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1d
answered Automorphisms of finite order in $Out(\widehat{F_2})$
2d
comment fake $S^{2k}\times S^{2k}$
If you assume that the image of $S^{Diff}(M)\to KO(M)$ is a subgroup, you can get some interesting conclusions. You could use the fiber bundle construction in either direction; you can declare either sphere to be the base. This gets you two different $Z$s. They are related by $Aut(S^{4n}\times S^{4n})$, of course. But inside $KO(M)$ they are two different subgroups. So they generate a whole $Z^2$. But that was a very big assumption at the beginning. It's hard to imagine proving that it's a subgroup without proving the whole surgery theorem and thus being able to compute $S^D(M)$ completely.
2d
comment fake $S^{2k}\times S^{2k}$
There is a natural map $S^{Diff}(M)\to KO(M)$ -- what is the tangent bundle [actually the difference of new minus old, or else the image isn't a subgroup]. And the Pontrjagin character gives an approximate isomorphism $KO(M)\sim \bigoplus H^{4*}(M)$. In your case, this is just $H^{4n}(S^{4n}\times S^{4n})\oplus H^{8n}(S^{4n}\times S^{4n})=Z^2\oplus Z$. [The total Chern class turns multiplication into addition. The Chern character is a homomorphism. But it has denominators. Same for P.] But it doesn't hit the whole $Z^3$. Anything in the image must satisfy the Hirzebruch signature formula.
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 $SL_2(\mathbb Z_\ell)$.
Jul
28
comment Automorphisms of finite order in $Out(\widehat{F_2})$
If the element of $GL_2$ is diagonal, then the obvious lift to $Out(F)$ (even $Aut(F)$) is still finite order. But if you want to lift an arbitrary torsion element, you have to be more careful.
Jul
28
comment “Spec” of graded rings?
@user40276 I did my exercise and I must restrict the claim: The free CDGA on an acyclic complex supported in degrees $2n$ and $2n+1$ is equivalent to the ground ring, as expected, but the the free CDGA on an acyclic complex supported in degrees $2n-1$ and $2n$ has nontrivial homology. I am using a homological differential that lowers degree; with the cohomological convention, the two cases switch.
Jul
28
comment fake $S^{2k}\times S^{2k}$
Yes, there are other examples. The group structures mentioned in the question and this answer are compatible, so this $\mathbb Z$ cannot fill up the whole $\mathbb Z^2$. The way to see the compatibility is to interpret both via the Pontrjagin class. This construction only changes the Pontrjagin class along the base, but it is also possible to change it along the fiber.
Jul
16
comment “Spec” of graded rings?
@user40276 a good exercise is to see that the free functor from chain complexes to CDGAs does not preserve weak equivalences. I think that it fails even with the simplest acyclic complex. Tyler Lawson uses a different failure of a Quillen adjunction to show that a particular model structure does not exist.
Jul
15
comment Euler characteristic of a curve
Why do you call it HRR, rather than just RR? Because of the Chern classes? I think most people reserve Hirzebruch for the generalization to higher dimensions and don't distinguish 1d variants.
Jun
26
comment Next steps on formal proof of classification of finite simple groups
Here is a complementary suggestion. Serre complains that people do not distinguish between theorems conditional on CFSG and those conditional on CFSG+ATLAS. Formalizing the statement and (completely omitted) proof of ATLAS sounds pretty easy to me, but of a very different flavor than formalizing the proof of CFSG (and different is good if your goal is to learn about formal math). Moreover, one should check that the formalization is usable by using it to formally prove conditional theorems.
Jun
8
comment What are some examples of non-commutative $\mathbb{Q}$-monoids and/or $\mathbb{R}$-monoids?
For the $\mathbb Q$ version to work it is important that the group to be nilpotent, so that the exponential preserve rational matrices. . . There is a smaller, though perhaps not simpler, non-commutative Lie group, the affine group $ax+b$. The exponential is still a bijection, so it gives an $\mathbb R$-group, but the exponential does not preserve rational elements. It contains $\mathbb Q$-monoids, but they are more complicated.
Jun
8
comment What are some examples of non-commutative $\mathbb{Q}$-monoids and/or $\mathbb{R}$-monoids?
If you restrict to rational coefficients, you get a $\mathbb Q$-monoid. If you work formally subject to $x^n=0$, this example still works, but now it is a finite-dimensional nilpotent Lie group, quite like André's example.
May
28
comment How does the solenoid structure of $\mathbb{A}/\mathbb{Q}$ lift to $PGL(2, \mathbb{A})/ PGL(2, \mathbb{Q})$?
Before considering $G(\mathbb Q)\backslash G(\mathbb A)$, consider $G(\mathbb Z[\frac12])\backslash G(\mathbb R\times\mathbb Q_2)$. Also, before $G=PGL_2$, consider $G=\mathbb G_a,\mathbb G_m,SL_2$.
May
24
answered Alexander duality for non-manifolds
May
4
awarded  Good Answer
Apr
28
comment Intuition behind the definition of quantum groups
A quantum group is a new tensor product on the category of representations. I think that something along the lines of the KZ equations give a coordinate-free construction. At first glance, they just give the braiding, but that's a good start.
Apr
10
comment The image of the Hurewicz map for rational loop spaces
I expect that is a counterexample, but if she didn't specifically point you to it, I doubt that it is the simplest counterexample.
Apr
10
comment Reference request: linearly independent cycles in a manifold
Is this a topology question or a linear algebra question? There are two ingredients to reduce to linear algebra: that the manifolds give homology and cohomology classes (so that you have a maps from $k^n$ to $H_j$ and $H^j$); and that transverse intersection is cap product. Now you have a linear algebra question about the rank of a bilinear form, specifically that a bilinear form with invertible determinant has full rank. So it detects every element of $k^n$, yet the form factors through $H_j$, so there was no kernel.
Apr
6
comment Topological $n$-manifolds have the homotopy type of $n$-dimensional CW-complexes
Actually, Tom knows from Milnor that a topological manifold has the homotopy type of a CW complex, which gives the comparison theorem.
Apr
6
comment Topological $n$-manifolds have the homotopy type of $n$-dimensional CW-complexes
Wall's theorem requires singular cohomology as input, so you need another theorem, probably also in Bredon, comparing singular cohomology to sheaf cohomology, just using the manifold hypothesis.