Reputation
4,921
Next privilege 5,000 Rep.
Approve tag wiki edits
Badges
11 37
Newest
 Nice Answer
Impact
~90k people reached

1d
awarded  Nice Answer
Jan
26
awarded  Nice Answer
Jan
21
comment Inducing up the group homomorphism between mapping class groups
@SamNariman Let me try that again. The homeomorphism is canonical, so it will preserve be a diffeomorphism if the smooth structure is canonical. I claim that a ramified cover of a surface inherits a smooth structure from the smooth structure downstairs. Let me specialize to a double cover. A function upstairs is the sum of a function that descends to downstairs and an “odd” function that takes negated values on the two sheets. Define the smooth structure by saying that an odd function is smooth if its square (which descends) is smooth.
Jan
12
awarded  Revival
Jan
12
answered Inducing up the group homomorphism between mapping class groups
Dec
30
comment Algebraic K theory, Karoubi completion and splitting
Your premise is false: the map on $K^0$ is not a rational isomorphism. Just take the subcategory of objects with trivial class in $K^0$. See Thomason's Classification of triangulated subcategories.
Dec
17
awarded  Notable Question
Nov
2
awarded  Pundit
Oct
13
awarded  Nice Answer
Oct
11
comment Calculation-free proof of the Weyl Integral formula for U(n)
Why polynomial rather than rational? Is that what the last couple comments are about?
Sep
28
comment Can relative flatness of a sheaf be tested using (faithfully) flat morphisms?
No, you cannot drop the "faithfully." For example, if $X$ is the empty scheme, it is flat, but it detects nothing. The empty scheme is a good heuristic for testing whether faithfulness is relevant. In fact, given the characterization of faithfully flat morphisms among flat morphisms as the surjective ones (ie, the ones with nonempty fibers), it is the whole story.
Aug
29
comment Homology generated by lifts of simple curves
Just draw some pictures and see that the OP's curves generate the rational homology. I think integral, too, but I'm not sure. Are you sure Boggi and Looijenga are talking about the same group?
Aug
28
comment Homology generated by lifts of simple curves
It doesn't look like a counterexample to me.
Aug
28
comment Motivic fundamental group of the moduli space of curves?
I'm not familiar with the moment method, but it sounds pretty difficult to apply. Doesn't it require you to know about the cohomology of all curves? Does it amount to: if most curves have the maximal MT group, then the monodromy is the MT group? But don't other results say that you just need a single curve with the maximal MT group?
Aug
11
comment Automorphisms of finite order in $Out(\widehat{F_2})$
Most of my comments were about the proof that the profinite completion of a free group is torsion-free. About virtually free groups, all I said is that they reduce to the free case. Here is a weaker statement: if a $G$ is virtually free, an extension $F\to G\to Q$, where $F$ is free and $Q$ is finite, then a torsion element of $\hat G$ has nontrivial image in $Q$. This does not rule out the possibility of torsion in $\hat G$ not conjugate into $F$, but the possibilities are uniformly bounded by $Q$, a finite group accessible before completion. In that sense it cannot be "new."
Aug
7
comment Automorphisms of finite order in $Out(\widehat{F_2})$
For every element $g$ of a profinite group $G$, there exists a finite index subgroup $H$ containing $g$ so that the image of $g$ in $H^{ab}$ is nontrivial... Indeed, if an element is not the identity, then there is some finite quotient $Q$ in which it is not the identity. The subgroup of $Q$ generated by the image of $g$ is an abelian group. Let $H$ be the inverse image of that subgroup, a finite index subgroup of $G$. This group is defined by the property that it maps to an abelian group not killing $g$, so it does not die in $H^{ab}$... If $g$ were torsion, then $H^{ab}$ would have torsion.
Aug
6
comment Manifold approximations to $BO(3)$
$BSpin(3)$ has the very nice representation $\mathbb H\mathbb P^\infty$. The fiber sequence $BSpin(3)\to BSO(3)\to K(\mathbb Z/2,2)$ suggests that $BSO(3)$ is on par with $K(\mathbb Z/2,2)$. ($BO(n)$ reduces to $BSO(n)$ in a way independent of $n$.)
Aug
5
comment Automorphisms of finite order in $Out(\widehat{F_2})$
Shatz's book has some discussion of free profinite groups. The key point is that the finite index subgroup of a profinite completion is the profinite completion of the corresponding finite index subgroup of the discrete group. That reduces to the free case. Also, an element of a profinite group is detected by the abelianization of a finite index subgroup. So if a profinite group has a torsion element, it would have a finite index subgroup with torsion abelianization. But by the key point, that subgroup is also the profinite completion of a free group, hence no.
Jul
30
answered Automorphisms of finite order in $Out(\widehat{F_2})$
Jul
28
comment Automorphisms of finite order in $Out(\widehat{F_2})$
What is going on in the paragraph starting with "Second"? Since $SL_2(\mathbb Z)$ is virtually free, so is its profinite completion, and thus it has no new torsion. But that very freedom gives it many quotients with many other orders of torsion, like $SL_2(\mathbb Z_\ell)$.