Is it true that there is no homomorphism from a semisimple Lie group without compact factor to a compact Lie group?
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3$\begingroup$ Wouldn't such a map be a compact factor of the original Lie group? $\endgroup$– LSpiceCommented Oct 8, 2020 at 3:18
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$\begingroup$ @LSpice I don't see why the image would be normal. However, the non-existence of a nontrivial/injective homomorphism (there's always the trivial homomorphism) follows, for instance, from the fact that the affine group $\mathbf{R}\ltimes\mathbf{R}$ has no faithful finite-dimensional unitary representation. $\endgroup$– YCorCommented Oct 8, 2020 at 5:56
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2$\begingroup$ This question is strictly covered in this one which was answered in comments. I have now provided the detailed answer there. $\endgroup$– YCorCommented Oct 8, 2020 at 9:39
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$\begingroup$ @YCor, maybe I don't know the terminology. I thought a compact factor just meant a quotient with compact image. I believe that there are obstructions (which is why I just asked in a comment); I just don't see why it matters whether the image of that quotient is normal in some larger compact group. Is the issue whether the kernel is an almost direct factor? $\endgroup$– LSpiceCommented Oct 8, 2020 at 11:41
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1$\begingroup$ But the problem is that it's not a priori obvious that the image is closed. So the quotient by the kernel might be noncompact. $\endgroup$– YCorCommented Oct 8, 2020 at 12:27
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2 Answers
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See below a detailed version of the comment of @LSpice. (Edited taking into account a comment of @YCor.) This is an answer to the question on homomorpisms of real algebraic groups.
Proposition. Let $\varphi\colon G\to K$ be a homomorphism of real algebraic groups, where $G$ is a connected semisimple real algebraic group without compact factors and $K$ is a compact real algebraic group. Then $\varphi$ is trivial (identically 1).
Proof. Such a homomorphism $\varphi$ induces an isomorphism $$ G/{\rm ker}\,\varphi\,\overset\sim\longrightarrow\, {\rm im\,}\varphi.$$ The image ${\rm im\,}\varphi$ is closed in $K$. Therefore, ${\rm im\,}\varphi$ is compact. On the other hand, from the theory of real semisimple algebraic groups we know that since $G$ is a connected semisimple real algebraic group without compact factors, it has no nontrivial compact quotients. It follows that $G/{\rm ker}\,\varphi=\{1\}$, and hence $\varphi$ is trivial.
Note that any compact real Lie group is algebraic. However, a noncompact real semisimple Lie group might be non-algebraic. For example, any nontrivial cover of ${\rm SL}(2, {\Bbb R})$ is non-algebraic.
answered Oct 8, 2020 at 6:54
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1$\begingroup$ I don't think this is a proof. It's a trivial reduction to showing that there's no injective homomorphism from a [nontrivial!] semisimple group with no compact factor into a compact Lie group. The remainder of the proof (which I mentioned in a comment, or alternatively doing the case when $G$ is locally isomorphic to $\mathrm{SL}_2(\mathbf{R})$) is fairly standard (and probably quite a duplicate here), but contains all the substance of the proof, which is not apparent here. Just to keep in mind, there's an injective hom from the noncompact Lie group $\mathbf{R}$ into some compact Lie group. $\endgroup$– YCorCommented Oct 8, 2020 at 9:02
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$\begingroup$ @YCor: I agree. I had in mind real semisimple algebraic groups. $\endgroup$ Commented Oct 8, 2020 at 10:35
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1$\begingroup$ ... with algebraic homomorphisms. (Which is automatic among Lie group homomorphisms between semisimple algebraic groups, which is not hard but strictly harder this whole argument). $\endgroup$– YCorCommented Oct 9, 2020 at 12:58
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I think an elementary argument is that if such a homomorphism exists, one can pull back the Cartan-Killing form (or an extension of that, in case the compact group has positive dimensional center) to an Ad-invariant definite inner product on the Lie algebra of a quotient of the original group. This quotient is then compact.
answered Oct 9, 2020 at 12:56
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1$\begingroup$ Indeed. This argument works in the broader setting of this older question. $\endgroup$– YCorCommented Oct 9, 2020 at 13:04