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7:51 AM
@MikeMiller: Thanks! I actually wrote the stuff you explained (about the homotopy type of the operators of fixed kernel/cokernel) already for a future paper. But this is still helpful. Especially the remark that "being of finite dimensions" is only a constraint in infinite dimensions. This is a nice viewpoint.
 
 
3 hours later…
10:57 AM
@ThomasRot I would also be interested in knowing the homotopy type of $\text{Fred}^I_{[k_0, k_1]}$, the space of Fredholm operators with index $I \geq 0$ and $0 \leq k_0 \leq \dim \text{ker}(A) \leq k_1$. This would help me better understand this process of trading kernel for cokernel, even for $k_1 = k_0+1$. But I didn't see how to do this.
 
 
1 hour later…
12:22 PM
@MikeMiller: That is a nice question. I should probably wait longer before typing anything, but...
What about finite dimensions? If we take an n dim vector space, and look at the linear transformations of rank at least k. Then as k decreases the space becomes more and more connected.
So if k=n, we have the invertibles, but if k=n-1, then the space becomes connected (there is a path from diag(-1,1,...,1) to diag(1,....)). if k=n-2 it becomes simply connected (even 3 connected I guess). I don't know what happens with higher homotopy groups.
But let me think a bit longer on this
 
One can see I think without too much difficulty that $\text{Fred}^I_{[k_0, k_1]} \to \text{Fred}^I_{\geq k_0}$ gets highly connected as $k_1 \to \infty$ just from transversality (the subspace of operators with large kernel has high codimension). This applies just as well in finite dimensions, though you can't say anything about really high homotopy groups
 
right
 
I don't really know if there's a good recipe for taking the homotopy type of pieces of a stratified space (and some sort of relationships between them) and reconstructing the homotopy type of the whole thing
 
12:43 PM
I guess transversality should be the main tool
but maybe you only get very weak statements that way
 
12:55 PM
@ThomasRot I would have guessed there's some statement for "linearly stratified spaces" like these that says that X[i,j] is maybe the adjunction of X[i] with X[i+1,j], glued along the link of the singularity
So you describe X[i,j] as a mapping cone (not great for homotopy groups but something)
 
@MikeMiller: I don't think I understand this statement fully
link of the singularity means what in this context?
 
To be clear I'm not stating anything that I know to be true, just a guess. Let X be a stratified space, with locally closed strata X[j] indexed on the naturals, and such that the closure of X[j] contains X[i] if and only if i >= j. I guess there is some subspace Lk(X[i,j]) < X[i] which is the boundary of a neighborhood of the singular subspace X[i+1,j] inside of X[i,j], and a map from the link to X[i,j]. Then I hope that one can describe X[i,j] as the (homotopy) union X[i] cup_{Lk} X[i+1,j].
 
Ok, and in this context you might be able to describe these neighborhoods, hence the thing you call the link, as the total space of some tautological bundle over the lower dim stratum
 
Right, this was my hope
 
Yeah, I think it is possible to get explicit bundles in this case.
 
1:21 PM
If $\Phi_k^l$ denotes the space of Fredholm operators with index k, coker l, then it is homotopy equivalent to BO(k+l)\times BO(l) as you said. The externel tensor product of the (dual of the first, non-dual of the second) tautological bundles is isomorphic to the normal bundle of \Phi_k^l in the space of all fredholm operators \Phi_k of index k. You can also see this bundle as the homomorphisms from the kernel to the cokernel.
Hmmm, the rank (at most) 1 homomorphisms do not really form a vector subbundle, do they? That is sort of the link where you are gluing right?
I should sit down and do some more computations.
 
2:24 PM
@MikeMiller Cough cough
 
2:35 PM
@DenisNardin: That is a lot of words that I need to learn.
 
@ThomasRot Well, maybe the whole theory is a bit of overkill. But the point is that you also need to remember the homotopy type of the link of a stratum inside the next one
(also, if it wasn't obvious, I was linking to two separate papers)
 
 
4 hours later…
6:56 PM
is it true that $\mathbb{Q}_p \cong \mathbb{Q} \otimes_\mathbb{Z} \mathbb{Z}_p$? i am 95% convinced, but i can't find any references that say so, which seems surprising for something that if true would be so basic
 
7:12 PM
@AaronMazel-Gee Yes, that should be true. I mean, you can get $\mathbb{Q}_p$ from $\mathbb{Z}_p$ by just inverting p. If you are after a formal proof, note that \mathbb{Q} is the localization of \mathbb{Z} at the multiplicative closd subset of non-zero elements in \mathbb{Z}. So $\mathbb{Q} = S^{-1}\mathbb{Z}.$ Then $\mathbb{Q} \otimes_\mathbb{Z} \mathbb{Z}_p \cong S^{-1} \mathbb{Z}_p.$
On the right hand side of this isomorphism localize with respect to the image of S under the map $\mathbb{Z} \to \mathbb{Z}_p.$ So I invert $p$ in $\mathbb{Z}_p,$ so I win. (Since every non-zero element in $\mathbb{Q}_p$ can uniquely be written as $p^n u$ for $ n \in \mathbb{Z}$ and $u$ a $p$-adic unit)
(Sorry if this was too long, but I got the impression that you wanted to make your strong conviction more rigorous)
 
 
4 hours later…
11:11 PM
@Dedalus right, thanks for confirming -- that was my argument as well. maybe the reason this isn't mentioned in basic introductions to p-adics is just that tensor products aren't necessary to discuss them otherwise.
 

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