Given a triangulated category $D$, there is a Chern character from the Grothendieck group to the Hochschild homology: $$ch:K_0(D) \to HH_0(D).$$ Consider a pair of projective Calabi-Yau threefolds $X$ and $Y$ which satisfy homological mirror symmetry. This means that there is an equivalence of triangulated categories $$D Fuk(X) \to D^b Coh(Y)$$ This induces isomorphisms $K_0(X)\to K_0(Y)$ and $HH_0(X)\to HH_0(Y)$. In terms of the Hodge decomposition, $$HH_0(DFuk(X)) \simeq H^{0,3}(X)\oplus H^{1,2}(X) \oplus H^{2,1}(X) \oplus H^{3,0}(X)$$ $$HH_0(D^bCoh(Y)) \simeq H^{0,0}(Y) \oplus H^{1,1}(Y) \oplus H^{2,2}(Y) \oplus H^{3,3}(Y).$$ The Chern character for $Y$ is the standard Chern character of coherent sheaves.

Question 1: Is the chern character for $X$ given by taking the (Poincare dual of the) fundamental class of a Lagrangian submanifold?

Question 2: Is the resulting square of Chern characters commutative?

A related question is whether the isomorphism on $HH_0$ respects the direct sum decompositions from Hodge theory. This seems unlikely because the fundamental class of a Lagrangian is defined over $\mathbb R$, so it is invariant under the complex conjugation which swaps the summands of $HH_0(X)$. On the other hand, sky scraper sheaves on $Y$ have Chern character supported on $H^{3,3}(Y)$.

If indeed the isomorphism does NOT respect the Hodge decompositions, then what do they each correspond to on the other side?

  • $\begingroup$ Unless I'm missing something the construction* of chern characters for dg categories is obviously functorial, so the associated square most be commutative? (I don't know how to identify the chern character on the A-model side btw.) *ie using Morita invariance to note HH(perf) =k[0], noting there is a unique map perf--->C taking k to an object x and letting the induced map on HH_{0} define ch(X). nb this further endows the chern character with a cyclic structure ie it descends to HC $\endgroup$
    – EBz
    Sep 17, 2019 at 22:30

1 Answer 1


Well, you pointed out a difficult problem in HMS and your guess is correct, equivalences between categories do not preserve the Hodge decomposition in general. Let me give you an example : let $X$ be an abelian threefold and $X^{\vee}$ the dual abelian variety. The well-known Fourrier Mukai transform between $D^b(X)$ and $D^{b}(X^{\vee})$ sends $H^{3,3}(X)$ to $H^{0,0}(X^{\vee})$.

In an appropriate situation, you can construct Kümmer threefolds $Z$ and $Z^{\vee}$ from $X$ and $X^{\vee}$ which are still derived equivalent and for which the derived equivalence still sends $H^{3,3}(Z)$ to $H^{0,0}(Z^{\vee})$. Now, let $Y$ be a CY3 mirror to $Z$. The equivalences (mirror maps):

$$ D^b(Z) \longrightarrow DFuk(Y)$$ and $$D^{b}(Z^{\vee}) \longrightarrow DFuk(Y)$$ can not both preserve the Hodge decompositions since the equivalence $D^b(Z) \longrightarrow D^b(Z^{\vee})$ does not.

In fact, this problem is certainly one of the main issue when you want to compare the Hodge structures of two derived equivalent varieties.

As far as your last question is concerned, it is difficult to have a guess a priori on what part of the Hodge decomposition of $\mathrm{HH}_0(X)$ is send to which of the decomposition of $\mathrm{HH}_0(Y)$ (it far from being even clear that pure part are sent onto pure parts).

To alleviate this problem, the notion of homological unit has been introduced in there. The homological unit is a graded subalgebra of the Hochschild cohomology (or a graded sub-module of the Hochschild homology) of a "well-behaved" category and is defined following the choice of a rank function. It is a categorical substitute of the algebra $H^{\bullet}(X,\mathcal{O}_X)$ when $X$ is a smooth projective variety.

The algebra structure on the homologial unit has been proved to be independant of the rank function for varieties of dimension less or equal to $4$, showing in particular that it is a derived invariant in dimension $\leq 4$. More recently, it has been proved to be derived invariant for any CY3 category (not necessarily of geometric origin) provided that the category contains enough "nice" objects (condition which is satisfied for all known examples of CY3 categories). So it seems that this unit is the right object to study (at least on the B-side) when willing to compare Hodge structures on both sides of an equivalence.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.