Gradient of the Lie exponential map on SO(n)

Question

I am interested in computing the gradient of $f(e^A)$ when $A$ is a skew-symmetric matrix.

If we write $e^A = B$ and we denote the gradient of the function $f$ on the ambient space evaluated at $B$ as $G$, using the formula for the derivative of the exponential map we get that $df(e^A)$ is

$$ df(e^A)_A(H) = \frac{1}{2} \langle B^\intercal G - G^\intercal B, \frac{1-e^{-\mathrm{ad}_A}}{\mathrm{ad}_A} H\rangle $$ where the scalar product is the canonical scalar product on $\mathbb{R}^{n \times n}$ and $H$ is a skew-symmetric matrix.

Now, I don't see how to get an expression for the gradient of $f(e^A)$ from this formula.

In particular, I would be interested in a formula that depends on the matrix $T=B^\intercal G - G^\intercal B$.

Answer 1

I'm answering my own question with a not-completely-satisfactory answer.

If we write $K_A = \frac{1-e^{-\mathrm{ad}_A}}{\mathrm{ad}_A}$ then $K_A$ is a linear operator, and as such we can write $$ \nabla(f \circ \exp)(A) = \frac{1}{2}K^*_A(B^\intercal G - G^\intercal B) $$ where $K^*_A$ denotes the adjoint of $K_A$.

Edit: Now, we can compute $K^*_A$, which gives $K_{A^\intercal}$.