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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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  • $\begingroup$ Perhaps a more interesting question: is every measurable cocycle cphomplogous to a smooth one? $\endgroup$
    – ThiKu
    Commented Feb 4, 2013 at 12:38
  • $\begingroup$ mathoverflow.net/questions/57426/… $\endgroup$
    – Steve Huntsman
    Commented Feb 4, 2013 at 13:33
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    $\begingroup$ Whatever theorem you want to prove about Lie groups should hold at least for the Lie group $\mathbb R$, right? $\endgroup$
    – Gerald Edgar
    Commented Feb 4, 2013 at 14:09
  • $\begingroup$ Tom, did you mean to say function, or did you mean homomorphism? $\endgroup$
    – Todd Trimble
    Commented Feb 5, 2013 at 17:05
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    $\begingroup$ handslap. Of course, homomorphism. $\endgroup$
    – Tom LaGatta
    Commented Feb 5, 2013 at 18:02

4 Answers 4

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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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    $\begingroup$ See here for example: mathoverflow.net/questions/64116/… $\endgroup$
    – Marc Palm
    Commented Feb 4, 2013 at 9:32
  • $\begingroup$ The counterexample works if X is not open. Taking X as the connected component of the identity wouldn't work. $\endgroup$
    – Marc Palm
    Commented Feb 4, 2013 at 13:19
  • $\begingroup$ 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. $\endgroup$
    – Tom LaGatta
    Commented Feb 5, 2013 at 18:04
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    $\begingroup$ 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/… $\endgroup$
    – Tom LaGatta
    Commented Feb 5, 2013 at 18:06
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    $\begingroup$ And André Henriques provides the very nice intuition: mathoverflow.net/questions/64116/… $\endgroup$
    – Tom LaGatta
    Commented Feb 5, 2013 at 18:07
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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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    $\begingroup$ This is more a comment than an answer. It's a great comment, but a comment nonetheless. $\endgroup$
    – Todd Trimble
    Commented Feb 4, 2013 at 16:13
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    $\begingroup$ Dear Todd, you are right. I fact, it was the first time I made a comment, so sorry for the mistake. $\endgroup$
    – Yanqi QIU
    Commented Feb 5, 2013 at 14:24
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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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