7
$\begingroup$

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?

$\endgroup$
0

1 Answer 1

15
$\begingroup$

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.

$\endgroup$
10
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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, 2023 at 9:58
  • 1
    $\begingroup$ @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, 2023 at 11:02

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.