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 Dec 7 accepted Whitehead double Dec 7 asked Whitehead double Nov 28 accepted evaluate two sums in analytic number theory Nov 28 accepted what is the definition of “two parallel copies of a surface S” Nov 27 revised evaluate two sums in analytic number theory fix typos Nov 27 asked evaluate two sums in analytic number theory Nov 27 comment what is the definition of “two parallel copies of a surface S” thanks, so it is kind of disjoint union, the resulting surface has genus 2g? Nov 27 asked what is the definition of “two parallel copies of a surface S” Nov 18 awarded Citizen Patrol Oct 23 comment question on subgroup of compact group thanks a lot for your answer! The subgroup I am interested in is closed, coming from fixed point of a group element in some group acting on this $G$. My knowledge on topological groups, especially lie groups is almost zero..., I apologize for the inaccuracy. Any books on this stuff to recommend? Thanks again! Oct 23 accepted question on subgroup of compact group Oct 23 revised question on subgroup of compact group add one more tag Oct 23 comment question on subgroup of compact group @CameronWilliams, I mean $K\leq G$ is of finite index if the left coset $G/H$ is finite. Equivalently, if the Haar measure of $K$ is nonzero. Maybe there is better name for this? Oct 23 asked question on subgroup of compact group Oct 16 awarded Yearling Sep 10 comment Extending a measurable map $f: G \to \mathbb{R}/\mathbb{Z}$ to a continuous group homomorphism Assume $f$ is a genuine group homomorphism, I.e.,$f(x+y)=f(x)+f(y)$ holds for all $(x,y)\in G\times G$, so in your definition of $F$, $f(x+y)+f(-y)=f(x)$, then $F(x)=f(x)$ for all $x\in G$, so how do you know $F$, which is now $f$ is continuous? To be honest, whenever I left some comments, you kept ignoring what I am talking about, I do not know why. Sep 10 comment Extending a measurable map $f: G \to \mathbb{R}/\mathbb{Z}$ to a continuous group homomorphism Suppose $f$ is a group homomorphism, then by definition of your $F$, $F(x)=f(x)\forall x\in X$, but we only assume $f$ is measurable, so you can not say that $F$ is continuous. Sep 10 comment Extending a measurable map $f: G \to \mathbb{R}/\mathbb{Z}$ to a continuous group homomorphism .. more details?I suspect we would run into trouble of some suttle issues that Kleppner faced. Sep 10 comment Extending a measurable map $f: G \to \mathbb{R}/\mathbb{Z}$ to a continuous group homomorphism Here are a couple of things I want to say. 1, when I see $a.e., x,y\in G$ I would understand it as for some conull subsets $V,W\subset G$, $x\in V, y\in W$ instead of what you mentioned above. 2, suppose some statement P holds for $(x,y)\in Z$ where $Z\subset G\times G$ is conull set. In general, you CAN NOT find conull $V,W\subset G$ such that P holds for $x\in V,y\in W$, since in general $Z$ is complicated, say $Z=G\times G-diagonal$. 3, In your answer $F(x)=f(x) a.e., x\in G$ is true, but I do not see how to show $F$ is continuous, a group homomorphism. If you have time, could you provide.. Sep 9 comment Extending a measurable map $f: G \to \mathbb{R}/\mathbb{Z}$ to a continuous group homomorphism Yes, you can show that $f(x+y)+f(-y)-f(x)=0$ for a.e. $(x,y)\in G\times G$. But the point is that you claim it holds for a.e. $x\in G$ and a.e. $y\in G$, i.e., you claim that you can find $V, W$, both are conull subset in $G$ such that the above equality holds for $x\in V, y\in W$. Could you tell me why? How do you argue that from $Z$ you can get $V, W$?