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 $Hess (f).(V, W)=\nabla ^ {\nabla f}\_{V}.W$



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> **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 $G$?
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> 
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This question is motivated by the following post.


[Is there an explicit formula for the hessian of "Determinant"?](https://mathoverflow.net/questions/249219/is-there-an-explicit-formula-for-the-hessian-of-determinant)



The Hessian of invariant functions on a Lie group
Yes. 


In what follows, I use standard notation for the derivative; see, e.g., $\S$2.3 of *Banach Spaces and Differential Calculus* (Chapter 2) of the book referenced below.


*First Derivative of $f$.*


As the OP stated, a function $f: G \to \mathbb{R}$ is $G$ invariant means that 
$$
f(A) = f(g A g^{-1}) \quad \forall A, g \in G \tag{$\star$}
$$ If we differentiate this relation once we obtain
$$
D f(A) \cdot B\_1 = \frac{d}{d\lambda} \left. f( g (A+\lambda B\_1) g^{-1} ) \right|\_{\lambda=0} = D f( g A g^{-1} ) \cdot g B\_1 g^{-1} \tag{$\star \star$}
$$ where $B\_1 \in T\_A G$. In the special case of the determinant on $G = Gl(n,\mathbb{R})$, this is simply saying that: 
$$
D \det(A) \cdot B\_1 = \det(A) \operatorname{trace}(A^{-1} B\_1) = \det( g A g^{-1} ) \operatorname{trace}(g A^{-1} g^{-1} g B\_1 g^{-1} ) 
$$ which follows from basic properties of the determinant and trace of a matrix.


*Second Derivative of $f$.*


Now differentiate ($\star \star$) to obtain: 
$$
D^2 f(A) \cdot (B\_1, B\_2) = D^2 f( g A g^{-1} ) \cdot ( g B\_1 g^{-1}, g B\_2 g^{-1} ) \tag{$\star \star \star$}
$$ where $B\_1, B\_2 \in T\_A G$. In the special case of the determinant on $G = Gl(n,\mathbb{R})$, \begin{align\*}
D^2 \det(A) \cdot (B\_1, B\_2) &= \frac{d}{d \lambda} \left. D \det(A+\lambda B\_2) \cdot B\_1 \right|\_{\lambda=0} \\
&= \frac{d}{d \lambda} \left. \det (A + \lambda B\_2) \operatorname{trace} ( (A+\lambda B\_2)^{-1} B\_1 ) \right|\_{\lambda=0} \\
 &= \det(A) \operatorname{trace} (A^{-1} B\_2) \operatorname{trace}(A^{-1} B\_1) - \det(A) \operatorname{trace}(A^{-1} B\_2 A^{-1} B\_1 )
\end{align\*} which is clearly $G$-invariant in the sense of relation ($\star \star \star$).


*Higher Derivatives of $f$.*


This result for the first and second derivative seems to be true for higher derivatives of a $G$-invariant function. In other words, differentiation preserves $G$-invariance in the sense given above.


**Reference**


For background info see, e.g., *An Introduction to Lie groups* Chapter 5 of


Jerrold E. Marsden, Tudor Ratiu, and Ralph Abraham. **Manifolds, Tensors, and Applications.** Second Edition. Vol. 75. Springer Science & Business Media, 2012.


