Let $X$ be a smooth projective variety (say over $\mathbb{C}$). An object $F \in D^b(X)$ is said to be rigid if $\mathrm{Ext}^1(F,F)=0$. I was wondering if we can always find a rigid object on a projective variety of dimension bigger or equal to $2$ (see the edit below for comments on the dimensional hypothesis). Ideally, I would also like the Chern character of this object to be non-zero.

In case $H^1(\mathcal{O}_X) =0$, any line bundle will do the job. On the other hand, if $H^1(\mathcal{O}_X) \neq 0$, the existence of the trace maps shows that the rank of such an object must be zero. I have some specific examples in mind (mostly structure sheaves of rigid subvarieties of some special varieties), but I would like to know if such objects exist in general on any smooth projective variety.

Edit: as Johan elliptically points out in the comments, the Grothendieck-Riemann-Roch Theorem shows that $\chi(F,F) =0$ for $F \in D^b(X)$, when $X$ is an elliptic curve. In particular, if $F$ is a coherent sheaf, the non vanishing of $\mathrm{Hom}(F,F)$ implies necessarily that $\mathrm{Ext}^1(F,F) \neq 0$, as there are no higher Ext's. On the other hand, we know that that an object in the derived category of an elliptic curve is quasi-isomorphic to the direct sum of its shifted cohomology sheaves. From this, we can deduce that all objects have non vanishing $\mathrm{Ext}^1$.

This seems however a very specific phenomenon related to curve (as the category $Coh(X)$ is then hereditary and any object in the derived category is quasi-isomorphic to a sum of shifted coherent sheaves). This is why I will make an assumption on $\dim X$.

  • 3
    $\begingroup$ Hint: elliptic curve. $\endgroup$
    – Johan
    Jul 26, 2022 at 19:18
  • $\begingroup$ The zero object is rigid. $\endgroup$ Jul 26, 2022 at 21:13
  • $\begingroup$ @JasonStarr : good point! This is why I asked that, ideally, the Chern character would be non zero. $\endgroup$
    – Libli
    Jul 26, 2022 at 21:19
  • $\begingroup$ If you consider the "filtration" by good truncations, I believe that every rigid object on an elliptic curve is quasi-isomorphic to zero. $\endgroup$ Jul 27, 2022 at 0:12
  • $\begingroup$ @JasonStarr : That may be. On the other hand, I know prove in the edit that any non zero object on an elliptic curve has non vanishing Ext^1. This seems however very specific to the case of curve ( because then $Coh(X)$ is hereditary) and I have added an extra hypothesis on $ \dim X$ to improve my question. $\endgroup$
    – Libli
    Jul 27, 2022 at 6:05

1 Answer 1


I am writing up as one answer the comments by @Johan, by @Libli, and by myself. If either of them prefers to write an answer, I am happy to delete this answer.

Let $A$ be an Abelian variety. For every scheme $S$ and every $S$-valued point $x\in A(S)$, denote by $\mu_x$ the associated translation automorphism of the $S$-scheme $S\times A$, i.e., $\mu_x(y) = x+y$.

Denote by $\widehat{A}$ together with the invertible sheaf $\mathcal{P}$ on $\widehat{A}\times A$ the relative $\text{Pic}^0$ of $A$, normalized so that $\mathcal{P}|_{\widehat{A}\times\{0\}}$ is the structure sheaf on $\widehat{A}$. Of course $\widehat{A}\times A$ is a commutative group scheme with its structure as the product of two commutative group schemes. Denote by $G$ the noncommutative group scheme structure on $\widehat{A}\times A$ defined by $$ ([\mathcal{L}],x)\bullet([\mathcal{M}],y) = ([\mathcal{L}\otimes \mu_x^*\mathcal{M}],x+y).$$ There is an "action" of $G$ on the bounded derived category of coherent sheaves on $A$ that associates to each $([\mathcal{L}],x)$ in $G$ and each bounded complex $C^\bullet$ of coherent sheaves on $A$ the associated bounded complex of coherent sheaves, $\mathcal{L}\otimes \mu_x^*(C^\bullet).$ (According to an article of Orlov, this action identifies $G$ with the identity component of the group of autoequivalences of the bounded derived category of coherent sheaves on $A$.)

If $C^\bullet$ is not quasi-isomorphic to the zero complex, i.e., if it is not an exact complex, there there exists an integer $p$ such that the cohomology sheaf $h^p(C^\bullet)$ is nonzero. In general, a "flat deformation" of the complex $C^\bullet$ does not necessarily give rise to a flat deformation of the coherent sheaf $h^p(C^\bullet)$, since base change is not left exact. However, for a connected, smooth group scheme $G$, for a $G$-equivariant family of deformations over $G$, the coherent sheaves $h^p$ are compatible with base change: this holds over a dense open of $G$ (since $G$ is reduced), and this dense open is $G$-invariant, thus it is all of $G$. Therefore, the deformations of $C^\bullet$ arising from the action of $G$ give rise to a deformation of $h^p(C^\bullet)$.

By hypothesis, the coherent sheaf $h^p(C^\bullet)$ on $A$ is nonzero. If the rank is positive, then the action of the normal subgroup $\widehat{A}\times\{0\}$ of $G$ on this sheaf is nontrivial by considering "det" of the coherent sheaf. If the rank of the sheaf is zero, i.e., if the support of the sheaf is a proper closed subscheme of $A$, then the action of the subgroup $\{[\mathcal{O}_A]\}\times A$ of $G$ on the sheaf is nontrivial since it "moves" this proper closed subscheme. Either way, the action of the group scheme $G$ on the sheaf is nontrivial.

Since the action of $G$ already produces nontrivial deformations of the sheaf $h^p(C^\bullet)$, it also produces nontrivial deformations of the complex $C^\bullet$. Thus, the only rigid complexes in the bounded derived category of $A$ are quasi-isomorphic to zero.


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