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I would like to do Tao's exercise 6 (i) but before I can even attempt it I need to be clear about his terminology.

Exercise 6

Show that the Lie algebra $gl_n(\mathbb{C})$ of the general linear group $GL_n(\mathbb{C})$ can be identified with the space $M_n(\mathbb{C})$ of $n \times n$ complex matrices, with the Lie bracket $[A,B] := AB - BA$.

When he writes "identified", does he just mean there is an isomorphism? So the exercise would be to find an explicit isomorphism from $M_n(\mathbb{C})$ to $gl_n(\mathbb{C})$? Is that correct?

Many thanks for your help.

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Yes. Depending on how much you know, the proof of this can be quite long. –  Matt Sep 19 '11 at 14:25
    
I know nothing. I started to read about Lie groups yesterday. I read that the exponential map maps elements in the Lie algebra to elements in the Lie group. I wonder if that is useful. Probably not because the exponential map is not surjective, I suppose.... –  Rudy the Reindeer Sep 19 '11 at 14:38
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Well, for one thing, a Lie algebra is an algebra -- which means that it is also a vector space over the underlying field $\mathbb{C}$. So you might first show that there is an isomorphism respecting the vector space strucure. (Very easy.) Being a Lie algebra, it also has a binary operation, the Lie bracket. In this case you want to argue that the commutator on $M_n(\mathbb{C})$, under the natural vector space isomorphism, actually obeys the properties you want the Lie bracket to obey. –  Craig Sep 19 '11 at 15:14

3 Answers 3

up vote 4 down vote accepted

The group $G:=\text{GL}_n(\mathbb C)$ is an open subset of $M_n(\mathbb C)$. A vector field on $G$ is a map $$ X:G\to M_n(\mathbb C). $$ The left action of $x\in G$ on such vector field is given by $$ (xX)(y):=xX(x^{-1}y). $$ Thus, $X$ is invariant under this action if and only if $X(x)=xX(1)$ for all $x$ in $G$. Such a vector field is said to be left invariant. In particular, for each $a\in M_n(\mathbb C)$ there is a unique left invariant vector field $\widetilde a$ such that $\widetilde a(1)=a$. We have $\widetilde a(x)=xa$ for all $x$ in $G$, and $\widetilde a$ is $C^\infty$. We must check $$ \left[\ \widetilde a\ ,\ \widetilde b\ \right]=\widetilde{[a,b]} $$ for all $a,b\in M_n(\mathbb C)$.

Let $e_{ij}\in M_n(\mathbb C)$ be the matrix with a one in the $(i,j)$ position and zeros elsewhere. It suffices to verify that the above display holds for $a=e_{ij}$ and $b=e_{pq}$.

We have $$ X_{ij}(x):=\widetilde{e_{ij}}(x)=x\ e_{ij}=\sum_{p,q}\ x_{pq}\ e_{pq}\ e_{ij}=\sum_p\ x_{pi}\ e_{pj}. $$ If we write $$ X_{ij}=\sum_k\ x_{ki}\ \ \frac{\partial}{\partial x_{kj}}\quad,\quad X_{pq}=\sum_r\ x_{rp}\ \ \frac{\partial}{\partial x_{rq}}\quad, $$ then the Lie bracket $[X_{ij},X_{pq}]$ is just the commutator of the differential operators $X_{ij}$ and $X_{pq}$, and we get $$ [\widetilde e_{ij},\widetilde e_{pq}]=[X_{ij},X_{pq}]=\delta_{jp}\ X_{iq}-\delta_{qi}\ X_{pj}=\widetilde{[e_{ij},e_{pq}]}, $$ as was to be shown.

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Note that the above proof uses only the very definition of the Lie algebra of a Lie group. –  Pierre-Yves Gaillard Sep 19 '11 at 18:25

My answer is a bit lacking, I omit calculations and take certain facts for granted.

Whenever a manifold $M$ is an open subset of some vector space $V$, there is, for every $m\in M$ a simple identification $T_mM\approx V$, which comes from the inclusion $\varphi :M\to V$ which is a smooth chart. Since $\mathrm{GL}_n(\mathbb{C})$ is an open subset of $\mathrm{M}_n(\mathbb{C})$, this gives you a natural identificaiton $$T_{Id}\mathrm{GL}_n(\mathbb{C})\approx \mathrm{M}_n(\mathbb{C}).$$ Explicitely, you map $A\in\mathrm{M}_n(\mathbb{C})$ to the class of paths $\lbrace t\in(-\epsilon,+\epsilon)\mapsto Id+tA\rbrace$ ($\epsilon$ depends on $A$, and is small enough so that $Id+tA$ is invertible). To compute the bracket you can use $[X,Y]_{Lie}=[X,Y]_{Vector~Fields}$ where $X,Y$ are in the Lie algebra and are identified (in the second bracket) to their left invariant extensions. The right hand side is explicitely computable if you know that the flow of $X\in\mathfrak{g}\approx\Gamma^{left~inv}(G)$ has $\mathrm{Fl}^t_X(g)=g\times\mathrm{Fl}^t_X(Id)$ and in the case of $\mathrm{GL}_n(\mathbb{C})$ you have $\mathrm{Fl}^t_X(Id)=\exp(tX)$ where $\exp$ is the usual matrix exponential. Then you only have to differentiate a path of matrices.

EDIT It's probably easier to go back to the definition of the Lie Bracket as $[X,Y]_{Lie}=ad(X,Y)=d_{Id}Ad(X)(Y)$, since for any $g\in\mathrm{GL}_n(\mathbb{C})$, $Ad(g):\mathrm{M}_n(\mathbb{C})\to\mathrm{M}_n(\mathbb{C})$ is conjugation by $g$.

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"Identify" usually means "exhibit a nice isomorphism." Making that rigorous in general is a bit tricky, but you know it when you see it.

In this case, you could make it rigorous in the following way. Note that both $\mathfrak{gl}_n(\mathbb{C})$ and $M_n(\mathbb{C})$ have a natural action of $\mathrm{GL}_n(\mathbb{C})$ (by the adjoint action and conjugation respectively), so a "nice" (or "natural" or "canonical") isomorphism should respect this action.

But Tao is almost certainly using "identify" informally. He just means find an isomorphism, with the subtext that you shouldn't need to make any "arbitrary choices" along the way.

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