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 Mar31 comment approximate vanishing in Pontryagin dual thanks for your interest in this problem, hope you can obtain some nice results on it! Right now, i have to I work on something else, but if I can help you in any way, please let me know. Thanks again!!! Mar29 comment approximate vanishing in Pontryagin dual thank you very much!! Indeed, this question arose from a failed attempt to solve a research problem. The original problem is still not solved, maybe it is still interesting for you, so let me mention it below. Jan27 comment When two projections in a C*-algebra are “almost” Murray-von Neumann equivalent, they are equivalent Intuitively, maybe we can first show that when $aa^*$ is close to $p$, then the range projection of $aa^*$, denote it by $r_1$, is close to $p$, maybe the assumption is strong enough to imply $||r_1-p||<1$, so they are unitary equivalent, do the same thing for $q$, note that $r_1, r_2$ are unitary equivalent. Jan25 comment one end group with positve first Betti number$\beta^{(2)}_1(G)>0$ Thanks again! It is good to know that... Jan24 comment one end group with positve first Betti number$\beta^{(2)}_1(G)>0$ Thanks, to see the surface groups have one end, if my understanding is right, just look at the Cayley group(some tiling of the plane); is there any way to see directly that the surface groups could not be of the form predicted by the Stalling theorem? Btw, how to see in (2), $B^1$ is not closed?... Nov27 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? Oct23 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! Oct23 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? Sep10 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. Sep10 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. Sep10 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. Sep10 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.. Sep9 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 $V, W$? Sep6 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? Sep6 comment Extending a measurable map $f: G \to \mathbb{R}/\mathbb{Z}$ to a continuous group homomorphism "...iff the measure of the slices of $G\times G\Z$ is zero a.e., by Fubini". I still do not see how to use the assumption that $f(x+y)=f(x)+f(y), \forall (x,y)\in Z" to get$f(x+y)+f(-y)=f(x)$for a.e. x, a.e. y in G. Sep6 comment Extending a measurable map$f: G \to \mathbb{R}/\mathbb{Z}$to a continuous group homomorphism$Z\subset G\times G$is not necessarily of product type. Aug12 comment a measurable function on a LCA group coincide with an mulitplicative character almost everywhere Hi, could you tell me where you have met this fact? I asked an essentially same question recently, and I want to understand this stuff better. Aug11 comment Existence of proper I.C.C. subgroup my fault, I should mention that$G$is inner amenable group if there exists a finite additive probability measure on the power set of$G$which is invariant under conjugate action of$G$.(Recall for amenable group, we require that measure is invariant under left action of$G$). Aug11 comment induction exercise doubt Hint: When do induction, you can not change the assumption, in your case, the assumption,$x_1x_2... x_{n+1}=1$is the assumption, but you could translate your assumption into different form. Try set$x_i'=x_i for 1\leq i\leq n-1, x_n'=x_nx_{n+1}$. Aug11 comment What is the closed form for this sequence, powers of$4\$? try oeis.org, but it seems does not work...