10
$\begingroup$

I have an assignment problem that is coming from Brian Hall's book Lie Groups, Lie Algebras and Representations: An Elementary Introduction.

Suppose $G \subseteq GL(n_1;\Bbb{C})$ and $H \subset GL(n_2;\Bbb{C})$ are matrix Lie group and that $\Phi:G \to H$ is a Lie group homomorphism. Is the image of $G$ under $\Phi$ a matrix Lie group?

The definition of a matrix lie group that I have is a closed subgroup of $GL(n;\Bbb{C})$ for some $n$. I do not know about things like differentiable manifolds and the like. By a Lie group homomorphism, we mean a continuous group homomorphism from $G$ to $H$ (continuity is with respect to the usual topology coming from the euclidean metric).

Now an example I have in mind is the Heisenberg group $G$ that consists of all matrices of the form

$$\left\{\left(\begin{array}{ccc}1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{array}\right) : a,b,c\in \Bbb{R} \right\}.$$

I can get a surjective group homomorphism (which I believe is also continuous) from $G$ to $\Bbb{R}^2$ under addition simply by sending

$$\left(\begin{array}{ccc}1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{array}\right) \to (a,c) \in \Bbb{R}^2.$$

This would be a counterexample to the problem above if only I were to know how to show that $\Bbb{R}^2$ does not embed as a matrix lie group in $GL(n;\Bbb{C})$ for any $n \in \Bbb{N}$. The difficulty in showing this is it is not obvious if such an embedding exists or not.

For example $\Bbb{R}$ under addition surprisingly is a matrix lie group. I have a copy of $\Bbb{R}$ sitting inside of $GL(2;\Bbb{C})$, namely as the set of all matrices

$$\left\{ \left(\begin{array}{CC} 1 & a \\ 0 & 1 \end{array}\right) : a\in \Bbb{R} \right\}.$$

How do I know if $(\Bbb{R}^2,+)$ can or cannot be given the structure of a matrix Lie group? If only I were to know representation theory, I believe this is asking if there exists a finite dimensional complex faithful representation of $\Bbb{R}^2$.

Thanks.

  • 3
    What about this as an embedding of $(\Bbb{R}^2,+)$ into $GL(4;\Bbb{C})$?: $$\left\{ \left(\begin{array}{CCCC} 1 & a & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & b \\ 0 & 0 & 0 & 1 \end{array}\right) : (a,b) \in \Bbb{R}^2 \right\}$$2012-08-04
  • 3
    To get an example of a continuous homomorphism whose image isn't a closed subgroup, identify the torus $T = \mathbb{R}^2/\mathbb{Z}^2$ with a Matrix Lie group and consider the quotient map $\mathbb{R}^2 \to \mathbb{R}^2/\mathbb{Z}^2$. A line $L$ in $\mathbb{R}^2$ with irrational slope is a closed subgroup of $\mathbb{R}^2$ but the image of $L$ under the quotient map will be dense in $T$ (it's a line that wraps around the torus without closing itself up). Here's [an animation](http://www.youtube.com/watch?v=YfKxzKM-MNc).2012-08-04
  • 0
    @t.b. We identify the torus with $\left(\begin{array}{cc} e^{i\theta}&0\\0& e^{i\theta} \end{array}\right)$?2012-08-04
  • 0
    Yes, this is one possibility (but you should take *two* different $\theta$'s)2012-08-04
  • 0
    @t.b. So from what I get, such a line will be dense in $T$ but can never be the whole torus so is not a closed subgroup of it.2012-08-04
  • 0
    Yes. Exactly. Do you see why what you say is true? How do you prove that?2012-08-04
  • 0
    @t.b. Just look at the unit square. Now draw the line $y = \sqrt{2}x$. Then it touches the right side of the square when $x = 1$, $y = \sqrt{2}$. Now because the left and right sides of the square have been identified, the line comes out of the left edge at $(0,\sqrt{2})$ and continues on till it hits $((1-\sqrt{2}/\sqrt{2},1)$. Since the top and bottom edges have been identified, the line comes out of the bottom to hit the right edge at some irrational point. I don't know how to prove this formally, but the idea is that because the slope is irrational you can never ever get back to where2012-08-04
  • 0
    @t.b. you started. I believe it is dense in the whole of the unit square (eventually), and it does not fill up all of it because the rationals on any edge are never hit by the line.2012-08-04
  • 0
    @t.b. In short the unit square is filled up with lines of the form $y = \sqrt{2}x + a$ where $a$ is some irrational.2012-08-04
  • 0
    Well, it's not very hard to prove that if any line closes up the slope must be rational: if $(a,b) + t(1,\lambda) = (a,b) + (p,q)$ with $(p,q) \in \mathbb{Z}^2$ and $t \neq 0$ then $\lambda \in \mathbb{Q}$. For density see [irrational rotation](http://en.wikipedia.org/wiki/Irrational_rotation) and [this thread](http://math.stackexchange.com/q/70201).2012-08-04
  • 0
    @t.b. What I meant is that I would like to prove that the line only hits irrationals on the edges and not rationals.2012-08-04
  • 0
    This can be seen with essentially the same argument. Assume that the image of the line $t\mapsto t(1,\lambda)$ passes through the image of $(p/q,0)$. Select appropriate representatives and deduce that $\lambda$ is rational.2012-08-04

1 Answers 1

16

In fact $(\mathbb R^n,+)$ is a matrix Lie group for all $n$. Note that the map $$\begin{pmatrix} x_1\\ \vdots\\ x_n \end{pmatrix}\mapsto \begin{pmatrix} 1 & 0 & \cdots & x_1\\ & \ddots & & \vdots\\ 0 & \cdots & 1 & x_n\\ 0 & \cdots & 0 & 1 \end{pmatrix}$$ which sends a vector $x$ to the identity matrix with the zeroes of the last column replaced by coordinates of $x$, is an injective homomorphism from $(\mathbb R^n,+)$ to $GL(n+1,\mathbb C)$.

  • 0
    Thanks for your answer. As I suspected my guess would fail.2012-08-04