Why Hessian is a (0,2) symmetry tensor?

Question

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.

Answer 1

There is a more general property at play here that is worth noting: if $\varphi : M \to N$ is a diffeomorphism of smooth manifolds, then the pullback map $\varphi^{*} : T^k N \to T^k M$ on bundles of covariant k-tensors will preserve symmetric and anti-symmetric properties of your covariant k-tensors.

For example, consider the case when $g$ is a symmetric covariant 2-tensor on $N$, i.e., $g$ is a Riemannian metric on $N$. Then the pullback $\varphi^{*}g$ becomes a Riemannian metric on $M$ defined on vector fields $X, Y \in \chi\left(M\right)$ by $$ \varphi^{*}g \left(X, Y\right) := g\left(\varphi_{*}X, \varphi_{*}Y\right). $$ Note that from the symmetry of $g$ on $N$ it follows that \begin{align*} \varphi^{*}g \left(X, Y\right) &= g\left(\varphi_{*}X, \varphi_{*}Y\right)\\ &=g\left(\varphi_{*}Y, \varphi_{*}X\right)\\ &=\varphi^{*}g\left(Y, X\right). \end{align*}

And more to the point of your question, this implies that the symmetric or anti-symmetric properties of a covariant k-tensor are preserved under Lie derivatives. For example, suppose now that $g$ is a symmetric covariant $k$-tensor on $M$, that $X$ is a vector field on $M$, and $\psi_{t}$ is the flow of $X$ (i.e., the one-parameter family of diffeomorphisms of $M$ generated by $X$).

Then by definition we have that the Lie derivative of $g$ at a point $p \in M$ in the direction of $X$ is $$ \left(\mathcal{L}_{X}g\right)_{p} = \lim\limits_{t \to 0} \frac{\left(\psi^{*}_{t}\right)_{p}\left(g_{\psi_{t}\left(p\right)}\right) - g_{p}}{t}. $$

The numerator of the above expression is merely the difference between the pullback of the symmetric tensor $g$ at $\psi_{t}(p)$ to $p$ and the symmetric tensor $g$ at $p$. But since pullbacks of diffeomorphisms preserve symmetry properties of tensors, the numerator is the difference between symmetric tensors at $p \in M$ and is thus symmetric. The resulting Lie derivative will thus be symmetric as well.

Answer 2

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.