@LeakyNun the normal basis theorem says this just the regular representation and that's how you prove additive Hilbert 90

HNY @Mathein

happy new year! @Ted

Rehi Ted, hey Mathein!

I just greeted you a fine lines ago, Demonark.

Hey @Daminark

Oh snap I didn't see that, I had been off my phone. But yeah how are things going for you?

@Ted did you cook something fancy for the holidays?

I cooked several days for a party for 35 people about 2000 miles from my home, @Mathein. Others cooked some too.

Oh wow, nice, where was this?

Ann Arbor, Michigan.

---If you're still there can you put in a good word? --- :P

Rip

I didn't see anyone in the math department, although I know one or two slightly.

But yeah I actually just got an email from them because one of my letter writers still didn't send his thing. I already had 3 letters for them so I said to just proceed without if waiting any longer is gonna cause processing delays

well, lectures resume on monday, so I'm enjoying the free time and going out with friends etc. (that's why I'm still awake at 1:45 am)

That's good, @Mathein. I'm glad to hear you're being social!

Nice. Is it the start of a new term?

yeah, I've had insight that doing only math isn't healthy long-term

nope, it's still the winter term

we have a semester system

Ah I see

but there are already some courses anounced for the next semester, it sounds pretty nice

when do you get feedback for your grad school applications? @Daminark

I think February and March are when most answers come back

I think that's a fair statement, Demonark.

Hopefully more February because the angst is draining

Okay so I have a differential geometry question about this excerpt above from Lee's Smooth Manifolds book. Let me denote the chart $(U, (x^i))$ by $(U, \phi)$ so that $\phi(p) - (x^1(p), \dots, x^n(p))$. So firstly for the left hand side of the highlighted part, I'm sure it should be $$v^i\frac{\partial \hat{ x^j}}{\partial x^i}(p)$$ where $$\hat{x^j} = x^j \circ \varphi^{-1}$$. Now I don't see why $x^j = \hat{x^j}$ which would be necessary for the highlighted part on the left to be obtained

What are those silly hats?

It's supposed to be the co-ordinate representation of $x^j$

I don't know why you're doing that.

Oh, I see. Yes, officially $\partial/\partial x^i$ acts on a function $f$ by acting on $f\circ \phi^{-1}$.

But no one writes that ...

Yeah that's what I meant

Does Lee use that notation earlier?

Actually disregard what I said about $x^j$ not equaling $\hat{x^j}$ because that's nonsense since they don't have the same domains

I'm sure he says he's dropping it, because once the definitions are made, everyone says "we're just going to write ..."

When he writes $x^j$ he means the function $x^j\circ\phi^{-1}$. That's what everyone means.

@TedShifrin Yep, I think he said he was dropping it somewhere

It seems though like we have $$\frac{\partial \hat{x^j}}{\partial x^i}(p) = 0$$ for $i \neq j$, and I don't see why that's true because formally $\hat{x^j}$ in this case is actually a function $\hat{x^j} : \hat{U} \subseteq \mathbb{R}^n \to \mathbb{R}$ so I don't see why we'd have $$\frac{\partial \hat{x^j}}{\partial x^i}(p) = 0$$ for $i \neq j$

The point is that you think of the function in local coordinates.

That's exactly the point.

In $\Bbb R^2$, consider $f(x,y)=y$. What is $\partial f/\partial x$?

$0$ in that case

But your case is exactly the same.

Hmm are you saying then that $\hat{x^j}(a^1, \dots, a^n) = a^j$ for $(a^1, \dots, a^n) \in \hat{U}$?

Yes, of course.

That's what the $j$th coordinate function does. :)

Ohh so that's how the component functions are defined? (I'm actually kinda shocked)

$x=(x^1,\dots,x^n)$ is the identity function on $\Bbb R^n$, @Perturb.

Ooooh wait I think I got it, the coordinante functions are the $\hat{x^j}$'s but they very different from the component functions $x^j : U \to \mathbb{R}$ which comprise $\phi(p) = (x^1(p), \dots, x^n(p))$

comprise :)

No, you're confusing yourself. Everything is the same.

Yeah woops

@TedShifrin Sorry have to power down my PC, it's lightning over here, thanks for all your help!

Well, not quite the same. Be safe.

Last night dream is some weird maths. (More details in another room as it will fill in the screen)

> Back home, I discuss with Akiva in maths chat about that problem ,and wondering what will you call a particle the size of a cow. Akiva said that will be the ox

Link here if interested

Can somebody tell if the answer of this question is correct? math.stackexchange.com/questions/2986579/…

In my opinion, it is correct.

hmm. chat's dead again.

hi @Ted

Didn't realize I was still logged in here ...

A dead chat is a sleepy chat

A dead cat is not

Well, I'm almost a dead chatter ... not quite yet.

lol

Meanwhile, I am thinking whether the literature actually has examples of limits similar to the dream where the sequence in question are geometric objects and not just numbers or functions. Perhaps if I recall stuff from category theory correctly, the notion of direct limits might provide a way to construct such things

by treating each geometric shape connected by some homomorphism as categories

Interestingly, the real life counterpart of what is shown in the dream is very young, a paper in 2010.

sciencedirect.com/science/article/abs/pii/S000926141000967X

> A conjecture on the energies of particles in other regular polyhedra is proposed.

It will be very cool as we approach 2020, reality started to resemble more of my dreams

@TedShifrin so apparently if charpoly(A) = f then charpoly(A^2) = f(x^1/2) f(-x^1/2)

Oh wow

First question, is there an easy way to see that this expression also gives a polynomial?

Oh wait ofc lol

Err actually less sure now. I can kinda buy it but I don't see it in my soul just yet

Do these look like girl shoes?

I need an honest opinion

To me they definitely don't

I was trying them on at the store and the sales lady said they did

But it's rarely good to take me at my word for that

I didn't buy them

:/

I just need fancy shoes

I give up, I'm just going to wear sneakers every where

Um, no, @CaptainAmerica. Stupid sales lady. Maybe they don't look good on you, but I've owned shoes like that.

@Leaky: Of course. $A^2-xI = (A-\sqrt x I)(A+\sqrt x I)$.

Demonark: Yes, it's a polynomial. Think of it this way. $g(t)=f(t)f(-t)$ is an even polynomial, hence a polynomial in $t^2$.

@TedShifrin I might try for a darker color

It's a matter of what you're wearing them with and how they look on you. Maybe you don't want suede (which is what that is). They're more easily damaged in rain and snow than other things that you can wax and protect.

I would think your parents would tell you this stuff ...

@Daminark: I tried to edit my thing to Demonark to ping you and instead it added a $t^2$ at the end. :(

Ah that's slick, nice

My mom told me they were suede, but I don't really dress up.(she thought they were nice...)

Oh weird, the edit messed up everything.

I think they are nice, @CaptainAmerica, but we know I have no taste. :P

lol

I guess Leaky isn't responding ...

Rest in rip

Rest in rest in peace

Is that a meme

Well, g'night, fellas. See you later in the weekend.

See ya later

@Daminark Wut. Isn't the joke rip in pieces?

See you Ted! And @Dair I have many variants

but rest in rip isn't a recursive acronym tho...

agh my brain hurts.

it looks like I keep missing active chat by that much.

I mean we weren't too active lmao

A few messages here and there

@Daminark Sounds more active than my social life

Rip in dead

(As I said, many variants)

So what've you been doing recently? You have graduated if I remember right?

Yeah.

I've been doing some web dev for now. I'm applying to CS grad school. Trying to do some stuff in evolutionary computation or something related.

Nice

I've been putting the math to the side for a bit... I've been working on some personal CS projects. Even if the idea feels relative simple, it can take a while to actually program it. Computers are definitely pretty picky compared to math TAs lol.

not only that but some of these machine learning/AI algorithms are pretty inconsistent and it's like one day they work, the other day they decide to flop completely.

I've been mosty active on codereview either indirectly asking for help with the website i'm working on or asking for help with my personal projects haha.

Lol yeah coding can be kinda annoying because of all the micromanagement

If I see a segfault ever again I'm gonna cry

lol it's been forever since i've seen a segfault.

I don't think Rust usually gets those.

and if Python gets a segfault it's a bug in the interpreter and largely out of your control i.e. really rare.

but you probably won't see a segfault again since you're doing math :D

wait, lol were you programming in C? If so, why were you programming in C? I don't think math people need to do that...

hi chat

@Daminark are you here?

@Jacksoja Hi!

Person i don't think i've ever seen

:)

I have question about this

can the map from S_4 to S_3 be given in an explicit form ?

I took intro to CS for two quarters

One was Racket which I enjoyed

One was C which has really really made me hesitant to return to coding ever again

@Daminark :(

I mean you can in principle give it an explicit form by just writing out elements but I don't think it'll be something especially nice

so it is just a computational question right?

there is no way to solve it without trying the products

i mean once the kernel is given, we can find what each of the 4 elements of S_4 act in the same way

You can say it's just gonna be S_4 mod the normal subgroup of identity+ double transpositions

@Daminark For me, I had this weird thing where I came to school with a lot of prior software engineering experience. I did usually (with the exception of one project) really well on the projects in school. Got wrecked on the math-puzzle stuff for exams. Decided I should focus more on the math-puzzle stuff and switched my major to math lol.

what I mean is, the kernel is { 1, (12) (34) , (13) (24) , (23) , (14 ) )

and if we want to know what elements of S_4 act the same way , we take the coset gK

Yeah that sorta thing, you can just write it out really

okay thanks

neat example

@Dair I def like the puzzle side more than implementation. Hell, I'm taking algorithms next quarter

@Daminark I definitely agree lol. I'm just better at the implementation haha.

@Jacksoja fun fact: that's the only surjection S_n to S_m for m > 2

@Daminark what do you mean ?

Or like, n ≠ m of course

i feel like this example should be studied further by me

seems very important

C can be ruff though... I remember doing SSE stuff and that was brutal.

@Daminark Ever do any programming in Coq, Idris, Agda, Metamath, or Lean?

Nope. Are those the automatic theorem proving stuff?

Basically. Formal verification languages. I don't think Metamath has any automated stuff

I haven't really had the energy/time to invest in it like I probably should, but it's kind of cool. A lot of the introductory material (well for Idris at least) is kind of like a discrete math course. You don't end up being confused about what is consider "obvious" and what "isn't" because really it either compiled or it didn't.

Well, I was pretty confused at least... I had trouble understanding a lot of basic stuff like "The remainder is always between 0 and n-1" when I took discrete math lol.

good times. probably should head out soon. cya.

hi @AlessandroCodenotti

Hi

it'sa mazi ngho wth isi shar der tor ead thanifIdothis

it'samazinghowthisiseasiertoread

Finally, my creativity rivals my dream self

anyone tell me prerequisites for bilinear pairing in elliptic curves ?

Let $<a_{n}>$ be a sequence of positive numbers such that $a_{n} = \sqrt{a_{n-1}a_{n-2}}$, its given that $<a_{n}>$ converges, but i dont see how. Also where does it converge to? Help.

@Shobhit well $a_n=1$ satisfies your conditions

actually I think any sequence of positive numbers satisfying $a_n = \sqrt{a_{n-1} a_{n-2}}$ converges

so the "given" is superfluous

also it should converge to $a_1^{1/3} a_2^{2/3}$

Thought: People from second semester of college refers to their first semester as "last semester"

@GaloisintheField jmilne.org/math/Books/ADTnot.pdf

CFT is in I.8

@MatheinBoulomenos was denkst du das? math.stackexchange.com/q/3060308/328173

kennst du den "Ordnungsatz"?

nope

@BalarkaSen how does man overcome the horrendous symbols in dealing with manifolds?

but I agree with Jyrki

@LeakyNun what horrendous symbols

the conductor of a quadratic number field is really easy to understand

@BalarkaSen I mean, the integrations

forms?

yes

thats the simplest way to globalize local calculations

Still unclear what symbols in that process ate awful

I'd argue forms simplify notation

I mean, too many variables, I can't see what is going on

That's multivariable calculus. You can't really avoid that. Once you have some basic technical results part of the point of forms is that you can stop writing coordinate expressions like this

yeah I don't really like multivariable calculus

That theorem is literally saying that everything is coordinate independent

@LeakyNun Tough luck studying manifolds then

right

thing is, multivariate calculus isn't optional

Every field has technical results. This is like a page of symbol pushing. That's pretty tame for what it could be.

I don't like symbol pushing either but the content of this is pretty much the formula at the top of the page...

@BalarkaSen Should I try to go back to bed or answer this

Yes, I'd like to see your answer to that

Which one!

Lol

The MSE question

Aight

is the empty set a manifold?

is the empty function smooth?

You tell me

It seems saving an offline copy of the DLMF paid off. I knew a government shutdown would happen some time during my hiatus.

If $\phi :\Bbb R\to \Bbb R$ is smooth, cancels on $]-\infty,0]$ and is non-zero on $]0;+\infty[$, how do I solve the equation $\phi D =\delta_0$ for $D$ in the space of ditributions ($\delta_0$ being the dirac at 0) ?

More explicitely I'm interrested in the case $\phi(x) = e^{-1/x}$ for $x\gt0$

@LeakyNun I think I resolved the confusion

Im trying to find a proper dense subset of $\Bbb R ^{[0,1]}$ that is sequentially closed. any ideas? (the topology is the usual - product topology)

@MatheinBoulomenos what the

@LeakyNun ?

I mean, that's unexpected, lol

it's just the $p=2$ case being weird again

for $p \neq 2$ you have $1+p\Bbb{Z}_p \cong \Bbb{Z}_p$, so any finite index subgroup is indeed of the form $1+p^n\Bbb{Z}_p$

@LeakyNun The empty set is a manifold

An open subset of a manifold is a manifold

The empty set is open

QED

Nevermind i got it

Actually I didn't

Is there a method or online tool to check answers to line integral problems? For the line integral of [xy, x^2]; C: x = y^2 from (0, 0) to (1, 1), I keep getting 3/4 but my answer key says 3/5, and I'm not sure if it's a typo.

@AkivaWeinberger I mean you can argue pretty directly. A manifold is a topological space such that for any point blah blah blah. Empty set has no points, so it's an auto-manifold by vacuous logic

True

@BalarkaSen You must argue directly. Akiva's argument is implicitly this.

I think the following should be true and easy but I'm missing something

For every cardinal $\kappa$ there exist a cardinal $\lambda>\kappa$ with $\lambda^{\aleph_0}=\lambda$

I can prove this under GCH but I'm pretty sure it should work unconditionally

@AlessandroCodenotti Where does Loh prove it? The book has gotten quite a bit larger since I last saw it!

Which version of the book are you looking at?

I was just looking at the draft on her website(with the white and gray backgrounds)

I don't think you need AC, but maybe there is some weird generality where you do...

@Alessandro can't you take any cardinal $\lambda'$ with $\lambda' > \kappa$ and then put $\lambda = \lambda'^{\aleph_0}$? then you'll have $\lambda^{\aleph_0}=\lambda'^{\aleph_0^2}=\lambda'^{\aleph_0}=\lambda$

@PaulPlummer Proposition 7.2.9

@MatheinBoulomenos Oh, of course

Nice one

The first part of that proposition to be precise

When deriving the jump condition for grens functions a lot of lecture notes say that since $\del _x G$ and G are bounded they don’t make any contribution to the integral $\int_{\zeta-\epsilon}^{\zeta-\epsilon}\alpha{\partial_xG }^2 + \beta \partial_xG + \gamma G = \delta (x-\zeta)$ as you make $\epsilon \rightarrow 0$but since the first derivative is discontinuous at $\zeta$, how does this not contribute?

I’m just gonna retype that

@TedShifrin sorry you have to witness my terrible latex

gimme until my masters I’ll have figured it out

there we go

@AlessandroCodenotti Yah i don't think you need AC. it seems like her proof might need some form of AC to get an actually function, but you can do a similar construction where you give a sufficient partition $[0,L]$(in terms of the quality of $f$ and quasigeodesic) and do a sort of piecewise/discrete path where each partition only gets sent to one point.

But I guess with her proof you don't really have to go through these "extra" steps

Hmmm it's bit late here now, I'll think about what you said carefully tomorrow

Thanks for thinking about it though!