4
$\begingroup$

Let $X$ be a smooth projective variety over $\mathbb{C}$ and $E$ a slope-stable vector bundle on $X$ with regard to some ample line bundle $H$.

Question: What can we say about the algebra structure of $Ext^{\ast}(E,E)$?

Since this is a fairly general question, let me be more precise.

Let us for simplicity assume that $E$ is a smooth point in the moduli spaces of stable sheaves $M_H(v)$ with $v=v(E)$ its Mukai vector. Are there some general results on the algebra structure $Ext^{\ast}(E,E)$? In particular, do we know how the (Yoneda) product looks like or what properties it has?

For curves and Fano surfaces for example the product is trivial, whereas for K3 surfaces the pairing $Ext^1(E,E) \times Ext^1(E,E) \to Ext^2(E,E)\cong Hom(E,E) \cong \mathbb{C}$ is perfect and skew-symmetric. I was wondering if there are any known structural results in higher dimensions?

For example, is the product (e.g. on Calabi-Yau varieties, as suggested by the K3 case) graded commutative? Since $E$ is assumed a smooth point in moduli, the obstructions to deform $E$ vanish and so the Maurer-Cartan equation gives that the product $Ext^1(E,E) \times Ext^1(E,E) \to Ext^2(E,E)$ is skew-symmetric

More generally, one can consider $RHom(E,E)$ as a differential-graded Lie algebra and as such one can associate to it a deformation functor which in this case controls the local deformation theory of $E$. This can be quite complicated, so for simplicity let us assume that this differential-graded Lie algebra is formal, i.e. quasi-isomorphic to its cohomology which is precisely $Ext^{\ast}(E,E)$. Are there on a smooth projective $X$ any a priori constraints on the algebra structure of $Ext^{\ast}(E,E)$?

$\endgroup$

1 Answer 1

2
$\begingroup$

There have been recently various results revolving around this question. Let me quote a few:

$\bullet$ For any line bundle $L$ on $X$, the graded algebra $\mathrm{Ext}^*(L,L)$ is always graded-commutative. More generally, for any autoequivalence $\Phi$ of $\mathrm{D}^b(X)$, the graded algebra $\mathrm{Ext}^*(\Phi(\mathcal{O}_X),\Phi(\mathcal{O}_X))$ is graded commutative. See for instance this short proof by Suarez-Alvarez.

$\bullet$ If $\Delta_X$ is the diagonal in $X \times X$, then the graded algebra $\mathrm{Ext}^*(\mathcal{O}_{\Delta_X},\mathcal{O}_{\Delta_X})$ is graded commutative. Hence, the Hochschild cohomology algebra on $X$ is graded commutative. This is equally proved in the paper by Suarez-Alvarez I mentionned above.

$\bullet$ If $\mathrm{rank}(E) \neq 0$, then the trace map shows that $\mathrm{Ext}^{*}(E,E)$ has the structure of a faithful $H^{*}(\mathcal{O}_X)$-algebra. This algebra structure is conjectured to be a derived invariant (in car $0$). This is proved in dimension $\leq 4$ (and is some other situations related to moduli theory). It will be disproved in car $p>0$ in a forthcoming paper of Addington and Bragg.

$\bullet$ Hochenegger and Krug proved that for any $E \in \mathrm{D}^b(X)$, if $\mathrm{Ext}^*(E,E) = k[t]/t^{n +1}$ with $\deg(t) \geq 2$, then the DG-algebra $\mathrm{RHom}(E,E)$ is automatically formal.

$\endgroup$
4
  • $\begingroup$ Thank you very much for this great answer! May I quickly ask how you apply the result of Suarez-Alvarez to obtain that $Ext^{\ast}(E,E)$ is graded-commutative, i.e. what is the suspended monoidal category such that the endomorphism ring of the unit object equals $Ext^{\ast}(E,E)$? $\endgroup$ Jun 14, 2021 at 12:31
  • $\begingroup$ @NicoBerger : you should have a look at section 2 of Suarez-Alvarez paper (and more precisely, section 2.1). $\endgroup$
    – Libli
    Jun 14, 2021 at 20:20
  • $\begingroup$ I am sorry, I still have troubles understanding this in general. If we take in section 2.1 $\mathcal{C}=Coh(X)$, then the unit object is $\mathcal{O}_X$ and one concludes that $Ext^{\ast}(\mathcal{O}_X,\mathcal{O}_X)$ is graded-commutative. So this also holds for objects in the orbit of $\mathcal{O}_X$ under $Aut(\mathrm{D}^b(X))$. But for a general slope-stable vector bundle $E$ what is the category $\mathcal{C}$ to which we apply the discussion of section 2.1? $\endgroup$ Jun 15, 2021 at 7:36
  • $\begingroup$ @NicoBerger : you are right. I have had a wrong recollection of Suarez-Alvarez result. I used it some time ago to highlight some properties of Hochschild cohomology (where it applies, see section 2.5) and I wrongly remembered it was correct in a larger context. I will edit my answer. $\endgroup$
    – Libli
    Jun 15, 2021 at 13:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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