The Hessian of invariant functions on a Lie group

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"?

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.

• Dear Nawaf Thank you very much for your complete and interesting answer. (As I gave +1 to your answer, already) – Ali Taghavi Sep 21 '16 at 20:07