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Dedalus
Dedalus
09:03
@DenisNardin Is there a way to get to the higher homotopy groups of the homotopy pushout in the example I gave above? Blakers-Massey doesn't seem to apply... Are there some LES to use?
Qiaochu Yuan
Qiaochu Yuan
09:35
@Dedalus: there's no simple description of the higher homotopy groups of a homotopy pushout. think of S^2 as a pushout of pt -> S^1 <- pt; we know all the higher homotopy groups of all the spaces involved but certainly don't know those of S^2...
Dedalus
Dedalus
10:06
@QiaochuYuan True. However, maybe one can sometimes get something about the say 2-type by say crossed modules?
But that would probably involve annoying algebra
 
1 hour later…
Mr. Chip
Mr. Chip
11:14
Say Y is a non-singular positive-dimensional subvariety of X = P_k^n, k alg. closed. It's possible to prove that k coincides with the global sections of the formal completion of X along Y. Can anyone suggest why that could help to prove the following: if f: X \to Z is a morphism sending Y to a point, where Z is a k-variety, then f sends X to a point. Thanks.
I guess my basic problem is a lack of intuition for what the global sections ("formal-regular functions") are or what they should tell us.
 
1 hour later…
Denis Nardin
Denis Nardin
12:36
@Mr.Chip The intuition you should have is that a formal completion behaves somewhat like a tubular nbd
Mr. Chip
Mr. Chip
Denis: I'm afraid I don't know what you mean by a tubular nbd.
@DenisNardin, sorry
Denis Nardin
Denis Nardin
So what your condition is saying is that there are no nonconstant functions on some nbd of Y. Now take f:X→Z a function that is constant on Y, with value some z_0. Then there some affine nbd U of z_0 such that f restricted to a tubular nbd of Y is contained in U. But then by the hypothesis f must send all U to z_0. Hence f is constant in an open, and so it is constant (if X is connected)
Tubular neighborhood
In mathematics, a tubular neighborhood of a submanifold of a smooth manifold is an open set around it resembling the normal bundle. The idea behind a tubular neighborhood can be explained in a simple example. Consider a smooth curve in the plane without self-intersections. On each point on the curve draw a line perpendicular to the curve. Unless the curve is straight, these lines will intersect among themselves in a rather complicated fashion. However, if one looks only in a narrow band around the curve, the portions of the lines in that band will not intersect, and will cover the entire band without...
The description I gave above is very impressionistic, but it is not too hard to turn it into a true proof
Mr. Chip
Mr. Chip
Hmm, okay, that makes a lot of intuitive sense. Thanks.
Would the right way to do it be to assume for contradiction no such U existed?
I mean, we're looking at an infinitesimal thickening. I'm not sure I see how to pass from that to a true neighbourhood.
By U I meant something other than you notated. I meant to refer to the tubular nhd.
Denis Nardin
Denis Nardin
You don't "pass" from one to the other, in general the nbd U does not actually exist. It is only that the infinitesimal thickening behaves similarly enough to a nbd that you can make these arguments work
For example in the rigorous proof instead of letting U be a nbd of z_0, you let it be the completion of Z at z_0
Mr. Chip
Mr. Chip
Ah, yes, okay. This is one of the things I thought of. f: X to Z should yield a map between the completions, no?
Hartshorne doesn't mention the process in functorial, but it seems like it should be.
Functorial in the pair (X,Y) I guess.
process is*
Eric Peterson
Eric Peterson
13:27
what saul said is about as much as i can say. there are some remarks at the end that are of a different flavor, where he talks about comparing his two "periods" \theta and \vartheta, but i lost the thread too far back to be able to understand
 
5 hours later…
Tyler Lawson
Tyler Lawson
18:36
so morava's paper does this thing where he takes a FGL F(x,y) over a local ring and replaces it by pi^{-1}(pi x, pi y) where pi is a uniformizer, and it turns into something additive. i'd wish i'd been able to zero in on what he's saying about how that comes into play

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