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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.
Apr
5
comment Topological $n$-manifolds have the homotopy type of $n$-dimensional CW-complexes
That is overkill and probably circular.
Apr
4
comment The image of the Hurewicz map for rational loop spaces
Jeff knows that it is not finitely generated in the commutative dga case. mathoverflow.net/questions/182437/…
Apr
3
comment Pseudomanifolds and Poincaré duality
No, that doesn't work. It is necessary that the parameterizing variety have $\chi=0$.
Mar
31
comment Pseudomanifolds and Poincaré duality
Here's a modification of the algebraic example that feels less "finite." You can interpret the elliptic curve example saying that elliptic curves have a canonical polarization, so the canonical bundle of curves on $M_{1,1}$ has a projective structure, thus there is an associated bundle of affine cones. Same for $M_g$. But there are more interesting complete curves in $M_g$, so those support a bundle of affine cones, so the singularities twist more.
Mar
28
comment Pseudomanifolds and Poincaré duality
(but if you are only interested in homology with $\mathbb Q$ coefficients, 2,3,4 are fine)
Mar
28
comment Pseudomanifolds and Poincaré duality
@DavidC maybe this example can be made algebraic. Replace the the hyberbolic automorphism with the order 6 (not 2,3,4) automorphism of the right elliptic curve. And replace the cone on an elliptic curve with an affine cone. The link of its singularity is not the elliptic curve, but the $S^1$ bundle over it. And replace the circle with a complete variety, say, another elliptic curve. But the surgery is hopeless, so twisted coefficients are out.
Mar
27
comment Pseudomanifolds and Poincaré duality
Also, the third bullet point in David's definition of pseudomanifold is common, but I think it's a silly axiom. It amounts to the normalization being connected. I guess the point is to reduce the number of orientations, just as a connected manifold has at most one. Connectedness hypotheses are usually a bad idea, but this one is also unwieldy.
Mar
27
comment Pseudomanifolds and Poincaré duality
I continue to doubt that there are any algebraic examples, but this one has a nice dualizing sheaf, so my earlier suggestion was inadequate to eliminate them.
Mar
27
answered Pseudomanifolds and Poincaré duality