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Dec
18
comment Lefschetz fixed notation
The main question is whether one notation is winning, which doesn't depend on whether you see a difference between the choices, or even if you don't care about consistency across papers. If no convention is winning, then there are subsidiary questions about what is better. Maybe those were a mistake.
Dec
18
asked Lefschetz fixed notation
Dec
17
comment Which mapping class group representations come from algebraic geometry?
This question already has nontrivial answers, but trivial answers are useful, too. (1) The tensor product of the symplectic representation with itself is a new PVHS. (2) Every finite index subgroup gives a permutation representation, which is a (boring weight 0) PVHS. Even if they factor through $Sp_{2g}(\mathbb Z)$, they don't come from the defining representation. And the group is residually finite, so many don't.
Nov
18
comment Reference request for cohomology of coverings
@Aleksey There are examples with higher cohomology when the fundamental group is just $\mathbb Z^n$.
Nov
17
answered Reference request for cohomology of coverings
Nov
16
comment Is there a category whose isomorphisms are precisely the simple homotopy equivalences?
It may be worth knowing that if you take the nerve of the simple homotopy equivalences (ie, abandoning interest non-equivalences, retreating from categories to groupoids), you get an interesting space, which is different from the nerve of the homotopy equivalences, and not just in its set of components.
Nov
5
comment Which mapping class group representations come from algebraic geometry?
Yes, it works. There are two difficulties. The first is that there are lots of choices; you just have to make all of them. That is: take the universal curve over the moduli stack of double covers of a genus $g$ curve ramified in $n=4g-4$ points. This maps to $M_g$, so its cohomology is a local system there. The second problem is the stackiness: when you take cohomology, it's like taking $\mathbb Z/2$ invariants, but the interesting cohomology all has action by $-1$. But you can just take the tensor square, or some other ad hoc option.
Nov
5
answered Which mapping class group representations come from algebraic geometry?
Oct
27
comment Are there nontrivial involutions of $S^7\times S^7$ with fixed point set homeo to $S^7$?
Yes, the method is surgery, so it has the drawbacks of surgery. We'd like to distinguish involutions by the isomorphism type of the quotient of the complement of a regular neighborhood of the fixed set, but that's hard so we mark so that we can apply surgery. And even if there are isomorphism once we drop the marking, that doesn't mean that there are automorphisms of $M$ that induce them.
Oct
26
comment Does every smooth, projective morphism to $\mathbb{C}P^1$ admit a section?
@Marty, that's a great idea. It suggests a generalization: that a smooth proper map to a curve contains a complete curve which maps to the target by an etale map.
Oct
25
answered Are there nontrivial involutions of $S^7\times S^7$ with fixed point set homeo to $S^7$?
Oct
24
comment Doing some homological algebra in triangulated categories
I am skeptical that you can iterate the construction in the third paragraph beyond $n=2$.
Oct
23
comment Does every smooth, projective morphism to $\mathbb{C}P^1$ admit a section?
Is there any reason for "projective" rather than "(algebraic) proper"? (Of course, for analytic proper maps, it's false.)
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.