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?