Example of a semisimple Lie group which is not an algebraic group?

Question

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?

Answer 1

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.

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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.

Answer 2

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 $G$.