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Mike Miller
Mike Miller
12:16 AM
@ArunDebray I think that this is a useful subtlety to point out, though. I think the right statement for (1) is a stable trivialization, though I think (1) is true as stated for any bundle of rank 3 (using obstruction theory to see every oriented bundle over a 2-complex is isomorphic to a stabilization of an oriented plane bundle, and that rank 3 bundles over a 2-complex are classified by $w_2$). I guess you knew this, but I thought it might be stating anyway.
 
Mike Miller
Mike Miller
12:27 AM
bundles over a 3-complex* (the map BSO(3) -> B^2 Z/2 coming from classifying SO(3) -> BZ/2 -> BSU(2) is a 3-equivalence), might be worth* stating
 
Arun Debray
Arun Debray
1:16 AM
@MikeMiller I knew the idea, but it is good to have the details, which I did not know. Thank you!
 
 
5 hours later…
Saal Hardali
Saal Hardali
6:21 AM
Is there a Galois correspondence between certain purely inseparable extensions and formal groups?
I know there's something for purely inseparable of exponent 1 and restricted lie algebras (although i have no idea how it works) and was wondering whether there's a nice general story here.
There's an obvious formal group associated to every purely inseparable extension which is just the scheme of automorphisms. For infinite algebraic extensions it will be a fornal scheme of course.
So the question is whether the usual statements of galois theory hold in this setting...
 
 
1 hour later…
Simon Pepin Lehalleur
Simon Pepin Lehalleur
7:46 AM
@dhy If by "ayoub learning thing" in "Oberwolfach", you mean the summer school in Berlin about the conservativity conjecture :-) then you are welcome to drop by.
 
Thomas Rot
Thomas Rot
8:11 AM
@MikeMiller: thanks a lot for your explanations. There is definitely a lot to learn for me. I never really got into the nitty and gritty of spectral sequences, and this is a good reason to do so.
I actually want to do something extremely easy: To understand the equivariant cohomology of X and X\times S^1, when the group is acting freely on the second factor. This means that the group acts freely on X\times S^1, so the equivariant homology would be that of the quotient. But I'm having some troubles visualizing this. Is it just the homology of X?
 
 
1 hour later…
Denis Nardin
Denis Nardin
9:19 AM
@ThomasRot Borel homology of Y is just ordinary homology of the homotopy quotient $Y_{hG}$, so in the case of an induced G-space $X×G$ it is just the homology of $X=(X×G)_{hG}$
 
 
2 hours later…
Sean Tilson
Sean Tilson
11:38 AM
@ThomasRot Given various statements about the equivalence between Borel and Bredon cohomology, I might point you towards the only reference I know of for the K\"unneth theorem with respect to Bredon cohomology. Mandell and Lewis have a paper on this. I would imagine that we could chat about it a bit and determine if there is some collapse result in the case you are interested in.
 
Thomas Rot
Thomas Rot
@DenisNardin: thanks, that is what I thought in the end
@SeanTilson: Thanks, I'll have a look. I was chatting with Christian on some equivariant analogues of the stuff we did last week. The map I came up with is maybe the ordinary product, because of the iso of Denis.
 
user1732
user1732
12:46 PM
@AaronMazel-Gee i left you a message :-)
 
 
2 hours later…
user2236
user2236
2:25 PM
Starts at 4:30
 
user131753
2:56 PM
Let $\mathbf{CSp}$ be the category of closure spaces, i.e., of spaces of the form $(X,\text{Cl}_X)$ where $\text{Cl}_X$ is the closure operator on $X$. The morphisms of $\mathbf{CSp}$ are the maps $f: X\to Y$ such that $(f\circ \text{Cl}_X)(A)\subseteq (\text{Cl}_Y\circ f)(A)$ for all $A\subseteq X$.
 
user131753
Then there seems to exist an embedding from $R$-$\mathbf{Mod}$ (the left category of $R$-modules) to $\mathbf{CSp}$.
 
user131753
For if $(M,\circ)$ be a $R$-module then we can define a closure operator $\text{cl}_{M}:\mathcal{P}(M)\to \mathcal{P}(M)$ by letting $\text{cl}(S)=\langle S\rangle$ for all $S\subseteq M$ (where the $\langle S\rangle$ denotes the submodule generated by $S$).
 
user131753
Furthermore if $f$ is a $R$-module homomorphism between $(M,\circ_M)$ and $(N,\circ_N)$ then the same $f$ can be viewed as a closure space $(M,\text{Cl}_M)$ and $(N,\text{Cl}_N)$ where $\text{Cl}_M$ and $\text{Cl}_N$ are defined as above. This is so because $$f(\langle S\rangle)\subseteq \langle f(S)\rangle$$ for all $S\subseteq M$.
 
user131753
So if we define a functor $\mathscr{F}:R\text{-}\textbf{Mod}\to\textbf{CSp}$ which associates to each $R$-module $(M,\circ)$ the corresponding closure space $(M,\text{Cl}_M)$ and to each $R$-module morphism $f$ the same $f$, we get an embedding. Isn't it?
 
user131753
3:12 PM
But that would mean $R\text{-}\textbf{Mod}$ to be isomoprhic to a subcategory of $\textbf{CSp}$. My question is: what is the relation of this category with $\textbf{Top}$?
 
Denis Nardin
Denis Nardin
3:54 PM
@user170039 My understanding from the Kuratowski description of topological spaces is that $\mathrm{Top}$ is the subcategory of $\mathrm{CSp}$ spanned by those closure spaces such that $\mathrm{cl}\varnothing=\varnothing$ and $\mathrm{cl}(A\cup B)=\mathrm{cl}A\cup \mathrm{cl}B$ (i.e. $\mathrm{cl}$ preserves finite unions). It's quite clear that the span operator does not satisfies those axioms
 
user131753
@DenisNardin Indeed.
 
user131753
Any other relations?
 
user131753
Just to be clear by a closure operator I was referring to the following,
 
user131753
Closure operator
In mathematics, a closure operator on a set S is a function cl : P ( S ) → P ( S ) {\displaystyle \operatorname {cl} :{\mathcal {P}}(S)\rightarrow {\mathcal {P}}(S)} from the power set of S to itself which satisfies the following conditions for all sets X , Y ⊆ S {\displaystyle X,Y\subseteq S} Closure operators are...
 
Denis Nardin
Denis Nardin
It's unclear to me why should there be any other relations... There might be adjoints to the inclusion, but they're probably on the uninteresting side. Is there anything you expect?
 
user131753
4:10 PM
@DenisNardin Actually I was hoping that to show that a subcategory of $R$-$\textbf{Mod}$ to be equivalent to $\mathbf{Top}$ using this embedding.
 
Denis Nardin
Denis Nardin
I wouldn't count on that
 
user131753
@DenisNardin I know. I was told this earlier while I asked a related question here.
 
Mike Miller
Mike Miller
5:26 PM
@ThomasRot @ThomasRot So $G$ is cyclic or $S^1$. In the latter case indeed the quotient is just $X$ via the mantra "anything x G is anything else x G"; you can write down the equivariant homeomorphism by hand, and in paritcular, this is the same as $X_{triv} \times G$.
If $G$ is some finite cyclic group this is more interesting: if $\varphi$ generates the $\Bbb Z/n$ action on $X$, equip $X \times \Bbb R$ with the $\Bbb Z$-action $(x, t) \mapsto (\varphi x, t+1)$. Quotienting by the action of $n\Bbb Z$ gives rise to the space $X \times S^1$, with a leftover $\Bbb Z/n$ action that is precisely yours. Quotienting by the full $\Bbb Z$ is by definition the mapping torus of $\varphi$. So you are computing the homology of the mapping torus of $\varphi$.
The action of $H^*(B\Bbb Z/n)$ is induced by the map $\text{MT}(\varphi) \to B\Bbb Z/n$ classifying the covering space.
 

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