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165147
bio website math.berkeley.edu/~ianagol
location Berkeley
age 45
visits member for 5 years, 7 months
seen 3 hours ago

I'm a professor at UC Berkeley working predominantly in the field of low-dimensional topology.


1d
comment Injectivity of the Dehn-Nielsen-Baer map?
I guess they address the issue of homotopy/homotopy equivalence vs. diffeomorphism/isotopy in Chapter 1.
2d
comment Penrose tiling substitution is bijective
This paper and it's references might shed some light on this question: ams.org/journals/tran/1996-348-11/S0002-9947-96-01640-6
2d
comment Injectivity of the Dehn-Nielsen-Baer map?
The mapping class group is usually defined as diffeos. modulo isotopies. To show equivalence with $Out(\pi_1(M))$, first one may show equivalence with self-homotopy equivalences up to homotopy, and then show that homotopy equivalences up to homotopy are equivalent to diffeomorphisms up to isotopy.
2d
comment Injectivity of the Dehn-Nielsen-Baer map?
As for homotopy equivalence implies homeomorphism, one may prove this by induction on a hierarchy, a la Waldhausen. Similarly for homotopy equivalent homeos. are isotopic.
2d
comment Is there a non right-orderable torsion-free factor of the Braid group on 3 strands?
I don't understand your logic - why can't there be finitely many relators?
2d
comment Is there a non right-orderable torsion-free factor of the Braid group on 3 strands?
Yes, those two relations (I didn't compute m and l, but this should be easy).
May
26
comment Trigonal loci in Teichmueller spaces
I think the result is essentially due to Hilden - look at the top of p. 8 of the paper, where the reference is given.
May
26
comment Is there a non right-orderable torsion-free factor of the Braid group on 3 strands?
Yes, it is easy to obtain such representations. One adds a relation of the form m^a*l^b, where r=a/b and m is the longitude, l the meridian.
May
23
answered How to show whether a given knot and its mirror image are the same or not?
May
23
revised Degrees of self-maps of aspherical manifolds
deleted 6 characters in body
May
20
answered hyperbolic structure on Figure–8 knot complement
May
18
comment When can a class in $H^1(M;\mathbb{Z})$ be represented by a fiber bundle over $S^1$
In dimension 4, it might suffice that $ker(\pi_1(M)\to \mathbb{Z})$ is a PD(3) group; if it were, conjecturally this group would be an aspherical 3-manifold group, hence there ought to be a homotopy equivalent 4-manifold with the same fundamental group. However, the Borel conjecture is still open for 4-manifolds. I'm not sure if there is a natural conjecture in the case that $\pi_2(M)\neq 0$.
May
15
awarded  Nice Answer
May
6
comment Kleinian groups containing an isomorphic copy of itself
Yes, Scott (and Shalen) actually prove that the manifold is homotopy equivalent to a compact 3-manifold. In fact, even with torsion, it is homeomorphic to the interior of a compact orbifold by tameness - of negative euler characteristic if it is non-elementary. In higher dimensions, this strategy at least should apply to geometrically finite Kleinian groups with non-zero Euler characteristic.
May
6
comment Kleinian groups containing an isomorphic copy of itself
In 3-dims. (discrete groups in $PSL_2(\mathbb{C})$), either a finitely generated group has finite covolume, in which case @YCor's comment applies, or it is infinite covolume, in which case if it is non-elementary, the Euler characteristic is $<0$ (by the Scott core theorem, finitely generated groups are finitely presented, so Euler characteristic makes sense), and therefore it cannot be isomorphic to a finite-index subgroup of itself. So Rivin's answer is overkill at least in 3D (co-Hopfian is much stronger than what you're asking for).
May
5
comment Do unit quaternions at vertices of a regular 4-simplex, one being 1, generate a free group?
Ok, excellent, that shows it is a lattice!
May
5
comment Do unit quaternions at vertices of a regular 4-simplex, one being 1, generate a free group?
Nice description. One thing I still don't quite get is why $\Gamma_0$ is generated by the 4 given elements? I don't see how you can deduce this without a bit more information (a free group on 4 elements has many 4 generator subgroups).
May
4
comment Can every $\mathbb{Z}^2$ disk be pinball-reached?
It seems that you're searching for something like "Sinai billiards".
May
4
comment Avoiding mean-curvature flow dumbbell neck-pinch by inflating a surface
@JosephO'Rourke: Did you mean to discuss Ricci flow? I was confused by the title of your question, since you actually seem to be looking for something related to mean curvature flow.
May
3
comment Do unit quaternions at vertices of a regular 4-simplex, one being 1, generate a free group?
@QiaochuYuan: Ok, I was using the term "Bass-Serre tree" incorrectly. I really meant the tree associated to $SL_2$ of a local field, which is a special case of a Bruhat-Tits building. Serre's book Chapter II section 1 is the usual reference. Shalen's notes (linked in my other comment) Chapter 3 also has a nice discussion.