Mathematics

12:35 AM

Alright finally Laplacian time

So to see the change of variables going on

$r^2 = x^2 + y^2$, take partial derivatives wrt $x$ and get $2rr_x = 2x$

So $r_x = \frac{x}{r}$, and then $r_{xx} = \frac{r - r_x x}{r^2} = \frac{r - \frac{x^2}{r}}{r^2} = \frac{r^2 - x^2}{r^3} = \frac{y^2}{r^3}$

Similarly $r_y = \frac{y}{r}$ and $r_{yy} = \frac{x^2}{r^3}$

Or lemme see exactly what I need to do here

So chain rule gives $f_x = f_r r_x+ f_{\theta}\theta_x$ and $f_y = f_r r_y + f_{\theta} \theta_y$

And $f_{xx} = (f_r)_x r_x + f_r r_{xx} + (f_{\theta})_x \theta_x + f_{\theta}\theta_{xx}$

$f_{yy} = (f_r)_y r_y + f_r r_{yy} + (f_{\theta})_y \theta_y + f_{\theta} \theta_{yy}$

So now we have to compute $(f_r)_x$ and $(f_{\theta})_x$

$(f_r)_x = f_{rr}r_x + f_{\theta r} \theta_x$

$(f_{\theta})_x = f_{\theta r} r_x + f_{\theta \theta} \theta_x$

And similar for $y$

So putting that all together

$f_{xx} = (f_{rr}r_x + f_{\theta r}\theta_x)r_x + f_r r_{xx} + (f_{\theta r} r_x + f_{\theta\theta}\theta_x)\theta_x + f_{\theta}\theta_{xx}$

$f_{yy} = (f_{rr}r_y + f_{\theta r}\theta_y)r_y + f_r r_{yy} + (f_{\theta r}r_y + f_{\theta\theta}\theta_y)\theta_y + f_{\theta}\theta_{yy}$

Edward Evans

gross

Amin Idelhaj

So those are the quantities I want. I have $r_x$, $r_{xx}$, $r_y$, $r_{yy}$

So now $\tan(\theta) = \frac{y}{x}$, take partials wrt $y$

$\sec^2(\theta) \theta_y = \frac{1}{x}$, so $\theta_y = \frac{\cos^2(\theta)}{x} = \frac{x}{r^2}$

And $\theta_{yy} = \frac{-2xy}{r^4}$

Now taking partials wrt $x$ is gonna look different, we get $\sec^2(\theta) \theta_x = -\frac{y}{x^2}$

So $\theta_x = -\frac{y}{r^2}$, and $\theta_{xx} = \frac{2xy}{r^4}$

Conveniently $\theta_{xx} + \theta_{yy} = 0$

And $\theta_x^2 + \theta_y^2 = \frac{1}{r^2}$

So now putting everything together, help me God

I was called now I'm back

$f_{xx} = f_{rr}r_x^2 + 2f_{\theta r}\theta_xr_x + f_{\theta\theta}\theta_x^2 + f_rr_{xx} + f_{\theta}\theta_{xx}$

$f_{yy} = f_{rr}r_y^2 + 2f_{\theta r}\theta_yr_y + f_{\theta\theta}\theta_y^2 + f_rr_{yy} + f_{\theta}\theta_{yy}$

Now we have some convenient facts

Well first let's see

$r_x^2 + r_y^2 = 1$

$r_{xx}+r_{yy} = \frac{1}{r}$

$\theta_{xx}+\theta_{yy} = 0$

$\theta_x^2 + \theta_y^2 = \frac{1}{r^2}$

$r_x\theta_x + r_y\theta_y = 0$

So finally

$f_{xx}+f_{yy} = f_{rr} + \frac{1}{r}f_r + \frac{1}{r^2}f_{\theta\theta}$

So now in polar coordinates, the hyperbolic Laplacian is $(1-r^2)^2(f_{rr} + \frac{1}{r}f_r + \frac{1}{r^2}f_{\theta\theta})$

Ted Shifrin

1:14 AM

Yes, the answer is right. I can get there in mere lines with forms :)

Amin Idelhaj

Lmao, maybe I wish I had done that. Though I guess since we're using a different metric it's a bit trickier for me to reason about Laplacian on forms

It should be $\star d \star d f$ right?

Ted Shifrin

If you think briefly like a physicist, unit analysis checks out. Also you can see $\log r$ is a solution, as cx analysis tells you it must be.

Yes. Well, you didn't take the metric into account when you did yours. You just had a final formula. Here $\star$ builds in the metric if you want to.

Amin Idelhaj

Well the metric is what brought in $(1-r^2)^2$ factor

I guess it's probably notable in this case that even though the Laplacian is different, it seems like the harmonic functions are the same?

Anyway let's try with forms

So the trick here is to ask what the star operator looks like

So $df = f_r dr + f_{\theta}d_{\theta}$

Ted Shifrin

But you need the metric to figure out $\star$

What's the orthornormal basis for $1$-forms?

Amin Idelhaj

Hyperbolic angles should be the same as Euclidean ones so I guess it's just taking an orthonormal basis for 1-forms in the Euclidean case and then multiplying by $(1-r^2)^2$

Basically $(1-x^2-y^2) dx$ and $(1-x^2-y^2) dy$

Ted Shifrin

1:28 AM

What's $dx\otimes dx + dy\otimes dy$ in polar?

You're right that you will just stretch from the Euclidean case.

Amin Idelhaj

Hmm, this might be where I need to cry uncle a bit since I haven't really ever thought about tensor (as opposed to exterior) products of forms

My naive approach to the overall problem would be to say that $\star dx = (1-x^2-y^2)dy$ and $\star dy = -(1-x^2-y^2)dx$

Amin Idelhaj

1:53 AM

So now let's take the Poisson kernel

$\frac{1-r^2}{1-2r\cos(\theta - \phi) + r^2}$, where $\phi$ is just some fixed guy

So let's compute

$$\frac{1-r^2}{1-2r\cos(\theta)\cos(\phi)-2r\sin(\theta)\sin(\phi) + r^2}$$

Normally it seems like this is a thing when $\phi = 0$ but Helgason wants it with a possible $\phi$ in there for what he's doing

So let's show this guy is a Laplace eigenfunction

Ted Shifrin

The Euclidean metric is $dr\otimes dr + r^2 d\theta\otimes d\theta$. Work it out,

You know tensor product from algebra, just forget skew-symmetry.

Amin Idelhaj

2:17 AM

Alright, I will do that. Though for now I'll finish up the Poisson kernel

So let's take the derivative of that monster wrt r

Let's call that guy $P$

$$P_r = \frac{-2r+4r^2\cos(\theta-\phi)-2r^3 - (-2\cos(\theta-\phi)+ 2r)(1-r^2)}{(1-2r\cos(\theta-\phi)+r^2)^2}$$

Simplify that to get

$$P_r = \frac{-4r+(2r^2 + 2)\cos(\theta - \phi)}{(1-2r\cos(\theta-\phi) + r^2)^2}$$

Yeah Ted was right about this being revenge for me talking smack about diffgeo lol

$$P_{rr} = \frac{-4+4r\cos(\theta-\phi)}{(1-2r\cos(\theta-\phi)+r^2)^2} - \frac{2(-4r +(2r^2+2)\cos(\theta-\phi))(-2\cos(\theta-\phi)+2r)}{(1-2r\cos(\theta-\phi) + r^2)^3}$$

If this is wrong I'm going to cry

$$P_{\theta} = \frac{-(1-r^2)(2r\sin(\theta-\phi))}{(1-2r\cos(\theta-\phi)+r^2)^2}$$

$$P_{\theta\theta} = \frac{-(1-r^2)(2r\cos(\theta-\phi))(1-2r\cos(\theta-\phi)+r^2)^2 +2(1-2r\cos(\theta-\phi)+r^2)(2r\sin(\theta-\phi)(1-r^2)(2r\sin(\theta-\phi))}{(1-2r\cos(\theta-\phi)+r^2)^4}$$

$$P_{\theta\theta} = -\frac{(1-r^2)(2r\cos(\theta-\phi))}{(1-2r\cos(\theta-\phi)+r^2)^2} + \frac{8r^2\sin^2(\theta-\phi)(1-r^2)}{(1-2r\cos(\theta-\phi)+r^2)^3}$$

Okay this has a chance of working

I really hope I didn't drop a sign somewhere

So $\frac{1}{r}P_r + \frac{1}{r^2}P_{\theta\theta}$ first

$$\frac{1}{r}P_r + \frac{1}{r^2}P_{\theta\theta} = \frac{4r\cos(\theta-\phi)-4}{(1-2r\cos(\theta-\phi)+r^2)^2} + \frac{8\sin^2(\theta-\phi)(1-r^2)}{(1-2r\cos(\theta-\phi)+r^2)^3}$$

Thorgott

3:03 AM

oh my lord

Balarka Sen

4:32 AM

@Thorgott im become dead

Amin Idelhaj

Lmao

I'm also worried this doesn't even work but I had office hours

skullpatrol

4:49 AM

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Amin Idelhaj

6:52 AM

Alright finally

Verified it

So this guy is harmonic for sure

Call this guy $P(z,b)$ for Helgason reasons

And then apparently $\Delta_z(P(z,b)^{\lambda}) = 4\lambda(\lambda-1)P(z,b)^{\lambda}$

Mathphile

7:34 AM

$\lceil \log_b (x+1) \rceil = \lfloor \log_b (x) \rfloor +1$ right?

Alessandro Codenotti

8:52 AM

Suppose that $X$ is an homogeneous topological space. It's easy to see that it is not true that every (necessarily continuous) function between finite subspaces of $X$ extends to an automorphism of $X$, but is true that for every pair of points there is an automorphism of $X$ swapping them? Maybe I'm missing something obvious

Tomasz

9:17 AM

Hi

Can someone tell me, which form of Riemann's Hypothesis needs to be proved? Is the one for 0 > c > 1 enough?

Edward Evans

9:35 AM

"Good morning"

Alessandro Codenotti

morning

Edward Evans

What happened that made someone star all of @Amin 's computations lol

Tomasz

9:51 AM

Anyone alive here?

Edward Evans

@Tomasz what do you mean by "the form with $0 < c < 1$" ?

Alessandro Codenotti

Any other good music to suggest? @Edward

Edward Evans

Sybreed - Slave Design

timeless album

Tomasz

The one that's extended from the basic n^-s

Edward Evans

I don't know what you mean by that lol

Tomasz

9:53 AM

Sorry I don't know these symbols. I'm learning now to get those million bucks.

Edward Evans

Good luck

Tomasz

Thanks

Edward Evans

Where does your question come from?

Tomasz

I found a solution and now want to know if it's ok

Edward Evans

you haven't even said what the question is

Tomasz

10:04 AM

There's the basic form of SUM (n ^ -s) and then there's the one that has alternating signs. And there's the function equation. So I want to know if the one with alternating signs is enough for a proof, as from what I've been reading that should do.

Edward Evans

Depending on where you read that, it's probably true

lol

Tomasz

I must be a funny carrot.

Edward Evans

The zeta function is related to the alternating zeta function (also called Dirichlet's eta function) by $\eta(s) = (1 - 2^{1-s})\zeta(s)$ so a zero of the zeta function is a zero of $\frac{\eta(s)}{1 - 2^{1-s}}$ shrugs

Analytic number theory is "not yet" my thing

Mike Miller

@AlessandroCodenotti Take Z with the topology consisting of intervals $(n, \infty)$ for $-\infty \leq n \leq \infty$. Then f: Z -> Z is continuous iff it is nondecreading, I believe

So f(n) = n+k is continuous for any k and so this is homogeneous

But f can never swap two points

Alessandro Codenotti

but the inverse of f(n) is not continuous then

I was also playing around with orders to find a counterexample

Mike Miller

11:48 AM

@AlessandroCodenotti No, I disagree, translation is always continuous

Nondecreading meaning $x \geq y \implies f(x) \geq f(y)$

Not $f(x) \geq x$

So translation is a homeomorphism

Alessandro Codenotti

Ah of course, sorry for the brainfart

Mike Miller

You're good

Balarka Sen

12:04 PM

@MikeMiller I think it's obvious that the next thing I am supposed to learn is Banach representations of noncompact groups

user2103480

12:44 PM

@EdwardEvans someone wanted to push away your awful "norm wildberger" joke

Mike Miller

@BalarkaSen Bad

Balarka Sen

Eigenfunctions of the hyperbolic Laplacian I assume

Edward Evans

1:22 PM

@user2103480 and lo, it hath returned

Alessandro Codenotti

Good (deleted) question mike

Mike Miller

Am I banned

Looks like someone deleted it for me, I thought I got flagged

Balarka Sen

I decided its better to try to understand Weierstrass preparation

Alessandro Codenotti

In the message history it says that "Feeds" deleted it

chat.stackexchange.com/users/-2/feeds this bot

Maybe it automatically filters some swear words? I have no idea

Balarka Sen

weird

fucking weird

what the fuck is this weirdness

lets see

@EdwardEvans Rivers of Nihil is good

Edward Evans

1:31 PM

ye man

Faded Giant

@AlessandroCodenotti No. The message was flagged.

Alessandro Codenotti

I see, thanks for clearing up the confusion

So whoever mods deletes a message it shows as if deleted by the bot?

Faded Giant

If the net flag count of the message reaches six or if the message is flagged by a moderator, the message is deleted by Feeds.

If a moderator manually deletes the message, it will show this moderator instead.

Balarka Sen

@EdwardEvans what the hell this is death metal jazz

that should not exist

Edward Evans

hahaha

Alessandro Codenotti

1:40 PM

@BalarkaSen Can I introduce you to between the buried and me

@FadedGiant Makes sense, thanks

Balarka Sen

oh i have heard of colors but never listened to them

Alessandro Codenotti

youtube.com/watch?v=dM8WjgRA6o8 try this one

Balarka Sen

thanks will do

Alessandro Codenotti

@Edward you might like it too if you don't know them already

Alessandro Codenotti

2:04 PM

What are interesting examples of spaces such that any (necessarily continuous) bijection between finite subspaces extends to an automorphism of the space? Does $\Bbb R^2$ have this property? Does this property imply trivial $\pi_1$?

Balarka Sen

Homeomorphism group of any manifold acts $n$-transitively on the manifold, for any $n$.

Alessandro Codenotti

That's different from what I'm asking

Balarka Sen

Confused.

How is that different

Thorgott

surely $\mathbb{R}^2$ does

Alessandro Codenotti

No wait, sorry

Thorgott

2:07 PM

I can't argue it without handwaving, but surely it does

Mike Miller

@BalarkaSen You can take any n-point set to any other, he's asking to permute them with specified permutation

But the proof of n-transitivity already proves the latter

Balarka Sen

Oh. Yeah

Alessandro Codenotti

I don't remember, in the definition of say $2$-transitive can you swap two points?

Balarka Sen

No, but you can do this easily by modifying the proof.

user193319

If $a \in [0,1)$, does $\sum_{k=1}^\infty \frac{a^k}{k}$ converge?

Alessandro Codenotti

2:09 PM

Ok but now I'm confused

Balarka Sen

Eh, $n$-transitive does mean that, you can just take a permutation of the original collection of points.

Whatever

Alessandro Codenotti

Pick 4 points on $S^1\subseteq\Bbb R^2\setminus\{0\}$ in clockwise order $a,b,c,d$, how can the permutation swapping $b,d$ and fixing $a,c$ be extended to the whole space?

Balarka Sen

$a, c$ are antipodal, right? Imagine a reflection of the circle.

Fixing $S^0 = \{a, c\}$

Thorgott

and if they're not directly antipodal, you just do some squishing

Balarka Sen

Ya

Alessandro Codenotti

2:13 PM

Ah of course I see

Ok I'm going back to complete sequences of covers, I can't work with spaces you can actually see without getting confused apparently

Balarka Sen

But wait, there's no self-homeo of $\Bbb R$ realizing $(13)(24)$, right?

Order-preservation

user2103480

@AlessandroCodenotti mood

Balarka Sen

It has to be either increasing or decreasing

Alessandro Codenotti

@BalarkaSen right

Balarka Sen

So what is this $k$-transitivity I am talking about? I don't know. I mean $\text{Homeo}(M)$ acts transitively on the configuration space of $k$-points, $(M^k \setminus \{\text{various diagonals}\})/S_k$.

Alessandro Codenotti

2:17 PM

That was my example too for homogeneous does not imply this property

Maybe all $n$ up to the dim of the manifold?

@BalarkaSen Hmm what does that mean?

Balarka Sen

My statement forgets about the order, it doesn't answer your question.

One should be able to write some proof for $\Bbb R^2$

Tobias Kildetoft

You might be talking about $k$-homogeneous actions (transitive on subsets of size $k$)

Thorgott

is it actually true for $\mathbb{R}^2$, I'm not sure anymore

Alessandro Codenotti

I need to leave for a while, sorry

We can talk about this later if you're still interested

Balarka Sen

@Thorogtt Yes, I think so.

Thorgott

2:26 PM

I'm convinced you can take any $n$-element subset of $\mathbb{R}^2$ to any other $n$-element subset via homeomorphism, but I'm starting to be worried about permutations

Balarka Sen

Let $1, 2, \cdots, n$ be points on the real axis in $\Bbb R^2$. Any permutation is a product of transpositions, so it suffices to extend transpositions without affecting the other points.

Now given two points you can draw a topological disk around it which doesn't hit any of the other points

On a disk $D^2$ say $p, q$ are two points on the same diameter

Push the diameter so that $p$ and $q$ reverse order

I'll draw a picture

You can take 1st picture to 2nd picture by an isotopy, not just a homeomorphism.

Doing this for each transposition extends your permutation

I think this is what they call a braid diagram

@Alessandro This proves it for $\Bbb R^2$ hence for $\Bbb R^n$ for all $n \geq 2$, hence for all manifolds of dimension $\geq 2$ because you can always take finitely many points to be inside a chart by a homeomorphism (by $k$-homogeneity if you wish, thanks @Tobias)

(The braid diagram connection is as follows; the $n$-stranded braid group $B_n$ is MCG of $D^2$ with $n$ punctures; draw the essential curves between the punctures and see the image of these curves under the homeomorphism corresponding to the braid element - this configuration is reminiscent of what I am drawing and determines the braid uniquely)

geocalc33

2:52 PM

can you build a manifold based on information on the boundary?

sonicboom

3:05 PM

I have an $n\times n$ matrix where the diagonals are equal to one and the off diagonals all have magnitude smaller than one. Is this matrix invertible? What is the quickest way to prove it?

Tobias Kildetoft

At least not if you allow negative entries. Haven't thought too much about if they are all non-negative

Balarka Sen

It's not true

You need stronger conditions, look up being diagonally dominant

Alessandro Codenotti

It is true for diagonally dominant matrices

I got sniped :/

Balarka Sen

:}

Alessandro Codenotti

I'm going to read what you wrote earlier now