3
$\begingroup$

I wonder if there are any examples of semisimple Lie groups that couldn't correspond to any algebraic group, or actually every semisimple Lie group is algebraic?

$\endgroup$
1

2 Answers 2

7
$\begingroup$

The smallest example is the following. $SL_2(\mathbb{R})$ is a semisimple Lie group which has the same representation theory as its Lie algebra $\mathfrak{sl}_2(\mathbb{R})$. On the other hand, its maximal compact is $SO(2)$ so its fundamental group is $\mathbb{Z}$, and so it has $n$-fold covers for any $n$; each of these groups has the same representation theory as $\mathfrak{sl}_2(\mathbb{R})$ also, which means all of their finite-dimensional representations factor through the quotient to $SL_2(\mathbb{R})$.

This means none of these groups admit faithful finite-dimensional representations, so none of them can be the real points of a linear algebraic group. The $2$-fold cover is a metaplectic group as David Loeffler links to in the comments.

In the positive direction, for compact Lie groups see this MO answer.


Edit, 9/1/20: This is maybe a little terse so I'll be more explicit what facts I'm using about Lie groups and coverings and so forth. Probably I don't need the full strength of some of what I say below but it's useful context and stuff worth knowing anyway.

  • Every finite-dimensional Lie algebra $\mathfrak{g}$ is the Lie algebra of a unique (up to isomorphism) simply connected Lie group $\widetilde{G}$. Every connected Lie group $G$ with Lie algebra $\mathfrak{g}$ is covered by this simply connected group.
  • If $G$ and $H$ are two connected Lie groups, then the differentiation map $\text{Hom}(G, H) \to \text{Hom}(\mathfrak{g}, \mathfrak{h})$ is injective, and it is bijective if $G$ is simply connected.
  • Setting $H = GL_n(\mathbb{R})$ or $GL_n(\mathbb{C})$, it follows as a corollary of the previous point that a Lie algebra $\mathfrak{g}$ and its simply connected Lie group $\widetilde{G}$ have the same finite-dimensional representation theory (real or complex), in the sense that the differentiation map $\text{Rep}_f(\widetilde{G}) \to \text{Rep}_f(\mathfrak{g})$ is an equivalence of categories.
  • Given a representation of $\mathfrak{g}$, the corresponding representation of $\widetilde{G}$ can be recovered using the exponential map $\exp : \mathfrak{g} \to \widetilde{G}$. The exponential map is functorial in the sense that if $f : G \to H$ is a homomorphism of Lie groups then the obvious diagram commutes; this means that to compute the action of $\widetilde{G}$ on an $n$-dimensional representation $\rho : \mathfrak{g} \to \mathfrak{gl}_n$ we can exponentiate the corresponding matrices $\rho(X)$ from $\mathfrak{gl}_n$ to $GL_n$.
  • $G = SL_2(\mathbb{R})$ is a semisimple Lie group with simple Lie algebra $\mathfrak{g} = \mathfrak{sl}_2(\mathbb{R})$. The finite-dimensional representations of $\mathfrak{sl}_2(\mathbb{R})$ are completely reducible, and the irreducible representations of $\mathfrak{sl}_2(\mathbb{R})$ are precisely the symmetric powers $S^n(\mathbb{R}^2)$ of the defining representation. A priori these representations need only exponentiate to representations of the universal cover $\widetilde{G} = \widetilde{SL_2(\mathbb{R})}$, but in fact all of them exponentiate to $SL_2(\mathbb{R})$ (since the defining representation exponentiates to $SL_2(\mathbb{R})$ and the others are obtained from it), so every irreducible representation of any nontrivial cover of $G$ factors through the covering map to $G$.
  • It follows that nontrivial covers of $G = SL_2(\mathbb{R})$ (which exist) have no faithful finite-dimensional representations.
$\endgroup$
2
  • $\begingroup$ Thanks a lot!But I am relatively unfamiliar with the mixture of covering space theory and lie/representation theory mentioned in your answer. Do you have any good reference that discusses this mixture? $\endgroup$
    – No One
    Sep 1, 2020 at 22:48
  • $\begingroup$ Not off the top of my head. I can edit the answer and sketch in some more detail what facts I'm using exactly. $\endgroup$ Sep 2, 2020 at 0:06
0
$\begingroup$

In plainer terms:

Let $p\colon \tilde G\to G=SL(2,\mathbb{R})$ be a cover map. Consider $\phi\colon \tilde G \to H$, a morphism of Lie groups, where $H$ is a complex Lie group. The map $d \phi \colon \tilde{\frak{g}} \to \frak{h}$ is a morphism of real Lie algebras. Extend it a morphism of complex Lie algebras $\tilde{\frak{g}}_{\mathbb{C}}=sl(2,\mathbb{C})\to \frak{h}$. Now, the group $SL(2,\mathbb{C})$ is simply connected, so there exists $\psi\colon SL(2,\mathbb{C})\to H$ with the given map between Lie algebras. We conclude $\phi = \psi \circ p$, that is, the morphism $\phi$ factors through $G= SL(2,\mathbb{R})$. As a conclusion, there does not exists an injective morphism of Lie groups $\tilde G\to GL(n, \mathbb{C})$ (or $GL(n, \mathbb{R})$).

So: $G\subset G_{\mathbb{C}}$ simply connected, $\tilde G\to G$ cover, then any map from $\tilde G$ to a complex Lie group factors through $G$.

$\endgroup$

Your Answer

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

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