Mathematics - 2021-11-08 (page 2 of 2)

anakhro

18:00

There is tonnes of cool history behind the foundations of functional analysis and the notion of spectrum there, too.

leslie townes

"tonnes" - he's a spy! get him.

or she.

my understanding is that it's basically an accident that the spectrum defined in linear algebra turned out to be something that could, with appropriate operators, coincide with the emission spectrum of atoms and maybe molecules. maybe i'm wrong about that.

i don't think the hydrogen atom was settled until the 1950s and people were calling it 'spectrum' in functional analysis before that.

anakhro

Yes, the coincidence of the naming was completely by chance.

leslie townes

what a great coincidence, though.

we should reword history so that it was on purpose

one day in the 1920s, von neumann was sleeping under a hydrogen tree and an atom fell on his head. something like that.

anakhro

There were some great comments in Dieudonne's history of functional analysis book on the physicists of that time:

>>>These were the hectic years during which quantum mechanics was developing at breakneck speed, with a new idea popping up every few weeks from all over the horizon. The theoretical physicists who were developing the new theory were groping for adequate mathematical tools, trying in sucession infinite matrices without any consideration of convergence (as late as 1924, most physicists did not even know what a finite matrix was!), differential operators, “continuous” matrices (w…

Do we not have quote formatting in this chat. I thought it was >

gives up

leslie townes

i've never read that book. i'm suspicious of bourbakists, particularly as to analysis. but a history book ought to be fine.

anakhro

18:12

It's actually really good if you are into history of math stuff. He keeps his Frenchness to a minimum when it comes to math.

You'd almost never guess his mathematical opinions.

leslie townes

albrecht pietsch's history of banach spaces and linear operators is good. have you seen it?

also, i love "he keeps his Frenchness to a minimum." that's a very funny turn of phrase.

anakhro

I have not seen Pietsch's book, I will have to have a look, thanks!

leslie townes

lax includes some historical information in his functional analysis book but it is mostly about how various mathematicians died in the WW2 era and not the world's most light reading.

anakhro

There was a good book about mathematicians during the nazi era in Germany. I forget who wrote it.

Also not light, in the light-hearted sense.

Ah, it was Segal who wrote it.

leslie townes

oh yeah, it was reviewed all over the place.

not a great look for a lot of those guys (and they were all guys).

i haven't read it, i expect i know at least some of it from other sources. and the older i get the more i try to stay away from doom and gloom.

anakhro

18:21

Mathematicians seem to have a good share of doom and gloom.

copper.hat

just looking forward to the rapture

anakhro

Why? St. Augustine already doomed us mathematicians to hell.

copper.hat

hell, heaven, its just a matter of sign

anakhro

copper.hat is banking on being able to change a sign convention if he gets placed in the bad place.

leslie townes

we know where copper is going.

copper.hat

18:30

i will confound them at the gates with my axiom of cheese

leslie townes

i'll see him there, it'll be great.

the bar is open later.

copper.hat

i've seen the hell demo, its incredible.

anakhro

silly leslie, bar usually means closed.

maths researcher

why is $X^4-YX^2+1$ irreducible in $\mathbb{R}(Y)[X]$?

user193319

Let $f$ be a nonconstant entire function with $f(z) \neq 0$ everywhere. Show that the set $\{z \in \Bbb{C} : |f(z)| < 1\}$ is unbounded.

I could use some help on this problem. Call the set $U$. First, note that $U \neq \Bbb{C}$, otherwise Liouville's theorem would tell us that $f$ is in fact constant.

anakhro

18:39

Sometimes people like to create a new function called g = 1/f

user193319

But I don't see how to argue that $U$ is unbounded. The fact that $f$ is never $0$ made me immediately think to define $g(z)=1/f(z)$ and then try to argue by contradiction that $g$ is bounded.

But I couldn't see it. Probably being dumb.

@anakhro Will that work?

Oh, yeah...I think I am being dumb...

copper.hat

Is the representation of a convex function as the $\sup$ of dominated affine functions considered a well known result?

I mean, would a 3rd year cs student know it.

anakhro

@copper.hat I wouldn't, but maybe your cs students are super smart.

@mathsresearcher remind me of the notation, R(Y)...is that rational polys?

maths researcher

yup @anakhro

anakhro

What have you tried so far?

maths researcher

18:49

i mean there is an analogue for the rational roots test on $\mathbb{F}(Y)$

right?

anakhro

Do you have that in your toolset? I was thinking of just doing something with the degrees in x.

maths researcher

so can I say that if $f=a_0+.....a_nx^n\in \mathbb{F}[x]$ where $a_0\neq 0$ and $a_n\neq 0$ then if $\theta \in \mathbb{F}(X)$ is a root of $f$, $\theta = \frac{f(x)}{g(x)}$ so that $f(x)|a_0$ and $g(x)|a_n$

?

@anakhro is that statement true? it should essentially be like $\mathbb{Z}$ and $\mathbb{Q}$

anakhro

I assume you are doing this for a class?

copper.hat

@anakhro its my daughter, she's smart, but i don't know what is considered standard any more.

maths researcher

@anakhro yeah

copper.hat

19:00

anyhow, i infer that it is probably not starndard.

anakhro

@copper.hat Let's just say, I didn't learn that when I did my undergrad in math.

@mathsresearcher What tools do you have in your tool set from class?

maths researcher

just gauss's lemma and eisenstein

copper.hat

thanks! she was learning stuff in year one of undergrad that i learned in grad school, so you never know.

maths researcher

he did write that there is an analogue of gauss's lemma for polynomials

but he hasn't stated it

so I assume it should be similar to what i've written above

anakhro

@mathsresearcher So then I would say avoid trying to invent new theorems on top of that. Your professor probably wants you to use the tools you have been provided.

The rational part is the tricky thing, but that's only for stuff in terms of y that you see anything in the denominator.

maths researcher

19:04

but if i've only been provided with eisenstein and gauss's lemma, how in the world can I use those for irreducibility since they're statements on $\mathbb{Q}$ and $\mathbb{Z}$

even the degree 2 and 3 condition hasn't been stated

anakhro

What does it mean that it's irreducible or not irreducible?

maths researcher

that is, $f$ being degree $2$ or $3$ irreducible iff has no roots

irreducible means degree $\geq 1$ for which if $f=gh$ then $degg=0$ or $deg h =0$

anakhro

So what would it mean if x^4 - yx^2 + 1 were reducible (for the sake of contradiction)?

Leaky Nun

20:06

@LukasHeger dass eine Komplexmannigfaltigkeit von Dimension d Realdimension 2d hat ist topologisch nicht algebraisch; nun verstehe ich nicht, warum eine Mannigfaltigkeit uber einem algebraisch gescholossen Korper gleichlich Dimension 2d haben scheint

Akiva Weinberger

21:05

Here's a neat thing to think about

You may have heard of the Kolmogorov complexity

Given a universal programming language L, the Kolmogorov complexity of a natural number n is the length of the smallest program in L that specifies n

(measured in bits, I suppose. Or megabytes etc)

Suppose we switch from language L to language L'. What's the worst that can happen to our complexities?

Can we bound $K_{L'}(n)$ given $K_L(n)$ and knowledge of $L'$ and $L$?

Another thing to think about. This has close ties to incompleteness. Can you figure out how? What sorts of statements about the Kolmogorov complexity cannot be proven? Given a theory T (say, ZFC or PA or something) and a programming language L, can you given me a statement involving $K_L$ that cannot be proven in T?

PM 2Ring

21:40

@AkivaWeinberger One difficulty here is dealing with the Berry paradox en.wikipedia.org/wiki/Berry_paradox

Akiva Weinberger

@PM2Ring This is true. But we've made our construction precise by specifying a language. It's informative to figure out what happens when you try to do Berry's paradox in this setting

maths researcher

22:08

Suppose $\theta$ is a root of $X^3+X^2+1\in \mathbb{F}_{2}[X]$. Then, $X^3+X^2+1=(X^2+(1+\theta)X+\theta(1+\theta))(X+\theta)$

how do I factorize (X^2+(1+\theta)X+\theta(1+\theta)) more in $\mathbb{F}_{2}(\theta)[X]$

PM 2Ring

@AkivaWeinberger A precise language helps, but I don't think it avoids the paradox. Also, there's time complexity to consider, and the halting problem. The shortest program for a given n may execute a ridiculous number of loops.

Leaky Nun

@mathsresearcher you just need to find the roots by trial and error

Akiva Weinberger

@PM2Ring I think the halting problem is precisely what prevents the paradox!

maths researcher

I subsituted $X=a+b\theta$ for $a,b\in \mathbb{F}_2$, but I kept getting $b^2+b+1=0$

which is a contradiction

PM 2Ring

@AkivaWeinberger Maybe. I'll have to think about that... Obviously, there's a trivial finite length program to produce any finite n. But we can't just search for a shorter program to produce it by simply executing all syntactically valid programs in lexicographical order because there may be no upper bound on the runtime of a program.

But i guess you're already aware of that stuff. :)

Akiva Weinberger

22:18

Right. So while we can define Kolmogorov complexity, it seems we can't compute it.

Astyx

@mathsresearcher did you figure this out?

maths researcher

nope @Astyx

PM 2Ring

Exactly. It's a nice concept, but (so far) it's been of limited practical use.

Astyx

@mathsresearcher You can find the roots. none of them are in $\mathbb R(Y)$ so if it's irreducible, it's deg 2 * deg 2. Write it as such and find equations on the coefficients

and you should get your answer

maths researcher

oh, this is the other problem

I figured that one out

@Astyx

Ted Shifrin

22:29

Hi @Astyx

Astyx

Hi

Howdy

maths researcher

i'm now trying to show that if $\theta$ is a root of $X^3+X^2+1\in \mathbb{F}_2[X]$ then $X^3+X^2+1$ factors into 3 linear factors in $\mathbb{F}_2(\theta)[X]$

Lukas Heger

@LeakyNun was genau meinst du?

maths researcher

so far, i've expresssed $X^3+X^2+1=(X^2+(1+\theta)X+\theta(1+\theta))(X+\theta)$

Lukas Heger

@mathsresearcher do you know about the Galois theory of finite fields?

Leaky Nun

22:34

@LukasHeger dass die Kohomologiedimension von einer Alebraischen Mannigfaltigkeit von Dimension d 2d ist

maths researcher

@LukasHeger not yet, no

Lukas Heger

@LeakyNun well, it's true over complex numbers and a lot of alg geo doesn't depend on which algebraically closed base field you take

maths researcher

this is supposed to be an introductory problem to Galois theory of finite fields

Leaky Nun

@LukasHeger ja aber...

Lukas Heger

@Leaky wait are you even sure that the cohomological dimension is 2d?

Astyx

22:38

That sounds wildly false

Lukas Heger

I think it's at most d by Grothendieck's vanishing theorem

stacks.math.columbia.edu/tag/02UZ

Leaky Nun

die hochste KD dass es haben kann oder

Ted Shifrin

@maths Did you divide the polynomial by $X-\theta = X+\theta$?

Astyx

Oh wait, cohomological dimension = dimension of highest nonzero cohom group?

Ted Shifrin

Working in $F_2(\theta)[X]$, of course.

Lukas Heger

22:39

@mathsresearcher once you get to Galois theory it's obvious that $X^3+X^2+1=(X+\theta)(X+\theta^2)(X+\theta^4)$

@Astyx yes

Ted Shifrin

It's not unusual to assign problems like this long before that stuff, @Lukas. I've done it myself.

maths researcher

@TedShifrin I divided $X^3+X^2+1$ by $X+\theta$, yes and obtained $X^3+X^2+1=(X^2+(1+\theta)X+\theta(1+\theta))(X+\theta)$

Lukas Heger

@Ted sure

Ted Shifrin

OK. @mathsresearcher Now factor your quadratic.

Akiva Weinberger

@PM2Ring You can use it to get a new proof of Gödel's first incompleteness theorem :)

as well as a surprising connection to fractal dimension (Hausdorff dimension)

One day I will remember how to properly spell that man's name

Astyx

22:41

Akiva: do you know about D-finite functions?

Akiva Weinberger

No

Astyx

welp

maths researcher

yes, that's what i'm trying to do @TedShifrin. Can't seem to guess a priori that $-\theta^2$ and $-\theta^4$ are roots

Lukas Heger

@LeakyNun the theorem by Grothendieck I linked to that an algebraic variety of dimension d has at most cohomological dimension d. I think your confusion may stem from the fact that GAGA does not work for the cohomology of constant sheaves

If $X$ is an irreducible complex variety of dimension $d$ and we consider the cohomology of the constant sheaf $\Bbb Z$, then we get $H^{2d}(X,\Bbb Z)=0$, but $H^{2d}(X^{an},\Bbb Z) = \Bbb Z \neq 0$

comparison of sheaf cohomology groups between the algebraic and the analytic category only works for coherent sheaves

Leaky Nun

@LukasHeger ich beziehe mich auf der Etalekomologietheorie, sehe zbw die Poincaré-Dualität

Ted Shifrin

22:46

Using the equation that $\theta$ satisfies, can you find $\alpha,\beta$ with $\alpha+\beta = 1+\theta$ and $\alpha\beta = \theta(1+\theta)$?

Lukas Heger

@LeakyNun there's a comparison theorem with singular cohomology so it has to be 2d...

also it's just something that you get of the calculations

Astyx

@Akiva Do you know about polynomial factorization algorithms?

Lukas Heger

I don't have a conceptual explanation other than comparison over $\Bbb C$...

Akiva Weinberger

I can guess at what they do, but beyond that, no

Astyx

There are some fields over which polynomial factorization is not decidable

Akiva Weinberger

22:52

Oh weird!

And multiplication is computable over these I assume?

Leaky Nun

@LukasHeger auch hat eine Abelsche Varietät Rang 2g, die Dimension g hat

das verstehe ich auch nicht

Lukas Heger

@LeakyNun maybe look at the computation for the étale cohomology of curves, that's not too hard and it gives you some explanation for why the second étale cohomology doesn't vanish

Leaky Nun

haben diese 2 Tatsachen eine Verbindung?

Astyx

Not so weird once you know why, the example I know of is $\mathbb Q[\sqrt{p_i}, \text{the i-th program halts}]$. So deciding if $X^2-n$ can be factored would give a program that decides whether a program halts

The fun thing is that we have very fast algorithms for computing the factorization of polynomials over finite fields, but they are not deterministic

Lukas Heger

@LeakyNun not really no

Astyx

22:56

And we don't know if there exist deterministic algorithms with similarly low complexity

Leaky Nun

die geheimnisvolle 2

Lukas Heger

@LeakyNun the Picard group does appear in étale cohomology though

and that's related to the Jacobian of a curve, which is an abelian variety

Leaky Nun

@LukasHeger hangt es zussamen mit die Surjektivität der Funktion uber Pic^0(X), deren multipliziert mit n?

H2(X,μn) = Pic(X)/nPic(X) = Z/nZ?

Lukas Heger

yes

Leaky Nun

diese ist die Berechnung, auf deren du dich beziehst?

Lukas Heger

23:00

yeah

Leaky Nun

und was Erklarung gibt es mir?

Lukas Heger

you also have H1(x,μn)=Pic(X)[n]=Pic_0(X)[n]

@LeakyNun well, it's an explanation for why the cohomological dimension of a curve is 2

because it's a proof for that

Leaky Nun

heh...

@LukasHeger ich glaub dass es ist eine geometrische Erklarung, die ich finden probiere

dies Beweis gibt mir keine Intuition

Lukas Heger

23:17

@LeakyNun I'm not sure I'd even call étale cohomology "geometric". The computations of étale cohomology groups I saw always reduce the computation to some Galois cohomology group and the hard part is calculating that Galois cohomology group

it seems a bit much to ask for a geometric explanation if the theory is not geometric

Leaky Nun

dan was ist es dir?

Lukas Heger

I'd say it's an algebraic theory

Leaky Nun

und die Dimensionen hat dir keine Meinung?

Lukas Heger

the dimension is something that comes out of the computation and over C it agrees with singular cohomology (as it should)

that's enough meaning for me

let's say you have Spec(K) of a field K, that can have various dimensions depending on K . What's the geometric interpretation of that?

Leaky Nun

also ist die Moral des Gesprach, dass ein algbebraisch abgeschlossen Korper wie C ist?

Lukas Heger

23:24

over char 0 yeah

the whole motivation of étale cohomology was to construct a cohomology theory that behaves like singular cohomology over C

Leaky Nun

@LukasHeger weisst du, was ist das Λ, das in Seite 3 des Dokument ersheint, dessen ich dich sendete?

Lukas Heger

nein

Leaky Nun

23:44

@LukasHeger i've looked at the adjective ending table, which said that it's -en for masculine singular genitive adjectives with no determiner in front of it

but i'm struggling to construct a sentence out of it; kannst du mich helfen?

Slate

My new reading material arrived today, very excited.

Lukas Heger

@LeakyNun der Genuss deutschen Weines ist zu empfehlen

Leaky Nun

@LukasHeger danke!

user435118

I'm confused. I've read that in definite integration, the constant of integration cancels out. Why is there a constant of integration in the first place? For example, what function integrates to x^2 + 4? I certainly can't think of one.

Lukas Heger

@Xnero 2x integrates as much to x^2+4 as it does to x^2

Leaky Nun

23:55

@Xnero the "indefinite integration" and "definite integration" are two different things that happen to be related by a formula, so don't treat the "constant cancels out" thing too seriously; it's just a mnemonic

basically indefinite integration gives you a constant of integration and definite integration doesn't

user435118

@LukasHeger Even in definite integration?

Lukas Heger

@Xnero a definite integral gives you just a number

user435118

@LeakyNun I thought that, I was just confused why it cancels out if it doesn't exist in the first place.

Leaky Nun

@Xnero "it cancels out" is a mnemonic; there's no why; it's just that indefinite integration gives you a constant of integration, and "it cancels out" is just how the formula relating the two integrations works

it's like you're asking me why i need to divide by 2a in the quadratic formula; sure i can explain it but i would just be proving the formula; the "divide by 2a" has no special meaning

user435118

@LeakyNun So technically, "it cancels out" is wrong?

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$Mathematics$ Mathematics