@AkivaWeinberger Yep, peasy. Now let's say $(X, A)$ is a pair of topological spaces, with $r : X \to A$ being a retract. Note that the LES becomes a SES $0 \to H_n(A) \to H_n(X) \to H_n(X, A) \to 0$.
This is a useful tool. E.g., you can prove that $D^n$ doesn't retract onto it's boundary using this.
So you know how to prove the Brouwer fixed point theorem in arbitrary dimensions now.
(by SES I meant split exact sequence there, not just short. So $H_n(X) \cong H_n(A) \oplus H_n(X, A)$).
@MikeMiller i figured out where i was confused on an elementary point re: symplectic manifolds, and i'm torn between being glad i finally know better and being annouyed at how silly i was being.
the silliness was that i somehow had got it into my head that expressions like $X_F$ for vector fields had some meaning besides that of "vector field dual to dF"
which is to say, that $X_F$ has any meaning whatsoever prior to saying "Hey, i'm on a symplectic manifold"
which meant in particular i kept trying to write $X_q$ on the symplectic manifold $(q,p)\in\mathbb{R}^2$ as $X_q=\partial_q$, and getting confused at how the right version seemed to know about $\omega=dq\wedge dp$
On a different note, I can compute things. This stuff is good, not at all tedious.
I want to read the proof polar coordinates though, so I think I'll skip physical applications and read the change of variables theorem and then come back later.
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@BalarkaSen About the cyclic dunce cap… That can't be embedded in $\Bbb R^3$, can it? But for some reason I got it in my head that it could be embedded in $S^1\times S^2$, but whether that's actually true remains to be seen
I am not sure why. Those are equivalent pictures. And I am not sure how to think of identifying all 3 edges by doing a pair first and then doing the other.
I don't know how to prove that something isn't embeddable in $\Bbb R^3$. Or $S^1\times S^2$, for that matter. I suspect it has something to do with Euler characteristic (like proving that certain graphs are nonplanar).
If $[G:S]=n$ then we have to prove that $<g^n> \in S$. If I say that $[G:S]=n$ implies that $g^{|S|n}=e$ and hence $<g^n>$ is a generator such that it belongs in $S$ would that be correct?
I can prove that your triple dunce cap doesn't embed in any manifold whose homology doesn't surject onto $\Bbb Z/3$. I can prove it doesn't admit a non-pathological embedding into $S^2 \times S^1$.
It probably also doesn't admit a pathological embedding but that's harder for me to see.
Replace the complement with $H_*(M,M\semtinus X)$.
Works for arbitrary closed oriented manifold.
After computing the homology of the complex my first statement is a near triviality in the presence of Alexander duality. There's some more work in the second (hence a demand about non-pathological)
We have $\langle \widehat{x}, \phi \rangle= \langle x, \widehat{\phi} \rangle$ How do we continue? Do we now use the definition of the fourier transform?
Usually when one practices something you become better at what you are doing. My development at this forum is quite the opposite. I used to ask well received questions. Now my questions are so bad that they get closed.
Let $x>1$ be a real variable.
$$f(x) = \lim_{n = \infty} \sum_1^n \ln(n)^2 + n \ln(x)^2 - \sum_1^n \ln(x+n)^2$$
Is there an integral representation for $f(x)$ that uses no $\sum$ , named polynomials nor non-analytic functions ?
( so no " cheap" Floor function or Sum in the integrand )
Notice t...
@AkivaWeinberger If you can just embed the prism the bottom of which is the cap and the top is just plain vanilla triangle inside $S^1 \times S^2$, why can't I glue an interior of the triangle on the top after it embeds to get an embedding of the cap inside $S^1 \times S^2$ (which, as Mike said, is not possible)? That seems wrong.
@AkivaWeinberger That's fine, I was just skeptic about what happens when you embed the mapping cylinder of the degree 3 map inside, i.e., why can't you fill the easy end with a disk to contradict what Mike said. Your nontrivial loop explanation helps.
@BalarkaSen I just thought of another way to think about this.
What space is the solid torus quotiented by its boundary torus?
Pretty sure that it's $(S^1\times S^2)/S^1$.
And there's an embedding of the mapping cylinder of degree 3 into the solid torus, in such a way that its boundary lies entirely on the boundary torus.
Because you first pinch the boundary circles of the meridianal disks, which gives $S^1 \times S^2$ and then identify all those pinch points which squishes a longitudinal $S^1$.
There is a lottry of 20 different balls numbered 1 to 20. what is the probability to get in 10 in a row lotteries numbers from 8 to 20 ? (only 7 balls can be drawn).
Hello again. I posted my question regarding on probability mass functions on Stats.SE: http://stats.stackexchange.com/q/204202/67220 If anyone would like to take a look at it, I'd be happy to chat with you right here. I'd also be happy to give you points if you leave a response, of course.
@PVAL I guess that makes sense. I see how that could be useful in other areas of math. But saying "I want to prove fixed point theorems" doesn't easily point you in the direction of thinking of cycles modulo boundaries.
@BalarkaSen Remind me — two subspaces are in cobordism with one-another if their union is the boundary of a manifold-with-boundary. Or something like that?
now, for most points $z$ near $\eta$ i expect that integral to be complex. but (if i'm remembering right) there should be a line of $z$ passing through $\eta$ for which this integral is purely real
(i have to say locally to avoid worrying about branch points)
i also know (from stuff i did a while ago and don't remember as well) that if i pick $\eta=\pm a$, then one instead has three curves emanating from this branch point
@AkivaWeinberger Bordism is dual to cobordism, I think. $X$ be an arbitrary topological space, then given two $n$-folds $M, N$, maps $f : M \to X$ and $g : N \to X$ are bordant if there is a cobordism $W$ between $M, N$ and a map $h : W \to X$ such that $h$ extends $f$ and $g$.
i should also point out that the standard terminology for this is (evidently) Stokes lines and anti-Stokes lines, depending on whether you pick purely real or purely iimaginary
i.e. lines is the standard language, despite them really being curves.
@AkivaWeinberger Homology between cycles (by homology I mean the equivalence relation) is essentially the same as this, with manifolds replaced by simplicial complexes.
@PVAL But Poincare's original idea of homology was cobordisms, right? I read that somewhere. Someone criticized his ideas weren't rigorous enough so he used a different approach or something.
@pval yeah, should be. there are some possible complications due to the branch cut, but that shouldn't be a problem if one stays in the upper half-plane.
I think the higher dimensional analogue to the existence of these lines is the "h-principle for totally real submanifolds" developed by Eliashberg which my adviser has tried to tell me things about.
@MaryStar There are infinite groups with each element of finite order. This (or at least the way to do this that is is obvious to me) should be about the order of the elements. Not about the order of the whole group.
@Danu: When I say context I mean like the surrounding words. But yes I would assume he means that $\overline U$ is compact and contained in the interior of $V$.
@BalarkaSen It wasn't that his ideas weren't rigorous enough, those darn pedants, it's that they actually did not make sense. Everything was submanifolds, not "maps out", and nothing worked; not to mention there were no abelian groups, he was just counting numbers of "homologically independent" things.
Like submanifolds being homologous meant there was something mutually bounding them, but this does not make any sense if your submanifolds eg intersect.
The idea was there but it was still not actually correct. So he realized the way to do everything was simplices.
@PVAL: Is that actually useful? I thought h-principles were just cool.
Certainly Eliashberg's existence theorems for symplectic/Stein structures have been the theorems (maybe outside Donaldson's and Freedman's) with the most widespread applications in GT.
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Let $x>1$ be a real variable.
$$f(x) = \lim_{n = \infty} \sum_1^n \ln(n)^2 + n \ln(x)^2 - \sum_1^n \ln(x+n)^2$$
Is there an integral representation for $f(x)$ that uses no $\sum$ , named polynomials nor non-analytic functions ?
( so no " cheap" Floor function or Sum in the integrand )
Notice t...
