Mathematics - 2020-08-09 (page 1 of 2)

Balarka Sen

01:17

@Thorgott @Lelouch Yeah that's what I meant, the general case follows from Cartan's magic formula

Much better than doing coordinates

That circle of ideas is known as the Moser trick; you have an isotopy taking a closed form $\omega_0$ to another closed form $\omega_1$ and you want to say that means the forms differ by an exact form.

Thorgott

ah, makes sense

Balarka Sen

Technically speaking Moser trick goes backwards. Here's an application; say $M$ is a closed manifold, $\omega_0$ and $\omega_1$ are two volume forms on $M$ such that $\text{vol}(M, \omega_0) = \text{vol}(M, \omega_1)$.

Then there is a diffeomorphism $f : M \to M$ such that $f^* \omega_1 = \omega_0$

Thorgott

that's cool

Ted Shifrin

re @Thor, hi, a @Balarka

Balarka Sen

Hi @Ted!

@Thorgott Want to try and prove this?

Thorgott

01:29

rehi Ted

Balarka Sen

I have to work this out everytime but the main idea is, write the concordance of forms $\omega_t := t \omega_0 + (1 - t) \omega_1$, and then try to solve for an isotopy $f_t : M \to M$ such that $f_t^* \omega_t = \omega_0$.

Thorgott

it sounds like a good exercise

but I'll save it for later

Balarka Sen

Mmk

Thorgott

I think I'm currently getting close to finally figuring out the sphere bundle

Ted Shifrin

What is so difficult for you?

Balarka Sen

01:34

$f_t^* \omega_t = \omega_0$ implies $\mathcal{L}_{X_t} \omega_t = 0$ where $X_t$ is the time-dependent vector field that integrates to $f_t$... Cartan's magic formula implies $0 = \iota_{X_t} d\omega_t + d\iota_{X_t} \omega_t$...

i think thats how it goes anyway; you differentiate a differential equation to solve it lol

Oh I messed it up

$\partial_t(f_t^* \omega_t) = \mathcal{L}_{X_t} \omega_t + \partial_t \omega_t$

So I need to solve for $\partial_t \omega_t + \iota_{X_t} d\omega_t + d\iota_{X_t} \omega_t= 0$

If $\omega_t$ is already closed as in our case the middle term goes away

And $\partial_t \omega_t = \omega_0 - \omega_1$ which is actually some exact form because $\omega_0$-volume and $\omega_1$-volume of $M$ are the same.

If $\alpha$ is the potential we just need to solve for $\iota_{X_t} \omega_t = \alpha$ which is always possible since the volume forms are nowhere vanishing

Akiva Weinberger

02:44

Let $\langle x\rangle$ be the integer nearest $x$, defined for $x$ not a half integer. It equals $\lfloor x+\frac12\rfloor=\lceil x-\frac12\rceil$.

Then the $n$th nonsquare number is $n+\langle\sqrt n\rangle$.

The $n$th nontriangular number is $n+\langle2n\rangle$.

In particular, those are never square (resp. triangular).

Swarup

Hey, can we divide a straight line into $m:n$ ratio, using ruler and compass only? You know, with the aid of Euclidean Geometry, and not with the Cartesian one?

Akiva Weinberger

$m$ and $n$ are integers? Yes.

Swarup

Uh yes, $m$ and $n$ integers. So what do I look up for? I mean, what keywords should I use? (^^")

Akiva Weinberger

Here's one way: mathopenref.com/constdividesegment.html

You can search "divide a line segment into equal parts using compass and straightedge" or something similar

Once you've divided into $m+n$ pieces you're golden

Swarup

Yup :3 Thank you, this works!

Akiva Weinberger

02:53

Here's another way, but it assumes you know how to draw parallel lines using compass and straightedge (not shown in the video):

Dividing a line segment into equal parts (Thales theorem)

►

@Swarup

Swarup

Ah huh, yes! Tysm! I'll check it out too!

I actually had this one in high school... sigh, I forgot everything...

*the first one

Akiva Weinberger

It's interesting how you can divide a line in thirds but not an angle

(Theorem: You cannot construct a 20 degree angle with compass and straightedge, but you can construct a 60 degree angle.)

The proof involves linear algebra. It's to do with statements like this: the set of numbers of the form $a+b\sqrt 2$ where $a$ and $b$ are rational (written $\{a+b\sqrt2:a,b\in\mathbb Q\}$) is 2-dimensional vector space over $\Bbb Q$

Very rough proof sketch: Let $\alpha=\cos(20^\circ)$. It turns out that $8\alpha^3-6\alpha-1=0$; it's the root of a cubic equation.

Swarup

Yeah... I'm technically still in high school...

I have only done calc 1 & 2 by myself out of enthusiasm xD

ronak jain

Hi everyone

Can someone tell me what will be the general formula for tan(a+b+c+d......+n)

Akiva Weinberger

OK @Swarup

Swarup

03:06

@AkivaWeinberger But actually, I heard of this too! I heard it was a famous problem for a while. But isn't it actually a group theory problem?

Akiva Weinberger

It's to do with a field called Galois theory. A lot of Galois theory is about connecting things called fields to groups. But for this I don't think you need groups

Do you know what fields are?

They're sets of numbers that are closed under addition, subtraction, multiplication, and division (except division by zero)

For example: It turns out that, if $a$ and $b$ are rational, then $\dfrac1{a+b\sqrt2}$ equals $c+d\sqrt2$ for rational $c,d$

Swarup

Yep, I know what fields are. But I don't know things beyond that (rings, and... and... well things like that).

Akiva Weinberger

So $\{a+b\sqrt2:a,b\in\Bbb Q\}$ is a field

Swarup

I see

Akiva Weinberger

When I say it's a "vector space over $\Bbb Q$", I mean (roughly) I can specify its elements by lists of rationals (in this case, two rationals, $a$ and $b$)

It's a two-dimensional vector space over $\Bbb Q$

Similarly, $\{a+b\sqrt[3]2+c\sqrt[3]4:a,b,c\in\Bbb Q\}$ is also a field

(division isn't 100% obvious)

It's a three-dimensional vector space over $\Bbb Q$

Puzzle: write $1/(1+\sqrt[3]2)$ in that form

@Swarup Here's some notation. Write $\Bbb Q(\sqrt[3]2)$ to mean the smallest field containing the rationals and $\sqrt[3]2$

Swarup

03:16

@AkivaWeinberger Yep, (roughly) it sounds like a function...

Akiva Weinberger

It turns out that $\Bbb Q(\sqrt[3]2)=\{a+b\sqrt[3]2+c\sqrt[3]4:a,b,c\in\mathbb Q\}$

and so $\mathbb Q(\sqrt[3]2)$ has dimension $3$ over $\mathbb Q$.

So, back to geometry

Swarup

That links a tuple of two numbers to another number... but I guess the interesting details really lie in the abstraction that comes from more advanced subjects like LA, Number Theory, Analysis...

Akiva Weinberger

I mentioned that $\alpha=\cos(20^\circ)$ satisfies $8\alpha^3-6\alpha-1=0$. It's the root of a cubic equation. (I'm skipping details, but it ultimately comes from the angle addition formula)

So it turns out that$$\mathbb Q(\cos(20^\circ))=\{a+b\cos(20^\circ)+c\cos^2(20^\circ):a,b,c\in\mathbb Q\}$$

It's dimension 3

(this comes from how it's a root of a cubic equation; again, skipping details)

(It turns out that it's important that the cubic equation is irreducible: it's not the product of two equations of smaller degree)

(I won't prove this)

So. About geometry

Theorem: Say you start with a line segment of length 1, and you have a straightedge and compass

and you manage to construct a line segment of length $x$.

Then the dimension of $\Bbb Q(x)$ is a power of 2.

Corollary: You cannot construct a line segment of length $\cos(20^\circ)$

ronak jain

Did someone look at my problem ?

Akiva Weinberger

Corollary: You cannot construct a 20 degree angle (if you could, you could use it to make a segment of length $\cos(20^\circ)$)

@ronakjain I don't know the answer

Balarka Sen

03:22

Suppose $X_1, X_2, \cdots$ are independent random variables and $X_1', X_2', \cdots$ are independent copies such that $X_i'$ has the same distribution as $X_i$. Then variance of some function $Z = f(X_1, X_2, \cdots)$ is bounded by $\sum_{i = 1}^\infty \Bbb E[(Z_i - Z)_+]^2$ where $Z_i = f(\cdots, X_i', \cdots)$, replacing the $i$-th random variable by it's independent copy.

That is to say, variance of the function can be bounded by computing how much $f$ depends on each individual variable

Swarup

Why can't I, solely because $\alpha$ is the solution of a cubic equation?

Akiva Weinberger

Because 3 is not a power of 2

Ted Shifrin

@AkivaWeinberger This is an (easy) exercise in my algebra book. :)

Akiva Weinberger

I didn't prove the theorem

Swarup

Oh... I see

Akiva Weinberger

03:24

To prove the theorem, it's helpful for me to bring up some more machinery

Swarup

Thanks for giving me a rough outline though :)

Akiva Weinberger

Let's look at $\Bbb Q(\sqrt2,\sqrt3)$. This is the smallest field containing $\sqrt2$, $\sqrt3$, and the rationals

It's things of the form $a+b\sqrt2+c\sqrt3+d\sqrt6$

Every element can be represented by four rationals, so it has dimension 4… over $\Bbb Q$.

But! We can also write its elements in this form:

$(a+b\sqrt2)+(c+d\sqrt2)\sqrt3$

Every element can be represented by two things in $\Bbb Q(\sqrt2)$.

So what this means is, $\Bbb Q(\sqrt2,\sqrt3)$ has dimension 2 over $\Bbb Q(\sqrt2)$.

This is the first time I talked about dimension over something other than $\Bbb Q$.

Some notation: write $[\Bbb Q(\sqrt2,\sqrt3):\Bbb Q(\sqrt2)]$ to mean the dimension of $\Bbb Q(\sqrt2,\sqrt3)$ over $\Bbb Q(\sqrt2)$.

And in general, $[G:H]$ for the dimension of $G$ over $H$.

Then:$$[\Bbb Q(\sqrt2,\sqrt3)]=[\Bbb Q(\sqrt2,\sqrt3):\Bbb Q(\sqrt2)]\cdot[\Bbb Q(\sqrt2):\Bbb Q]$$

$$4=2\cdot2$$

In general, $[G:K]=[G:H][H:K]$, where $G,H,$ and $K$ are fields

Now we're getting pretty deep into Galois theory

But I want to tell you this to tell you where the "powers of 2" come from

Think of a ruler and compass construction. Think of the x- and y-coordinates of those points, and think of the field generated by all of those coordinates

Every time you draw a new line or circle, you construct new points (the intersections with the lines and circles you've previously drawn)

So you can consider $\Bbb Q($all the (coordinates of the) points you've constructed$)$

and we want to know its dimension over $\Bbb Q$

It's simpler to ask for its dimension over $\Bbb Q($all the points you constructed until a step ago$)$

(until you drew that last line or circle)

Let $S_n$ be all the (coordinates of the) points you created up to and including the $n$th step of the construction.

Then it turns out the field $\Bbb Q(S_n)$ either equals $\Bbb Q(S_{n-1})$, or it has dimension $2$ over $\Bbb Q(S_{n-1})$

$[\Bbb Q(S_n):\Bbb Q(S_{n-1})]=1\text{ or }2$

(The main reason is, when you intersect a circle with a line or a circle, to get the coordinates of the new points, you usually have to solve a quadratic equation)

So, assuming we accept that, then$$[\Bbb Q(S_n):\Bbb Q]=[\Bbb Q(S_n):\Bbb Q(S_{n-1}][\Bbb Q(S_{n-1}):\Bbb Q(S_{n-2})]\dotsb[\Bbb Q(S_1):\Bbb Q]$$is a product of a bunch of $1$s and $2$s, so it's a power of 2

If you could construct $\cos(20^\circ)$, then $\Bbb Q(S_n)$ would contain $\Bbb Q(\cos(20^\circ)$, so we could ask what $[\Bbb Q(S_n):\Bbb Q(\cos(20^\circ)]$ is.

But$$[\Bbb Q(S_n):\Bbb Q]=[\Bbb Q(S_n):\Bbb Q(\cos(20^\circ)][\Bbb Q(\cos(20^\circ):\Bbb Q]$$

$$\text{power of 2}=[\Bbb Q(S_n):\Bbb Q(\cos(20^\circ)]\cdot 3$$

$3$ isn't the factor of any power of 2

So contradiction. The conclusion is that $\Bbb Q(\cos(20^\circ))$ isn't a subfield of $\Bbb Q(S_n)$ (where $S_n$ is the numbers we constructed with a straightedge and compass), so $\cos(20^\circ)$ isn't in $S_n$, so we can't construct $\cos(20^\circ)$.

Swarup

03:49

fuck...

Akiva Weinberger

@Swarup It's OK if you didn't get all that

Swarup

I'll try understanding it!

Akiva Weinberger

Also, it's not obvious how to define "dimension"

Swarup

No, we actually had a load shedding :( so I couldn't get here for a while...

I'll go on and read it

Akiva Weinberger

The main idea for that section is, I kept on saying "dimension over $\Bbb Q$", can you have dimension over something other than $\Bbb Q$? (Yes)

$\Bbb Q(\sqrt2,\sqrt3)=\{x+y\sqrt3:x,y\in\Bbb Q(\sqrt2)\}$

(I did not prove that $[G:K]=[G:H][H:K]$. This is not obvious and requires proof.)

abhas_RewCie

03:57

Good Morning

How many of you all giving JEE this year?

Swarup

Oh man... all of this sounds pretty tough :(

abhas_RewCie

Also, I'm not totally mentally stable in lockdown

Pretty sure won't make it through JEE this year.

May need to take a year off for it :-(

Swarup

If I manage to construct a line segment of length $x$, how does that imply that the dimension of $\mathbb{Q}(x)$ is a power of 2?

Ahh forget it, don't answer me :( a FAIR amount of grip in fields is prolly needed to grasp all that, which I don't think I have...

Akiva Weinberger

@Swarup By induction

abhas_RewCie

@AkivaWeinberger Is that something related to topology or what?

Akiva Weinberger

04:02

Well hold on

First we prove $\Bbb Q($_everything_ we've constructed so far$)$ is a power of 2, by induction. I called that $S_n$

Then, since $x\in S_n$, I know the field $\Bbb Q(x)$ is a subset of $\Bbb Q(S_n)$

so I can ask what the dimension of $\Bbb Q(S_n)$ over $\Bbb Q(x)$ is

By the multiplication formula I stated but didn't prove, $[\Bbb Q(S_n):\Bbb Q]=[\Bbb Q(S_n):\Bbb Q(x)][\Bbb Q(x):\Bbb Q]$

On the left side of that equation we have a power of two

so the two things on the right side of it have to multiply to that

so they're factors of a power of two

If you're a factor of a power of two, you're a power of two.

So $[\Bbb Q(x):\Bbb Q]$, the dimension of $\Bbb Q(x)$ over $\Bbb Q$, is a power of two.

@Swarup In any case, hopefully you at least have a rough idea of how it goes. Eventually you'll learn enough to fill in the gaps. With experience comes understanding

uhoh

0

Q: Is a 2:1 Lissajous orbit possible in the Circular Restricted Three Body Problem?

Was Queqiao in a halo or Lissajous orbit? Why do sources disagree? says Proper halo orbits have the same period for their in-plane oscillations and out of plane oscillations, so they are closed orbits with roughy circular motion in the rotating frame, whereas Lissajous orbits are those where the...

Akiva Weinberger

04:51

Give a man a fish and he'll have food for a day. Give a fish a man and it'll have food for the rest of its life

Lelouch

@AkivaWeinberger <insert you had me in the first half ngl meme>

Balarka Sen

@BalarkaSen Idea for finitely many variables: Say $Z = f(X_1, \cdots, X_n)$. Look at the martingale differences $M_i = \Bbb E[Z|X_1, \cdots, X_i] - \Bbb E[Z|X_1, \cdots, X_{i-1}]$, so that $\sum_{i = 1}^n M_i = Z - \Bbb E[Z]$.

Lelouch

@BalarkaSen ooo ISI probability gawd

:P

Balarka Sen

probability is very good

Say $i < j$, then $\Bbb E[M_i M_j] = \Bbb E[\Bbb E[M_i M_j|X_1, \cdots, X_i]] = \Bbb E[M_i \Bbb E[M_j|X_1, \cdots, X_j]]$.

ronak jain

Hi everyone ! Can someone tell me that how can I proof that for two quadratic equations to have both roots same the only condition is that therecoefficient should be proportional ?

Lelouch

05:05

@ronakjain use factor theorem

ie if $P(a) = 0$ with mulitiplicity $m$ then $(x-a)^m$ divides $P(x)$

Balarka Sen

@Lelouch Actually surprisingly the probability courses here are terribad

ronak jain

@Lelouch can you proof ? I am not able to understand that

Lelouch

@BalarkaSen huh, didn't expect that

Balarka Sen

yeah i mean its just basic stuff

Lelouch

@ronakjain Well, if $\alpha, \beta$ are roots of $Q(x)$, then $(x-\alpha)(x-\beta)$ divides $Q(x)$. Now expand what divisibility means and you're done

@BalarkaSen CLT/Martingales were done ?

ronak jain

05:09

@Lelouch ok. I got that. Thanks

Balarka Sen

CLT is in the course, but since there's no measure theory you have to skip some details (i think the proof is optional, but thankfully was taught). there's nothing on martingales

there's a stochastic process elective that i'll take the next semester

ronak jain

But can you solve a one more problem. Can you tell me how to prrof the condition for one common root of two quadratic equations ?

Lelouch

can you recommend some books or something to read for probability, which is more into intuition than measure-theoretic stuff? I'm traumatized by the prob course done here

Balarka Sen

durrett

thats the holy bible

Lelouch

ooh, that looks cool. thanks

Balarka Sen

05:12

cmi has probability? damn

i thought you guys just do algebra lol

Lelouch

@BalarkaSen it has in second sem, but I didn't like the way it was done. Basically 80% was what I felt unnecessarily formalism instead of actual probabilistic intuition. Some parts I liked though (eg: Delhi scandal, probabilistic proof of Weirstrass)

Balarka Sen

whats the delhi scandal lmfao

there should be more measure theory in our courses

Lelouch

ok, so Karandikar taught the course. He's a pretty pro probabilistic

*probabilist

Balarka Sen

ah ok

Lelouch

so there was once a scandal in Delhi engineering enrance

the test were rigged or something

Balarka Sen

05:14

you mean he's pro-bability

Lelouch

he was on the team to detect if it was fraud or not

he told how he figured that out

I mean it was babied down but I liked it

Balarka Sen

ugh no stat in my town man

miss me with that crap

Lelouch

@BalarkaSen huh ?

Balarka Sen

i dont give a shit about stat lol

karandikar came here to give some stat talk, i fell asleep

Lelouch

yeah he dumbs down things too much for "intro" talks

Balarka Sen

05:16

the stat courses here are so annoying

Lelouch

are they mandatory ?

Balarka Sen

its like a gigantic waste of time

yeah

Lelouch

can you take elective in third sem ?

Balarka Sen

yes

Lelouch

what did you take ?

Balarka Sen

05:17

oh wait no, we can take elective in third year; so that'd be 5th and 6th sem

Lelouch

@BalarkaSen oh rip

@BalarkaSen not even in 4th sem lol ?

Balarka Sen

nah

Lelouch

enjoy stat then :P

Balarka Sen

but im in 5th sem now (or will be, whenever this thing starts)

Lelouch

@BalarkaSen they should make some exception for you ?

Balarka Sen

05:18

no why should they

i mean i dont care much i can just read something

Lelouch

@BalarkaSen Well here normally people can take electives in fourth sem, but if you have ok-ish performance, you can take in third sem, and if you're really pro you can take in second sem (Mohan took diff geo)

Balarka Sen

Oh like that, yeah they asked me if I want to skip some courses, but I'd have to take the exam anyway

So I said screw it

Either let me skip the entire course or I'll just sit for it

Not going to prepare for the exam if I am skipping the course

Lelouch

I mean, isn't it a huge waste of time to sit through the course if you can test out with little prep ?

is attendance mandatory or smth

Balarka Sen

I am not going to read for shit courses if I don't sit for the course; like sitting for classes is half the preparation done anyway

and I didn't want to skip serious math courses because why not make a solid foundation

I learnt a lot from sitting in really simple courses like linear algebra even

Lelouch

How do you manage to properly learn through the courses you already know everything about?

I know like 70-90% of the stuff that's happening a priori, but the remaining 30-10% is important but I end up getting bored so I miss them.

ronak jain

05:24

@Lelouch can you tell me proof for a common root of two quadratic equations ?

Balarka Sen

i dunno, there's always new things to learn

Lelouch

@ronakjain The resultant would be zero

compute the resultant

ronak jain

@Lelouch I did not get this. Can you please give a explanation

Lelouch

Resultant

In mathematics, the resultant of two polynomials is a polynomial expression of their coefficients, which is equal to zero if and only if the polynomials have a common root (possibly in a field extension), or, equivalently, a common factor (over their field of coefficients). In some older texts, the resultant is also called the eliminant.The resultant is widely used in number theory, either directly or through the discriminant, which is essentially the resultant of a polynomial and its derivative. The resultant of two polynomials with rational or polynomial coefficients may be computed efficiently...

speaking of linear algebra, how will you motivate the proof of spectral theorem ?

Balarka Sen

just do it for normal operators

ronak jain

05:27

@Lelouch I am just preparing for JEE. And all that you posted seem to be non understandable to me. Can you explain thst in your language ? I just want the condition for two quadratic equations

Balarka Sen

@Lelouch Quickest proof of spectral theorem for normal operators is to show that generalized eigenvectors of normal operators are eigenvectors, so the Jordan canonical form is actually diagonal, and checking by hand that eigenvectors are all orthogonal

Normal operators is the optimal condition, so everything becomes clear if you use just that much hypothesis

Lelouch

@ronakjain The main idea is that if $f(x)$ and $g(x)$ have a common root, then you can find linear $p(x), q(x)$ with $f(x)p(x) + q(x)g(x) = 0$ (why ?). This can be stated in terms of solving linear equatoins (how/why ?). Now rest is using a system of linear equation has a nontrivial zero iff the determinant is zero (you can blackbox this fact ig)

Akiva Weinberger

Round $n+\sqrt n$ to the nearest integer. It is never square

Balarka Sen

OK man @Akiva

Akiva Weinberger

In fact, $n+\langle\sqrt n\rangle$ is the nth nonsquare number

Balarka Sen

05:34

Nobody cares

Akiva Weinberger

I'm surprised

Lelouch

@BalarkaSen um, yeah but how do you a priori get that all eigenvalues do lie in $\mathbb{R}$ without mimicking the proof of real spectral theorem ?

Balarka Sen

What? Having eigenvalues in $\Bbb R$ is obvious

Lelouch

wait why ?

nevemrind, ofc

Balarka Sen

If $T$ is normal, $T^* T$ is symmetric, which has real spectrum

Lelouch

05:38

yeah, sorry I realized it was obvious immidiately after asking it

Manan

I need help formatting plain text with spaces in an exponential. Can anyone help?

Balarka Sen

There are also deformation proofs; normal operators which have multiplicity in the spectrum can always be wiggled a little so they have a simple spectrum

for which spectral theorem is obvious

and these kind of proofs work for arbitrary fields if you use Zariski topology

Lelouch

@BalarkaSen Wiggled as in, the set of the such operators are dense ? (like an way to prove CH is to show diagonalizable operators are dense in Zariski topology)

Balarka Sen

yeah

Akiva Weinberger

@Manan $a^{\text{blah blah $b$ blah}}$

Like that?

a^{\text{blah blah $b$ blah}}

Balarka Sen

05:42

I would say the most important thing I have learnt in my linear algebra course is how to think about Hermitian inner products

that shit is nonobvious (TM)

Lelouch

@BalarkaSen How are you going to generalize the notion of a bilinear form to arbitary fields ? Matrices like $\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$ are nilpoitent in $\mathbb{F}_2$

Balarka Sen

yeah nonzero characteristic is a problem; but sometimes you can pass to the algebraic closure

the details depend on what you want to do

Lucas Henrique

@AkivaWeinberger take a look at $\left \lfloor{n + \sqrt n + \dfrac{1}{2}} \right \rfloor$

Lelouch

@BalarkaSen btw, this is a nice problem, you can try if you don't know about it: Determine all possible values of $[\overline{K}:K]$

the answer is $1, 2, \infty$

Lucas Henrique

i don't have a clue on how artin proved that

if it's proper and finite, then we have the complex numbers from bizarro world o_o

Balarka Sen

05:47

@Lelouch yeah believable

Lelouch

@BalarkaSen How do you think about it? The way I think of it is having a matrix $M$ with $\langle v, w \rangle = v^T M w$

@BalarkaSen it's not very easy to proof though, atleast the (only) proof I know is quite involved

Balarka Sen

yeah i guess this has to do with formal reality and so on

i can believe this is hard

Lelouch

formal reality ?

Balarka Sen

@Lelouch the real part is an actual inner product, the imaginary part is a symplectic form

Formally real fields are where -1 is not a sum of squares; it should be possible to prove that formal real closures have infinite degree over the base

whereas formally real closed fields have degree 2 algebraic closure

formal reality is also the same thing has having an order structure, but maybe this is not relevant

Lelouch

These are all relevant. It's not hard to finish from here assuming A-S.

Balarka Sen

05:54

whats A-S

Lelouch

I'm bit surprised how did you think of formally real fields so quickly, I thought that was the hard part

@BalarkaSen Artin-Schreier

Balarka Sen

this is some relevant exercise in dummit foote

Lelouch

oh

Balarka Sen

@Lelouch oh ok i dunno this, x^p - x - a shit?

i thought that was char p garbage

Lelouch

@BalarkaSen no, Artin-Schreier is literally what I asked to prove

if you assume some other characterization of formal real field

Balarka Sen

05:57

Um, huh?

Artin–Schreier theory

See Artin–Schreier theorem for theory about real-closed fields.In mathematics, Artin–Schreier theory is a branch of Galois theory, specifically a positive characteristic analogue of Kummer theory, for Galois extensions of degree equal to the characteristic p. Artin and Schreier (1927) introduced Artin–Schreier theory for extensions of prime degree p, and Witt (1936) generalized it to extensions of prime power degree pn. If K is a field of characteristic p, a prime number, any polynomial of the form X p −...

Oh there are two Artin-Schreiers

lol

Lelouch

see theorem 2: sa-logic.org/sajl-v2-i2/10-Luciano-SAJL.pdf

Balarka Sen

Ok I didnt know these nutcases did formally real fields

Manan

Thanks @AkivaWeinberger - I found on TeX SE $\textsuperscript{text goes here}$

Balarka Sen

E[E[Z|F]|G] = E[Z|G] if G is a sub sigma-algebra of F, right? lol

Of course

Too hard to keep track of garbage

Lucas Henrique

06:15

@Manan doenst look too right

Lucas Henrique

06:58

if $K/L$ is Galois and $L/F$ is Galois, then $K/F$ is Galois, right?

separability is ok, and I'm too sleepy to write down normality

Edward Evans

07:16

Take $F = \Bbb Q$, $L = \Bbb Q(\sqrt{2})$ and $K = \Bbb Q(\sqrt[4]{2})$

Lucas Henrique

that's a good one. thanks.

Manan

07:32

@LucasHenrique It's as good as $^{\text{}}$

Specter Prophet

$\pi=2.34129...$

Specter Prophet

08:31

en.m.wikipedia.org/wiki/Laplace%E2%80%93Beltrami_operator

Alessandro Codenotti

09:08

By going through the Artin-Schreier stuff you should also get a almost completely algebraic proof of the fundamental theorem of algebra

Balarka Sen

09:52

@Alessandro Here's what I was confused about earlier. If you throw away a $1/n$-ball around $1/n \cdot \Bbb Z^2$ in $\Bbb R^2$, and take the pointed GH limit, it converges to $\Bbb R^2$ because the metric on $\Bbb R^2 - \Bbb Q^2$ and $\Bbb R^2$ are comparable; the infimum distance between two points on $\Bbb R^2 - \Bbb Q^2$ is the same as that of $\Bbb R^2$

But take some other measure 0 set, let's say even a straightline segment $L$ in $\Bbb R^2$. $\Bbb R^2 - N_{1/n}(L)$ does not pointed GH converge to $\Bbb R^2$

Because the metric on $\Bbb R^2$ and $\Bbb R^2 - L$ are not comparable. The GH limit introduces a bulge near $L$

This is an interesting example to me

I think the GH limit is what you get if you cut $\Bbb R^2$ along the the segment $L$ and "separate" the two flaps

Alessandro Codenotti

10:25

@BalarkaSen wait why?

Which metric are you using on $\Bbb R^2\setminus N_{1/n}(L)$?

Balarka Sen

Take two points on the opposite sides of $L$, the distance between them in the interor metric of $\Bbb R^2 - L$ is not the same as the one on $\Bbb R^2$

@Alessandro The length metric, not the subspace metric. Should have made that clear.

I only care about geodesic metric spaces

Alessandro Codenotti

Ah ok, makes sense then

Balarka Sen

You agree with my picture of the GH limit, yeah?

slit the plane along the segment $L$ and separate the flaps

like you do in branch cuts to construct riemann surfaces

Alessandro Codenotti

So the point is that the first set is made of isolated points, you don't waste any time walking around them

Balarka Sen

yeah

it doesn't locally disconnect the ambient

Thorgott

11:32

I don't think A-S helps with the FTA

or am I missing an obvious a priori reason why $\overline{\mathbb{R}}/\mathbb{R}$ should be finite?

Specter Prophet

(removed)

user271418

12:01

Hello there

Specter Prophet

(removed)

after u say hello u post question

JessicaK

Hi I have kind of a dumb question, that I don't think is meaningful to make a post for. In this question math.stackexchange.com/questions/2837202/… I am having a lot of trouble understanding what $\partial f(x) = \{\nabla f(x) \}$ means. The squiggly parenthesis are especially confusing me because it makes it look like $\partial f(x)$ is now a set with a single element?

Thorgott

looks like nonsense to me

JessicaK

littleO's answer is also confusing me because it says if f is differentiable, but I don't see how you get meaning out of \nabla without the function being differentiable in the first place?

Specter Prophet

please how I read latex (ಥ_ಥ)

Thorgott

12:11

ah, I see, it does make sense

$\partial f(x)$ is the subderivative and, as such, a set

@Specter see top right corner

JessicaK

Ah I've never heard of a subderivative before. I will have to figure that out then

I think I see what is going on now. So a subderivative always lies under the graph of f which isn't necessarily unique. But if the function is differentiable and convex, then it is unique and equal to the gradient?

Thorgott

yeah

JessicaK

Thank you so much for the help

Specter Prophet

12:26

it works it works! it fails to work

why don't sem create chat like this

mathim.com

Alessandro Codenotti

12:39

@Thorgott because it is real closed

The algebraic closure of a real closed field is always a degree 2 extension, that's part of what A-S gives you

Thorgott

I don't know the real closed stuff, but from which property of $\mathbb{R}$ does this follow?

Balarka Sen

@Alessandro @Thorgott The analysis lies in showing R is real-closed

Alessandro Codenotti

The proof I know needs that odd degree polynomials with different sign values on a and b have a root in [a,b], which is why I said almost algebraic

@BalarkaSen or in showing an equivalent characterization of real closed, depending on your definition R could be real closed by definition :P

Balarka Sen

being real-closed is equivalent to every odd deg poly having a root in the field, and every element to be either a square or - a square

@Alessandro give me one definition of real closed where its obvious that R is real closed without any analysis

Thorgott

I mean, I know the standard Galois-theoretic proof of the FTA, but I don't see how to do it with A-S

Balarka Sen

12:50

the point is real closed is equivalent to not being algebraically closed by immediately being algebraically closed once you adjoint square root of -1

Alessandro Codenotti

@BalarkaSen a field is called real closed if it is elementary equivalent to the real numbers

Balarka Sen

FTA is baked in there

@Alessandro oh not a logic definition man

come on

you'll do some model theory trick on me

Alessandro Codenotti

Lmao

Thorgott

yeah, but why is $\overline{\mathbb{R}}/\mathbb{R}$ finite

once we have that, A-S does give us the FTA

Balarka Sen

F real closed implies F(sqrt(-1)) is algbebraically closed man

Alessandro Codenotti

12:53

That's equivalent to being real closed actually

Balarka Sen

overline F = F(sqrt(-1))

Thorgott

so why is $\mathbb{R}$ real-closed then, whatever that means

substituting tautologies isn't gonna give us a proof

Balarka Sen

because every odd degree polynomial has a root in R

that implies real closed

just read the real closed stuff man its in dummit foote as an exercise

Thorgott

that sounds like it's basically the standard Galois proof of the FTA

Balarka Sen

yeah it is

Thorgott

12:55

ok, so A-S doesn't really give us a different proof of the FTA

Balarka Sen

thats right

there is no interesting algebra proof of FTA

Thorgott

the standard one is nice tho

Balarka Sen

meh

just do loopy loop

Thorgott

ok, can't argue with that

Balarka Sen

lmao

Thorgott

12:57

but still better than the elementary proof

Balarka Sen

the loopy loop proof actually gives proof for quaternions as well

so its very superior

Thorgott

does it work on real closed fields?

Alessandro Codenotti

Isn't there a proof for quaternions using that they are basically a union of slices that look like C?

Balarka Sen

yeah

by Cullen-regularity of polynomial functions

lmao

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