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

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.

• It's just a non-Euclidean generalisation of en.wikipedia.org/wiki/Hessian_matrix – J.G. Oct 10 '18 at 14:24
• do you mean gradient by divergence? – peter Feb 20 '20 at 19:25

## 2 Answers

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.

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.