You know it doesn't bode well when the professor emails us to make sure we are mentally prepared for a tough exam :(

hi @TedShifrin

hi @Shobhit

how are you?

hi @Tobias

@TedShifrin Hi

@Semiclassical That's only one of its countably many values.

Doing OK, Shobhit, thanks, and you?

@TedShifrin Hey ! Good morning :)

hi @Tanuj

Afternoon here now, but close enough :)

i am good, have complex analysis exam tommorow :(

Oh , okay. Have a good afternoon then .

There are worse subjects than complex analysis, Shobhit :)

@TedShifrin

Any thoughts about this ?

Why should that be interesting, @Tanuj?

@TedShifrin because , when I go for termination of $a$ , I make the question too lengthy everytime

i am more upset because i think i wont be able to take number theory as elective subject, very few students. @ted

It just seems like a problem that's made up to be a mess.

@Shobhit: That's how I ended up teaching topology as a reading course a few years ago.

@Tanuj: What do they mean, "independent of $a$"? If you vary $a$, you'll get a different function $y=y(x)$, so I actually don't think the question makes sense.

i dont think there is anything that you can't teach @TedShifrin

@TedShifrin $a$ is a constant here , the $dy/dx$ they have expressed is entirely in terms of $x$ and $y$

Yes, @Tanuj, but the $y$ will be different when $a$ is different. So this is crap.

@Shobhit: I taught topology a lot of times during my career. But there are things I never taught (some I wish I had).

@TedShifrin idk , this does have an answer according to my book

@Tanuj: I think they're wrong. Just giving an answer with no $a$ in it does NOT make it independent of $y$. You're on different curves, so $(x,y)$ will be different even with the same $x$.

I mean you can implicitly differentiate and eliminate $a$. You'll have an expression just in $x$ and $y$. That doesn't make $dy/dx$ independent of $a$.

@TedShifrin I mean I think they just mean the first part. Can you tell how to do that ?

I'm still saying the question is just wrong. Just solve for $dy/dx$ and then plug in $a$ in terms of $x$ and $y$. It's a mess, but ...

Who's going star-crazy?

lol

Anyhow, Shobhit, complex analysis is beautiful material. What is your exam covering?

i was noticing that too

XD

whats up with the random starring ?

oh , great !

@TedShifrin yeah, fair

but even that one value is goofy

Whoever's doing the stars, please stop.

No more constellations.

Why is that any goofier than $e^{i\pi} = -1$, Semiclassic?

headdesk

(at the ironic starring)

It may be ironic — It's certainly not iconic.

eh, "real to the imaginary can be real" is not so weird to me as "imaginary to the imaginary can be real"

though maybe that's just a sign of how comfortable I am with Euler's formula

yes, it is very good and easy to understand (most of it). Mostly its, complex integration, analytic functions, linear fractional transformations, conformality, and symmetry. I think the symmetry part is related to or is inversive geometry, not sure. @TedShifrin

Whoever keeps starring unnecessarily: It's not funny and it's not helpful. Stop.

OF course imaginary to imaginary can be real (among other things).

It’s just algebra , it don’t mean anything

True enough.

Oh, that's cool, Shobhit — yes, inversive geometry is precisely what that is.

That said, I can also think of $e^{i \pi}=-1$ in terms of $z\mapsto z^{1/2}$

Are fiber bundles a type of a tensor bundle ?

and that's weird, but only finitely weird

Not necessarily, @Zee. Vice versa, yes.

What about vector bundles ?

Right, bivalued is less intimidating than $\Bbb Z$-times valued.

on the other hand, doing $i\mapsto i^i$ has infinitely many possible values (which, I know, you mentioned before)

In which place, Zee?

right

On a smooth manifold ?

But that's all because of the first thing, Semiclassic. Log is multivalued just because $e^{2\pi i}$ had the afrontery to come out $1$.

No, Zee. Rewrite your question. I don't know where you're putting vector bundles — in place of tensor bundles or in place of fiber bundles. Again, tensor bundles are specific examples of vector bundles, and vector bundles are specific kinds of fiber bundles.

Oh I see

@TedShifrin hah

What kind of an algebraic structure is associated with a fiber bundle ?

@Tanuj: I still say the question is garbage, but if my algebra is right, the derivative simplifies rather nicely to $\sqrt{\dfrac{1-y^2}{1-x^2}}$. :P

@Zee, none is necessarily associated. The fibers might just be topological spaces.

is a fiber bundle an instance of a sheaf ?

@TedShifrin yea , but how did you simplify that ?

No, the topological structure works wrong, @Zee.

Why ?

It's not interesting, @Tanuj. I substituted for $a$, got a mess, then totally cleared denominators and simplified, factoring out what I could and then noticing what was left was the same in numerator and denominator.

The topological definition of a sheaf makes the projection map a local homeomorphism. So that works for a fiber bundle only when the fiber is discrete. @Zee

Greeting @Balarka

Ted!!

Alright thanks shifrin

LOL, you're welcome.

I wish we could get someone banned for over-starring. Seriously, stop it.

I'll accept the ironic stars on that one.

I don't understand this childish behavior

But I am starring your post nonetheless

Semiclassic and I have both asked nicely before.

Our local mods aren't around so much anymore. I miss them.

Same :(

Especially robjohn, Pedro, and DanielF. Even anon is a rarity now.

Maybe I should vaporize, too.

noooooooooooooooooo

NO

i will fail in my exams if u left

Nonsense.

@PedroTamaroff @DanielFischer @robjohn Please come back

@TedShifrin in which course will i study this manifold thing?

What do you mean by manifold thing?

I don't know your curriculum.

Manifolds are in my honors multivariable math class. They're standardly taught in multivariable analysis, differential topology, graduate courses in manifolds and differential geometry, etc.

oh, we have them as electives

Well, Balarka, I guess DogAteMy has gotten occupied with other things (like maybe school), so I haven't seen/heard any Riemannian geometry in a few days.

@Tanuj: Did you get it to work out?

He asked me why $d\omega(X, Y) = X\omega(Y) - Y\omega(X) - \omega([X, Y])$ earlier today

I pointed to the Cartan's magic formula

Unfortunately I didn't have time to explain

@TedShifrin yup , I did , thanks a ton ! :)

Any JEE people here ? Can someone tell me some website where I can find all previous year questions ( starting from 80's )

Cool, @Tanuj. But you understand why I say their wording is just wrong?

@TedShifrin Yup ! Totally.

@Balarka: I just answered a question precisely doing that a week ago. Tell DogAteMy to find it. I still typically check it in local coordinates (having observed that the RHS is a tensor).

Good point, that's one possible way to do it

@AkivaWeinberger ^^

@Akiva: DogAteMy — check my comment but also read this.

@Balarka and DogAteMy: At some point we should discuss how first-variation formulas are just Stokes's Theorem plus the Cartan magic formula.

Ah right that problem

So what's the best way to prove the Cartan magic formula? :)

Stokes's Theorem? Local coordinates?

I have a highly rated MSE question on this

Arnold suggests a proof by magic

I do it in an indirect way by showing the left and right hand sides are cochain maps on the deRham complex that agree on degree 0

Probably not the most enlightening way

That’s the best way.

@BalarkaSen how old are you ? ( Too random , but I was curious)

In less fancy terms, induction on $k$.

It suffices to prove many things on 1-forms. It’s a great trick to know.

Just 0-forms here.

@Tanuj Somewhere between 13 to 27

Yes, that's the proof I usually give that $d$ commutes with pullback. Just the chain rule and we're done.

Not sure precisely where

flunks Balarka on English

Somewhere on the interval [13, 27]

Is that better?

Ah , @BalarkaSen you're still too young if you're 27

No.

between ... and ... ... or in the interval ... :P

I remember when Balarka was 13, but thankfully those days are past.

Larkÿ, did you see the pdp video about the tanks

@BalarkaSen I never get the 'high level stuff' you're always talking about . What exactly do you do ? (as in do you study or some job)

@TedShifrin That's not what a man living in that interval would say!!

On the Earth, not in the Earth

He'd be too thin to have a voice, Balarka.

How old are you seriously ?

Very pointillius vocal chords

@Tanuj I'm 18, I was just joking about

@0celo7 Not yet. Give me an hour or so

I'll look then

Did you see the article I pinged you and Bernardo with?

It contains glorious symbolism from the fatherland

Problem: Let $X$ be compact; let $f_n \in C(X,\Bbb{R}^k)$. If the collection $\mathcal{F} = \{f_n\}$ is pointwise bounded and equicontinuous, then the sequence $f_n$ has a uniformly convergent subsequence. Proof: Since $\mathcal{F}$ is pointwise bounded and equicontinuous, $\mathcal{\overline{F}}$ is compact as well as a metric space, which means it is sequentially compact. Since $\{f_n\}$ is a sequence in it, it has some convergent subsequence.

Call it $\{g_n\}$ for simplicity and let $g$ be its limit. Then for every $\epsilon > 0$, there exists $N \in \Bbb{N}$ such that $d(g_n(x),g(x)) \le \rho (g_n,g) < \epsilon$ for every $n \ge N$ and every $x \in X$, which means $g_n \to g$ uniformly in $\Bbb{R}^k$.

I meant the Soviet part, not the Nazi part...

I'm 18 too . Where did you learn all of this ? I mean you're like waaay ahead in terms of what we are actually taught.

How does that sound?

@BalarkaSen which one

@Tanuj: He's been learning a lot with help from a few of us here and also from university faculty ... for probably 3 or 4 years or more. Akiva is our American version :)

Yeah I did

It’s a prank

@user193319: The point is to make the metric explicit at the outset. You didn't do that. The metric gives the topology of uniform convergence immediately.

Cool

@Tanuj I got interested and sidetracked into a rabbithole of mathematics. I still don't claim I'm good at it, I just happen to have an encyclopedic knowledge on some parts. Honestly I value creativity/problem-solving more than knowledge

Akiva is the real man

I plan to steal his brain at some point

@0celo7 This one: stereogum.com/1986899/…

I hate my life.

DogAteMy has tremendous intuition and also learns very quickly.

@TedShifrin Right???

@BalarkaSen Akiva is a boy?

I've had a few exceptional university students over the years. It's a treasure.

@Abcd: Well, at 18 we probably say young man :)

Whereas I'm an antique relic.

@TedShifrin The metric on $\Bbb{R}^k$? By not specifying the metric I was just doing what Munkres did. He started all proofs by saying "Let $d$ denote the standard or square metric on $\Bbb{R}^k$." So my proof is not right?

@TedShifrin Nah, I thought Akiva was a female name!

@Tanuj: It's not a race.

@Abcd: it's a Hebrew name.

No, the metric on the function space, @user193319.

@TedShifrin yea right , Indian education system is though , like the 'definition of race'.

@Rick okay thanks again!

Better to have passion and really understand things, @Tanuj.

@Tanuj Very true. It's quite draining.

hi @Leyla

@TedShifrin Oh. That metric is given by $\rho (f,g) = \sup\{d(f(x),g(x)) \mid x \in X\}$.

Hi, how are you @TedShifrin ?

@TedShifrin yup :) This gives me a reason not to give up !

which is why I wrote "$d(g_n(x),g(x)) \le \rho(g_n,g) < \epsilon$..." in my proof.

Hi @Ted

Comment vas-tu ? Mieux j'espère ?

@BalarkaSen How were boards?

@user193319: I know. I was just suggesting that if you recall the topology/metric at the beginning, then it's pretty obvious.

@Abcd I still have 2 weeks before them

Un peu, @Astyx. On verra, merci.

@TedShifrin @TedShifrin Is Balarka studying topology since he was 13?

@BalarkaSen Ohkay.

I think I've been chatting with Balarka here for about 4-5 years, so maybe even earlier.

I didn't learn any topology until I was 19 or 20.

he is so ahead of his times!

@LeylaAlkan you're welcome

I reiterate that I value being creative and being able to solve problems more

The two qualities I lack extremely

Oh please

damn

@Balarka: We've yelled at you before for the false self-deprecation. You are always doing problem-solving of one sort or another. Not the tricky Indian exam stuff. But mathematics, yes.

@Leyla I think Rick has misled you, ${F(x+h)-F(X)\over h}$ is not generally equal to $F'(x)$, you need to take the limit as $h$ goes to $0$ (provided it exists), And what's tricky is actually proving $$\lim_{h_1\to 0}{{\lim_{h_2\to 0} {F(x+h_2) - F(x)\over h_2} -\lim_{h_2\to 0} {F(x+h_2-h_1) - F(x-h_1)\over h_2}}\over h_1} = \lim_{h\to 0}{F(x+h) -F(x)-(F(x)-F(x-h))\over h^2}$$

That's not the way to do that, @Astyx :P

What do you mean ?

@Astyx porcupine tree are awesome, I prefer some of their stuff to the solo works of Wilson (those are real good too), but Balarka is more on the metal side on prog than the rock one

Hi @Ted

Oh right I forgot about the Porcupine reco

Hi @Alessandro

Let me bookmark it

I suggest fear of a blank planet from porcupine tree and harmony korine by Steven Wilson (watch the video of the latter)

The whole album "The raven that refused to sing" by Wilson is very good too

@TedShifrin It's not false from my perspective. I think it's two things I should seriously put attention to, and I'm learning from how Akiva thinks (eg, the recent conversations with you and Akiva about the torsion was very enlightening)

Not really self-deprecating as much as pointing out what I think one should focus on more than "learning lots"

@Astyx we kept it short actually

I'm a happy boi. I'm good with what I know.

As I said, exploration and depth of understanding are more important than encyclopedic knowledge, Balarka.

Short ?

@TedShifrin I agree, that's what I said!

@Astyx: I vote for Taylor's Theorem, not double limits.

Yes I did too

Hey people

Hi dami

@AlessandroCodenotti Thanks, also bookmarked!

Hi Demonark.

@Ted here : chat.stackexchange.com/transcript/message/43380369#43380369 :)

Gotcha.

@AlessandroCodenotti I have been feeling very Italian today. Lunch was left over risotto from yesterday, and while I ate I occasionally went to stir the bolognese on the stove.

But then Rick said this : chat.stackexchange.com/transcript/message/43380369#43380369 which is why I'm trying to re-explain

@Balarka About metal prog I like youtube.com/watch?v=dQw4w9WgXcQ a lot

FUCKING HELL

JESUS GOD

:)

When we do variable separation in the context of series solutions and have a positive semi-definite operator acting on the separated functions like $-\dfrac{\Theta''(\theta)}{\Theta(\theta)} = \text{constant}$, is it often a good idea to set the constant equal to a square say $n^2$ so that we may sum over $n \in \mathbb{Z}$?

My question probably made no sense

It's only going to be boundary conditions that tell you the possible eigenvalues, @Lozansky.

But $\mathcal{A}=-d_{\theta \theta}$ is a positive semi-definite operator so we know $\mathcal{A} u = \lambda u$ only has non-negative eigenvalues??

Right.

@Astyx I found a fantastic French meme recently that you'll appreciate

I can't wait to see it

@TobiasKildetoft that's great! How did you make the risotto?

Here it is: www.imgur.com/a/G6rU8

Heh

@AlessandroCodenotti In a very lazy and probably not very Italian way actually

I guess I deserved it

didn't really stir much while it cooked, and used completely ordinary rice for it. Still tasted great though

@Astyx rekt

@Tobias: You're a charlatan!

get 1-up'd son

It's 1-1

Yes but say I want a complex Fourier solution $\sum_{n \in \mathbb{Z}} c_n e^{in\theta}$, then I should have to assume $\text{constant} = n^2$?

used fennel rather than celery and added shrimp, peas and carrots

When the "coefficient of determination" is the pearson correlation times 2. Then the formular is r^2 = coefficient of determination. But why is the coefficient of determination R^2 and not r^2 ? Because then r = R ?

@Lozansky: Boundary conditions will have to dictate the period.

Sounds good, I eat a lot of different kinds of risotto because I often eat with my classmate and it's a comfortable dish to prepare for a lot of people

@TedShifrin Sorry, I know the period is $2 \pi$

Well, that's what tells you the eigenvalues are $n^2$, then.

I used to find it really tiring to make, until I found out that it really did not need to be stirred anywhere near as much as I was doing

I am unable to prove that the cube roots of unity represent vertices of an equilateral triangle. Attempt:

@TedShifrin This is trivial?

@Lozansky: What are the solutions of $x''(t) + k^2 x(t) = 0$?

For angle proof, $\arg(\omega^2 - \omega)- \arg(\omega -1)= 120^\circ$

@Ted

Uh

@Abcd Use sides, not angles

@BalarkaSen I want to prove for both

First for angles then for sides.

They are $x(t) = e^{kt}$

Not quite, @Lozansky.

Well minus as well

Or just trig

Well, what is it?

On safe side I say $a \sin(kt) + b \cos(kt)$

So what values of $k$ will give you period $2\pi$?

@Abcd Ugh, that's tedious unless you use some kind of trick (eg, two cube roots of unity makes an angle of 120 with the origin)

@TedShifrin Only $\pm 1$

Well, remember that $\sin(2t)$ also has period $2\pi$ (just not the smallest period).

Hm right, so all integers

So that's where your $n^2$ comes from. If you have different boundary conditions, the period will change accordingly.

@BalarkaSen Angle between lines joining $z_1,z_2, z_3, z_4 $ is given by $\theta = \arg(\dfrac{z_4-z_3}{z_1-z_2})$ right? Then angle here between line joning $\omega ,1$ and $\omega, \omega^2$ should be $\arg{w}= 120^\circ$

Why doesn't this work?

@TedShifrin If I have something like $R(r) = ar^{\lambda} + br^{-\lambda}$ and I want $|R(0)|<\infty$, that's equivalent to saying $R(r) = cr^{|\lambda|}$, no?

OK.

@TedShifrin can you please see why its not working?

@Abcd You need to be careful when manipulating arg like that

I haven't been following, @Abcd.

What are you trying to do?

@TedShifrin Finding the angle between lines joining $1, \omega$ and line joining $\omega, \omega^2$

Where $\omega$ is any nonzero complex number?

No, cube root of unity

Schoen and Yau are going to drive me crazy. Upon closer inspection, there's some illegal integration by parts going on.

The cube root of unity. He doesn't want to use a side argument to prove that they form an equilateral triangle

Oh, and what's the definition of angle between two lines? There are two possible angles.

The switched from an "defined somewhere" derivative to a distributional derivative and I don't buy it

It's actually not clear that this function is differentiable a.e.

Got my mistake.

Can a harmonic function have large swaths of critical points?

$\omega$ lies in 2nd Quadrant

So $\arg(z)= \pi - \alpha$, where $\alpha = \arctan\dfrac{y}{x}$ (acute)

Be careful.