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Apr
5
reviewed Approve Sturmian subword whose reverse is not a subword
Mar
28
comment A symmetric embedding of manifolds
@SebastianGoette: do you mean that M cannot be a $4k$-dimensional Kähler manifold?
Mar
23
comment No matter how many algebraic invariants we attach to topological spaces, there will always be nonhomeomorphic spaces agreeing on all their invariants
Anton Fetisov has given an excellent and interesting answer, and it seems quite possible that this is what the OP's professor meant. Therefore I would vote against closing this question.
Mar
18
comment Grothendieck says: points are not mere points, but carry Galois group actions
You're probably right, I wasn't thinking about compact support or not, but the distinction is relevant here. It does still seem you need Artin-Verdier duality to promote simple connectivity to contractibility. I'm not at all a number theorist though!
Mar
18
comment Grothendieck says: points are not mere points, but carry Galois group actions
No, as far as I understand, Artin-Verdier duality means that $\text{Spec}\mathbb{Z}$ has cohomological dimension at least 3 (in fact equal to 3, if 2-torsion is ignored). This kind of thing is precisely why I made my original comment.
Mar
18
comment Grothendieck says: points are not mere points, but carry Galois group actions
Another problem with this "suggestion": these covers are not actually covers, but ramified covers! So perhaps the claim should be that $\text{Gal}(\mathbb{Q}^{\text{ur}}/\mathbb{Q})=0$ suggests that $\text{Spec}(\mathbb{Q})$ is a point (but of course it still isn't).
Mar
18
comment Grothendieck says: points are not mere points, but carry Galois group actions
@QiaochuYuan: Sorry, I don't mean to pick on you too much; it's just that this particular misconception for this particular example is unusually persistent. I'm not sure why certain people continue to say "suggests that X is Y" when they mean "suggests that X and Y have the same fundamental group", especially since they don't extend the same suggestibility to other fields: I haven't heard anyone say this suggests $\text{Spec}(\mathbb{Q})$ should be a profinite $K(\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}),1)$.
Mar
18
comment Does the “holomorphic spheres-to-continuous spheres” forgetful function respect the mixed Hodge structures on homotopy groups?
What does the subscript 0 mean in $\Omega^2 X_0\sim [(S^2,0),(X,x)]_0$? If it means "basepoint component", then why do you say "for any connected component"? If not, why is the isomorphism between $\pi_k(\Omega^2 X_0)\otimes \mathbb{Q}$ and $\pi_{k+2}(X)\otimes \mathbb{Q}$ canonical?
Mar
18
comment Grothendieck says: points are not mere points, but carry Galois group actions
@QiaochuYuan: Thanks for the edit. But the claim that "you can think of $\text{Spec}\,\mathbb{F}_q$ as behaving like a 'profinite circle'" still seems to rely on the fact that $\pi^{et}_i(\text{Spec}\,\mathbb{F}_q)=0$ for $i\geq 2$, or at least that $H^i_{et}(\text{Spec}\,\mathbb{F}_q;\,\mathbb{Z}_{\ell})=0$ for $i\geq 2$. These facts are true, as far as I understand, but not trivial, and in any case are not related to the position of Grothendieck the OP asked about.
Mar
18
comment Grothendieck says: points are not mere points, but carry Galois group actions
The fact that the finite covering theory of $\text{Spec}\,\mathbb{F}_q$ is the same as the finite covering theory of $S^1$ is certainly evidence that they have the same étale fundamental group. But in what sense could this be enough information to determine the étale homotopy type? After all, the finite covering theory of $\text{Spec}\,\mathbb{C}$ is the same as the finite covering theory of $S^{13}$, but that doesn't mean they have the same étale homotopy type.
Mar
15
reviewed Approve Reduction to some physical interpretation of this formula
Mar
1
comment Evaluating an infinite sum related to $\sinh$
Thank you for this enlightening connection!
Feb
16
awarded  Good Answer
Feb
11
reviewed Approve discrete-mathematics tag wiki excerpt
Jan
2
comment Why are abelian groups amenable?
@Misha: There was a small mistake in my proof (it didn't handle well the case when $g_i$ has finite order), which I've fixed now. It seems that this may have been your objection. If not, could you clarify what the issue is?
Jan
2
revised Why are abelian groups amenable?
corrected proof in case when g_i has finite order
Jan
2
comment Principal bundles that can't be detected by spheres
Any other circle bundle over the 2-torus will work as well, won't it?
Jan
2
revised Why are abelian groups amenable?
add proof of main claim
Jan
1
answered Rings that inject in all p-adic integers
Nov
12
answered equivariant embeddings from the k-th configuration space to the k+1-th configuration space