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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.

edited Feb 7 at 17:46

asked Feb 4 at 8:54

Tom LaGatta
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Perhaps a more interesting question: is every measurable cocycle cphomplogous to a smooth one? – unknown (google) Feb 4 at 12:38

mathoverflow.net/questions/57426/… – Steve Huntsman Feb 4 at 13:33

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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 at 14:09

Tom, did you mean to say function, or did you mean homomorphism? – Todd Trimble Feb 5 at 17:05

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handslap. Of course, homomorphism. – Tom LaGatta Feb 5 at 18:02

show 2 more comments

Marc Palm · Accepted Answer · 2013-02-04 13:14:24Z UTC

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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.

edited Feb 4 at 13:14

answered Feb 4 at 9:30

Marc Palm
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See here for example: mathoverflow.net/questions/64116/… – Marc Palm Feb 4 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 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 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 at 18:06

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And André Henriques provides the very nice intuition: mathoverflow.net/questions/64116/… – Tom LaGatta Feb 5 at 18:07

Christoph Wockel · Answer 2 · 2013-02-06 20:13:26Z UTC

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.

Yanqi QIU · Answer 3 · 2013-02-04 09:20:23Z UTC

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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 $\mathbb{R}$.

answered Feb 4 at 9:20

Yanqi QIU
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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 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 at 14:24

Eugene Lerman · Answer 4 · 2013-02-04 16:50:40Z UTC

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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.

answered Feb 4 at 16:50

Eugene Lerman
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Is a measurable homomorphism on a Lie group smooth?

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