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Oct
12
comment The quotient stack $[\mathbb{A}^n / \mathrm{GL}_n]$
This stack classifies vector bundles of rank $n$ together with $1$ global section. You see it just like you see that $A^1/G_m$ classifies line bundles with a global section.
Oct
11
comment Are there nontrivial involutions of $S^7\times S^7$ with fixed point set homeo to $S^7$?
This involution is conjugate to the swap by $(x,y)\mapsto (xy,y)$.
Oct
7
comment Why the Dold-Thom theorem?
@მამუკა ჯიბლაძე, I agree that is the right perspective, but you need a little bit more. $\mathbb Z[X]$ is a simplicial abelian group. There are two things you can do with it: turn it into a chain complex and take homology; or turn it into a space and take homotopy groups. You need to know that they correspond. This is sometimes included in Dold-Kan.
Oct
5
accepted Is homology finitely generated as an algebra?
Oct
5
comment Finite complexes whose homotopy groups are not “finitely generated”
Given the answers to my question, I've changed my mind and think the answer is no: that there are rational examples that are not finitely generated. But I don't have any.
Oct
4
comment Is homology finitely generated as an algebra?
@john Wow! I expected the opposite. I think you're right: the whole algebra is finitely generated as a module over the kernel of $d$, so the kernel of $d$ is finitely generated as an algebra.
Oct
3
comment Is homology finitely generated as an algebra?
The homological convention is that $d$ lowers degree, while the cohomological convention is that it raises degree. You can switch between them by negating degrees, but that switches whether a DGA is positive or negative.
Oct
3
comment Is homology finitely generated as an algebra?
Do you have citations for Bhatt and Halpern-Leistner?
Oct
3
comment Is homology finitely generated as an algebra?
Your last parenthetical sentence implies that $R$-modules are a reflective subcategory of $\mathbb Q[x,y]$-modules, thus an example of a smashing localization. This is generally true about open subschemes. This is familiar algebraic localization in codimension 1. Codimension higher than 2 isn't much different than 2.
Oct
3
comment Is homology finitely generated as an algebra?
This has the advantage over Akhil's example of being generated in positive degrees. It is positive with respect to the cohomological convention, but if you move $z$ to degree 4, it is positive with respect to the homological convention. . . . Can't you make a free one by $dz=abc$, with $a,b,c$ odd?
Oct
3
awarded  Nice Question
Oct
3
asked Is homology finitely generated as an algebra?
Oct
2
comment Finite complexes whose homotopy groups are not “finitely generated”
I meant that I believe it's true, but I don't have an explanation.
Oct
2
comment Finite complexes whose homotopy groups are not “finitely generated”
By Quillen's model of rational homotopy theory by DGLAs, the rational question becomes: if we have a differential graded Lie algebra that is finitely generated (and free?) as a graded Lie algebra, is its homology finitely generated as a graded Lie algebra? I think so.
Sep
19
comment $H^4(BG,\mathbb Z)$ torsion free for $G$ a connected Lie group
I went back to Deligne and he gives an easy proof using just the one spectral sequence. Since everything on the 4 line is torsion-free, $H^4(BT)$ is the sum of those pieces, including $H^4(BG)$. So the image must saturate over $\mathbb Z$. In particular, if you know the rational result, the integral result follows. But identifying the image doesn't seem very closely related to knowing that there is no kernel.
Sep
18
comment vanishing higher cohomology group for property T group?
The paper says that the group has an aspherical presentation complex. This is a finite classifying space for the group. It gives a resolution of $\mathbb Z$ by $\mathbb Z[G]$-modules.
Sep
14
comment on Brieskorn Manifolds
The question of whether a the subset "is" a smooth manifold is problematic. That a topological space is a manifold is a property of that space, a sensible question. But a smooth manifold is extra data. It is sensible to ask whether a subset of a smooth manifold is a smooth submanifold, hence inherits a smooth structure.
Sep
14
comment vanishing higher cohomology group for property T group?
Ken Brown, Cohomology of Groups, VIII 6.7. Or his notes, 4.3: math.cornell.edu/~kbrown/papers/cohomology_hangzhou.pdf
Sep
10
comment In what sense is the classification of all finite groups “impossible”?
(1) what? the arxiv link? the MO link? (2) It is not clear to me that the standard tame question, classifying matrices up to conjugation, is itself in P. Rational canonical form is not, because it requires factorization, but I suspect that there is a way around that.
Sep
10
comment $H^4(BG,\mathbb Z)$ torsion free for $G$ a connected Lie group
That makes sense, though I am still very confused. That answers your question, right? It identifies the group with a group that is obviously torsion-free.