Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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
add comment

1 Answer

up vote 3 down vote accepted

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 $g\in GL_n(\mathbb{R})$, $\det(g)\neq 0$).

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.