@AlessandroCodenotti That's what I was thinking. For instance, $\langle \bigcup i_k(G_k) \rangle$ can contain the element with a nonidentity element in every component, and $\prod^w G_i$ doesn't contain any of these elements.
I agree, but the exercise says thay $\prod^w G_i$ is the weak internal product of the $i_k(G_k)$ (while $\prod G_i$ would be the internal product of $i_k(G_k)$)
Hi. I'm looking for sufficient conditions of $\lim_{K\to\infty}K^{-1}\sum_{i=1}^K a_{i,K}=0$, where $0\leq a_{i,K} \leq \ell < \infty$. Am I right that I get it by the dominated convergence if I assume $a_{i,K}\to 0$ for every $i$ as $K\to\infty$?
So the trajectory is I think just $\phi(t;x_0)$. And the $\omega$ limit set consists of all the points for which $\phi(t;x_0)$ comes near that point as $t$ goes to infinity (ie, there is a sequence of times such that $\phi(t_n;x_0)$ converges to a point in the omega limit set)
How to draw an object and rotate it in oblique frontal (dimetric) projection properly ?
An illustration of projection:
I've already made a program (Pascal with Graph unit) which does it, but I think that it draws an object incorrectly.
program p7test;
uses PtcCrt, PtcGraph;
type
TPixel ...
@TedShifrin Akiva brought up another good point on the holonomic approximation theorem that stumped me for a while. The answer turns out to be rather surprising.
I standardly put the $x$-axis coming out of the paper, the $z$ axis vertical, and the $y$ axis making something like a $100^\circ$ angle (to the right) with the $z$-axis. @V.7
I definitely don't like $x$ pointing completely to the left. More like $120^\circ$ counterclockwise from the $z$-axis.
@TedShifrin Yeah, think about the section of the 1-jet bundle over an annular neighborhood of $U$ of the unit circle in $\Bbb R^2$ that's given by $s : U \to U \times \Bbb R\times \Bbb R^2$ given by $s(x, y) = ((x, y), 0, (-y, x))$. How can you perturb the the circle in $U$ to get a new neighborhood $V \subset U$ such that there is an "exact section" $((x, y), f(x, y), \nabla f(x, y))$ that approximates $s$ arbitrarily close?
@TedShifrin Well, no, we were dealing with integrability. Also over intervals, where this problem does not occur; the key point is $U$ is not simply connected.
The argument you'd do is that integrating over $(-y, x)$ around that perturbed circle in $V$ would give you $2\pi$, but if you integrate $\nabla f$ around that circle that's always... 0. The resolution is the approximation is $C^0$, not uniform or $C^1$ or anything.
@TedShifrin This is the relevant theorem, quoting Eliashberg-Mishachev: Let $K \subset V$ be a polyhedron of codimension $\geq 1$ and $\omega$ a p-form. Then there exists an arbitrarily $C^0$-small diffeotopy $h_\tau : V \to V$ such that $\omega$ can be $C^0$-approximated near $\tilde{K} = h_1(K)$ by an exact p-form $\tilde{\omega} = d\alpha$.
@V.7, if you know some linear algebra, I can send you a section of my linear algebra book that talks about how to do perpective projection the way programs like Mathematica do it.
Let $\{G_i \mid i \in I\}$ be a family of groups, then $\prod^w G_i$, the external weak direct product, is the internal weak direct product of the subgroups $\{i_k(G_k) \mid k \in I\}$, where $i_k : G_k \to \prod G_i$ is the canonical embedding.
I have already shown that the $i_k(G_k)$ are n...
@Balarka: I'm guessing that the key for DogAteMy's (or my) example is for your diffeotopy to rotate the circle, turning $x$ into $y$ and $y$ into $-x$, and then the form becomes exact.
But you're trying to then use a Cantor-diagonal trick to prove that $y^\infty$ lies in the $\omega$-limit set, @Sha.
You're supposing $y^j$ is a sequence in the $\omega$-limit set which converges to some point $y^\infty$. You want to prove that $y^\infty$ is also in the $\omega$-limit set.