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Let $G$ be a Lie group, and let $\mathcal B(G)$ its Borel $\sigma$-algebra. Suppose that $f : G \to G$ is a Borel-measurable homomorphism. Is $f$ smooth?


Edit: My original question said "measurable function" instead of the more accurate "measurable homomorphism." Marc Palm and other people answered both questions very nicely:

  • there are obviously non-smooth measurable functions on Lie groups, and
  • all measurable homomorphisms on Lie groups are smooth.
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Perhaps a more interesting question: is every measurable cocycle cphomplogous to a smooth one? –  ThiKu Feb 4 '13 at 12:38
    
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Whatever theorem you want to prove about Lie groups should hold at least for the Lie group $\mathbb R$, right? –  Gerald Edgar Feb 4 '13 at 14:09
    
Tom, did you mean to say function, or did you mean homomorphism? –  Todd Trimble Feb 5 '13 at 17:05
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handslap. Of course, homomorphism. –  Tom LaGatta Feb 5 '13 at 18:02

4 Answers 4

up vote 8 down vote accepted

No, take a proper closed set $X$ and define $f|X = 1$ and $f|G-X = g \neq 1$.

If you consider $f$ to be a group homomorphism, then the answer is yes in general.

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See here for example: mathoverflow.net/questions/64116/… –  Marc Palm Feb 4 '13 at 9:32
    
The counterexample works if X is not open. Taking X as the connected component of the identity wouldn't work. –  Marc Palm Feb 4 '13 at 13:19
    
Thanks @Marc Palm. I meant homomorphism but I foolishly neglected to say it. Why is the answer yes in general? Intuitively, the Borel $\sigma$-algebra contains all the topological information of the space, hence the differential information since $G$ is a Lie group. However, this is just heuristic speculation and I don't know a good argument which makes this fact apparent. –  Tom LaGatta Feb 5 '13 at 18:04
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Excellent, I just looked at your other question. Alain Varette provides a nice reference to Adam Kleppner's paper. A measurable homomorphism of locally compact groups is continuous. Since Lie groups are locally compact, this proves it. Direct link to Alain Varette's answer: mathoverflow.net/questions/64116/… –  Tom LaGatta Feb 5 '13 at 18:06
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And André Henriques provides the very nice intuition: mathoverflow.net/questions/64116/… –  Tom LaGatta Feb 5 '13 at 18:07

Perhaps it is interesting to know that for group cocycles the answer is in general no (even up to coboundaries). For instance the extension $\mathbb{Z} \to \mathbb{R} \to S^1$ is described by a 2-cocycle $S^1 \times S^1 \to \mathbb{Z}$ that is measurable, but this cannot be chosen to be smooth or continuous.

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Probably you impose some algebraic structure on your function $f$, like $f$ is a homomorphism, no? Since your question reminds me similar results on the additive measurable function on $\mathbb{R}$.

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This is more a comment than an answer. It's a great comment, but a comment nonetheless. –  Todd Trimble Feb 4 '13 at 16:13
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Dear Todd, you are right. I fact, it was the first time I made a comment, so sorry for the mistake. –  Yanqi QIU Feb 5 '13 at 14:24

Not every measurable function on the real line is continuous, let along smooth. Real line is a perfectly nice Lie group.

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