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1d
comment Isomorphism between $\mathbb R^3$ and the the Heisenberg group
This question isn't of research level. I'm voting to close.
May
2
comment Link between the hairy ball theorem and the fundamental theorem of algebra
You probably want the slightly more sophisticated version of the Hairy Ball theorem which says that the sum of the indices of the zeros equals 2.
Apr
26
comment Why do people study representations of 3-manifold groups into $SL(n,\mathbb{C})$?
One last comment. You might also be interested in the discussion under this blog post: ldtopology.wordpress.com/2013/04/23/… .
Apr
26
comment Why do people study representations of 3-manifold groups into $SL(n,\mathbb{C})$?
... I would certainly regard that as strong motivation. But actually $SL(2,\mathbb{C})$ seems to do quite well: arxiv.org/abs/1510.08493 .
Apr
26
comment Why do people study representations of 3-manifold groups into $SL(n,\mathbb{C})$?
Let's stick to the irreducible case to keep things simple. Then I have in mind the JSJ decomposition, together with the Seifert invariants of the toroidal pieces, the hyperbolic structures of the atoroidal pieces, and the gluing maps. Certainly this is hard to compute in general, but it often works quite well in practice. I agree that this approach is unlikely to, say, find a polynomial-time algorithm to recognize knottedness. Is it conjectured that invariants coming from higher representation varieties might recognize knottedness quickly, say? (cont'd)
Apr
26
revised Are braid groups conjugacy separable?
Edited to give a much fuller answer.
Apr
24
comment Applications of space filling curves
Also, this question should be community-wiki.
Apr
23
comment Applications of space filling curves
The Cannon--Thurston map very naturally associates a space-filling curve to any fibred hyperbolic 3-manifold.
Apr
22
answered Are braid groups conjugacy separable?
Apr
22
comment Why do people study representations of 3-manifold groups into $SL(n,\mathbb{C})$?
So the answer is that you might hope to generalize well known existing invariants? It would be nice to know what people then hope to do with those invariants. Since geometrization provides a very nice complete set of invariants for any 3-manifold, the business of finding new invariants for their own sake seems poorly motivated to me.
Apr
21
comment Big list - Equivalent descriptions of Hodge conjecture?
This should be community-wiki.
Apr
19
comment Is there a smooth manifold which admits only rigid metrics?
@studiosus -- that seems to do it! Thanks!
Apr
18
comment Is there a smooth manifold which admits only rigid metrics?
I'm missing the other direction of the equivalence. It seems like you need an argument to show that an isometry is necessarily finite order... Or is it obvious?
Apr
18
comment Limits of quotient groups in the space of marked groups
In your example, isn't $K$ just the intersection $\bigcap_iK_i$, since you can pass to a nested subsequence?
Apr
17
comment Properties of a special finitely presented groups
It's a standard fact (I should of course have said all torsion-free one-relator groups are left-orderable). The only proof I know is to use the ideas of arXiv:1410.2540 to prove that they're locally indicable, and then to quote the well known theorem that locally indicable groups are left-orderable.
Apr
17
comment Properties of a special finitely presented groups
All one-relator groups are left-orderable.
Apr
11
comment Some question on haar measure for sumsets of closed subsets of profinite groups
@user89334, sorry, you're quite right, I did miss that hypothesis. I withdraw my objection! (As Paul Garrett notes above, the first incarnation of the question wasn't especially clear.)
Apr
10
comment Some question on haar measure for sumsets of closed subsets of profinite groups
Well, the claim that there's a unique invariant measure is clearly false, since we can scale. There should be lots of examples in which the Haar measure of a double-coset is zero. For instance, the double coset of two pro-cyclic subgroups of a profinite free group should have infinitely many disjoint translates, and hence have zero Haar measure.
Apr
10
revised Some question on haar measure for sumsets of closed subsets of profinite groups
edited tags
Apr
10
comment Some question on haar measure for sumsets of closed subsets of profinite groups
This argument isn't right. The restriction of $\mu_H$ could be the zero measure (and frequently is, I think).