Mathematics - 2019-02-07 (page 1 of 3)

chandx

00:00

daym, that would mean that in the same manner i could produce known identity $\sin (2x) = 2\sin x \cos x$

Pedro Tamaroff

Hi Ted.

Ted Shifrin

@Pedro !!!! I was just talking about you the other day :)

Yup, @chandx. For sure.

Lucas Henrique

@KasmirKhaan I flagged this. I mean... he's a doctor. I talk with guys that aren't even undergraduates and they're absurdly intelligent, makes me feel I'll never reach such "intellectual nirvana". Don't be disrespectful; know your place, please.

Pedro Tamaroff

@TedShifrin What was the subject?

Ted Shifrin

@Lucas: I know kasmir well. It's OK.

Pedro Tamaroff

00:02

I was dropping by here cause I saw a flag waving in the air.

Ted Shifrin

LOL @Pedro — your saying "toasts" and "fruits" :P

We don't need flagging, guys.

Pedro Tamaroff

French toasts?

Ted Shifrin

and also "fruits," I recall.

@Lucas: There's a problem that too much flagging goes on in here. It makes people in the rest of Stack Exchange upset. There's no need to flag. I can handle this fine.

Lucas Henrique

Sorry, @Ted. I'll speak to the room owners before. Just thought that the comment was just... too much. And we know it's recurrent.

Pedro Tamaroff

Wait, "my saying toasts and fruits"?

Ted Shifrin

00:04

Yes, @Pedro ... that's about the only non-standard English I've noticed from you.

But I just ran into someone else (not a native English speaker) who referred to "fruits" rather than "fruit" and it reminded me of you :P

Jacksoja

Hi chat

Ted Shifrin

hi @Jacksoja

Pedro Tamaroff

@TedShifrin I'll take that as a compliment. It's Thrusday in Dublin, gotta go. Bye!

Ted Shifrin

Bubye ;P

Leaky Nun

is every holomorphism (between complex manifolds) orientation preserving?

Lucas Henrique

00:07

Sorry for the trouble, @Pedro. Goodbye :)

Ted Shifrin

@Leaky: They're called holomorphic maps. You mean biholomorphic?

Yes, figure out why that's true for $\Bbb C$ first.

Leaky Nun

I mean holomorphic map

Ted Shifrin

You mean with holomorphic inverse ... or at least, you mean locally so.

Leaky Nun

why?

chandx

wait i did some mistakes in raisin to power 5, should be ${\sin}^5 \phi - 10{\cos}^2 \phi {\sin}^3 \phi + 5 {\cos}^4 \phi \sin \phi$

Ted Shifrin

00:09

Because you need an invertible linear map to discuss whether or not it's orientation-preserving

Leaky Nun

well I have yet to figure out why every complex manifold is real-orientable

Ted Shifrin

Right.

@chandx: I don't have it memorized, but that looks right.

Pascal's triangle really is your friend.

Lucas Henrique

Binomial theorem

In elementary algebra, the binomial theorem (or binomial expansion) describes the algebraic expansion of powers of a binomial. According to the theorem, it is possible to expand the polynomial (x + y)n into a sum involving terms of the form a xb yc, where the exponents b and c are nonnegative integers with b + c = n, and the coefficient a of each term is a specific positive integer depending on n and b. For example (for n = 4), ( x + y ) 4 = x ...

You can just substitute the terms :)

Leaky Nun

@TedShifrin $J = \begin{bmatrix} u_x & u_y \\ v_x & v_y \end{bmatrix} = u_x v_y - u_y v_x = u_x^2 + u_y^2 > 0$

is that an alternate way to memorize CR equations :P

Ted Shifrin

Oh, if you haven't done that linear algebra exercise, here it is.

The real derivative of a complex-differentiable function is a scalar times a rotation matrix. That's why holomorphic maps (with nonzero derivative) are conformal.

Generalize to $\Bbb C^n$. :P

Leaky Nun

00:14

indeed, follows from CR

Ted Shifrin

QED :)

Leaky Nun

they're all harmonic functions of dimension 2n aren't they

is everything related to symplectic manifolds?

Ted Shifrin

Kähler manifolds are a special case of symplectic manifolds.

chandx

ye, so it would be ${\cos}^5 + 5i{\cos}^4 \phi {\sin} \phi - 10 {\cos}^3 \phi {\sin}^2 \phi - 10i{\cos}^2 \phi {\sin}^3 \phi + {\cos} \phi {\sin}^4 \phi + i{\sin}^5 \phi $ so imaginary is ${\sin}^5 - 10{\cos}^2 \phi {\sin}^3 \phi + 5{\cos}^4 \phi {\sin} \phi $

now its clearer i guess

Ted Shifrin

Yup.

chandx

00:18

wow, so using complex numbers u can derive known trig identities

Ted Shifrin

yuppers!

Leaky Nun

yes!

Lucas Henrique

You lose order but you get algebraic closure and some cheats :P

Jacksoja

$E \in R^{k}$ is convex if $\lambda x +(1-\lambda) y \in E$

@TedShifrin in common language, what does this tell us ?

Ted Shifrin

\subset

It tells you that the line segment joining any two points of $E$ is entirely contained in $E$.

Jacksoja

00:21

first one yes

but (1-lambda )

why that multiple ?

Ted Shifrin

Regroup so that the $\lambda$s are together. What do you have?

Oh, and you should have said: for $0<\lambda<1$.

Lucas Henrique

@TedShifrin "teacher! teacher! I know how to do this one! here!!"

Ted Shifrin

LOL, you might have done some of my homework exercises, @Lucas? :P

chandx

its also cool that u can see a quadratic form in 2-order curve, do diagonalization on quadratic form and in eigen vector base u can see much clearly wherer curve is hyperbola or smth else

Jacksoja

lambda ( x-y) + y

Ted Shifrin

00:24

Draw a picture of that, @Jacksoja.

one of the main applications of eigenvalues and the spectral theorem, yes, @chandx.

Lucas Henrique

@Jacksoja: as you approach 0 and 1, you get closer and closer to $x$ and $y$, respectively.

Mike Miller

@LucasHenrique Did you finish up that argument from earlier?

Jacksoja

I see but it is a smart way to stating convexity

alrigh thanks !

Ted Shifrin

@Jacksoja: That's the standard way of writing the points on the line (segment) ... If you limit $0\le \lambda\le 1$, then you get the segment. Otherwise, you get the entire line. And you can generalize to more points.

Hmm, @Eric's back. I wonder if those 11 1/2 rolling eyes Demonark rolled hit him.

Jacksoja

@TedShifrin it works for nbhoods and that is what I wanted to show

Érico Melo Silva

00:32

nary a rolling eye has hit me

Ted Shifrin

OK, Eric ... I wasn't sure how crazy Demonark was gonna be with 'em.

Érico Melo Silva

lmao

Ted Shifrin

oh oh @Fargle is back

Fargle

Wait what?

Kasmir Khaan

@TedShifrin I have no idea how you did that but am very mad !

-.-

Ted Shifrin

00:35

I didn't do it.

You got other people mad and they flagged you and the mods did it. But seriously ...

However, I can do it ... so can other room-owners.

But you shouldn't come in here contradicting me when I tell you simple things. It's just rude and it's obnoxious, and enough ...

Kasmir Khaan

I really did not mean anything rude

I told you i saw that compactness lecture and that is now what I needed for my course

not only in R^n but needed it in the general settings

Ted Shifrin

The appropriate reaction should have been, "Oh really? Let me go look!" and LOOK CAREFULLY.

Kasmir Khaan

BUT I watched ur lectures

for multivar

Ted Shifrin

You obviously didn't understand that lecture. Or you wouldn't ask me why it's useful to know about convergent subsequences.

Kasmir Khaan

I was not trying to be rude

Ted Shifrin

00:38

If you understand it for $\Bbb R^n$, then you can answer your own questions.

I know, but you often are anyhow.

Kasmir Khaan

R^n is not my intrest

how am rude?

Ted Shifrin

You're missing my point. If you understand that case, you can generalize to metric spaces pretty easily.

Kasmir Khaan

okay thanks

Ted Shifrin

The only thing that's different is that it's special to say that in $\Bbb R^n$ compact sets are closed and bounded. In a general metric space you need to say totally bounded.

Kasmir Khaan

that is what I seeked when ask you

compact defined as open covers

it seemed a bit strange as defintion for me

Ted Shifrin

00:41

Right. It developed after a lot of history. That's not where mathematics started.

Kasmir Khaan

our course is just real analysis

and this seem to serve more in topology

so I wanted to know the meaning of compactness in these settings =p

Leaky Nun

I probably shouldn't say this but I'm still curious as to what the flagged messages was :P

Ted Shifrin

I think Lucas totally overreacted.

Kasmir Khaan

Ted kicked me out for 30 mins

Ted Shifrin

No, I did not.

Kasmir Khaan

00:43

or lucas?

Ted Shifrin

I could have ...

Kasmir Khaan

I assumed only could do that

mainly becasue

Ted Shifrin

The mods can kick people out for months. I can only do it for 30 minutes. (Same for the other room-owners, of whom there are now 5 or 6.)

Kasmir Khaan

you said warned me about it seconds before

Leaky Nun

@KasmirKhaan maybe... don't point your fingers if you don't know how the system works?

Ted Shifrin

00:44

Yes, I said I could. And you should not contradict me like you were doing.

Kasmir Khaan

omg am not pointing finger at any one

why am miss understood today ?

Ted Shifrin

You said, "Ted kicked me out for 30 mins" when I had already said I did NOT do it.

Leaky Nun

well i ain't seeing mr understood nearby so

Ted Shifrin

You have a tendency to do things like this.

Just chill out and get some sleep.

smacks Leaky

Kasmir Khaan

grrrrr kasmir is very mad a all of ya now -.-

yes smack that leaky again =p

Good night yall :)

Ted Shifrin

00:46

Night, Kasmir.

Dair

01:02

hi chat

Ted Shifrin

hi Dair

Dair

hi Ted

Looks like chat has been getting a bit heated.

Ted Shifrin

Nah, not really.

Some overreacting.

Dair

ah ok.

Mike Miller

@LeakyNun Complex manifolds are orientable because $GL_n \Bbb C \subset GL^+_{2n} \Bbb R$.

Leaky Nun

01:05

well that's a tautology

Mike Miller

Yeah.

Leaky Nun

that I'm still trying to prove

Ted Shifrin

Wait. Nothing's a tautology.

Mike Miller

@LeakyNun Great! Then so is orientability of complex manifolds.

Ted Shifrin

LOL

Dair

01:13

took me a while to realize that was a sort of a joke. my b.

chandx

01:25

anyways, i'll tell u smth interesting about mathematician Erdos

Got that Erdos had ADHD and so guess how the hell he was so unbelievably efficient in writin so many paper

He just took amphetamines

papers*

and he said that when he didnt take amphetamines "all he saw was a blank piece of paper" instead of productivity

nice right

its really hard to be professional mathematician without meds if u have adhd

Leaky Nun

@TedShifrin I can't come up with a more elementary proof than the fact that if you wedge the columns two by two then you get positive things, and then since they are 2-forms they commute (supercommutativity), so the overall 2n-form is positive

@MikeMiller ^

nvm that's ridiculous

Mike Miller

I would have said it in more old-fashioned language: the determinant of a block matrix can be computed as the determinant of what you get by taking the "determinant" of the block matrix to get a matrix out, and then taking the determinant of that - so long as all of the blocks commute.

Leaky Nun

hmm

chandx

do you know any professional mathematicians with ADHD?

i have adhd and wanna be research mathematician, thats not easy, but i hope with meds ill be able to

Monolite

01:42

Suppose the roots $r_1, \dots, r_p$ of the polynomial

$$x^p + a_1 x^{p-1} + a_2 x^{p-2} + \dots + a_p$$

all lie inside the unit circle. Is it true that the the roots of the polynomial

$$x^p + |a_1| x^{p-1} + |a_2| x^{p-2} + \dots + |a_p|$$

will also lie inside the unit circle?

dm__

all lies are inside a never ending circle. I agree.

Leaky Nun

r/im14andthisisdeep

Monolite

Did I make a typo? sorry english is not my first language

Leaky Nun

@Monolite no, your question is perfect

dm__

no, don't mind me, i'm being profound.

Monolite

01:46

Oh thanks! don't mind

Rumplestillskin

02:01

Is anybody here a fan of highly non-linear DE's? I have a DE given by $y'' + (1+C_1e^8y) y = C_2 e^8y + C_3 e^2y$ for some constants $C_1,C_2,C_3$. The only progress I've made is solving numerically! Any suggestions?

Leaky Nun

aha I finally figured out a proof

@MikeMiller @TedShifrin Using the JNF (for complex matrices) we can WLOG assume that the complex matrix is upper triangular (this doesn't change the determinant of the corresponding double-sized matrix because everything can be done on the level of real numbers); then this becomes a block matrix that is upper triangular, hence the determinant is just the product of the determinants of the 2x2 blocks, each of which is positive, qed

and bonus fact is that if $p(z) \in \Bbb C[z]$ is the char poly of $A$ then the char poly of the corresponding double-sized real matrix is $p(z) \overline{p(\overline z)}$

if $p(z) = z^n + a_1 z^{n-1} + \cdots + a_n$ then $p(z) \overline{p(\overline z)} = (z^n + a_1 z^{n-1} + \cdots + a_n) (z^n + \overline{a_1} z^{n-1} + \cdots + \overline{a_n})$

Mike Miller

02:17

@LeakyNun Nice. I don't know a non-computational proof of my general claim above about block matrices.

Leaky Nun

@MikeMiller I don't think your claim is true in general

Mike Miller

Sorry to hear that.

Silent

04:51

Let $f(x)=\frac{2x}{1+2x}+\ln (1+2x)$ and $g(x)=1$. I have to show that eventually $f(x)<g(x)$, i.e., for some $x_0>0$, $f(x)<g(x)$ for all $x>x_0$.

I know that $f'(x)<g'(x)$ for $x>\frac{\sqrt3}2$. Can I use that fact?

Iced Palmer

05:10

Off topic: By the Spectral theorem (linear algebra), I normal operator can be diagonalizable. Does this imply that a normal operator always has real eigenvalues?

In other words, when one ask if a normal operator $T$ can be diagonalized, are they asking if there exists a field $F$ for which the matrix of $T$ is diagonal, or are they assuming that it holds for any field $F$?

Zee

05:34

Does anybody get a weird feeling after doing algebra for hours continuously, like you can’t talk to people and your mind feels very heavy and rigid

I feel like a computer

I kinda like it but it’s painful

Jacksoja

@Zee Zee you are one hot potato

What kind of algebra did you do ?

Zee

Sheafs

Jacksoja

I don't even know what that is

Akiva Weinberger

Even the identity matrix doesn't work normally

Jacksoja

So I kinda understand what you went thru

@AkivaWeinberger I dont get the joke

or meme

Zee

05:38

I guess nobody understands love

Not surprised, knowing the dating skills of mathematicians

Dair

what is love?

Zee

Baby don’t hurt me

Jacksoja

Zee you seem like a cool guy

where are you from ?

Zee

Thank you. NY , you?

Jacksoja

africa

Never been to US

but am planning to visit some city, maybe cali or somehting

Zee

05:46

Colorado , NY , Cali , Washington are nice states to see

Akiva Weinberger

Multivariable calculus poster

Source

(Click to enlarge)

Dair

06:07

lol i saw this on reddit earlier today.

not using $\langle$ and $\rangle$ is a bet peeve of mine.

same goes for using $\emptyset$ instead of $\varnothing$.

Silent

06:38

Please help!

2 hours ago, by Silent

Let $f(x)=\frac{2x}{1+2x}+\ln (1+2x)$ and $g(x)=1$. I have to show that eventually $f(x)<g(x)$, i.e., for some $x_0>0$, $f(x)<g(x)$ for all $x>x_0$.

I know that $f'(x)<g'(x)$ for $x>\frac{\sqrt3}2$. Can I use that fact?

Akiva Weinberger

You want $f$ to be less than $1$?

Both $\frac{2x}{1+2x}$ and $\ln(1+2x)$ are increasing, no?

And it's already more than $1$ at $x=1$.

Silent

Sorry it should be $g(x)=x$

g'(x)=1

The solution manual to this problem says that f<g for $x>\sqrt3/2$, but i checked on desmos, this is not the case. It is true for f',g' that f'<g' for $x>\sqrt3/2$>

Akiva Weinberger

Ah

tarit goswami

Can anyone tell me how we can treat the operator $d^2/dx^2$ as $(d/dx)^2$ ?

Akiva Weinberger

Not literally squaring (multiplying it by itself), but applying it twice

Like the derivative of $f$ is $\frac d{dx}f$, the derivative of that is $\frac d{dx}(\frac d{dx}f)$

so people write it as $\frac d{dx}\frac d{dx}f=\frac{d^2}{dx^2}f$

It's just a notational device, though

You can run into trouble if you try to treat $\frac{d^2}{dx^2}$ like a fraction

Silent

06:51

@AkivaWeinberger, if you are free, can i ask please you another question?

*please ask :)

tarit goswami

@Akiva Ok, but in Physics, it is treated like this -

Akiva Weinberger

Sure

tarit goswami

$$-\frac{1}{2m}\big(\hbar^2\frac{d^2\psi}{dx^2}+(m\omega x)^2\big)=\frac{1}{\sqrt{2m}}\big(\frac{\hbar}{i}\frac{d}{dx}+ im\omega x\big)\times \frac{1}{\sqrt{2m}}\big(\frac{\hbar}{i}\frac{d}{dx}- im\omega x\big)$$

Akiva Weinberger

I think what you need to show is that eventually $f'-g'$ is negative and does not go to zero

Like, eventually you have $f'-g'<-.5$ for example

@Silent

Silent

OK! then apply mean value theoem to conclude, right?

Akiva Weinberger

06:54

@taritgoswami What happened to the $\psi$ on the right side?

@Silent That sounds right

Silent

ye!

tarit goswami

@AkivaWeinberger Yeah sorry, there will be $\psi$

Here, it is treated as multiplication of two operators

Silent

@AkivaWeinberger Why do we have for even function $f$ this equality $\int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx$?

tarit goswami

There is a $\psi$ at the end, let me write it in the way they have written -
$$-\frac{1}{2m}\big(\hbar^2\frac{d^2}{dx^2}+(m\omega x)^2\big)=\frac{1}{\sqrt{2m}}\big(\frac{\hbar}{i}\frac{d}{dx}+ im\omega x\big)\frac{1}{\sqrt{2m}}\big(\frac{\hbar}{i}\frac{d}{dx}-im\omega x\big)$$

Albas

07:16

Is there a way to find the minimum number of surface patches required to cover up a smooth manifold. For instance a sphere requires atleast two surface patches (stereographic projections) to cover it completely.

Mike Miller

07:27

@Albas When you say surface patches, do you mean charts?

The answer will be different depending on whether or not a chart is required to be diffeomorphic to $\Bbb R^n$ or to an open subset thereof.

I'm reasonably certain that in either case, for any $n$-manifold, you should need at most $(n+1)$ charts, and this bound is optimal on $\Bbb{RP}^n$. But I would need to look for references tomorrow, too sleepy now.

This should be closely related to the minimum number of critical points of an arbitrary smooth function, the "cup-length", and the "Lusternik-Schnirrelman category".

Albas

08:01

@MikeMiller yea I mean charts

Akiva Weinberger

Is there a 3D version of the inscribed angle theorem, involving spheres and solid angles?

I'm guessing yes

but I don't know how to prove it

8

Q: Can the inscribed angle theorem be generalized to solid angles in 3D? And beyond to n-dimensional space?

The "inscribed angle theorem" is a common 2-dimensional plane geometry fact. It states that for a circle the angle formed between any two points on the circumference with the center is twice the angle formed by those two points with any other point on the circumference. I will not elaborate on a ...

Oh

ÍgjøgnumMeg

08:19

Mooorning

Akiva Weinberger

@Silent That's true even if $f$ is not even. Let $x=a+b-u$. Then $dx=-du$, and $\int_a^bf(x)dx=\int_b^af(a+b-u)(-du)$.

Notice that the bounds of integration have changed, from $\int_a^b$ to $\int_b^a$.

Switching them back around gives us another minus sign, so we get${}=-\int_a^bf(a+b-u)(-du)$.

${}=\int_a^bf(a+b-u)du=\int_a^bf(a+b-x)dx$.

Compare this with addition: $\sum_{i=a}^bf(i)=\sum_{i=a}^bf(a+b-i)$, because the former is $f(a)+f(a+1)+\dotsb+f(b)$ and the latter is $f(b)+f(b-1)+\dotsb+f(a)$.

Silent

Thank you very much!!!

ant

@MikeMiller should this say connected or something? Shouldn't a disjoint union of n+2 copies of some n-manifold give a counterexample?

Leaky Nun

08:38

@ant no, the open subsets of R^n can be disjoint unions as well

ant

@LeakyNun why did you delete that

Leaky Nun

(this always gets me)

ant

He said in either definition

But yeah that's true in the second definition

Leaky Nun

I don't know what you mean, he didn't even say the word definition

Tobias Kildetoft

I hate working on my CV for the private sector. Was so much easier in academia where I could just point to my publications. Now I have to find some way to explain why I am great at programming even though I have nothing concrete to show for it.

Leaky Nun

08:39

oh, you mean charts

ant

`The answer will be different depending on whether or not a chart is required to be diffeomorphic to $\Bbb R^n$ or to an open subset thereof.`

`I'm reasonably certain that in either case, for any $n$-manifold, `

Leaky Nun

oh ok

then it should be connected

ant

Yep

Leaky Nun

come on you can understand that by context :P

ant

@LeakyNun I thought I might just be being dumb

Leaky Nun

08:41

I can cover $\Bbb RP^2$ in 2 charts though

ÍgjøgnumMeg

@Tobias same, even though I only have one "publication", I can just link my degree grades or smth

Tobias Kildetoft

@ÍgjøgnumMeg You are applying for PhD programs?

ant

@LeakyNun What does that counter or support?

ÍgjøgnumMeg

@Tobias I was supposed to start a masters in Heidelberg in April but I didn't get my app in on time, so I'm applying for the winter semester

but yes that's what I plan on doing

Tobias Kildetoft

@ÍgjøgnumMeg Then 1 publication is more than what most will have

ÍgjøgnumMeg

08:45

Indeed

Leaky Nun

@ant well he said you can only cover $\Bbb RP^n$ in $n+1$ charts

ant

What are your two charts?

Leaky Nun

draw the fundamental polygon

ant

done

Leaky Nun

remove the square edges, the first chart

remove the center, the second chart

@ÍgjøgnumMeg du gamla, du fria, du fjällhöga nord

ÍgjøgnumMeg

08:51

lol det får man inte sjunga på skolar

tror jag..

ah skolor* heter det

ant

@LeakyNun what is a square edge?

Leaky Nun

@ant remove the edges of the square, I mean

ant

Oh right

Leaky Nun

@ÍgjøgnumMeg really? why not?

ant

@LeakyNun The second chart you've removed the 2-cell?

Leaky Nun

08:55

no, I removed the single point at the center

ant

@LeakyNun Or you mean puncutre it

Oh right

Leaky Nun

yes

Tobias Shuxue Laoshi

Hello everyone! I have a problem in analytic number theory that I'm super stuck on. Anyone interested?

ant

@TobiasKildetoft I worry about that sometimes. I feel like I should be trying to find coding projects to work on to boost my CV if I move into the private sector

But then I'll end up spending too much time on that

Leaky Nun

@TobiasShuxueLaoshi just say it

ant

09:04

The A in my name is algebraic

Tobias Shuxue Laoshi

I'll write it up in case anyone would like to have a go. Let $\alpha\geq 0$ and $\delta<1$ be fixed real numbers. Let $f$ be an arithmetic function with $f(n)=O(n^\alpha)$ and $A_f(x)=\sum_{n\leq x}f(n)=O(x^\delta)$. Let $g=f\ast 1$ where $1(n)=1$ for all $n\geq 1$ and $\ast$ is the Dirichlet convolution.

That's the setup.

Prove now that $\sum_{n\leq x}g(n)=cx+O(x^\frac{(1-\delta)(1+\alpha)}{2-\delta})$.

ant

@LeakyNun What do the edges of the square correspond to on the real projective plane? Shouldn't they correspond to two loops on the space, representing the two homotopy classes?

Tobias Shuxue Laoshi

I can get a bound using the hyperbola method, but it doesn't have a main term and also doesn't have any $\alpha$ in it.

Leaky Nun

@ant I don't know

ant

Is it open though?

Your claimed chart (removing the edges) I mean

Tobias Kildetoft

09:09

@ant I have plenty of code written, but it is all garbage stuff written solely for my own use in determining KL cells in small ranks.

I am also working on a small Android app, but that was mainly because I felt weird not knowing how those worked

ant

is the android app a game?

Tobias Kildetoft

No

Leaky Nun

@ant sure, it's homeomorphic to the open unit square

(you just drew it)

Tobias Kildetoft

For now it is just going to contain a single useful tool to help calculate how to do things evenly when taking in while knitting

ant

That's open in R^2, not RP^2?

Don't you want it to be open in the subspace topology coming from R^3?

Leaky Nun

09:11

@ant RP^2 is homeomorphic to quotient of the unit square, as fundamental polygon shows

a quotient map is open

ÍgjøgnumMeg

@Leaky something about nationalism

Leaky Nun

there's no subspace topology coming from R^3

you can't embed RP^2 in R^3

ant

Right quotient topology from R^3

Leaky Nun

no, quotient of R^3\{0}

or quotient of S^2

Tobias Kildetoft

@LeakyNun Right, because that would imply that there was a real-life example of torsion in the fundamental group (and that would be weird)

Leaky Nun

09:12

@TobiasKildetoft lol

@TobiasKildetoft hey you're close to sweden

@ÍgjøgnumMeg that's stupid

ja alskar dett sang xd

Tobias Shuxue Laoshi

Den är tråkig tycker jag.

Leaky Nun

@TobiasShuxueLaoshi ah I just realized what your username means xd

(I speak Chinese btw)

Tobias Shuxue Laoshi

我也会说中文。不过水平比较低。

Leaky Nun

it's ok lol

Chinese is hard!

@TobiasShuxueLaoshi du gamla, du fria, du fjällhöga nord

Tobias Shuxue Laoshi

@LeakyNun Haha, when it comes to national anthems, I think it's pretty boring, at least harmonically.

Anyway, are you good at analytic number theory?

ant

09:20

ba'ax yaan Maax ku t'aan u maaya xan

Leaky Nun

you don't like your own national anthem?

no I'm not good at it

ÍgjøgnumMeg

@Leaky du älskar den låten*

Leaky Nun

what's wrong with sång?

ÍgjøgnumMeg

låt is more common

but there's nothing wrong with sång lol

Leaky Nun

I see

Tobias Kildetoft

09:53

Awesome, Jango just decided to play Kaizers Orchestra (even though they really don't fit into the station I had chosen). Been forever since I listened to them.

Tobias Shuxue Laoshi

@LeakyNun Nah, not really.

Leaky Nun

@TobiasShuxueLaoshi :o

ÍgjøgnumMeg

@TobiasShuxue min ex-flickvän sa att man inte får sjunga den på skolan, stämmer det?

Tobias Shuxue Laoshi

10:41

@ÍgjøgnumMeg Nja, det får man nog. Men det kan säkert anses lite provokativt.

Leaky Nun

ja, jag vil leva, jag vil dö i norden

@TobiasShuxueLaoshi I’ve heard that immigrants is a huge problem

refugees, that is

Rudi_Birnbaum

Hej vänner

Leaky Nun

hej

Rudi_Birnbaum

taler du ochså svensk

?

Leaky Nun

nej :p

Rudi_Birnbaum

10:49

lite

?

Leaky Nun

I happen to have understood you

I know like < 10 words

Rudi_Birnbaum

If you speak English you know more (and German on top)

ÍgjøgnumMeg

@TobiasShuxueLaoshi ah okej, det är konstigt, man sjunger det alltid i England o.O

den*

den engelska*

hahaha

Rudi_Birnbaum

@AkivaWeinberger Wow thats really cool! Yes that what I had in mind.

Leaky Nun

@ÍgjøgnumMeg what do we sing in england?

Rudi_Birnbaum

10:53

its coming home

ÍgjøgnumMeg

@Leaky nationalsången

Tobias Shuxue Laoshi

@LeakyNun It's not. Sweden is just unused to immigration. It'll take time to adapt to becoming a more international country.

Leaky Nun

@ÍgjøgnumMeg din sol, din himmel, dina ängder gröna!

abenthy

Hi

Rudi_Birnbaum

@ÍgjøgnumMeg Hur e läget?

Tobias Shuxue Laoshi

10:55

Hi

ÍgjøgnumMeg

@Rudi det e bra, jobbar bara :(

Rudi_Birnbaum

Sorry my swedish is declining ...

abenthy

I've got a question. Is there a description of the ideals of $\hat{Z} = \prod_{p} \mathbb{Z}_p$? I'm asking because I want a neighborhood basis of the finite adele ring of $\mathbb{Q}$.

Leaky Nun

@ÍgjøgnumMeg do most swedes know their anthem?

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$Mathematics$ Mathematics