Mathematics

12:24 PM

Observation: D(e^{n 2\pi i x})=n 2\pi i (e^{n 2\pi i x}$ and $E(e^{n 2\pi i x})=e^{n 2\pi i (x+1)}=e^{n 2\pi i x})$ if $n$ is an integer

So that's a simultaneous eigenfunction of $D,E.$

In fact, it's also an eigenfunction of $E^{1/2}.$

SteamyRoot

12:39 PM

@LeakyNun Right, I was bored enough to actually think about that puzzle.
$$\sum_{r=0}^m(-1)^r \begin{pmatrix}m\\ r\end{pmatrix}(x+r)^n = (-1)^m \sum_{r=0}^m(-1)^{m-r} \begin{pmatrix}m\\ r\end{pmatrix}(x+r)^n = (-1)^m \Delta^m x^n = (-1)^m \Delta^{m-n} n! = 0$$

Liad

12:51 PM

i have $f: S \ ^1 \to \Bbb R $ continuous i need to prove that there exists $x \in S \ ^ 1 $ s.t $f(x) = f(-x)$.
So, suppose there isn't such a point. so define $g(x) = f(x) - f(-x) $ , $g \ne 0$ so we can assume w.l.o.g that $g \gt 0$ . so if we define $h(x) = \dfrac{g(x)}{||g(x)||}$ we have a map $h : S \ ^ 1 \to S \ ^ 1$. how can i get to a contradiction from here? or maybe im not in the right way?

(it is a question in topology course, we learned about retracts maybe i need to use it somehow.)

Akiva Weinberger

We also know that $g(-x)=-g(x)$

and $h(-x)=-h(x)$

SteamyRoot

What is $-x$ for $x \in S^1$ ?

Opposite point

Antipodal?

Yeah

12:53 PM

i thought $-x = -(x_1,x_2) = (-x_1,-x_2)$ doesn't it?

Akiva Weinberger

Yeah, that's the antipodal point

SteamyRoot

It depends how you define $S^1$...

Liad

huh. ok

Sha Vuklia

@Semi you have time?

Leaky Nun

@SteamyRoot $S^n = \left\{ x \in \mathbb{R}^{n+1} : \left\| x \right\| = r \right\}$ from Wikipedia

SteamyRoot

12:54 PM

A common definition is a subset of $\mathbb{C}$, where $-x$ would have a different meaning.

Leaky Nun

@SteamyRoot not really

Liad

alright, didn't know that :)

SteamyRoot

@LeakyNun How so?

Liad

@AkivaWeinberger so $h$ must get $0$

Leaky Nun

@SteamyRoot because $-x$ would still be the opposite point.

SteamyRoot

12:55 PM

Oh, right.

In that sense

I was thinking of $e^{i\theta}$ where you consider $\theta \in \mathbb{R}/2\pi \mathbb{Z} \cong S^1$

Liad

@AkivaWeinberger if $g(-x) = -g(x)$ and $g$ is continuous then $g$ must get a $0$, is that what you meant?

SteamyRoot

It's a question that is commonly asked as "prove there exist two antipodal points on the equator having the same temperature".

Balarka Sen

$h$ is a map $S^1 \to S^0$ though, is it not?

Akiva Weinberger

@Liad Wait ignore all that

I got confused

Sorry sorry sorry

SteamyRoot

It tends to be easier here if you work with a map $T: [0,2\pi] \to \mathbb{R}$ for the "temperature"

Leaky Nun

12:59 PM

@BalarkaSen what is $S^0$? Two points?

Liad

everything is ok :P

Balarka Sen

Yes.

Akiva Weinberger

@Liad You're right

I thought $g$ was from $S^1$ to $\Bbb R^2$ for some reason

or that $h$ was from $S^1$ to $S^1$

But yeah, $g$ is from $S^1$ to $\Bbb R$

Balarka Sen

Well, Liad wrote it wrong.

SteamyRoot

And you define a secondary map $T_a: [0,2\pi] \to \mathbb{R}: \theta \mapsto T(\theta + \pi) - T(\theta)$

Leaky Nun

1:01 PM

@BalarkaSen why would $h$ map $S^1$ to two points?

Akiva Weinberger

scrolls back up

Liad

yea, sorry $h: S \ ^ 1 \to \Bbb R -\{0\}$

Akiva Weinberger

Oh, so he did

Liad

:P

Balarka Sen

@LeakyNun $|h| = 1$

What are the norm 1 elements in R?

Leaky Nun

1:01 PM

10 mins ago, by Liad

i have $f: S \ ^1 \to \Bbb R $ continuous i need to prove that there exists $x \in S \ ^ 1 $ s.t $f(x) = f(-x)$.
So, suppose there isn't such a point. so define $g(x) = f(x) - f(-x) $ , $g \ne 0$ so we can assume w.l.o.g that $g \gt 0$ . so if we define $h(x) = \dfrac{g(x)}{||g(x)||}$ we have a map $h : S \ ^ 1 \to S \ ^ 1$. how can i get to a contradiction from here? or maybe im not in the right way?

Akiva Weinberger

In any case, $g(x)$ is either positive or negative

and $g(-x)$ is either negative or positive

In either case, IVT.

Balarka Sen

@Leaky Huh? $g$ maps to $\Bbb R$.

Leaky Nun

@BalarkaSen sorry, I see it now

you are right. $h:S^1 \to S^0$.

Balarka Sen

Borsuk-Ulam in dimension 1 is basically IVT, yup.

Liad

@AkivaWeinberger yea, thanks.

Balarka Sen

1:05 PM

@AkivaWeinberger We should write an expository on the Borsuk-Ulam theorem lol

it came up on a lot of stuff we've done recently

I actually don't even know all the various proofs of B-U.

Simply Beautiful Art

2:32 PM

WolframAlpha, convert united states to king

Akiva Weinberger

2:49 PM

Fun fact: If Quebec were to become independent, the US would become larger than Canada

@BalarkaSen Is Borsuk–Ulam still true if we replace the antipodal map with any degree -1 map?

Simply Beautiful Art

Balarka Sen

@AkivaWeinberger That's interesting. I think so.

Krijn

@Bala I have a question for you

Assuming you want to talk about alg geom

Balarka Sen

go ahead

Krijn

Say you have this friendly and nice curve $C$ over a finite field

And you want to map it to $\mathbb P^1$ over the same field

Take $d >> 0$ and $D$ an effective divisor of $C$ of degree $d$

Akiva Weinberger

2:58 PM

@SimplyBeautifulArt $\operatorname{km}\$^2$ is my favorite unit of measurement

Krijn

Can we find a (degree $d$) map $f: C \to \mathbb P^1$ with fiber $D$ above $0$?

I've been googling this for the past half an hour and I assume it has been answered or at least looked at, but can' t find too much

Simply Beautiful Art

@AkivaWeinberger I think $\$^2$ is almost like saying $\text{radians}^2$

Balarka Sen

@Krijn What's an "effective" divisor?

Krijn

@BalarkaSen No negative coefficients

Balarka Sen

Ah, ok. I was actually thinking about the exceptional divisor in blowups, or something of that sort.

But of course we're working with curves here.

Akiva Weinberger

3:02 PM

@SimplyBeautifulArt Radians squared is a legit unit of measurement, actually

Measures areas of parts of the sphere

Simply Beautiful Art

Hm, right... didn't think that through too well.

Akiva Weinberger

You'll see it for apparent size, I think

Simply Beautiful Art

So what does $\$^2$ mean?

Secret

but how can we be sure any function $f$ can be decomposed into a fourier transform, hence allowing us to exploit the linearity and that $D$ and $E^n$ have simultaneous eigenfunctions $e^{2\pi i x}$ thus allowing us to commute the $E^n$ and $D$ for each term in the expansion?

although I suspect for Akiva's question, I think we should be fine since the fourier transform of a polynomial always exists thus we might be able to use that to show the required commutation relations

although, I felt like when Akiva made that question, he might have a more elegant (read: simpler) way in proving that commutation relation, which probably requires theorems that I either don't know or I don't remember

Akiva Weinberger

I'm not actually sure if the proof is that simple

Secret

3:05 PM

NB. I also noticed that $\frac{d}{dx}$ is actually a shift operator in disguise...

Akiva Weinberger

By the way, note that it's important that these be over polynomials.

$f(x)\mapsto f(x)+\cos(2\pi x)$ commutes with $E$ but not $E^{1/2}$, for example.

(Luckily, there are no nonzero periodic polynomials)

@Secret How so?

Secret

$$\frac{df(x)}{dx}=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{E^hf(x)-f(x)}{h}=\lim_{h\to 0} \left(\frac{E^h-\textrm{id}}{h}f(x)\right)$$

Akiva Weinberger

Oh yeah

Krijn

@BalarkaSen Yeah, so we are in the easiest case

SteamyRoot

@AkivaWeinberger That's not a very linear operator though, is it?

Secret

3:12 PM

While yes under such form, $\textrm{id}$ and $\frac{1}{h}$ will definitely commute with any operator, the issue is that we need to deal with $E^h L$ where $L$ is some linear operator. But that will require proving that $[E,L]\implies [E^n,L]=0,\forall n \in \Bbb{R}$ which is not true for the general case as the periodic counterexample showed, but it might work on polynomials for reasons I need to check out how...

One issue with doing polynomial proofs is I often have no idea what is happening when a polynomial $f(x)$ becomes $f(x+1)$ until I expand it

and that's where things get messy...

Akiva Weinberger

@SteamyRoot Oh. You're right

Woops

Is $f(x)\mapsto f(x)\cdot\cos(2\pi x)$ good?

Secret

Meanwhile, the action by the derivative on polynomials is easy to see when we understood that all one variable differential operators have the form $D=\sum_{i=1^n}a_i(x)\frac{d^i}{dx^i}$, thus considering each powers of $\frac{d}{dx}$, each integer application of $\frac{d}{dx}$ on the polynomial effectively shift the polynomial "to the left" if we use the standard basis $\{1,x,x^2,x^3,x^4,...,x^n\}$, and scale its coefficients in a linear manner

Akiva Weinberger

Don't use $D$ for something other than $\frac d{dx}$, that's confusing

Use like $A$ or something

Secret

Ok, will do that next time since I timed out

Meanwhile graphically, $E^n$ will translate the graph of the polynomial (or in fact, any function) to the right by $n$ units, however how the coefficients are affected I think I need to work out its matrix representation