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This question follows up another question on this website.

Let $M$ be a Riemannian manifold and let $f : M \rightarrow \mathbb R$ be a smooth function. We assume that $\nabla$ is a connection over that manifold. The Hessian of $f$ is the symmetric tensor

$\operatorname{Hess} f = \nabla \nabla f$

The Levi-Civita connection seems to be symmetric, as can be seen from its definition in terms of Christoffel symbols. see Wikipedia article

Question Under which circumstances the covariant Hessian is a symmetric tensor? Is there an intrinsic way of seeing that the Levi-Civita connection is symmetric?

Rushabh Mehta
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shuhalo
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    The first link in your question does not work. Also, while $\text{Hess} f = \nabla \nabla f$ is correct, the equation $\text{Hess} f(X,Y) = \nabla_X \nabla_Y f$ is incorrect. The correct identity is $\text{Hess} f(X,Y) = (\nabla_X \nabla f)(Y)$, where we interpret $\nabla f = df$ as a $1$-form. – Jesse Madnick Dec 26 '20 at 19:33
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    The Hessian is only symmetric if the connection $\nabla$ is torsion-free. – Kajelad Dec 26 '20 at 19:48
  • I would add to @Kajelad comment that the main motivation in for the Levi-Civita connexion to be torsion-free is exactly because we wan Hessians functions to be symmetric. – Didier Dec 27 '20 at 10:25
  • The symmetry of the Hessian is a side note in most differential geometry classes but IMHO that property should be framed as the core motivation for the Levi-Civita connection. – shuhalo Dec 27 '20 at 16:04

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