Q:

$\mathsf{GL}(n, \mathbf{R})$ Mapping and Derivative

Q: I would like to find the derivative of the conjugation map
$$\psi: \mathsf{GL}(n, \mathbb{R}) \times \mathsf{GL}(n, \mathbb{R}) \to \mathsf{GL}(n, \mathbb{R}),~ (g,h) \mapsto ghg^{-1}$$
So far I have:

First consider taking the partial derivative of $h$. Formally, $\frac{\partial \psi}{\partial h}$. But since $\psi$ is linear we have that $h \mapsto ghg^{-1}$ is also linear. Thus,
  \begin{equation*}
\frac{\partial \psi}{\partial h} : X \mapsto gXg^{-1},
\end{equation*}
  where $X \in \mathfrak{g}$, and $\mathfrak{g}$ is the Lie algebra of $\mathsf{GL}(n, \mathbb{R})$.
Now, let $e^{X}$ be the matrix exponential. Then $g(t) = e^{tX}$ and 
  \begin{equation*}
g(t)h\left(g(t)\right)^{-1} = e^{tX}h\left(e^{tX}\right)^{-1} = e^{tX}he^{-tX},
\end{equation*}
  by properties of the exponential matrix (and inherently the exponential function). Furthermore, we can differentiate the exponential matrix with respect to $t$ such that
  \begin{equation*}
\frac{\partial}{\partial t} e^{tX} = Xe^{tX},
\end{equation*}
  and if we evaluate at $t=0$ we have
  \begin{equation*}
\frac{\partial}{\partial t} e^{tX} \Bigg\rvert_{t=0} = Xe^{(0)X} = X.
\end{equation*}
  Thus,
  \begin{align*}
\frac{\partial}{\partial t} g(t)h\left(g(t)\right)^{-1} & = \frac{\partial}{\partial t} e^{tX}he^{-tX}\Bigg\rvert_{t=0} \\
&= \left(Xe^{tX}he^{-tX} - e^{tX}hXe^{tX}\right)\Bigg\rvert_{t=0} \\
&= Xh - hX \\
&= 0~~~???.
\end{align*}
  Therefore, the derivative of the conjugate map is
  \begin{equation*}
\frac{\partial \psi}{\partial h} + \frac{\partial \psi}{\partial g} = gXg^{-1} + 0 = gXg^{-1}.
\end{equation*}

I know the derivative is the adjoint representation, Adj$(g)X = gXg^{-1}$, but I am unsure how to show that $Xh - hX = 0$, since matrix multiplication is not commutative. Am I mistaken to think $h$ is a matrix, when in fact it is a scalar?

A:

Quick reply: The mapping $\psi(g,h) = ghg^{-1}$ can be written as
$\psi(g,h) = C(g, h, i(g))$ where $C(A, B, C) = A B C$ is the composition tri-linear operator and $\imath(g) = g^{-1}$ is the inversion of the matrix. $C$ is smooth as (bounded) linear, and $\imath$ is smooth as we are working in a banach algebra. Now
$$\mathrm{D} C(A, B, C) \cdot (K, L, M) = KBC + ALC + ABM$$
$$\mathrm{D}\imath (g) \cdot h = -g^{-1} h g^{-1}$$
By composition rules if $g$ is invertible then $\psi$ is differentiable and,
$$\mathrm{D}\psi (g,h) \cdot (k,l) 
= \mathrm{D}C(g, h, g^{-1}) \cdot (k, l, \mathrm{D}i(g)\cdot k) = \ldots$$

