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21:00
Well, a wobbly parallelogram, where two of its sides are not wobbly
@0celo7 See the starred message
On second thought, don't
what the fuck
@BalarkaSen is this an elaborate trol
ley problem? Yes.
@BalarkaSen You use $\bigsum$ instead of $\sum$, right?
What's the correct code for that?
21:03
See, if you pull the lever, the trolley will switch from following the flow lines of the vector field $X$ and start following the flow lines of the vector field $Y$
Just \sum
@AkivaWeinberger Nah, in H bar Balarka said that he uses bigsum.
Which looks more neat.
Did I?
I use $\sum$
I use $\Sigma$
$\displaystyle \sum$
if you want
in The h Bar, Feb 7 at 17:02, by Balarka Sen
Also I'll write $\large{Σ}$ instead of $\sum$
Found it^
21:06
God this book is bad
I think Besse is one big erratum
Don't use $\Sigma$ lol
I was joking
taking balarka seriously leads to injury
2
$\varsigma$
In that game there are monsters called Endermen (among others), one of the friends I used to play with was a kid named Andrew
He quickly became Anderman
I guess that's not very relevant
21:09
I think we all know what endermen are
You used to play Minecraft?
@BalarkaSen didn't everyone?
No
Roblox >>> Minecraft
@BalarkaSen i think I have a Hint
I should probably try to patch together the two vector fields associated to the foliations
Is cross-posting "interdisciplinary" questions between boards legal?
21:15
with the vector field of the manifold itself (its time orientation, bc spacetime)
that should probably work
@Lozansky No, you might get shot for doing that.
Except if it's not time-orientable
@Slereah I have not been thinking about it but maybe
@0celo7 Not Nice
@0celo7 Or worse, expelled
lmao
Golden
@0celo7 Which doesn't look like a raccoon?
21:16
although I think even then that's fine, if I only consider the manifold and not the metric
reee
What's non-time orientable? Is that where, if you wait long enough, you end up where you started but upside-down?
@AkivaWeinberger Do you know a bit about Lorentzian manifolds
Non-time orientable is when, on a manifold with a metric of signature $(1, n)$, there is no vector field that is everywhere $g(X,X) < 0$
21:19
continuous
well, no continuous ones, anyway
I think, you need mod approval for that @Lozansky
Link please.
or else s h o g will delete you
@skullpatrol Here is the question math.stackexchange.com/questions/2671017/…
It's PDE so I figured it's 50/50 math/physics
21:26
@skullpatrol I had asked something in the periodic table?
Give it at least a couple of days here :-)
ok @Abcd
Salut, Anna alors tu as toujours pas pris ton Humex.. lol
What's a good book on foliations?
21:41
Is it ok to post the finish of this problem as an answer to my own question?
0
A: A collection of most of the properties about a linear operator and its trace.

Faust$( \Leftarrow ) $ Assume that $tr(T) = tr ( T^2) = \cdots = tr(T^n)=0 $ and that T is not nilpotent. Let $ m_i$ denote the multiplicity of the eigenvalue and let $\lambda_1, \cdots , \lambda_r $ denote all the distinct nonzero eigenvalues. using an above result and the assumption we have that ...

Oh my god
@BalarkaSen Look at this
So those are not homeomorphic
Multiply them by $I$
and they're homeomorphic
Oh
What the fuck
what the hell is that
That's surprisingly simple
Does the expression statistical probability have any sense? I have just seen a usage of it, where the term probability seems to suffice.
21:53
Depends on the context, but yes it does make sense in the long run.
Is there a Standard Trick to join two vector fields
If I have a vector field in some region and another vector field in another
Is there a trick to join them smoothly
Partition of unity, maybe?
@skullpatrol What's its meaning then?
Given many trials it is statistically probable.
@skullpatrol What's its exact relationship with the empirical probability?
The specific context I have just seen it being used is quantum mechanics, more precisely an article about the Stern-Gerlach experiment.
The sentence is:
> When we actually detect the atom, say in a spin-up state with statistical probability 1/2, two "collapses" or "jumps" occur.
22:03
A statistical probability is a probability found by using statistical information over many trails.
I guess I should take some compact support function around the submanifold where the vector field is like $\rho(x) X_2 + (1 - \rho(x)) X_1$
That way it will vary smoothly from one to the other
The only trouble for foliation being if it is ever $0$
@BalarkaSen Hey, on the vector field $x\hat\imath$, the flow line through $\hat\imath$ is $t\mapsto e^t\hat\imath$, right? Because the position is always the same as the velocity.
22:27
So the reason the square wasn't working before, was because I approximated "following $Y$ for $\epsilon$ amount of time" by moving from $p$ to $p+\epsilon Y(p)$.
So, for the vector field I just described, that would essentially move me from $1$ to $1+\epsilon$
I wasn't looking at the computation
where the true thing would be going from $1$ to $e^\epsilon$
and since we divide by $\epsilon^2$ at the end, linear approximations are not enough.
We need our approximations to be accurate to the second degree.
Second order?
What's the right word there
And maybe let's use $h$ instead of $\epsilon$ so as not to confuse it with $e$
so the point is that $1+h$ is not sufficiently close to $e^h$
I follow you but I'm not sure what you're trying to prove
22:31
I'm trying to explain why the computation before failed
I was computing the square thing where one of them was constant
and it should have given the directional derivative but it didn't
and the reason was because I was approximating badly
@AkivaWeinberger Pretty awesome
Fair enough
Did that make sense? @nbro
22:33
Btw, is this the new US policy model for equipping schools with guns?
Lol
It took me a few seconds
subscripts are bad for your health @Balarka
Seriously what the shit though. US has more legal arms count than population.
Not to mention the illegal stuff
You people are going to face an elaborate trolley problem if this goes on
Yeah, most everyone has two arms
5
22:35
Not a whole lot in cities though
Guns don't kill. People do.
Mostly in rural areas
@MikeMiller That one caught me off guard
@AlessandroCodenotti lmao
give it a star as a historical artifact in the pinnacle of mathematical notation
Yeah, the Las Vegas massacre lite the fuse...
@BalarkaSen what is it
22:42
Topological spaces
@AkivaWeinberger I have absolutely no idea what it's supposed to represent.
...everybody seems to be "aiming" at the younger crowd these days :(
As I explained, they are compact subsets of the plane satisfying $X\not\cong Y$ but $X\times I\cong Y\times I$
$\cong$? $\simeq$?
Sorry I don't see an explanation
$\approx$
$\cong$ is isomorphism and $\simeq$ is homotopy
$X \times I$ is a solid torus with two flaps sticking out
22:43
They're homeomorphic, which is isomorphism in the topological category
Rotate the flaps
I had no idea he meant that $I$
I thought it was algebraic geometry
and $I$ was an ideal or some shit
Oh, $I=[0,1]$
interval man
Yes, now I get it
22:44
@skullpatrol Honestly I'm glad that we had the arms act passed 150 years from now in India.
@AkivaWeinberger isomorphism should be restricted to algebra
Are there any locally compact hausdorff spaces that aren't paracompact, besides the long line? I haven't spent a whole lot of time studying the long line.'
No need to involve cats when only dogs are needed
I think if there's one nasty thing you should expect there to be a zoo of them
r/nocontext
22:46
@user193319 a cheating example is to take a large disjoint union of things
I think
@0celo7 What sort of things? How would you topologize it?
I might be forgetting what paracompactness means
Yeah @BalarkaSen the Brits did it right :)
I think he means, like, $X$ be a locally compact Hausdorff space and $D$ be an uncountable discrete set, take $X \times D$.
@AkivaWeinberger I see we follow the same page
Paracompactness does not refer to countability
Acc. to Bredon
so ignore that
> Another counterexample is a product of uncountably many copies of an infinite discrete space.
In mathematics, the Prüfer manifold or Prüfer surface is a 2-dimensional Hausdorff real analytic manifold that is not paracompact. It was introduced by Radó (1925) and named after Heinz Prüfer. == Construction == The Prüfer manifold can be constructed as follows (Spivak 1979, appendix A). Take an uncountable number of copies Xa of the plane, one for each real number a, and take a copy H of the upper half plane (of pairs (x, y) with y > 0). Then glue each plane Xa to the upper half plane H by identifying (x,y)∈Xa for y > 0 with the point (a + yx, y) in H. The resulting quotient space is the Prüfer...
According to the Wiki
22:49
sniped
Point set topology is hard
Tell that to Moore :P
What's a suitable Hilbert space when working with Bessel functions of the first kind?
$L_2(r, I)$ I guess
Well, do you know the orthogonality relation for Bessel functions of the first kind?
@Lozansky You want a weighted L^2 space is what Semic is getting at
23:02
Right.
Either that or absorb a factor of sqrt(x) into your basis functions
@Semiclassical then they won't be Bessel functions...?
Indeed not
@Semiclassical What I've come up with so far is that $\{ J_0(\alpha_{0k}r) \}_{k=1}^{\infty} $ form an orthogonal basis in $L_2(r, I)$ with the inner product $\int_{0}^{1} \overline{u(r)} v(r) r dr $
oh that's what you meant
23:06
Ugh
Though, is it L^2 in that case?
It’s definitely L^2 when the basis is $\{\sqrt{r}J_0(\alpha_{0k}r)\}$
What kind of basis is that?
The operator is $-\dfrac{1}{r}(ru')'$
I.e. a Sturm-Liouville operator with weight function $r$
Same kind of basis as yours, except that the inner product is just $\int_0^1 \overline{u(r)}v(r)\, dr$
23:11
Hm
Well I definitely don't want to change basis
i think I misunderstood your notation in retrospect
Well, I now know what space and inner product I want to work in
L^2(weight, domain) ?
Now I'm left with the integral $\int_{0}^{R} J_{0}(\alpha_{0k}r/R) r dr$
$L_2(r, [0,R])$
23:28
in my infinite wisdom I lost the notes for my talk.
I bet I left them on the projector last time.
@nbro they mean the same thing

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