Are complex semisimple Lie groups matrix groups?

Question

Actually, my question is a bit more specific: Does every complex semisimple Lie group $G$ admit a faithful finite-dimensional holomorphic representation? [As remarked by Brian Conrad, this is enough to prove that $G$ is a matrix group (at least when it's connected) because $G$ can be made into an (affine) algebraic group over $\mathbb{C}$ in unique way which is compatible with its complex Lie group structure, and under which every finite-dimensional holomorphic representation is algebraic. Furthermore, one can show that the image of a faithful representation would then be closed.]

Of course the analogous question for real semisimple Lie groups has a negative answer -- "holomorphic" having been replaced by "continuous", "smooth" or "real analytic" -- with the canonical counterexample being a nontrivial cover of $\mathrm{SL}(2,\mathbb{R})$.

For a connected complex semisimple Lie group $G$ I believe the answer is "YES." The idea is to piggy back off a 'sufficiently large' representation of a compact real form $G_\mathbb{R}$; here by "compact real form" I'm referring specifically to the definition which allows us to uniquely extend continuous finite-dimensional representations of $G_\mathbb{R}$ to holomorphic representations of $G$. I know (e.g. from the proof of Theorem 27.1 in D. Bump's Lie Groups) that such a definition is possible if we require $G$ to be connected (and I'd like to know if it's possible in general).

The details of the argument for connected $G$ are as follows. Consider the adjoint representation $\mathrm{Ad} \colon G \to \mathrm{GL}(\mathfrak{g})$. Since $G$ is semisimple, $\mathrm{Ad}$ has discrete kernel $K$. Consider next the restriction of $\mathrm{Ad}$ to $G_\mathbb{R}$. Observe that the kernel of this map is also $K$, for otherwise its holomorphic extension is different from the adjoint representation of $G$. Thus $K$ is finite, being a discrete, closed subset of a compact space. So by the Peter-Weyl theorem, we can find a representation $\pi_0$ of $G_\mathbb{R}$ that is nonzero on $K$. Extend $\pi_0$ to a holomorphic representation $\pi$ of $G$ and put $\rho = \pi \oplus \mathrm{Ad}$. Notice that $\rho$ is a holomorphic representation of $G$ with kernel $\ker\pi \cap K = 0$, which is what we were after.

What can we say if $G$ is disconnected?

Answer 1

As requested by Faisal, I am posting as an answer the observation that if $G$ has more components than the size of the complex numbers then G has no faithful finite-dimensional irreducible representation over the complexes, for cardinality reasons.

To be honest I thought Faisal's response to this would be "what happens if $G$ has only countably infinitely many components"? That is then a different question. Does every countable group have a faithful finite-dimensional complex representation? If this is well-known I'd be happy to hear the answer. If it isn't I'd be happy if someone else asked this as a separate question. If it's true that any countable group has a faithful representation then one might ask about complex Lie groups with countably infinitely many components.

Answer 2

In the spirit of the title of the question, the argument doesn't quite prove that $G$ is a matrix group, since more input is needed to prove that the faithful representation has closed image which is moreover algebraic. (A literature reference for finiteness of the center of a connected complex-analytic Lie group with semisimple Lie algebra -- I assume this is your definition of "semisimple" for the analytic group -- is Ch. XVII, Thm. 2.1 in Hochschild's "Structure of Lie Groups".)

Anyway, in fact the semisimplicity of the Lie algebra does imply algebraicity and closedness of the image. This follows by a graph argument, using input from the algebraic theory (see Cor. 7.9 in Borel's book). But the really nice part is that much more is true:

Theorem: The functor $G \rightsquigarrow G^{\rm{an}} = G(\mathbf{C})$ from linear algebraic groups over $\mathbf{C}$ to complex-analytic Lie groups restricts to an equivalence between the full subcategory of connected semisimple $\mathbf{C}$-groups and the full subcategory of connected complex Lie groups with semisimple Lie algebra. It also restricts to an equivalence between the full subcategory of connected reductive $\mathbf{C}$-groups and the full subcategory of connected complex Lie groups whose Lie algebra is reductive and whose center has identity component a power of $\mathbf{C}^{\times}$.

This implies much more than an affirmative answer to the initial question, since it implies that not only the objects but even the morphisms are all "algebraic" in a unique way.

Answer 3

The double cover of $GL(n,\mathbb{C})$ is not a matrix group.