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I recently began to study Lie group theory and I found the classical result:

Theorem: Every continuous homomorphism $\varphi: \mathcal{G} \to \mathcal{H}$ of groups is a homomorphism of Lie groups.

I was wondering the following: is the continuity of the homomorphism necessary? I was trying to find a discontinuous homomorphism between Lie groups but I'm unable to do so. There a fairly easy/non exotic example?

Cameron L. Williams
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jacopoburelli
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1 Answers1

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Take $G,H$ to be the group of positive reals under multiplication. Taking the logarithm transforms this question into finding a function $f: \mathbb{R}\rightarrow \mathbb{R}$ such that $f(a+b)=f(a)+f(b)$ for all real $a,b$ but $f$ is not continuous. This can be done with the Axiom of Choice. See here: Additive function $T: \mathbb{R} \rightarrow \mathbb{R}$ that is not linear..

As for if Choice is necessary to find a counterexample, I would be interested to know as well.

Terence C
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  • There are models of ZF where every subset of real numbers is measurable. – Moishe Kohan Apr 04 '23 at 16:12
  • @MoisheKohan certainly for this pair of Lie groups, it is well-known to require choice. But I'm curious if this can be shown for all such pairs. – Terence C Apr 04 '23 at 19:16
  • It is actually known that for, say, groups of real points of reductive algebraic groups the problem reduces to exotic homomorphisms of coefficient fields. – Moishe Kohan Apr 04 '23 at 19:33