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05:01
arxiv.org/abs/1707.01799 Nikolaus-Scholze
 
9 hours later…
14:08
hi, $$\text{Let } p \text{ prime, }G\text{ a group of order }2^p-1. \text{ Is true that }G\text{ abelian ? }$$
can you give a counterexample please
i suggest that you go find one yourself.
Well
here's a hint. consider p = 5
14:13
31 is prime
and so G is cyclique
is it 2^{p-1} or 2^p - 1, your mark up choices are confusing
My job is proposed enigma, I have propose them, and I believed, had a proof, but I do a mistake, now I don't know, if this result his true or not
if it's 2^{p-1} then p=11 gives examples: groupprops.subwiki.org/wiki/Groups_of_order_1024
2^{p}-1
sorry, for my bad english
just glancing at the table on wikipedia of composite mersenne numbers, it's not clear. The easiest thing to try would be finding divisors which are of the right size to form a semi-direct product, but I don't see any
14:31
Hello there. I'm reading Hatcher's AT and having trouble understanding the proof of Van Kampen's theorem in section 1.2
Can someone show me explicitly what the "compactness argument" is? How to explicitly construct the covering by the rectangles?
lines 5-7
@wilkersmon It is a bit painful to write down properly, but roughly the idea is as follows: for every point i of [0,1]×[0,1], you can find a small rectangle R_i, having i in its interior, such that F(R_i) is contained in one $A_\alpha$. Then the collection of the interiors of the R_i form an open cover of [0,1]×[0,1]. By compactness of the square, we can extract a finite subcover R_1,...,R_n. Then you just need to prove that (cont.)
you can find a sufficiently fine grid on [0,1]×[0,1] such that every square of the grid is contained in one of the R_i's. This follows, e.g., from Lebesgue covering lemma although there are more elementary ways.
I think that Hatcher wants to use the grid obtained by taking all the lines on the edges of the R_i's, which should work.
Thanks! Since each $A_\alpha$ is open and F is continuous, I think it suffices to take the preimages of each $A_\alpha$ as the open covering of I^2 then, right?
The other part I understood, thanks :)
15:21
I'm trying to pin down the abstract reason for the existence of the pontryagin-thom collapse map. The setting I think should be a relative finite type space embedded in a vector bundle. That is a map with finite type fibers $X $
$X \to Y$ s.t. $X$ comes with an embedding into some vector bundle.
Then I think the collapse map should be some connecting map in a homotopy fiber sequence obtained from one-point compactifying the vector bundle and stabilizing.
Or something along these lines.
Any help would be appreciated...
15:40
Isn't the Pontryagin-Thom collapse map, up to homotopy, the map from V into the cofiber of the inclusion $V \setminus M \hookrightarrow V$?
where $V$ is a vector bundle and $M$ is your space
 
1 hour later…
17:01
Probably something trivial that I'm not seeing right now. Is it true that [C^T, D] \cong [C, D]^{[T, 1_D]} for a comonad T? In the nlab page ncatlab.org/nlab/show/Eilenberg-Moore+category#Street72 , it says that this is true. On the other side, not every functor C^T --> D lifts to a functor C --> D since some morphisms in C are not algebra morphisms.
17:53
A personal production : $$ G_n \text{ the sub-group of order }2^n5^{100-n} \text{ of } (\mathbb Z/10^{100}\mathbb Z,+).\\ \text{ Calculate } \text{card}(\bigcup \limits_{n=0}^{100} G_n ).$$
 
2 hours later…
20:03
an indice : there are a simple solution which is not easy to find
 
2 hours later…
22:09
@SaulGlasman Yes! thanks for clearing this up! It's reassuring to see once again that the thom space is just a topological version of local cohomology.
I mean all the Thom business is basically doing local cohomology without choosing a cohomology theory so that you're treating all of them in one stroke. Is this a correct picture?
@SaalHardali well it's certainly not a historical picture.
yeah that much is clear to me...

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