Derivative of conjugate map The Wikipedia page for the adjoint representation has the following:

If $G$ is an immersed Lie subgroup of the general linear group $\mathrm{GL}_n(\mathbb{C})$ (called immersely linear Lie group), then the Lie algebra $\mathfrak{g}$ consists of matrices and the exponential map is the matrix exponential $\exp(X) = e^X$ for matrices $X$ with small operator norms.
Thus, for $g$ in $G$ and small $X$ in $\mathfrak{g}$, taking the derivative of $\Psi_g( \exp(tX) ) = g e^{t X} g^{-1}$ at $t = 0$, one gets
$$
  \mathrm{Ad}_g(X) = g X g^{-1}
$$
where on the right we have the products of matrices.
If $G \subset \mathrm{GL}_n(\mathbb{C})$ is a closed subgroup (that is, $G$ is a matrix Lie group), then this formula is valid for all $g$ in $G$ and all $X$ in $\mathfrak{g}$.
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What I don’t understand is how taking the derivative of $\Psi_g$ with respect to $t$ gives $\mathrm{Ad}_g$. I understand that $\mathrm{Ad}_g$ is the pushforward of $\Psi_g$, but don’t understand how the pushforward is related to the $t$ derivative in this case.
 A: I concede that it is a lot of circonlocutions in order to deliver a quite trivial result, but the idea is as follows : $g\in G$ lives in the Lie group, whereas $X\in\mathfrak{g}$ is an element of the associated Lie algebra. The great result of Lie theory constitutes the correspondence between a Lie group and its algebra through the exponential map, such that $A=e^{tX}\in G$ and $X = \left.\frac{\mathrm{d}A}{\mathrm{d}t}\right|_{t=0}$, i.e. $G\xrightarrow[]{\exp}\mathfrak{g}$ and $\mathfrak{g}\xrightarrow[]{\mathrm{d}/\mathrm{d}t}G$ in other words.
In your case, the conjugation map $\Psi_g$ is defined for the Lie group, but you want to "extend" it to its algebra, that is why $X$ is brought into the Lie group by exponentiation first, then $\Psi_g$ is applied to $e^{tX}\in G$ and, finally, differentiation permits to switch back to the Lie algebra. One has thus :
$$
\mathrm{Ad}_g(X) := \left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=0} \Psi_g(e^{tX}) = \left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=0} ge^{tX}g^{-1} = \left.ge^{tX}Xg^{-1}\right|_{t=0} = gXg^{-1}
$$
