# Adjoint action smooth (in Tapp's book)?

I'm reading "Matrix Groups for Undergraduates" by Tapp with a student. A "matrix group" means a subgroup $G$ of $GL_n(\mathbb R)$ which is (relatively) closed-- so if $(A_n)\subseteq G$ with $A_n\rightarrow A$, and $A$ is invertible, then $A\in G$.

In the book, Manifolds are treated in an adhoc way. Given $X\subseteq\mathbb R^m$ a map $f:X\rightarrow \mathbb R^n$ is "smooth" if for each $x\in X$ there is an open set $U\subseteq\mathbb R^m$ containing $x$, and a smooth map $g:U\rightarrow\mathbb R^m$ which agrees with $f$ on $U\cap X$. Then a manifold is defined in the obvious way.

We then have the adjoint action of a lie group $G$ on its lie algebra $\mathfrak g$. If $\mathfrak g$ is $d$ dimensional, then by taking a basis $A_1,\cdots,A_d$ of $\mathfrak g$, we can regard the adjoint action as a homomorphism $Ad:G\rightarrow GL_d(\mathbb R)$. So for each $g\in G$ there is a matrix $(X_{ij}(g))$ with $$g A_j g^{-1} =\sum_i X_{ij}(g) A_i.$$

How do we show that $Ad$ is smooth?

If we follow the definition from the book then we'd need to show that $G\rightarrow \mathbb R^{d\times d}; g \mapsto (X_{ij}(g))$ is smooth. So for each $g\in G$ I need an open set $U$ in $GL_n(\mathbb R)\subseteq\mathbb R^{n\times n}$ and a smooth function $f:U\rightarrow\mathbb R^{d\times d}$ such that $f(g) = (X_{ij}(g))$ for $g\in G\cap U$. This seems intractable...?

(If one has more Manifold theory, then this becomes sort of obvious, as $Ad$ is just the derivative of the conjugation action, which is smooth, and the derivative of a smooth map is smooth. But I want to stick to what the book has told us...)

Edit: Maybe I can actually argue as follows. The map $(A,B)\mapsto \operatorname{Tr}(AB)$ is an inner-product on $\mathbb M_n(\mathbb R)$; so I can find $B_1,\cdots,B_n\in\mathbb M_n(\mathbb R)$ with $\operatorname{Tr}(A_iB_j)=\delta_{i,j}$. Thus $$X_{ij}(g) = \operatorname{Tr}(g A_j g^{-1} B_i).$$ Thus I could take as my map $$f(h) = \big( \operatorname{Tr}(h A_j h^{-1} B_i) \big)_{i,j}.$$ This is now a composition of matrix inverse and multiplication, and trace, all smooth maps, hence $f$ is smooth.

Does this seem reasonable? Is this the easiest approach?

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I'm curious: How are you defining $\mathfrak g$? Matrices which are derivatives of paths "at the identity"? – Dylan Moreland Apr 12 '12 at 13:47
@Dylan: Yep, that's right. – Matthew Daws Apr 12 '12 at 13:56

The map $Ad:G\rightarrow GL(\mathfrak{g})$ extends nicely to a map $Ad:GL_n(\mathbb{R})\rightarrow Gl(\mathfrak{gl}_n(\mathbb{R}))$. In fact, since $GL_n(\mathbb{R})$ is an open subset of $\mathbb{R}^{n^2}$, we can take $U = GL_n(\mathbb{R})$ so that, by definition, if this extended map is smooth, then so is the original map.
This reduces the problem to checking smoothness in one concrete case. Here, one can the "standard" basis $E_{ij}$ of $\mathfrak{gl}_n(\mathbb{R})$. Using this basis, a direct computation shows that for each $i$ and $j$, each component of $gE_{ij} g^{-1}$ is a polynomial in the entries of $g$ and $g^{-1}$, and Cramer's rule shows the entries of $g^{-1}$ are given as rational functions of the entries of $g$ where the denominator is $\det(g)$.
So, overall, each component of $gE_{ij}g^{-1}$ is given as a rational function of the entries of $g$ with denominator $\det(g)$, so is smooth. (One might be worried about dividing by $0 = \det(g)$, but for $g\in GL_n(\mathbb{R})$, $\det(g)\neq 0$).
We identify $\mathfrak{gl}_n(\mathbb R)$ with $\mathbb M_n(\mathbb R)$ and so identify $\mathfrak{g}$ with a subspace of $\mathfrak{gl}_n(\mathbb R)$. So in the abstract setup we start with a map $f:X\rightarrow\mathbb R^d$, and then extend this to map $g:U\rightarrow\mathbb R^D$ (where $D>d$) such that $P g(x) = f(x)$ for $x\in X$, with $P:\mathbb R^D\rightarrow\mathbb R^d$ the natural projection. Yes, I guess that works-- is that what you meant by "extends nicely"...? – Matthew Daws Apr 12 '12 at 15:29
@Matthew: By "extends nicely", I just meant that it extends to a map that's much easier to understand. Sorry for the lack of clarity. I guess I'm also using the fact that "a function $f:X\rightarrow \mathbb{R}^d$ is smooth iff its composition with every smooth function is smooth" (Your $P$ above), to argue that the range doesn't affect smoothness. – Jason DeVito Apr 12 '12 at 15:49