What is the tangent space of a matrix subgroup at $A$

Question

I know that the Lie algebra $\mathfrak{g}$ of a matrix subgroup $G$ is its tangent space at $I$, but what would its tangent space be at $A \in G$? If I fix $B_\epsilon = \{A \in M_n \mid \|A\|_2 \le \epsilon \}$ then $\exp$ maps $\mathfrak{g} \cap B_\epsilon$ to a neighborhood of $I$ in $G$. Since $\exp$ maps from the tangent space to the space itself. I think in order for $\exp$ to map to a neighborhood of $A$, the source of my map should be something like $\mathfrak{g}A$, or should it be $A\mathfrak{g}$? It shouldn't matter because $\mathfrak{g}$ gets mapped to a neighborhood of $I$. But is there a more rigorous explanation that I can look to? Thanks in advance!

Answer 1

I am not sure what you are after exactly. The tangent space to $G$ at $A$ is a linear subspace of $M_n$ that can be defined as the space of all $c'(0)$ whether $c$ is a smooth curve from an open interval containing $0$ to $M_n$ which has values in $G$ and satisfies $c(0)=A$. This space coincides with $A\mathfrak g$ and with $\mathfrak gA$. This follows since left and right multiplication by $A$ are diffeomorphisms $G\to G$ that map $I$ to $A$, so their derivatives are linear isomorphisms from the tangent space to $G$ at $I$ to the tangent space to $G$ at $A$. Since left and right multiplication by $A$ are linear maps $M_n\to M_n$, also their derivatives are given by left respectively right multiplication by $A$.

If you are looking for a "replacement" for the exponential map, you don't even have to use the tangent space at $A$. Both $X\mapsto A\exp(X)$ and $X\mapsto \exp(X)A$ restrict to a diffeomorphism from some open neighborhood of $0$ in $\mathfrak g$ to an open neighborhood of $A$ in $G$.