Assume that $G$ is a Lie group with Lie algebra $\mathfrak{g}$. We fix an invariant Riemannian metric on $G$ and fix its corresponding $LC$ connection.

Consider the natural right action of $G$ on its Lie algebra $\mathfrak{g} \simeq \{X \in \chi^{\infty}({G}) \mid R_{g}^{*} X=X\}$, the space of smooth vector fields which are invariant under right multiplications.

In fact the right action is defined as follows:

For $g\in G$ and $X\in \mathfrak{g}$ define $X.g=L_{g}^{*} X$ where $L_{g}$ is the left multiplication by $g$.

So there is a natural (component wise) action of $G$ on $\mathfrak{g} \times \mathfrak{g}$

A smooth function $f:G \to \mathbb{R}$ is called $G$-invariant if $f(g^{-1}hg)=f(h)$ for all $g,h \in G$. For example $Det: Gl(n, \mathbb{R}) \to \mathbb{R}$.

A $2$- linear map $T: \mathfrak{g} \times \mathfrak{g} \to \mathbb{R}$ is $G$-invariant if $T((v,w).g)=T(v,w)$

Example The $2$-linear map $tr(u)tr(v)-tr(uv)$ defined on $M_{n}(\mathbb{R}) \times M_{n}(\mathbb{R})$ is a $Gl(n, \mathbb{R})$ invariant map, with the natural (conjugate) action of $Gl(n, \mathbb{R})$ on its Lie algebra $M_{n}(\mathbb{R})$ as described above.

Recall that for a Riemannian manifold with the corresponding $LC$ connection $\nabla$, the Hessian of a function $f$ defined on the manifold, is a two linear map on the tangent space with the formula $Hess (f).(V, W)=\nabla ^ {\nabla f}_{V}.W$

Question:Let $G$ be a Lie group and $f:G \to \mathbb{R}$ be a $G$-invariant smooth function. Is its Hessian $Hess(f)$ a $G$-invariant $2$-linear map on the Lie algebra $\mathfrak{g}$ of $G$?

This question is motivated by the following post.

Is there an explicit formula for the hessian of "Determinant"?