On a Remiannian manifold M, the Hessian of a smooth function $f$ on M is defined to be:
$$\operatorname{Hess}f=\frac{1}{2}\mathcal L_{\nabla f}(g)$$
where $\mathcal L$ stands for Lie derivative, $g$ is the metric of the manifold, and $\nabla f$ means the divergence of the function. It is said that Hessian is a symmetry tensor, but I am not sure why it's symmetric.
Best Answer
Note that, for any vector fields $X,Y\in\mathcal{T}(M)$, \begin{align*} &(\mathcal{L}_{\nabla f}g)(X,Y)\\ =&\mathcal{L}_{\nabla f}(g(X,Y))-g(\mathcal{L}_{\nabla f}X,Y)-g(X,\mathcal{L}_{\nabla f}Y)\\ =&\mathcal{L}_{\nabla f}(g(Y,X))-g(\mathcal{L}_{\nabla f}Y,X)-g(Y,\mathcal{L}_{\nabla f}X)\\ =&(\mathcal{L}_{\nabla f}g)(Y,X) \end{align*} by symmetry of $g$, so $\mathrm{Hess}\,f$ is indeed a symmetric $(0,2)$ tensor. Here I used the formula $$(\mathcal{L}_V\omega)(X_1,\dots,X_n)=\mathcal{L}_V(\omega(X_1,\dots,X_n))-\omega(\mathcal{L}_V X_1,\dots,X_n)-\dots-\omega(X_1,\dots,\mathcal{L}_VX_n)$$ where $\omega$ is a tensor field, and $V,X_1,\dots,X_n$ are vector fields.