A)$. $\square$

Proposition A.2

Given a Hochschild cochain classes $\varphi , \psi \in HH^{\bullet }\left( A \right)$ we have

$$\begin{aligned} D(\varphi \cup \psi )=(-1)^{|\varphi |d}\varphi \cap D(\psi ). \end{aligned}$$

In other words, D is a map of $HH^\bullet (A)$-modules (of degree d).

Proof

Since the Mukai pairing is non-degenerate it is enough to check

$$\begin{aligned} \langle D(\varphi \cup \psi ) , \alpha \rangle _{{\textsf {Muk}}} =\langle (-1)^{|\varphi |d}\varphi \cap D(\psi ) , \alpha \rangle _{{\textsf {Muk}}}, \end{aligned}$$

(26)

for all $\alpha \in HH_{\bullet }\left( A \right)$. By definition of D, we have

$$\begin{aligned} \langle D(\varphi ) , \alpha \rangle _{{\textsf {Muk}}} =(-1)^{|\alpha _0|'|\alpha _{1,n}|'}\langle \varphi (\alpha _1, \ldots , \alpha _n) , \alpha _0 \rangle , \end{aligned}$$

where $|\alpha _{1,n}|':=|\alpha _1|'+\cdots +|\alpha _n|'$. Therefore, using commutativity of $\cup$, the left-hand side of (26) equals

$$\begin{aligned}&(-1)^{|\varphi ||\psi |+|\alpha _0|'|\alpha _{1,n}|'} \langle (\psi \cup \varphi )(\alpha _1,\ldots , \alpha _n), \alpha _0 \rangle \nonumber \\&\quad = (-1)^{\Diamond } \sum \langle {\mathfrak {m}}_p(\alpha _1, \ldots ,\psi _a(\alpha _{i+1}, \ldots ), \ldots , \varphi _b (\alpha _{j+1}, \ldots ), \ldots \alpha _n), \alpha _0\rangle . \end{aligned}$$

(27)

where $\Diamond = |\varphi ||\psi |+|\alpha _0|'|\alpha _{1,n}|'+ |\psi |'|\alpha _{1,i}|'+|\varphi |'|\alpha _{1,j}|'$. On the other hand, using Proposition A.1, the right hand side of (26) equals

$$\begin{aligned}&(-1)^{|\varphi |d+ |\varphi |(|\psi |+d+1)}\langle D(\psi ) , \varphi \cap \alpha \rangle _{{\textsf {Muk}}}\nonumber \\&\quad = \sum (-1)^{\delta _1 } \langle D(\psi ), {\mathfrak {m}}_q(\ldots \varphi _b(\alpha _{j+1}\ldots ), \ldots \alpha _0, \ldots ) \alpha _{i+1} \ldots \alpha _{i+a}\rangle \nonumber \\&\quad = \sum (-1)^{\delta _2}\langle \psi _a(\alpha _{i+1} \ldots ) , {\mathfrak {m}}_q(\ldots \varphi _b(\alpha _{j+1}\ldots ), \ldots \alpha _0, \ldots )\rangle . \end{aligned}$$

(28)

where $\delta _1=|\varphi ||\psi |+|\varphi |'|\alpha _{i+a+1,j}|' +|\alpha _{0,i}|'|\alpha _{i+1,n}|'$ and

$$\begin{aligned} \delta _2\!=\!|\varphi ||\psi |+|\varphi |'|\alpha _{i+a+1,j}|' \!+\!|\alpha _{0,i}|'|\alpha _{i+1,n}|'+|\alpha _{i+1,i+a}|' (|\varphi |+|\alpha _{0,i}|'\!+\!|\alpha _{i+a+1,n}|'). \end{aligned}$$

Using cyclic symmetry of the pairing $\langle - , -\rangle$, a straightforward computation shows that the expressions in (27) and (28) are equal, which proves the desired result. $\square$

We can now prove Theorem 2.4. Indeed, Part (a) of the theorem is exactly Proposition A.2 proved above. In particular, for any $\varphi \in HH^{\bullet }\left( A \right)$, we have $D(\varphi )= (-1)^{|\varphi |d}\varphi \cap D({\mathbb {1}})= (-1)^{|\varphi |d}\varphi \cap \omega$. For part (b), it is well known that the cup product is associative and graded-commutative. The only condition left to check is the compatibility between product and pairing. We use Proposition A.1 to compute

$$\begin{aligned} D^*\langle \varphi \cup \psi , \rho \rangle _{{\textsf {Muk}}}&=(-1)^{(|\varphi |+|\psi |)d}\langle D(\varphi \cup \psi ), D(\rho )\rangle _{{\textsf {Muk}}} \\&= (-1)^{(|\varphi |+|\psi |)d+|\varphi ||\psi | + (|\varphi | +|\psi |+ |\rho |)d} \langle (\psi \cup \varphi )\cap \omega , \rho \cap \omega \rangle _{{\textsf {Muk}}}\\&= (-1)^{|\varphi ||\psi | + |\rho |d}\langle \psi \cap (\varphi \cap \omega ),\rho \cap \omega \rangle _{{\textsf {Muk}}}\\&= (-1)^{|\varphi ||\psi | + |\rho |d+ |\psi |(|\varphi |+d)} \langle \varphi \cap \omega , \psi \cap (\rho \cap \omega )\rangle _{{\textsf {Muk}}}\\&= (-1)^{ (|\psi | + |\rho |)d}\langle \varphi \cap \omega , (\psi \cup \rho )\cap \omega )\rangle _{{\textsf {Muk}}}\\&= (-1)^{ |\varphi |d} \langle D(\varphi ), D(\psi \cup \rho ) \rangle _{{\textsf {Muk}}}=D^*\langle \varphi , \psi \cup \rho \rangle _{{\textsf {Muk}}} \end{aligned}$$

Appendix B. Proof of Proposition 2.6

Let H be the map introduced in the proof of Proposition A.1 and extend it sesquilinearly to $CC_\bullet (A)[[u]]^{\otimes 2}$. We claim that for any negative cyclic chains $\alpha$ and $\beta$

$$\begin{aligned} \frac{{d}}{{d}u}\langle \alpha , \beta \rangle _{{\textsf {hres}}}&= \langle \nabla _{\frac{{d}}{{d}u}} \alpha , \beta \rangle _{{\textsf {hres}}} -\langle \alpha , \nabla _{\frac{{d}}{{d}u}} \beta \rangle _{{\textsf {hres}}}\nonumber \\&\quad + \frac{1}{2u^2}H\left( (b+uB)(\alpha ), \beta \right) +\frac{(-1)^{|\alpha |}}{2u^2}H\left( \alpha , (b+uB)(\beta ) \right) , \end{aligned}$$

(29)

which immediately implies the result.

Indeed, writing $\displaystyle \alpha =\sum _{n\ge 0}\alpha _n u^n$, $\displaystyle \beta =\sum _{n\ge 0}\beta _n u^n$ and using the definitions of the pairing and the connection to expand the above expression we see that the left-hand side equals $\sum _{n\ge 0} \sum _{k=0}^{n+1}(-1)^{n-k+1}(n+1)\langle \alpha _k , \beta _{n-k+1} \rangle _{{\textsf {Muk}}}u^n$. The right hand side equals

$$\begin{aligned}&\sum _{n\ge 0}\sum _{k=0}^{n+1} (-1)^{n-k+1}(n+1) \langle \alpha _k , \beta _{n-k+1} \rangle _{{\textsf {Muk}}}u^n \\&\quad +\sum _{n\ge -2}\sum _{k=0}^{n+2} \frac{(-1)^{n-k}}{2} \left( \langle b\{{\mathfrak {m}}'\}\alpha _k , \beta _{n-k+2} \rangle _{{\textsf {Muk}}} -\langle \alpha _k , b\{{\mathfrak {m}}'\}\beta _{n-k+2} \rangle _{{\textsf {Muk}}}\right) u^n\\&\quad +\sum _{n\ge -1}\sum _{k=0}^{n+1} \frac{(-1)^{n-k+1}}{2} \Big (\langle (\Gamma +B\{{\mathfrak {m}}'\})\alpha _k , \beta _{n-k+1} \rangle _{{\textsf {Muk}}} \\&\qquad \quad +\langle \alpha _k , (\Gamma +B\{{\mathfrak {m}}'\}) \beta _{n-k+1} \rangle _{{\textsf {Muk}}}\Big )u^n\\&\quad +\sum _{n\ge -2}\sum _{k=0}^{n+2} \frac{(-1)^{n-k}}{2} \left( H\left( b(\alpha _k), \beta _{n-k+2}\right) +(-1)^{|\alpha |} H\left( \alpha _k , b(\beta _{n-k+2})\right) \right) u^n \\&\quad +\sum _{n\ge -1}\sum _{k=0}^{n+1} \frac{(-1)^{n-k+1}}{2} \left( H\left( B(\alpha _k), \beta _{n-k+1}\right) -(-1)^{|\alpha |} H \left( \alpha _k , B(\beta _{n-k+1})\right) \right) u^n \end{aligned}$$

Therefore the claim follows from the following two identities

$$\begin{aligned}&\langle b\{{\mathfrak {m}}'\}x, y \rangle _{{\textsf {Muk}}} -\langle x , b\{{\mathfrak {m}}'\}y \rangle _{{\textsf {Muk}}} + H\left( b(x), y\right) +(-1)^{|x|} H\left( x, b(y)\right) =0, \\&\langle (\Gamma +B\{{\mathfrak {m}}'\})x , y \rangle _{{\textsf {Muk}}} + \langle x, (\Gamma +B\{{\mathfrak {m}}'\})y \rangle _{{\textsf {Muk}}} \\&\quad + H\left( B(x), y\right) -(-1)^{|x|} H\left( x , B(y)\right) =0, \end{aligned}$$

for arbitrary Hochschild chains $x=x_0|x_1 \ldots x_r$ and $y=y_0|y_1 \ldots y_s$. The first identity is exactly the content of the proof of Proposition A.1, since $|{\mathfrak {m}}'|=0$. For the second one, we first show by direct computation that

$$\begin{aligned}&\langle B\{{\mathfrak {m}}'\}x , y \rangle _{{\textsf {Muk}}}+ \langle x, B\{{\mathfrak {m}}'\}y \rangle _{{\textsf {Muk}}} + H\left( B(x), y\right) -(-1)^{|x|} H \left( x , B(y)\right) \nonumber \\&\quad = - \sum {\textsf {tr}}\left[ c \rightarrow (-1)^{\dagger }{\mathfrak {m}}'_* \left( x_j,\ldots ,x_0,\ldots ,{\mathfrak {m}}_*(x_i,\ldots ,c,y_n,\ldots ,y_0,\ldots ), y_m,\ldots \right) \right] \nonumber \\&\qquad - \sum {\textsf {tr}}\left[ c \rightarrow (-1)^{\dagger }{\mathfrak {m}}_* \left( x_j,\ldots ,x_0,\ldots ,{\mathfrak {m}}'_*(x_i,\ldots ,c,y_n,\ldots ,y_0,\ldots ), y_m,\ldots \right) \right] \end{aligned}$$

(30)

where $\dagger$ is as in (2). Then, counting the number of inputs as in the proof of Lemma 3.3, we see the right-hand side in (30) gives $(r+s)\langle x , y \rangle _{{\textsf {Muk}}}$. Finally we observe that

$$\begin{aligned} \langle \Gamma (x) , y \rangle _{{\textsf {Muk}}} +\langle x, \Gamma (y) \rangle _{{\textsf {Muk}}} = -(r+s)\langle x , y \rangle _{{\textsf {Muk}}} \end{aligned}$$

which therefore cancels with (30), proving the required identity.