Hi, is anyone present who is good with Lie groups?

I am a bit confused on the state of Hilbert's 5th problem.

My question basically is: is every topological group that is also a topological manifold, a Lie group?

If not, what were the additional constraints on the topological group?

.

Is there any technique for making a nice guess for the trial functions used in Galerkin method to solve BVP.

collection of problems

or it will come from practice i think!

Is there a standard notation for the mode of a distribution?

Hi, $$\text{Determinate } a\in \mathbb R, \text{ with } P(x)=x^5+x^4+x^3+x^2+x+a, \text{ with a double root in complexes}$$

Bwahaha, the Night King.

Hey @ALannister, @Alessandro, @PVAL!

Hi @Dami

How's it going?

Morning

I have an exam to study for, but quite well apart from that, what about you?

Hi @Steamy

I have an exam to supervise and to correct :P

On which topic?

Correcting exams sounds extremely boring

programming in python

Yo @Steamy!

Ah, nice. Do you also judge the clarity/efficiency of the code or just whether it gives the right result when run with some test cases?

I've got to pack over these next few days and then at home I'll have to get stuff for next year

But beyond that I've got few obligations so I'm just doing things randomly

How do the functions (or expressions) $\exp(\frac{x}{2})$, and $\sqrt{\exp(x)}$ over the complex numbers differ?

@AlessandroCodenotti Clarity not so much, I do give a few marks on efficiency

but, well, the level is so low that it's basically: if you didn't put any unnecessary loops, you get full marks for efficiency

@gxyd $\sqrt{z}$ is a multi-valued function and not a function in the usual sense, over the complex numbers

@SteamyRoot got it.

Like, if $z = 2\pi i$, then $\exp(z/2)$ would be $\exp(\pi i) = -1$, but $\sqrt{\exp(z)}$ would be both $1$ and $-1$, since their squares are $1$

Oh actually hmm

Okay so I'd like to ask exactly how Riemann surfaces arise out of multivalued functions

@Julius Yes, for a $\nu$ that works consider $\nu(E)=1$ if $0\in E$ and $\nu(E)=0$ otherwise, this is an outer measure and its $\sigma$-algebra of measurable sets it the whole powerset. The problem is that if you want a measure which is translation invariant and gives the "correct" measure to the intervals you can't have all subsets measurable. (Note however that the Lebesgue $\sigma$-algebra is much bigger than the Borel one)

Is there no chat-option for stack overflow?

If there's please send me the link.

I couldn't find it.

I think stack overflow chat is banned. That's surprising.

@Daminark Could you help me out?

I can only see 1 room here: chat.stackoverflow.com @TobiasKildetoft Can you see more rooms in this link?

I'm not terribly experienced in such matters

@Abcd Yes, I see a ton of rooms there

@TobiasKildetoft But I can't :"(

If they have banned me, it's really rude. I hate this rude behaviour of moderators.

Math. SE and Physics.SE are the best and most firendly!

@LeakyNun Is it possible to delete a stack exchange account? I want to delete my account and create a new one because I want a new beginning on this site.

@Abcd yes it is

@Abcd banned?

@LeakyNun Will I be able to retain my username Abcd?

@Abcd sure

but to make sure, you might want to change the name of this account first, to prevent collision

@Abcd But if you do this to get around a ban, that will result in a new, longer ban

Multiple accounts can have the same name

@LeakyNun Yes, long back (last year or last to last year) when I was new to this site I didn't know the appropriate manner of asking questions so I was banned I guess. (though I had asked hardly 5 questions) and now I agree they were poor quality questions but come on now I know how to ask/

@Abcd why did they ban you?

where did you ask the questions?

@LeakyNun Stack overflow. That was my first site.

@Abcd the main site or the chat?

@LeakyNun Both.

how long did they ban you?

@LeakyNun I am not sure but I think permanently

@TobiasKildetoft But see my new questions on Math.SE. The quality has improved. Shouldn't they remove the ban seeing the progress of user on other SE site?

@LeakyNun Farewell from this account :P.

@Abcd ok

@Daminark One way is through branch points and branch cuts

Basically, you're going to make a few copies of the complex plane, one copy per value the function can have at a point

So for the square root you'd have two copies, for the cube root 3 copies, and for the logarithm you'll have "$\mathbb{Z}$" copies

Of course, on each copy, the function will no longer be continuous but make a jump, so that'll be your branch cut.

and then you glue the copies together by connecting those cuts

@AlessandroCodenotti why can't it be translation invariant?

(the copies are usually called Riemann sheets)

Huh

@LeakyNun That proof basically constructs something called the Vitali set

@Daminark oh, ok

$\mathbb{R}/\mathbb{Q}$ is the culprit there

How does one proove that $$\sqrt{x}\in\Bbb{R}\text{\\}\Bbb{Q}\iff \nexists k \in \mathbb{Z} \text{ such that } k^2=x$$?

Like, if $x$ isn't a perfect square, how to prove that $\sqrt{x}\in\mathbb{R}\text{\\}\mathbb{Q}$?

@Mr.Xcoder consider its contrapositive

@LeakyNun Let me see what "contrapositive" is

Oh, ok

@LeakyNun So I should say that "if x is not a perfect square, then $\sqrt{x}$ is not rational", and assume that it is rational and get to a contradiction?

@Mr.Xcoder no, that isn't what I meant by contrapositive

Then what?

do you know what contrapositive means?

Switching the hypothesis and the conclusion and negating both.

you forgot to negate both

Oh, yeah right. Thanks... Let me think

My hypothesis is: "$\sqrt{x}\in\Bbb{R}\text{\\}\Bbb{Q}$", the conclusion is "$\nexists k \in \mathbb{Z} \text{ such that } k^2=x$". I switch them: "$\nexists k \in \mathbb{Z} \text{ such that } k^2=x\implies \sqrt{x}\in\Bbb{R}\text{\\}\Bbb{Q}$"

And now negate

"$\exists k\in \mathbb{Z}\text{ such that }k^2=x \implies \sqrt{x}\notin\Bbb{R}\text{\\}\Bbb{Q}$"

@LeakyNun Better now?

@Mr.Xcoder right

and try to simplify the consequent

antecedent $\implies$ consequent

Simplified: "$\exists k\in \mathbb{Z}\text{ such that }k^2=x\implies \sqrt{x}\in \mathbb{R}$".

@Mr.Xcoder no that isn't right

@LeakyNun Because of $\mathbb{C}$?

@Mr.Xcoder no

Oh no, it's in $\mathbb{Q}$

Sorry, typo

$\implies \sqrt{x}\in\mathbb{Q}$

can you prove it now?

Better?

@LeakyNun Let me think how

So I must prove that $\exists k\in \mathbb{Z}\text{ such that }k^2=x\implies\sqrt{x}\in\mathbb{Q}$... Well, that's quite obvious.

@LeakyNun Wait, if $\exists k\in \mathbb{Z}\text{ such that }k^2=x$, then isn't it that $\sqrt{x}\in\mathbb{Z}$...?

@Mr.Xcoder yes

And $\mathbb{Z}\in\mathbb{Q}\implies\sqrt{x}\in\mathbb{Q}\implies \sqrt{x}\notin \mathbb{R}\text{\\}\mathbb{Q}$.

@LeakyNun And that's it?

$\Bbb Z \in \Bbb Q$ is wrong

@LeakyNun What?

@Mr.Xcoder What??

How is $\mathbb{Z}\notin\mathbb{Q}$?

Oh, wrong sign

Should be $\mathbb{Z}\subseteq\mathbb{Q}$

\supseteq is the command you're looking for

Yep thanks

@Mr.Xcoder no that isn't right

that's the wrong way around though

@LeakyNun Done

(fixed)

\subset and \supset are the same symbol turned to the right or left, same with \subseteq and \supseteq

And that's just it?

@Mr.Xcoder right

o_O I didn't know of this technique...

@Mr.Xcoder it's called modus tollens

Fun and annoying fact: We learn basic mechanics (linear, without acceleration) in the seventh grade (Physics). Then we take a break in the eight, because we don't have the mathematical knowledge to move on to acceleration and get back to it in the ninth grade... Bad curriculum

@Mr.Xcoder I think you have now ended up with a different statement than you started with

@TobiasKildetoft you mean you have ended up

@LeakyNun the "t" was supposed to be a "w"

@TobiasKildetoft How come

You started with trying to show that if $x$ was not a perfect square then $\sqrt{x}$ was not a rational

but now the statement is that the squareroot of a perfect square is an integer

Forget about that

bye

$\exists n \in \Bbb N: \log_3 n \notin \Bbb N \land 2^{\log_3 n} \in \Bbb N$?

@LeakyNun Yes, of course

@TobiasKildetoft edited

Then no idea. My guess would be no

@TobiasKildetoft it's actually open :P

What in the world is ∧?

Logical and

Oh, I see

$\forall n \in \Bbb N: 2^{\log_3 n} \in \Bbb N \implies \log_3 n \in \Bbb N$?

@Blue what is the repeated limit?

it's defined to be the double limit

@Blue what is the definition of continuous?

Analysis on manifolds by munkres is amazing @MathmaticsAminP

Anyone here?

Hi @MathematicsAminPhysics

How r u?

Lucky

@BalarkaSen do you know Topology?

a little

Balarka knows everything, but if you have a question you should just ask and not ask to ask.

Yeah I have a question . Because I get peanut butter brain afterwards

The other thing about questions is that you should think for maybe a day on your own first before asking, if possible.

Oh .-.

But I'm trying to unless I'm missing a chunk of info in my book x.x

If you ask many questions without processing the information on your own first, you haven't really learnt anything.

So in the end, although you may get the answer to the specific problem, you haven't grown better as a result of the thinking process.

Just talking in general, not about this case here.

I know. I need to learn the material too but it's like my mind isn't processing because TOO MANY things going on in one sentence.

Good to see you around after so many years, lol.

@MathematicsAminP I dont think so.

Who me?

Yeah.

Haha thank s Anyway, A with a bar means closure right?

Yes, the closure of A.

And a circle means the interior of A.

Cool

My favourite book on general topology alone is Willard's General Topology.

It is also a very affordable Dover paperback.

Munkres had so much stuff going on at once

For some reason, I don't like Munkres.

I get overwhelmed sometimes so I read other books that step the brakes on

If you want a very simple book, it is Sutherland's Metric and Topological Spaces.

:O

Maybe I will delete my account again soon, lol.

No don't lol

@MathematicsAminPhysics yeah probably.

Does anyone know where I could find solutions to the exercises in Washington's Introduction to Cyclotomic Fields?

@user3343452 Hello, there are seldom exercise solutions given for books, only a few exceptions. =D

@WillHunting Damn, thank you anyway

Yes.

Well, assuming by compact you mean compact without boundary.

Morning

Good morning.

Good night.

Invariance of domain.

It's a nontrivial theorem.

Whereas if you just mean compact then it fails for the unit disc inside the real plane.

@TobiasKildetoft Technically the unit disk is not a manifold :)

wut

@BalarkaSen I suppose my idea of what a surface is is a bit hand-wavy

A manifold is locally homeomorphic to R^n; it's a manifold with boundary.

Oh, you mean the closed unit disk

@MathematicsAminPhysics Proof of invariance of domain requires some serious algebraic topology. Look in Hatcher.

There is no easy proof.

Well, uh, it's easy if you want to deal with differentiable embeddings. That is, if S is a C^1 submanifold of S'.

If you do not know anything you have no hope for understanding a proof. I don't have the patience for this.

@MathematicsAminPhysics Then you need to specify what definition of surface you mean

In case somebody with good knowledge of functional analysis and spectral analysis is around and has time to talk about this, there was this question about spectral decomposition of self-adjoint operators in another chatroom:

You need to know the bare minimum for understanding proof of a nontrivial theorem. If you want to prove that compact C^1 submanifold without boundary of a C^1 surface $S$ is the full $S$, it boils down to proving that for an injective C^1 map $f : U \subset \Bbb R^2 \to \Bbb R^2$, $f(U)$ is open for any open subset $U$ of $\Bbb R^2$.