Is there a "by hand" proof on the symmetry of the Atiyah class of $TX$?

Question

Let X be a complex manifold and $TX$ its tangent bundle. The Atiyah class $\alpha(E)\in \text{Ext}^1(E\otimes TX, E)$ for a vector bundle $E$ is defined to be the obstruction of the global existence of holomorphic connections on $E$. We can refer to M, Kapranov's paper "Rozansky–Witten invariants via Atiyah classes" or Sasha's answer to this mathoverflow question Atiyah class for non-locally free sheaf.

Now if we take $E=TX$ to be the tangent bundle of $X$ itself, we can prove that the Atiyah class $\alpha(TX)\in \text{Ext}^1(TX\otimes TX, TX)$ is symmetry, which means, $\alpha(TX)$ is in fact in $\text{Ext}^1(S^2(TX), TX)$.

The proof in Kapranov's paper is given in the language of torsor. In J. Roberts and S. Willerton's paper "On the Rozansky–Witten weight systems" (p. 29) they explain Kapranov's proof and mentioned that this symmetry "corresponds in differential geometry to the vanishing of the torsion."

This comment remind me about the similar result in Riemannian geometry that the curvature tensor is symmetric, which relies on the fact that the connection is torsion-free.

$\textbf{My question}$ is: Is there a direct "by hand" proof of the symmetry of the Atiyah class $\alpha(TX)$ without using torsors?

Answer 1

I'm not sure this answers your question since I think that "by hands" might have different meanings, but it seems to me that what you are asking for is precisely the content of Proposition 2.2 on page 14 of the paper of Roberts-Willerton you mentioned.

Let me explain what they do. The Atiyah class of a vector bundle $E$ can be represented by a Dolbeault cocycle: pick a smooth hermitian connection in $E$ and consider its curvature form $R$, which one views as a Dolbeault $1$-cocycle with values in $\Omega^1_X\otimes End(E)$. When $E=T_X$, Proposition 2.2 on page 14 of Roberts-Willerton's paper says that the skew-symmetrization of $R$ is a coboundary. More precisely, $R-R^{op}=\overline\partial(T^{1,0})$, where $T^{1,0}$ is the $(1,0)$ component of the torsion of the above hermitian connection.

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In any case, here is another proof that uses exact sequences... which you could consider being both conceptual and "by hands".

The Atiyah class of $E$ is the class of the following exact sequence: $$ 0\to E\to Diff^{\leq 1}_X(E)\to T_X\otimes E\to 0 \qquad(1) $$ where $Diff^{\leq 1}_X(E)$ is the sheaf of $E$-valued differential operators of degree $\leq1$.

In the case $E=T_X$ this gives $$ 0\to T_X\to Diff^{\leq 1}_X(T_X)\to T_X\otimes T_X\to 0\qquad(2) $$ and we have a morphism from it to the following one: $$ 0\to T_X\to Diff^{+,\leq 2}_X\to S^2(T_X)\to 0\qquad(3) $$ where $Diff^{+,\leq 2}_X$ is the sheaf of differential operators (with values in $\mathcal O_X$) of degree $\leq2$ and vanishing on constants. Moreover one can prove that
$$ T_X=\ker\left(T_X\oplus Diff^{\leq 1}_X(T_X)\to Diff^{+,\leq 2}_X\right)\qquad(4) $$ Hence the result.

From a more geometric point of view we have that:

• Exact sequence (1), resp. (2), splits if and only if there exists a holomorphic connection in $E$, resp. $T_X$.

• Exact sequence (3) splits if and only if there exists a torsion free holomorphic connection in $T_X$.

• (4) insures that (2) splits if and only if (3) does.

Answer 2

Here is a different answer, which is nevertheless the same as Damien's second proof written in a different language, and that can be found in Markarian's famous paper "The Atihay class, Hochschild cohomology, and the Riemann-Roch theorem". Look at Proposition 1 (i).

Answer 3

This seems to be a statement about the classification of connections without torsion. If $\nabla$ and $\nabla'$ are connections without torsion then $\varphi = \nabla - \nabla'$ is a symmetric, linear function $S^2(TX) \rightarrow TX$. If you want to classify connections without torsion you observe that they exist locally and that the data needed to glue local instances comes from $\mathrm{Hom}(S^2(TX),TX)$. Therefore the obstruction to gluing is a Čech 1-cocycle (i.e., a torsor) valued in $\mathrm{Hom}(S^2(TX), TX)$.