So I have $S^{n-1}$ as a Riemannian manifold with the metric and orientation it inherits from $\Bbb R^n$ with the standard ones. I want to verify that the volume form on $S^{n-1}$ is $\omega=|x|^{-n}\sum_{i=1}^n x^i dx^1\wedge\cdots\wedge\widehat{dx^i}\wedge\cdots\wedge dx^n$

If $S^n$ is the unit sphere, @Alessandro, why the $|x|$?

Or are you wanting the spheres of arbitrary radius?

Oh sorry, I messed up, $\omega\in\Omega^{n-1}(\Bbb R^n)$ and I want $i^\ast\omega$ on the sphere

Still, the norm is irrelephant unless you're doing arbitrary radius.

that's a contour plot of $\|g^{11}(x,y)\|^2$ where $g(x,y)=(y,x^2+2x-y)$ and I've chosen a variety of initial $(x,y)$

Oh, wait, I think I just figured out my issue

the idea being that, if most iterates escape to infinity, then they'll have large Euclidean norm

Anyhow, this is our usual $\iota_X dV$, where $X$ is the unit normal, @Alessandro.

Right, and when computing that I get the same expression but without $|x|$ factor, since the unit normal is $x^j\partial/\partial x^j$

But that doesn't matter, since $|x|=1$ on the sphere

Right, but do you see why it scales right for arbitrary radius?

(Because of the unit normal wink wink)

(The ranges on the outside are misleading: they should both be from 0 to 2)

For abritrary radius isn't the unit normal $|x|^{-1}x^j\partial/\partial x^j$? I think there's something wrong here

So the line of minimal norms should approximate the set of (x,y) which converge to zero

(If I used more numerical precision, then the line should get sharper and sharper)

It looks vaguely like $y=\sqrt(1+2x}-1$, I guess

But I probably can’t write home about that

@Alessandro Oh, you have $|x|^{-n}$. That does look off.

Yeah, that $\omega$ is invariant under the map $x\rightsquigarrow rx$. It gives solid angle.

Using the vector field I wrote above I get a factor of $|x|^{-1}$ in the form as well, which makes sense, but the $-n$ in the exponent seems suspicious to me

Yeah, you just have $R^{-1}$ in the volume form of $S^{n-1}_R \subset \Bbb R^n$.

The point of the $n$ is to get a closed form.

This is generalization of the inverse square force in $\Bbb R^3$. The fact that the integral over the sphere of radius $r$ is independent of $r$ is suggestive.

@TedShifrin Oh, I see, makes sense

Anyhow, is that issue resolved?

Yep

Next?

I'm computing the differential to check that it is closed now (not that I don't trust you but it looks like a good exercise :P)

OK. I'll be back after lunch :)

The fact Ted was hinting at was central force fields are conservative, which is the intuitive reason it should be closed

@TedShifrin Buon appetito!

now the question is

should i do math

or should i sleep

Sleep

or should i... watch anime

Sleep

lmao

Sleep while skillpatrol and mike miller make extensive advances in mathematics and leave you in the dust

i started watching this cool looking highly pretentious anime ergo proxy

It's not bad

@BalarkaSen sleep

it cant be if you have robots called Lacan and Derrida

I found a counter intuitive graph about the relationship in the US between video games and cime.

Observation: Correlation is not causation

tylervigen.com/spurious-correlations

observation: crime generally increases with temperature. (this of course does not mean it causes more crime)

the criminality is being channeled in to gaming violence

Observation: $\text{Corr}(X,Y)=\frac{\text{Cov}(X,Y)}{\sqrt{\text{Var}(X)\text{Var}(Y)}}$

Hi all

yo

Observation: the action or process of observing something or someone carefully or in order to gain information.

Greetings

anti sarkeesian which funnily enough is a pun

Observation: correlation correlates with causation

Gurten tage

Not a good question ? I got a -1

I like Thorgott's

I like throgots th ebest

Might be too many questions, mick. Too spread out and not focused enough, maybe?

I got -32

@mick it's those "find the next number in the sequence" question

there are infinitely many answers

Yeah but . Turning it into 10 separate questions seems silly

@Rithaniel observation: corr(X,Y)^2+corr(X,Z)^2+corr(X,Y)^2 never exceeds 1+2*corr(X,Y)*corr(X,Z)*corr(Y,Z)

Yes leaky. But the nature of the solutions I wonder about.

@mick how about g(1) = 2^2 and g(2) = 1^2 and g(n) = n^2 afterwards

or a good algorithm

Reaaally? That's a nifty inequality right there

observation: beached whales can explode at the speed of light

Hmm should include strictly nondeacreasing

@rithaniel u have fallen into Semiclassical's trap

@Semiclassical yeah, Sociology is such a soft science.

(implication: if X and Y are uncorrelated, then the correlations between X,Y and a third variable Z can’t be too strong)

sociology is the science of the patriarch

@BalarkaSen have you watched this

yup

oh yeah

@Rithaniel yeeep

oldest occurrence I know of that dates back to the 1890s

how do people cheat in blitz

observation: if we consider the conditional inequalities on several indedpended correlation coefficients...

he was making 50 queen moves in a couple seconds

@BalarkaSen well when the time gets low one can't cheat anymore

or maybe he expected to checkmate Aman before time gets low

okay I have exorcised my idiot demons

back to being somewhat sensible

@Rithaniel see page 10 here: zenodo.org/record/1432093#.XfVJRCRMHYU

Could you potentially have X and Z correlate really strongly but Y and Z be uncorrelated enough that the sum of the squares even out?

GM recognize patterns on the board

mostly global

@LeakyNun have u seen that one Hikaru vs Rybka game

some local

he trolled the computer

they queen walk

and then rage around the back

equipping those knights with 3 oclock snipe

and cleaning up with the castles

@BalarkaSen this?

yeah

As in: X,Y are uncorrelated and X, Z strongly correlated?

its like a totally locked drawn position but he has a knight and a bishop for two rooks so the computer thinks it has more chance of winning and eventually tries to break open the position to its disadvantage

In that case, then Y,Z can only be weakly correlated

@Ultradark that could be said of Psychology also, no?

I edited my question

you realize if the board is not square anymore it's still chess?

like hyperbolic warped chess

theres a geometric picture which can be made rigorous: think of X, Y, X as vectors in 3D space, and take their mutual correlations to be the dot product between them

on the tessallated bishop plane

i had a similar position (but equal pieces, just that it was a massive pawn block, so all we could do is maneuver until agreeing to a draw) so my friend recommended this to me

I play chess.

why we talk about chess ?

@skullpetrol I stated that sociology is the science of the patriarch and that is all i will say

then X,Y strongly correlated means X, Y nearly aligned (or anti-aligned)

I will never write about that subject again

My rating is about 1800 I think , yours ?

and X, Y uncorrelated means orthogonal

Balarka what is your rating

It's 1932

so if X,Y are perpendicular and Y, Z are nearly in the same direction, then X,Z must be nearly perpendicular

i dont have a rating, i dont play professional chess lol

you can still get rateddd

it's the elo

Then why are you talking about chess :)

i dont know how to respond to that question bruh

you talk cuz you can lol

kick him

My question edited math.stackexchange.com/questions/3476466/…

Everything clear now ?

snipe the grammar mistakes and you chillin

@BalarkaSen nice vid

v cool right

leaky nun If I had to guess I would say you are extremely good at chess

using knowledge about chess engine against it

"rf1-rf2-rf1-rf2-..."

hikaru is such a legend

I think he used a similar technique to beat komodo 19

Question: Is the Bayesian view of statistics more represented amongst minority teachers

all loogical moves

hit the pawn hit the queen hit everything

loogical

@Ultradark why would minority teachers prefer that view?

1:58:30

@BalarkaSen

I had a non-minority teacher who only presented it for like one class at the end of the semester

and i was like oh why didn't you tell us that earlier

nice

all of us were thinking, Dr. Prakesh down the hall would have put the two pedagogical views on equal footing

must have been a "purest"

yeah, he was lol

:-)

both of em are nice professors tho

We call him Dr. Probability

a platonist to the end

some real great learned men

who are doing a great service to the community

@Alessandro: Didn't want you to think I'd forgotten you.

did you see the graph I found Prof. @TedShifrin?

Nope.

I'm not sure there's any causality involved, even if there might be some correlation.

Hardly a firm statistical analysis.

yeah, just seemed counter intuitive to me

like watching violence on TV...

I hypothesize that you don't play many video games.

nope

As someone who does, this kind of correlation is actually closer to what I would have suspected

does anybody know of an algorithm to enumerate all the topologies of a finite point set?

@Rithaniel ok, thanks for the feedback

No problem. Also, remember, as people have indicated, correlation does not imply causation

indeed

@JoeShmo the number of such preorders on n points is enumerated here: oeis.org/A000798

Wow, I was expecting it to grow quickly, but that is faster than I thought it would have

Along with a lot of bibliographic references

If there’s a known algorithm, that seems like a good place to start your search

its right in wikipedia actually -- en.wikipedia.org/wiki/…

Yeah, that’s how I found the OEIS entry

and apparently there is no known formula for the number of topologies, and in particular no recurrence relation. i.e., there's no algorithm

Eh, not necessarily. It could be that there is an algorithm, but that it would take prohibitively long to run once n becomes large

Proper ideal means $I \ne R$ and $I \ne \emptyset$, right? Not $I \ne R$ and $I \ne \left\{0 \right\}$

no i dont care about the complexity of the algorithm. im just looking for one such algorithm, and its a little cryptic on wiki. it says "no known simple formula" (does there exist a complicated one?) im gathering that a recurrence relation doesn't exist

hi @JoeShmo

hi Ted!

Well, you can always enumerate all potential topologies and then check unions and intersections to see which ones fail

yeah i can always brute force

That is probably the least efficient algorithm possible, but it's still technically an algorithm

It can't be empty since to be an ideal it would have to be a subgroup of $(R, +)$; so really it's just $I \ne R$, I guess.

definitely, and it has runtime $2^{|X|}$

I mean, suppose that it takes one nanosecond to generate another preorder. That seems good, but there’s 519355571065774021 such preorders on 12 elements

That’s 20 years of runtime right there

And it only gets worse. So even if the algorithm generates them fast, the sheer number to construct is prohibitive

@TedShifrin do you have a problem set for typical questions in V.K. theorem / covering space theory?

Not really. I never actually taught algebraic topology so never developed problem sets. I would recommend Hatcher. He has superb exercises.

This seems like a nice question: math.stackexchange.com/q/3476550/137524