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V.7
V.7
18:00
So ... the main question is how to draw a line properly in oblique frontal projection.
As you see the Y divides by 2 ...
@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 $i_k(G_k)$)
Guys, what 'particular point' are they talking about?
they mean $x_0$?
I mean, the trajectory depends on $x_0$, so I don't really understand what they're trying to say
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 $K\to\infty$?
What's a trajectory?
18:10
@ShaVuklia How do you define the $\omega$-limit set? Usually I define it as the accumulation set of the trajectory.
The omega limit set of course depends on $x_0$.
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)
@Balarka, @Alessandro
Reading further, it seems like what doesn't matter is where to start along the trajectory
Yes, that is equivalent to my definition. I think your book defines it as the limit points of all iterates along points on the trajectory.
In which case (a) would make sense because of what Daminark said.
oh right, yea I agree with dami then
I still don't know what a trajectory is, what's the context, mechanics? (I can't help with your question just curious)
18:13
it's just a solution
with an initial condition
@Alessandro It's an integral curve of a vector field.
Or a solution to an ODE, like Sha says below (rip in rekt, by the way)
it's a solution to a differential equation
AGHH\
blyat
Ah, makes sense
what does a dashed line actually mean in the chat?
I have never understood that
18:16
donno if you can see it too :P
yeah I can
holy shit now I see more
That was the best snipe I've ever seen
ugh yea even tho I am the sniped one, I have to agree
The snipee?
18:17
ohh maybeeee it happens when someone edits something?
hm no
Who snipes the sniper?
Bertrand Russell
Mark Zuckerbergg
lol yea that can be his code name
18:19
@BalarkaSen Russell's sniper rifle does sound cooler than Russell's teapot
Nah Russell will rekk Zugg by asking him to friend every user that doesn't friend him/herself
good meme
my brain just did a big smoke "OHHH"
Hi @Ted
V.7
V.7
hey all :) how you would draw a line in 3 dimensions (oblique frontal (dimetric)) projection ?
Ah, glad to see you're discussing memes, not math :) Hi, Balarka.
@BalarkaSen it means the stuff after the dashed line occurred after you last switched tabs
18:26
@V.7: A line is a line is a line.
@anon Oh wow
Hi, @anon :)
V.7
V.7
Ofcourse, but as you see:
I don't see nothing.
V.7
V.7
There're lines which I guess I'm drawwing incorrectly
18:27
You need axes or other things in the diagram to get perspective.
V.7
V.7
0
Q: How to draw an object and rotate it in oblique frontal projection

V.7How 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 ...

If you're trying to experiment with computer graphics, draw a cube or a pyramid or something 3-dimensional.
V.7
V.7
@TedShifrin do you think that perspective should exist in oblique frontal projection ?
I don't know that terminology.
V.7
V.7
This projection looks like this:
18:28
@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.
Oh, really, Balarka?
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.
V.7
V.7
Yup, you're right ...
@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?
Hi Jasper.
18:31
@TedShifrin Ted :D
$(-y, x)$ is a non-conservative vector field... you cannot seemingly approximate it with conservative vector fields.
guys, any idea what $y^j$ and $y^\infty$ are?
That's been my complaint all along, @Balarka.
@TedShifrin Hello Ted. I hope you and your mum are well.
18:32
@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.
@TedShifrin Ted poor kasmir said hi :o
I don't see her regularly, Jasper, but will see her in the nursing home when I go back to ATL for a week in a few weeks.
I specifically brought up the $(-y,x)$ in one of my messages, @Balarka, but you probably were too distracted. :)
There was a shooting that killed 26, and a car ramming that killed 8.
A lot of killing in America.
@TedShifrin Ah, hm, maybe I missed it. But you were quite right to bring up that point, then. This is nontrivial.
Yes, because the National Rifle Association has all the politicians in their pockets.
Hi, poor Kasmir.
18:35
Finally :D
When you were making up switching $x$ and $y$, @Balarka, to turn $(0,1,x)$ into $(0,1,y)$, I specifically asked about what you'd do with $(-y,x)$.
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 Ahhh I remember now
@Balarka: A $C^0$ approximation is uniform on a compact set.
Or are you not meaning the usual norm?
@TedShifrin I mean it's pointwise
Not local pointwise?
Then I don't even know what it means.
V.7
V.7
18:37
So ... you think that this's right method to draw a point in 3 dimensions ( in this type of projection )
x1 := sp_t.x - sin( pi / 4 ) * sp_t.y / 2;
y1 := sp_t.z - sin( pi / 4 ) * sp_t.y / 2;
@Sha: Sorry, I meant to answer you but got waylaid. $y^j$ is just a sequence of points converging to the point $y^\infty$.
The approximation on the subsurface level is $C^0$; the 2-plane fields only approximate in the pointwise sense
I don't know what that means, @Balarka.
@V.7: Drawing perspective projections is very elaborate. It's not a simple 2-D rotation matrix at all.
Do you know linear algebra?
to prove that the kernel of an action of the Group G on a set A is the same as the kernel of the correspnding permutation G-->S_A
how does one starts?
V.7
V.7
@TedShifrin this's not a perspective projection
It's ortho ... isn't it ?
18:39
Well, I don't understand what the issue is. $(\partial f/\partial x, \partial f/\partial y)$ converge pointwise to $(-y, x)$ at each point.
I mean I kinda understand what it sais but dont know how to start the proof
The convergence is not uniform on $V$, so there is no reason the integrals should converge
@Kasmir: We said a month ago that that is the definition of a group action.
@Balarka: I'm lost. I thought you had $\epsilon$ control.
@TedShifrin Yes yes >< I know what it sais =p but I wonder how to write it in a mathematical language
I can say it in words
18:41
A group action is a homomorphism from the group to the permutation group of the set. Definition.
(or else immediately proved to be equivalent at the outset)
I accually got a Little more than that =p
each element of G, corresponds to a permuation of A
@Balarka: I don't see a sequence of functions in what you wrote. Do we have a sequence converging pointwise?
Yes, of course, we have a sequence of holonomic sections
I am being sloppy, sorry
But if you have any sort of dominated sequence, integrals still converge. Dominated convergence works without uniform.
so multiplication in G, g_1 *g_2 , is same as compostion of permutations associated with g_1 and g_2
18:43
That's what I'm saying, Kasmir. That is saying you a group homomorphism to the permutation group. Just get that in your head once and for all.
that being said, I have hard time coming up with what to write in the D.F examples
the exercices are so bad its crazy
I dunno.
V.7
V.7
So ... all in all, @TedShifrin is it right:
I've just rotated an object 20 degrees via Z axis
@TedShifrin Ted I know you told me this Before >< , I just wanted to show you that I really got it =p
@V.7: I honestly have no idea what you're drawing.
18:44
@TedShifrin Hm, I see. That makes it even more surprising.
Yeah, @Balarka: I totally do not understand this stuff.
ga =a ie g in the kernel of the action
V.7
V.7
@TedShifrin cube
means that sigma_g (a) = a
is that what I should write? @TedShifrin
for all $a$, Kasmir.
18:47
Ouos thanks
that would be comfused with stabilizer of a otherwise
You shouldn't draw a cube as a collapsed thing, @V.7. That's a terrible projection if you can't see that it's a 3-D object.
V.7
V.7
So I've sent you this screenshot with rotated cube.
@TedShifrin would the argument be just that?
V.7
V.7
This one is defaultly drawed
@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 $\tilde{\omega} = d\alpha$.
18:49
feels like most of the questions on dummit foot book are lame -.-
hate my teacher for choosing it -.-
@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.
It's a good book, Kasmir. Stop your bitching.
V.7
V.7
It would be awesome.
There are exercises on group actions all over the place as you use them ...
@V.7: You'll find my email in my profile. Send me an email and I'll send a .pdf of that section.
@TedShifrin okay -.- , but just for clarity, was that the only thing i should write? to prove that they are the same?
0
Q: Weak External Product is the Weak Internal Product of...

user193319 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...

18:51
You should explicitly write down the definition of what each of the things means, @Kasmir.
sup duds
Don't be calling me no dud, Eric.
hey @Eric
okay let phi : G--->S_A , ker (phi) is g in G such that phi(g) = id
id of S_A
18:52
@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.
OK, @Kasmir. And what about the other thing?
kernel of the action is g in G , such that ga=a for all a in A
@TedShifrin The diffeotopy is small
Rotation is very not small
@Balarka: Then my original complaint remains.
A nonzero $d\alpha$ stays nonzero.
V.7
V.7
@TedShifrin I've sent a message.
gives up
V.7
V.7
18:55
Hope you'll get it :)
The thing I was thinking was away from the negative real axis, think about $d\arctan(y/x)$
I got it, @V.7. I'm going to send just that one section. Obviously I cannot send a whole book.
there are too many interesting classes happening next term
And then near $(-1, 0)$ on the circle you have a small interval on which you want to holonomix approximate over
But you can't do that globally on the circle, obviously, @Balarka.
18:55
When you do the same trick in the slides
@Ted but it's not 'just some sequence' right? it's a sequence on $\omega(x_0)$?
phi(g) (a) = a but this is just the same thing as sigma_g (a) =a or ga=a @TedShifrin
@TedShifrin $\arctan(y/x)$ is defined outside an arc missing near $(-1, 0)$
I don't remember what it said, Sha. That was ages ago.
V.7
V.7
@TedShifrin ofcourse
So you have to destroy that dominated convergence in that tiny interval, @Balarka ... something has to get big.
Yeah...
I don't like that lol
What I don’t understand is why they talk about “limits of points converging to $y^j$”, because those limits should then be $y^j$ it seems
18:57
Oh yeah, $y^j$ is a sequence in the $\omega$-limit set you're trying to prove converges to a point of the set.
yea exactly
No, each point in the $\omega$-limit set is itself a limit ...
"the limits of points converging to $y^j$".. if the points converge to $y^j$, then their limit will be $y^j$
so I'm misreading something
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.

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