Let $E$ be a holomorphic vector bundle over a compact complex manifold (or projective algebraic variety) $X$.

The Atiyah class of $E$, $a(E)\in Ext^1(T_X,End(E))$, is defined to be the class of the extension $$ 0 \rightarrow End(E) \rightarrow \mathcal{D}(E) \rightarrow T_X \rightarrow 0 $$ where $\mathcal{D}(E)$ is the bundle of differential operators from $E$ to $E$ of order $1$ and scalar symbol, the map to the tangent being the symbol map.

It is a theorem of Atiyah that $a(E)$ generates the characteristic ring of $E$.

My question is: what can be said if $E$ is not a vector bundle, but just a coherent torsion free $\mathcal{O}_X$-module? Could a similar statement be true?

One has anyway the characteristic ring of $E$. To me looks like (although I may be wrong) that one can construct $\mathcal{D}(E)$ that fits the same exact sequence.

The problem is that in Atiyah's theory is essential that $E$ is locally free, since he proves the result through the curvature of connections on $E$, and these does not exist if $E$ is not locally free.

Is there any technique (from K-theory?) that would help? Or my problem is senseless?


1 Answer 1


It is better to define the Atiyah class as an element of $Ext^1(E,E\otimes\Omega^1)$. Then it is defined for all coherent sheaves, and even for all objects of the derived category. The most convenient definition is the following. Look at $X\times X$, let $\Delta:X \to X\times X$ be the diagonal, and $I$ --- the ideal sheaf of the diagonal. Then we have an exact sequence $$ 0 \to I/I^2 \to O/I^2 \to O/I \to 0 $$ on $X\times X$. Since $I/I^2 \cong \Delta_*\Omega^1_X$, it gives a morphism $\Delta_*O_X \to \Delta_*\Omega^1_X[1]$ in the derived category $D(X\times X)$. Now denote $p,q:X\times X \to X$ the projections, take any $E \in D(X)$, tensor this morphism by $p^*E$ and apply $q_*$. We will get a morphism $$ q_*(p^*E \otimes \Delta_*O_X) \to q_*(p^*E \otimes \Delta_*\Omega_X^1)[1]. $$ The projection formula shows that the first term is $E$, and the second is $E\otimes\Omega^1_X[1]$. So, we constructed an element in $$ Hom(E,E\otimes\Omega^1_X[1]) = Ext^1(E,E\otimes\Omega^1_X). $$

This Atiyah class has all the nice properties of the classical one. For example, one can express the coefficients of the Chern character as traces of its powers.

  • $\begingroup$ Thank you very much, looks like what I was looking for! Do you have any detailed reference for this? $\endgroup$ Feb 23, 2011 at 16:23
  • 2
    $\begingroup$ L. Illusie, Complexe cotangent et déformations $\endgroup$
    – Sasha
    Feb 23, 2011 at 16:41
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    $\begingroup$ Also take a look at papers of Caldararu, Makarian, and Roberts and Willerton. $\endgroup$
    – Chris Brav
    Feb 24, 2011 at 7:22
  • $\begingroup$ @Sasha late question: I think I can see the Atiyah class as well as a map $End(E) \rightarrow \Omega^1$. How can I express this in terms of the map you gave? Is it taking traces in some kind of way...? thanks! $\endgroup$
    – Simonsays
    Aug 29 at 9:58
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    $\begingroup$ @Simonsays: No, the Atiyah class is not a map, but rather an extension class; in particular, it can be understood as a class in $\mathrm{Ext}^1(\mathrm{End}(E), \Omega^1)$, or as a morphism in the derived category $\mathrm{End}(E) \to \Omega^1[1]$ (with a shift!). $\endgroup$
    – Sasha
    Aug 29 at 11:02

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