ougao
Reputation
1,418
Top tag
Next privilege 2,000 Rep.
 Apr 7 comment Infinite quotient with finitely many torsion elements @DerekHolt, oh yes, thanks a lot! Is it easy to see whether polycyclic groups satisfy this property I asked above? Apr 7 comment Infinite quotient with finitely many torsion elements @DerekHolt, thanks, I am not aware of this fact on nilpotent groups. btw, is it known that torsion elements in a polycyclic group also form a subgroup? Mar 21 comment angle" between two group elements @Rob, thanks! it seems different from what I expect... Mar 21 comment angle" between two group elements yes, so I include the question on defining the notion of cone" for a general countable discrete group. Mar 31 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!!! Mar 29 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. Jan 27 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. Jan 25 comment one end group with positve first Betti number$\beta^{(2)}_1(G)>0$ Thanks again! It is good to know that... Jan 24 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?... 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? 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 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? 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 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? Sep 6 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)$for a.e. x, a.e. y in G. Sep 6 comment Extending a measurable map$f: G \to \mathbb{R}/\mathbb{Z}$to a continuous group homomorphism$Z\subset G\times G\$ is not necessarily of product type.