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Of course the analogous question for real semisimple Lie groups has a negative answer -- "holomorphic" having been replaced by "continuous", "smooth" or "real analytic" -- with the canonical counterexample being a nontrivial cover of $\mathrm{SL}(2,\mathbb{R})$.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



Are complex semisimple Lie groups matrix groups?
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.


