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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 $Ext^{\ast}(E,E)$?

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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.

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  • $\begingroup$ 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

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