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seen Nov 20 at 14:49

Nov
18
awarded  Citizen Patrol
Nov
18
comment amenable + without $BS(m,n)$+finite $K(G,1)$implies virtually cyclic?
@QiaochuYuan, I am not sure it is suitable to be posted in MO, that's why I first asked it here.
Nov
18
comment amenable + without $BS(m,n)$+finite $K(G,1)$implies virtually cyclic?
@QiaochuYuan, yes.
Nov
17
asked amenable + without $BS(m,n)$+finite $K(G,1)$implies virtually cyclic?
Nov
11
revised torsion free for the 2nd cohomology group?
add a result
Nov
11
asked torsion free for the 2nd cohomology group?
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$?
Sep
9
revised Extending a measurable map $f: G \to \mathbb{R}/\mathbb{Z}$ to a continuous group homomorphism
added 37 characters in body
Sep
8
answered Extending a measurable map $f: G \to \mathbb{R}/\mathbb{Z}$ to a continuous group homomorphism
Sep
6
comment Extending a measurable map $f: G \to \mathbb{R}/\mathbb{Z}$ to a continuous group homomorphism
both are true. Is it related to my above comment?