Posted by: Steven Sam | July 10, 2009

## A Lie group which isn’t a matrix group

In this post, I want to give an example of a Lie group which is not isomorphic to a subgroup of a matrix group. This contrasts with the algebraic picture, where every affine algebraic group can be realized as a subgroup of a matrix group. I’ll just sketch the details.

The group we’ll use is the universal cover G of $\mathbf{SL}_2(\mathbf{R})$. One can realize $\mathbf{SL}_2(\mathbf{R})$ as a fiber bundle with fiber $\mathbf{R}$ over $\mathbf{R}^2 \setminus \{(0,0)\}$ via the map which sends a matrix to its first row, so its fundamental group is $\mathbf{Z}$. Similarly, we can use this map to show that $\mathbf{SL}_2(\mathbf{C})$ is a fiber bundle with fiber $\mathbf{C}$ over $\mathbf{C}^2 \setminus \{(0,0)\}$ and hence is simply connected.

Now let $r \colon G \to \mathbf{GL}_n(\mathbf{R})$ be any smooth map. It induces a map on Lie algebras $r_* \colon \mathfrak{sl}_2(\mathbf{R}) \to \mathfrak{gl}_n(\mathbf{R})$. We can use this to define a map $f \colon \mathfrak{sl}_2(\mathbf{C}) \to \mathfrak{gl}_n(\mathbf{C})$ by $f(A+iB) = r_*(A) + ir_*(B)$. It’s straightforward to check that f preserves the Lie bracket. Since $\mathbf{SL}_2(\mathbf{C})$ is simply connected, f is the derivative of some smooth map $F \colon \mathbf{SL}_2(\mathbf{C}) \to \mathbf{GL}_n(\mathbf{C})$.

To show that r cannot be injective, we will show that the composition $G \xrightarrow{r} \mathbf{GL}_n(\mathbf{R}) \to \mathbf{GL}_n(\mathbf{C})$ is not injective, where the second map is the usual inclusion. I claim that this composition is the same map as the composition $G \xrightarrow{\pi} \mathbf{SL}_2(\mathbf{R}) \to \mathbf{SL}_2(\mathbf{C}) \xrightarrow{F} \mathbf{GL}_n(\mathbf{C})$, where $\pi$ is the covering space map. By construction, these two compositions induce the same differentials on their Lie algebras. Now using the fact that both G and $\mathbf{GL}_n(\mathbf{C})$ are connected, this implies that the map on the level of Lie groups must be equal as well. But $\pi$ is not injective: we saw above that the fundamental group for $\mathbf{SL}_2(\mathbf{R})$ is nontrivial, so we are done.

## Responses

1. I really like this result. I don’t remember, though: is there a similar example over C?

So the interpretation is something like this:

Understanding mappings into matrix groups is precisely the same as understanding the category of finite-dimensional representations of a Lie group. So the subgroups of matrix groups are precisely the ones that can be distinguished by their finite-dimensional representation theory. For example, the algebraic groups can.

What this means, then, is that for understanding real Lie groups, we need to work with a larger category of representations. Do you have a favorite candidate? It shouldn’t be too large: it would be great if the representations have some analytic control.

2. Theo:

Your first question: by similar example, do you mean does there exist complex Lie groups which aren’t matrix groups? The answer is no if the group is semisimple since all such groups are algebraic. It seemed like a hard thing to search for, but there are examples of complex Lie groups which aren’t algebraic: certain 2-dimensional tori which are extensions of an elliptic curve by another elliptic curve. I found out about this from this paper:

Frans Oort and Yuri Zarhin, Endomorphism algebras of complex tori, Math. Ann. 303 (1995), no. 1, 11–29.

I guess this still doesn’t give an example of something which isn’t linear, oh well.

As for your second question: I don’t really know anything about Lie groups. Maybe taking all unitary representations is good enough? If that’s wrong, I’d guess taking all representations would be enough. I’m still tempted to think this is wrong since Tannaka-Krein duality is stated for compact groups. I’d like to know the answer if you figure it out though.

3. The existence of nonalgebraic complex tori has been known for quite a long time, and is basically a consequence of a theorem of Lefschetz concerning ampleness of line bundles. You can find a discussion near the beginning of Mumford’s Abelian Varieties, together with an example of a non-compact complex Lie group with two algebraizations, only one of which is linear.

There are some silly non-matrix Lie groups, like permutations of an infinite set. I think a slightly more interesting example (if it is in fact an example) is the quotient of 3×3 unipotent upper triangular matrices by a copy of the integers living in the center.

4. Hi Scott:

Thanks for the reference. I will take a look at that.

Your example of 3×3 unipotent upper triangular matrices modulo a copy of the integers in the center looks correct: if a representation is zero on this discrete subgroup, then it is zero on the whole center. Thanks!

5. I have nothing to contribute; I just wanted to say that this is a very nice post, short and sweet.