@DanielFischer Do you have any new thoughts about what we could possibly be overlooking about the parameter $a$? The $2 \pi$ restriction seems to apply to the entire family of infinite series.
Suppose that $R$ is a finite non-zero ring without zero divisors, and denote $R^{*}=R\setminus \{0 \}$. Consider the action of the multiplicative semigroup $R^{*}$ on itself by left translation; i.e., for $x \in R^{*}$, we have $x.y = xy$ by definition.
I just showed that $R^{*}$ acts on $R^{*}$...
@Semiclassical Numerical approximations for the sums using Wolfram Alpha for values of $a$ greater than $2 \pi$. But the formulas themselves suggest that a restriction of some sort is needed since they don't make sense for large $a$.
Guys any idea about this situation - "I want to read the book online from google books but for me internet service will not be there tomorrow , any technique how i can access the pages while offline..."
Let me summarize: Let $d \in D$ and $f = d(d,B) - d(d,A)$. $f > 0$ by definition. consider $Y$ an open-ball centered at $d$ with radius $\frac f 2$ we need to prove that $Y \subseteq D$ let $y \in Y$ we need to prove that $d(y,B) > d(y,A)$ so $d(y,A) < d(d,A) + d(d,y) < d(d,A) + \frac f 2 = d(d,B) - \frac f 2 < d(d,B) - d(d,y) < d(y,B)$ QED
@Mahmoud congratulations!
> Every increasing sequence in the long line converges to a limit.
What the hell
I'm quite interest in knowing that $(n,\sin n)$ converges to
Let $R$ be a finite non-zero ring without zero divisors. Denote $R^{*}=R \setminus \{0\}$. Consider the action of the multiplicative semigroup $R^{*}$ on itseslf by left translation - i.e., for $x \in R^{*}$, we have $x.y = xy$ by definition.
I already proved that $R^{*}$ acts on $R^{*}$ by bije...
sighs I suppose there is some argument among algebraists about that. But, as far as I'm concerned, to most human people, a ring is a semigroup under multiplication, not a monoid.
@Balarka is it true that every compact manifold is a CW-complex? We saw in class that this works for compact surfaces so I was wondering about the general case. I think you can cover it with a finite number of open $n$-balls and then do some fiddling with the borders where they overlap but I'm not sure how to formalize this last part (assuming it's correct)
@AlessandroCodenotti Homotopy equivalent to, yes. Homeomorphic to, open.
For smooth manifolds you're always triangulable so necessarily homeomorphic to a CW complex. But there are nasty topological 4-manifolds where things could go wrong.
Mike already answered to let me just say that for the fact upto homotopy equivalence you need the nerve theorem, once you get a good cover on your manifold (which is not too hard).
Also, my favorite proof that smooth manifolds are homeomorphic to CW complexes is through Morse theory @Alessandro. As a reference, look at the torus picture in wikipedia. For intro, there's a chapter in G&P but it doesn't tell you much.
@MikeMiller Ahh, sorry.
Out of curiosity, is dimension 4 the only place where things are open? I think there are 11 dimensional non-smoothable manifolds too (Kervaire manifold?); do we know if that's triangulable?
There are non-triangulable manifolds in every dimension greater than 5 (Galewski-Stern; independently Matsumoto; the final step by Manolescu), and plenty of non-smoothable manifolds. But we nonetheless still get CW structures in all dimensions other than 4.
This is all that Kirby-Siebenmann stuff I don't know well.
@BalarkaSen Ah, I see, maybe I'll read it at some point. Today the professor suggested us to look at Hatcher's book for the proofs of some facts we don't have the time to prove in this course so I guess I have no choice but to read it :P
Interesting: I'd never before heard of a Gauduchon manifold. But I just a few minutes ago got an email announcing a geometry conference in Rome on March 10, and one of the speakers is ... Gauduchon.
@DanielFischer Until I know what the issue is, I'm not going to be confident in any result I get using contour integration except in simple cases. That's what bothers me the most about this.
Is it nonstandard to define CW-complexes as having a finite number of $n$-cells for every $n$? Because that's included in the definition we saw in class but it doesn't seem to be case on wiki and Hatcher
Hm, I just checked and they are called finite CW complexes in the professor's notes, but I'm quite sure they where never called finite in class (or at least I never wrote it and I don't remember it)
@RandomVariable There's a problem letting $y \to \infty$. $\lvert J_k(z)\rvert$ grows like $\lvert z\rvert^{-1/2}\exp (\lvert \operatorname{Im} z\rvert)$, so replacing $\cot (\pi z)$ with $\pm i$ needs more justification than just uniform convergence. I don't see why that should work for small $a$ and stop working for $a \geqslant 2\pi$, though, but maybe looking in detail at the Bessel functions would reveal it.
@BalarkaSen infinite dimensional complexes are fine, just not with an infinite amount of cells apparently. I don't know why we're using this definition then
@DanielFischer So additional justification is needed for replacing $\cot(\pi z)$ with $\pm i$ because the magnitude of $J_{k}(z)$ does not remain bounded as $\text{Im}(z) \to \pm \infty$?
@AlessandroCodenotti Ya, the space admits a basis of s.c. open sets. It's also not semilocally simply connected, which says every point has a neighborhood such that any loop inside it is nullhomotopic inside the whole space.
Aka, there's a neighborhood $U$ around any $p \in X$ such that $i_* : \pi_1(U) \to \pi_1(X)$ is the zero map.
Ah, the point is any map $f : (X, x_0) \to (Y, y_0)$ between based topological spaces (aka $f(x_0) = y_0$) gives a map $f_*: \pi_1(X, x_0) \to \pi_1(Y, y_0)$ (how?). $i$ is the inclusion $U \to X$, and $i_*$ is the map induced from that
@RandomVariable Yes. We have $$I(y) = \int_{-\infty}^{+\infty} f_y(x)g_y(x)\,dx$$ with $g_y$ uniformly converging to $0$. To conclude $I(y) \to 0$ for $y\to \infty$, we need more. If the $f_y$ were dominated by an integrable function, that would suffice. But for $f_y(x) = J_k(x+iy)/(x+iy)$, we are very far from being able to appeal to the dominated convergence theorem.
Sure thing. In general understand how $X \times [0, 1]/X \times 0$ looks upto homotopy equivalence (the space is also denoted as $CX$ - C stands for ...?)
@DanielFischer What's $g_{y}(x)$ here? I thought we're talking about the top and bottom of the contour where $\cot(\pi z)$ is converging uniformly to $\mp i$ as $\text{Im}(z) \to \pm \infty$.
Hello, just want to confirm, I believe that $Aut_\mathbb {Q} \mathbb {Q} (\sqrt 2, \sqrt 3, \sqrt5)$ is $\mathbb {Z}_2 \times \mathbb {Z}_2 \times \mathbb {Z}_2$ since all automorphisms except the identity have order 2
@DanielFischer I wasn't even thinking about the ramifications of $J_{k}(z)$ not being bounded as $\text{Im}(z) \pm \infty$. Thanks for pointing that out. Would looking more closely at $J_{k}(az) \cot(\pi z)$ perhaps reveal something?
@LeGrandDODOM Yes, "abmalen" means you have a source/model (could be a picture, could be a bowl of fruit, a human) that you look at while you paint it (and you paint it much like it really looks). Plain "malen" is more general; you can paint after your imagination, or after a real-world model, if you paint after a real-world model, you can modify or distort it.
Assume I have a matrix $A$ that is isomorphic with a diagonal matrix $\Lambda$
Assume I have that matrix and I have that diagonal matrix
how can I compute the invertible matrix $P\in\mathbb R^n$ for which it holds that: $\Lambda=P^{-1}A P$?
I've jut gone through the trouble of finding that diagonal matrix, but I've never encountered a problem where I...
I wait
I think I know it
It's just the identity matrix of the change of basis (eigenvector basis to standard basis, I think)
right, never mind
oh wait
apparently it's the matrix of the eigenvectors
but that is of course what I described. I just didn't realise I'm going to plug in the coordinate vectors of the eigenvectors (w.r.t. the standard basis)
Giroux-Goodman says that I should be able to go between any fiberings of links (not necessairly fixing the link) by Hopf plumbing (or stabilization). I don't know of any non-trivial set of these operations
which arrives back at the same link with a different fibering
Locally simply connected means every point has a neighborhood which is simply connected. Semilocally simply connected means every point $p$ has a neighborhood such that any loop in that neighborhood based at $p$ can be nullhomotoped in the ambient space.
Hm, I guess I should try to understand why the whole space isn't simply connected and why those "bad" loops aren't contained in small enough nbhds of the point where the circles touch
Because the diagonal matrix is basically the matrix that is associated with the linear transformation $L_A$ w.r.t. a basis of eigenvectors, so you have to change the basis, which is done with the invertible matrix we were looking for
@ZachHauk The cover I defined was the set of all intervals on which the function $1/x$ is bounded. $(0,1)$ isn't in that cover, since $1/x$ isn't bounded on it