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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?

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    $\begingroup$ Why do you think Xh-hX=0 it is not true $\endgroup$
    – Elad
    Dec 10, 2017 at 9:36
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    $\begingroup$ I am confuse by your statement “I know the derivative is the adjoint representation, $\operatorname{Adj}(X)=gXg^{-1}$.” This cannot possibly be the derivative of your map. Perhaps that you ment to find the derivative of the map $h\mapsto ghg^{-1}$, for a fixed $g$. $\endgroup$ Dec 10, 2017 at 9:38
  • $\begingroup$ @Elad In general, unless X and h are both diagonal matrices, or equivalent, they wouldn't commute. $\endgroup$
    – jj8989
    Dec 10, 2017 at 10:37
  • $\begingroup$ @JoséCarlosSantos I believe my notation is off. I should have had $Adj(g)X$, not $Adj(X)$. I have edited. $\endgroup$
    – jj8989
    Dec 10, 2017 at 10:38

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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, $$\mathrm{D}\psi (g,h) \cdot (k,l) = \mathrm{D}C(g, h, g^{-1}) \cdot (k, l, \mathrm{D}i(g)\cdot k) = \ldots$$

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  • $\begingroup$ This seems like a more eloquent approach, but I do not follow in its entirety. Specifically, what gives us the ability to rewrite what we have using the composition tri-linear operator? $\endgroup$
    – jj8989
    Dec 10, 2017 at 10:51
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    $\begingroup$ Since you are working on a matrix group, the multiplication is the composition of matrices. You can ask the same question on a lie group for the function $\psi : G \times G \times G \longrightarrow G : (a,b,c) \longmapsto \psi(a,b,c) = abc$. $\endgroup$
    – yaqa
    Dec 10, 2017 at 22:55

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