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 $\Bbb{R}^2$.


  • 3
    $\begingroup$ What about this as an embedding of $(\Bbb{R}^2,+)$ into $GL(4;\Bbb{C})$?: $$\left\{ \left(\begin{array}{CCCC} 1 & a & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & b \\ 0 & 0 & 0 & 1 \end{array}\right) : (a,b) \in \Bbb{R}^2 \right\}$$ $\endgroup$ – Hugh Denoncourt Aug 4 '12 at 8:43
  • 3
    $\begingroup$ 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. $\endgroup$ – t.b. Aug 4 '12 at 10:13
  • $\begingroup$ @t.b. We identify the torus with $\left(\begin{array}{cc} e^{i\theta}&0\\0& e^{i\theta} \end{array}\right)$? $\endgroup$ – user38268 Aug 4 '12 at 11:03
  • $\begingroup$ Yes, this is one possibility (but you should take two different $\theta$'s) $\endgroup$ – t.b. Aug 4 '12 at 11:18
  • $\begingroup$ @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. $\endgroup$ – user38268 Aug 4 '12 at 11:20

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 $GL(n+1,\mathbb C)$.

  • $\begingroup$ Thanks for your answer. As I suspected my guess would fail. $\endgroup$ – user38268 Aug 4 '12 at 8:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy