Well I was just wondering if you have two generators $S: x \to 1/y$ and $T:y\to 1/x$ transforming points in the upper half plane

?

Is that the full question?

lol

oh

the rest of the question is... can I present this as a group maybe? And/or what's a conceptual way to think of the transformations?

Okay what do you mean by upper half plane?

I think it maps (x,y) points greater than $1$

to fractions

I only ask because $S$ and $T$ are the usual notations for the matrices that generate the full modular group

and that group acts on the (complex) upper half plane

$S : z \mapsto 1/z$ and $T : z \mapsto z + 1$

I would let it be complex, but maybe I should stick with real things

what does x mapsto 1/y even mean

Also ^

y isn't "in scope"

$(x,y)\mapsto (1/y,1/x)$

and what is T then

okay forget the first thing I wrote

this is probably the correct notation

$(x,y)\mapsto (1/y,1/x)$

that guy is an involution so.. $\langle T \rangle$ is just gonna be $\Bbb Z/(2)$

rigHT=?

wait remind me of that notation on the right?

integers quotiented by $2$?

yep

And is $S$ the same?

another involution

Sorry, with $T$ i meant $(x,y) \mapsto (1/y, 1/x)$

oh okay got it

what is S if T is $(x,y) \mapsto (1/y,1/x)$

they're not the same map right

idk

I mean should I just write it as this and forget the $S$ and $T$?

$(x,y) \mapsto (1/y,1/x)$

I honestly don't know the convention

okay so S and T were just the components?

yeah

If you're acting on the upper half plane then you want your inputs to be elements of the upper half plane lol

^

yeah they are

$y>0$

yes, in this notation, but with the S and T you wrote before they were just real numbers and the notation was weird

y is a complex number, what does it mean for y to be greater than 0

@Soham I think he's talking about a real upper half plane

or do you mean $x + iy \mapsto 1/y + i/x$?

ohh

well yeah that map is an involution so iterating it is kinda boring

it's an involution because of the inverting of each coordinate right?

it's an involution because it satisfies the definition of involution

:D

how?

Just use it on $(x,y)$ twice and see what you get

okay I'm performing the operation twice

Meg, what do people call you for short

They don't, my name annoys everyone in this chat

rofl

are you secretly another Ted

determined to be unique

hahaha, well Edward is my real name and Ted is a nickname for Edward sooo

got it!

TED 3 TED 3 TED 3

but nah, if you wanna @ me just @ and I should be the first name that comes up

Me or Érico

is the name from some kvlt song

looks like icelandic

It actually is hahaha

pietr stonne doiche!

although it's faroese

@Soham youtube.com/watch?v=X7lP4jwkB_A At around 3:00 he says "exit through me" and repeats that for the outro, and ÍgjøgnumMeg just means "through me" rofl

ya!

oooooh

i've heard of this album

it's not my favourite but it's good

youtube.com/watch?v=GUNOIHz2QQU best song from best album rofl

that is an absolute banger

aye

it's like demons are singing to me!

it feels really good to not rush through maths, i must say

for once

evening listening :)

I'm changing my name back to geocalc33

or hyperkolgomorovcomplexityupperbounds

where did Mathein and ryan

Mathein is super busy with his bachelor's dissertation atm

idk about Ryan

@Soham cool album

how does the modular flow relate to the modular group lol

not a clue

what if there was a math trivia app

but like pretty advanced

like: the trefoil knot is an example of a _______ knot

Never heard of a bachelor's dissertation before

a) nontrivial b) fibered c) unimodal d) semisimple

e) a and b

@Rithaniel we call it a dissertation in the UK lol

f) c and a

unimodal and semisimple dont sound like knot theory

then it's e)

ding ding ding ding ding

it would be too easy to answer that kind of question without actually knowing the topics in question :P

what's a fibered knot?

Well, we have masters and doctorate dissertations. I've heard of those before. But bachelors not so much

We say master thesis and doctoral thesis in the UK hahaha

A dissertation is where you have to defend something you've written before a council of professors, right?

yeah

I don't really know the literal meanings of the words; for a final undergraduate project we usually say dissertation, for masters and doctoral projects we say thesis lol

I'm not defending whether or not it's correct, I'm saying that's what we say :P

normally you don't have to defend a masters thesis I'm pretty sure

of course you do

I had to defend my undergrad dissertation too

lol

haha

well "defend" is a strong word, I had to give a talk on it and answer audience questions

I spoke too soon

Okay, cause I was about to say that might be a little bit tough on a bachelor's student. If it's just a project, that's more reasonable

well it's a dissertation

o.o

don't understand what the problem is rofl

what was it on

Algebraic number theory and Fermat's last theorem

(that's the title)

(Dangit. Chat didn't scroll for me and I missed some comments)

So yeah, this sounds similar to a graduate-level dissertation over here. There's no problem, though. It's just surprising to see something at a different level that I would have suspected it

I finished my final exam yesterday, for my bachelors, too. The thought of having to give a speech right now is a frightening thought

fair lol

tbf if you know your shit and you actually wrote the dissertation yourself, there's nothing that daunting about giving a talk on it

True. I suppose it's different if you've been expecting it throughout your undergrad degree

yeah, you also have a full academic year to write it

pop dissertation

due in one semester, must be at least 1000 pages

must be 95 percent original work

Do you take classes during that academic year too, or are you supposed to focus on the dissertation?

That'd be a nightmare, Ultra

yeah you take classes too

My cousin had to read 100 books in a semester and then get asked questions about them at the end of the semester

the books were in multiple languages

this is not for math though

comparative literature phd at Yale

I have question

Is my answer okay??

I feel like I kind of bent the rules

I extended the domain of $x$ to include $(1,\infty)$ and leveraged the projection that the function respects

the projection I was talking about earlier

@Balarka are you around?

@AlessandroCodenotti what is this about

Riemannian stuff

when did you start doing Riemannian stuff

When the diffgeo course I'm taking started talking about it

cool

@LeakyNun sir have you solved Riemann Hypothesis?

Namely I want to prove that the divergence of a vector field on an orientable Riemannian manifold $(M,g)$ is orientation independent, but it's not clear to me exactly what that means. I can write down a local coordinate expressions of $\mathrm{div} X$ that depends only on $X$ and $g$ but not on the volume form, is that enough to conclude its independence from the choice of orientation?

anyone like my drawing?

There's a sensei sitting cross-legged with his hood on and head bent down

he's meditating

@Ultradark Now, I realize how good Artist I'm...

evening chat

@AbhasKumarSinha! :(

it's okay lol :)

evening soham

$(\ln x)^2 - \ln x - 1 = 0$

so $\exp(\varphi)$?

you're quick lol

yeah it's the logarithmic version

also $\exp(-\phi^{-1})$

Hello, I can use some help with the pde u_x^2+u_y^2=u^2 with initial condition u(cos(phi),sin(phi))=1

what is the name for such an equation?

a transcendental quadratic equation?

because it yields two transcendental roots

eikonal equation?

@Ultradark JK, cool Modern Art.

Beauty is in the eyes of beholder...

Hi everyone

hi

Hello!

hello

When $x\to{2^{+}}$ what does $x^2$ goes to?

@ICCQBE 4

Doesn't it goes to $4^{+}$?

@ICCQBE what is $4^+$?

It goes to 4 from above, if that’s what you mean

I mean closer numbers to $4$ and greater than $4$ numbers.

@ICCQBE Read some stuff on real analysis....

That'll clear all your doubts...

I'm asking to learn, don't you know what is $a^+$ or $a^-$ notation?

@ICCQBE no, if there is any, I'm not aware...

I have a question about compactness of sets. Wikipedia says that a set X ist compact if every open cover of X has a finite subcover. But {(-1, 2)} is an open cover of [0, 1] (which is compact) but there is no finite subcover (I think).. Where am I making a mistake?

{(-1,2)} is a finite cover itself already

@AbhasKumarSinha Let me give you an example, $\lim_{x\to{0^+}}/frac{1}{x}=\infty$

the "finite" refers to the number of open sets in the cover, which, in this case, is 1

@ICCQBE nooooooooooo, that's only applicable under limits.

$\lim_{x\rightarrow 2^+}x^2=4$

@Thorgott yes now perfekt.

@ICCQBE the plus/minus superscript is used to denote whether you’re taking a right/left limit. But the value of the limit is just some number, so no superscript

@Thorgott So an open cover is not finite?

What you might want to say is that $x^2\downarrow4$ as $x\downarrow2$

@DominikSchmidt you need to be clear on what you mean by finite

@Semiclassical Thanks, that was the answer that I was looking for.

@AbhasKumarSinha Thank you for your helps.

a finite open cover is a cover by a finite number of open sets

{(-1, 2)} is a cover made of 1 set, 1 is a finite natural number

hence the cover you started with was already finite

@Semiclassical: Hi! how are you? I can use some help with the pde u_x^2+u_y^2=u^2 with initial condition u(cos(phi),sin(phi))=1

{(-1,2)} is a subcover of {(-1,2)}

you don't have to drop any open sets

think of how every set is a subset of itself

it doesn't say "proper subcover" or anything like that :)

Ah now I got it. Thanks for the help!

anytime

@Eran so you’ve got u=1 on the unit circle. What do you have for the characteristic odes?

Oh, wait. That’s nonlinear in derivatives

I'm not familiar with how method of characteristics works beyond the case $au_x+b_y=c$, tbh

I can look at the wiki page, of course

Okay then I'll try an easier one: u_x^2+u_y=0 with initial condition u(x,0)=0?

same issue, really. let me check something first tho

Thank you very much :)

okay: Since the boundary condition is u=1 on the circle, that suggests trying polar coordinates

Right, yes

in which case $(x,y)=(r\cos\theta,r\sin\theta)$

yes

and therefore $\partial_x = (\partial_x r)\partial_r+(\partial_x \theta)\partial \theta=(\cos \theta)\partial_r - (r\sin \theta)\partial_\theta$

and similarly $\partial_y = (\sin \theta)\partial_r+(r\cos \theta)\partial_\theta$

which, if you expand and simplify, yields $(\partial_x u)^2+(\partial_y u)^2 = (\partial_r u)^2+r^2 (\partial_\theta u)^2$

Moreover, our initial condition didn't depend on $\theta$ at all. that suggests looking for a solution of the form $u=f(r)$

in which case we have $f'(r)^2 = f(r)^2\implies f(r) = Ae^{\pm r}$

the condition $u(\cos u,\sin u)=1\implies f(r)=1$ then gives $f(r)=e^{\pm (r-1)}=e^{\pm (\sqrt{x^2+y^2}-1)}$

actually, that's a bad way f writing it

should be $u(x,y)=Ae^{\sqrt{x^2+y^2}}+Be^{-\sqrt{x^2+y^2}}$

Sorry I lost you with the first derivative. And why does the initial condition didn't depend on theta?

you had $u(\cos \phi,\sin \phi)=1$

yes

that's the same as $r=1$ in polar coordinates with no dependence on $\theta$

for the first derivative, I'm just doing the chain rule. (had a typo $\partial \theta\to \partial_\theta$ in the first step)

isn't the phi instead of theta?

yeah. but it's just a name

yes right. So the chain rule for with one? x?

you could just as well say polar coordinates are $x=r\cos\phi$, $y=r \sin\phi$

@Semiclassical yes, but then it does depend on theta

No, it doesn't. The output is the same regardless of the value of $\phi$

Right, my mistake.

$\phi$-dependence would mean the RHS was a function of $\phi$

in more detail, $$\frac{\partial u}{\partial x} = \frac{\partial r}{\partial x}\frac{\partial u}{\partial r}+\frac{\partial \theta}{\partial x}\frac{\partial u}{\partial \theta}$$

fixed

what bothers me, though, is that there's only one condition

I think you need another one, such as $u$ not blowing up at infinity

or the value of u at the origin

du/dx=(dr/dx)*(du/dr) Something's not right with the chain rule I think.

hmm. there's nothing wrong with what I wrote just now, but that's not what I had earlier

@Ultradark installing gap-system.org

what I had earlier was wrong, blah

I tried to solve this in a different way. I will right it down now.

should be $\partial r/\partial x = x/r$, $\partial \theta/\partial x=-y/r^2$, $\partial r/\partial y=y/r$, $\partial \theta/\partial y=x/r^2$

which I think gives... $u_x^2+u_y^2 = u_r^2 + r^{-2} u_\theta^2$

which ... gives the same answer as before.

I defined Gamma(s) the initial curve (cos(phi),sin(phi)).So u(Gamma(s))=1 So if you define f(s)=u(Gamma(s))=1 and differentiate both sides, I get 0=(p_1(0),p_2(0))*Gamma'(s) and they are perpendicular. That's equation 1. The second one: From the equation of the problem we get p_1(0)^2+p_2(0)^2=1 then we get: (p_1(0),p_2(0))=+-(Gamma_2(s)/|Gamma(s)|, -Gamma_1(s)/|Gamma(s)|

like I said, I don't know method of characteristics for fully nonlinear pdes

so I probably can't hep much

as in: $$\frac{\partial r}{\partial x} = \frac{\partial}{\partial x}\sqrt{x^2+y^2} = \frac{x}{\sqrt{x^2+y^2}} = \frac{x}{r}$$

and similarly with $\theta=\arctan(y/x)$

this is the derivative of what? Sorry I'm not following you.

$r$ in polar coordinates

$x=r\cos \theta$, $y=r\sin \theta \implies r=\sqrt{x^2+y^2},$ $\theta=\arctan(y/x)$

All I'm doing is changing to polar coordinates.

I'm familiar with changing to polar coordinates but unfortunately I don't understand your derivatives

I think I'll try solving it with method of characteristics but thank you very much anyway :)

Then I have no idea why. All I'm writing out is the chain rule.

I think I'll try the method of characteristics but thank you very much :)

see for instance en.wikipedia.org/wiki/…

mmkay. as I noted earlier, though, I don't see how there's a unique solution to the problem as written

there's just not enough conditions

Okay thank you! I think the math writing here confused me...