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Of course the analogous question for real semisimple Lie groups has a negative answer -- "holomorphic" having been replaced by "continuous", "smooth" or "real analytic" -- with the canonical counterexample being a nontrivial cover of $\mathrm{SL}(2,\mathbb{R})$.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

  • $\begingroup$ If you believe the answer is yes for connected G then why can't you just induce up a faithful rep of the conn component of 1 to get the result for general G? $\endgroup$ Feb 26, 2010 at 11:19
  • $\begingroup$ How would you carry out the induction if G has infinitely many components? $\endgroup$
    – Faisal
    Feb 26, 2010 at 11:38
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    $\begingroup$ If G is allowed to have infinitely many components, then why not make G a discrete group with more components than the cardinality of the complex numbers! This clearly has no faithful finite-dimensional complex representations. $\endgroup$ Feb 26, 2010 at 14:26
  • $\begingroup$ Okay, I guess that was pretty obvious. If you want to post that example as an answer, Kevin, then I'd accept it. $\endgroup$
    – Faisal
    Feb 26, 2010 at 16:52
  • $\begingroup$ I believe that a complex algebraic variety cannot have infinitely many disconnected components. At least, I'm pretty sure a group with infinite $\pi_0$ cannot be algebraic. In the algebraic group case, the connected component of the identity is an algebraic normal subgroup, and the quotient is algebraic and discrete, and I'm very confident that infinite discrete groups are not algebraic. $\endgroup$ Feb 26, 2010 at 17:12

3 Answers 3


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.

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    $\begingroup$ No, not every countable group has a faithful finite-dimensional representation over a field of characteristic 0. Necessary and sufficient conditions for finitely generated groups are given in "Interlude B" of Dixon et al.'s Analytic Pro-P-Groups, page 175. An easy example of a countable "non-linear" group is the direct sum of PSL(2,p) over the primes p. The minimal degree of a faithful representation of a single one of the PSLs is approximately p/2, so there is no single dimension large enough for all of them. $\endgroup$ Feb 26, 2010 at 21:54
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    $\begingroup$ Very nice Jack. So in fact the situation is clear: if there are finitely many conn components then there's always a faithful representation, but if there are countably infinitely many then there may not be. $\endgroup$ Feb 27, 2010 at 7:39

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.

  • $\begingroup$ Thanks for the answer, Brian. Unfortunately it's not quite what I'm looking for: I was mainly wondering about what happens for disconnected groups. I'm also curious about the availability of a ``nice'' compact real form for such groups. $\endgroup$
    – Faisal
    Feb 26, 2010 at 10:53
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    $\begingroup$ Once you have the equivalence over C in the connected case, it comes for free in the case of finite component groups, since the "algebraicity" is obtained by translation from the identity. So everything I said above holds with connnectedness relaxed to "finite component group". And Hoshchild's book proves in any Lie group with finite component group that maximal compacts meet every connected component, every compact is contained in one, and they're all conjugate. Thus, you get the compact real form from the connected case as well. $\endgroup$
    – BCnrd
    Feb 26, 2010 at 11:24

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

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    $\begingroup$ $GL(n,\mathbb C)$ is not semi-simple. $\endgroup$
    – Emerton
    Feb 27, 2010 at 2:18
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    $\begingroup$ @Scott: $GL(n,\mathbb{C})$ is simply connected! $\endgroup$ May 7, 2011 at 20:25
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    $\begingroup$ Well, that's not quite right; $GL_n(\mathbb{C})$ is homotopy equivalent to $U(n)$, which has fundamental group $\mathbb{Z}$. It's true that $SL_n(\mathbb{C})$ is simply connected and maybe it's true that $GL_n(\mathbb{C})$ is algebraically simply connected or something? (Not that I know what this means.) $\endgroup$ Aug 3, 2015 at 20:04
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    $\begingroup$ The double cover is indeed a matrix group $\endgroup$ Oct 2, 2018 at 2:39
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    $\begingroup$ Who knows what was happening when I wrote this, but perhaps I was thinking about the fact that the double cover of the identity component of GL(n, R) has no finite dimensional representations besides those coming from GL(n, R). (e.g. cf Lemma 5.23 of Lawson's Spin Geometry). $\endgroup$ Oct 2, 2018 at 11:43

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