Mathematics - 2019-08-23 (page 2 of 2)

Tuki

9:00 PM

they are perpendicular i.e inner product = 0

yes they were the basis vectors in this

you get quite easily confused by these

was thinking about $u$ is just any vector in $U$ but wasn't in this case

Semiclassical

yeah, that notation isn't doing you any favors

Alessandro Codenotti

@MatheinBoulomenos Ho iniziato a leggere del materiale per la mia tesi. Di che argomento stai scrivendo?

Semiclassical

anyways. so $\langle u_k,u_j\rangle=0$, except in one case

Tuki

yes the case where j=k

Semiclassical

Right. So when you run $k=1$ through $k=n$, you'll hit $k=j$ exactly once

and therefore the only term in that sum which will survive is $k=j$.

So what does my double sum look like afterwards?

Leaky Nun

9:05 PM

@AlessandroCodenotti after some unfruitful search and doubtful methods I was able to find "autarcia" that rhymes with "Lucia" including the consonant

MatheinBoulomenos

@AlessandroCodenotti Sto scrivendo della construzione delle representazioni de Galois dalle forme modulari. E tu? Stai leggendo delle algebre di operatori?

Alessandro Codenotti

@LeakyNun The usual spelling is autarchia (actually I had never seen autarcia before)

@MatheinBoulomenos leggendo*

Tuki

$$ P_U(v) = c_k \frac{ \langle u_k,u_k \rangle }{\langle u_k, u_k \rangle} = c_k $$

Leaky Nun

@AlessandroCodenotti yeah the dictionary listed autarcia as a variant

MatheinBoulomenos

grazie, era un errorino stupido

Leaky Nun

9:07 PM

do you know any such word then

Alessandro Codenotti

More or less, étale groupoids at the moment (to read about their groupoid C*-algebras later though) (switching to English because I don't know how those things are called in Italian)

Semiclassical

@Tuki You went from a double sum to a single value. But I only said that , as you sum from $k=1$ to $k=n$, you hit $k=j$ only once.

MatheinBoulomenos

I'm really unsure about the terms "representazione de Galois", "forma modulare" and "algebra di operatori", too

Semiclassical

You're having to do that sum once for every value of $j$ as well.

Alessandro Codenotti

rappresentazione

MatheinBoulomenos

9:08 PM

ah

Semiclassical

Moreover, once you've summed over $k$ and eliminated it, there is no way $k$ should remain in the problem.

MatheinBoulomenos

thanks

Tuki

oh yes so similiarly to nested for loops

MatheinBoulomenos

@Alessandro I'm a bit surprised that you don't do set theory or logic

Semiclassical

Right.

Tuki

9:10 PM

there should be $P_U(v) = c_1u_1 + c_2 u_2 + c_3 u_3 +\dots+ c_n u_n$

Alessandro Codenotti

@MatheinBoulomenos The only set theory professor is retiring, which will force the two postdocs in the logic group to leave Bonn as well, with no logic group remaining I had no choice but to find a different topic

MatheinBoulomenos

@Alessandro ah, I see

Semiclassical

~~should be $P_U$~~ ---

Tuki

fixed now

Semiclassical

okay, yes. or, in simpler terms: $P_U(v)=P_U(u+u^\perp)=u$.

That is: it maps the part of $v$ in $U$ to itself, and it annihilates the part of $v$ in $U^\perp$

(this also means that $P_U(P_U(v))=P(u)=u$. so applying it twice (or three times, etc.) is the same as applying it once. that's what the defining feature of a projection is to me: $P_U^2=P_U$)

Tuki

9:17 PM

Yes I think the formal description about projection states this $P_U^2=Pu$

Anyway @Semiclassical your help is much appreciated, so thanks for that

Semiclassical

just to complete the survey of what V.Ch was getting at

writing $v=u+u^\perp$ amounts to the decomposition $V=U\oplus U^\perp$ (i.e. every vector in $V$ can be written as a part from $U$ plus a part from $U^\perp$)

Tuki

well only problem was that if you take abstraction too far I cannot grasp the idea anymore

Semiclassical

And what we've seen is that $P_U$ takes $v=u+u^\perp\mapsto u$. That, by definition, is a vector in $U\oplus \{0\}$. (the zero vector is in every vector subspace by definition)

Tuki

Also for some unknown reason, I find proofs incredibly hard

Semiclassical

So the effect is that $P_U$ acts to map $V=U\oplus U^\top$ to $U\oplus\{0\}$

Alessandro Codenotti

9:22 PM

By the way @Mathei is there some notion of group bundle "better" than the groupoid $\bigsqcup G_i$? Where with "better" I guess I want some kind of nice compatibility of the fibers?

Tuki

this ball symbolised exactly what?

Semiclassical

$U\oplus W=\{u+w|u\in U,w\in W\}$

it's a pretty boring definition

Tuki

oh ok

Semiclassical

I'm not sure how much this abstraction helps, though. I'd probably stick with this:

Tuki

should be $w \in W$

Semiclassical

9:24 PM

yeah

Suppose $v\in V$. We know that we can write $v=u+u^\perp$ for $u\in U,u^\perp\in U^\perp$

and we further know that $P_U(v)=u$.

So (going to your original question): If you further know that $P_U(v)=v$, what can you conclude about $v$?

Tuki

that $P_U(v)=u \implies v \in U$

Semiclassical

ding

Tuki

For me that $v$ can be expressed as linear combination of some orthogonal basis in $U$ implies that $v \in U$

Semiclassical

Note that we didn't assume $v$ was in $U$. That's the point of $u^\perp$: You assume the possibility of that extra term, and show that if $u^\perp \neq 0$ then $P_U(v)\neq v$

Tuki

at intuitive level

MatheinBoulomenos

9:31 PM

@Alessandro if you have a morphism of $G$-sets $X \to Y$, you get an induced morphism on action groupoids $X/\!/G \to Y/\!/G$. Maybe if $X \to Y$ is surjective (e.g. $Y=*$ and $X$ is non-empty) this looks a bit like a group bundle?

ÍgjøgnumMeg

Hey @Alessandro and @Mathein again

lol

Tuki

since span of those orthogonal basis vectors cover all of $U$, but not all of $V$

Alessandro Codenotti

What does a morphism of $G$-sets mean again?

Ultradark

Does this sum converge? $$ S=\sum_{n=1}^\infty \ln(p_n^2)K_1(\ln(p_n^2)). $$ Here $p_n$ is the nth prime and $K_1$ is the modified bessel function of the second kind

MatheinBoulomenos

just $G$-equivariant map

ÍgjøgnumMeg

9:32 PM

@Mathein quick one: do you have a hint for why the degree $p$ subextension of $\Bbb Q(\zeta_{p^2})$ is unique? Or is it obvious af

hahaha

Alessandro Codenotti

So $g\cdot f(x)=f(g\cdot x)$?

Semiclassical

@Ultradark No. $\ln(p_n^2)\to\infty$ as $n\to\infty$ and $K_1(x)\to \infty$ as $x\to \infty$

MatheinBoulomenos

@Alessandro exactly

Semiclassical

and if the terms of a sum don't go to zero, the sum can't converge

MatheinBoulomenos

@ÍgjøgnumMeg the hint is Galois theory

@Alessandro in some sense I think that groupoids themselves are kind of like group bundles if you have a groupoid $G$, then I kind of think of it like a bundle of groups over the objects of $G$

Ultradark

9:34 PM

@Semiclassical oh yeah I see

Semiclassical

@Tuki sure, but that's just the definition of $u\in U$. it's not equivalent to $P_U(v)=v$.

ÍgjøgnumMeg

I mean the guy exists because $\operatorname{Gal}(\Bbb Q(\zeta_{p^2})/\Bbb Q) \cong \Bbb Z/(p) \times \Bbb Z/(p-1)$ but it's not just that that makes it unique right?

Alessandro Codenotti

Hmm I see what you mean but I was trying to see $G$ as a bundle over $G^{(0)}$ somehow

(because that's what happens with the disjoint union of groups)

MatheinBoulomenos

@ÍgjøgnumMeg if you have a cyclic group of order $n$ and a divisor $d$ of $n$, how many subgroups of order $d$ are there?

ÍgjøgnumMeg

just one for each divisor of $n$ right?

MatheinBoulomenos

9:37 PM

right

ÍgjøgnumMeg

bleh okay

oHOOOOOO

I was thinking "$\Bbb Q(\zeta_8)$ has $3$ quadratic subfields but only has degree $4$ so that can't be right"

but $V_4$ aint cyclic

Semiclassical

A linear algebra question of my own that I'm working on for an answer

ÍgjøgnumMeg

durr, thanks

MatheinBoulomenos

right, $2$ being the oddest prime again

ÍgjøgnumMeg

indeed

MatheinBoulomenos

9:41 PM

$2$ is so weird. we have explicit description of the absolute Galois groups of $\Bbb Q_p$ for $p \neq 2$ (i.e. generators and relations), but not for $p=2$

ÍgjøgnumMeg

yeah screw $2$

MatheinBoulomenos

in the book on local Langlands for $\mathrm{GL}_2$, there's a whole chapter devoted to char $2$, you get a whole new class of representations

Semiclassical

I have two $N$-by-$R$ matrices $\alpha,\beta$ and two other $N$-by-$R'$ matrices $\alpha_p$,$\beta_p$

ÍgjøgnumMeg

@Mathein nise

I'm enjoying cyclotomy atm

lol

Semiclassical

All four matrices have full column rank. In addition, I know $\alpha^\top \alpha_p=\beta^\top \beta_p=0$.

MatheinBoulomenos

9:44 PM

@ÍgjøgnumMeg yeah cyclotomic fields are cool

Semiclassical

and finally I know that $\alpha_p^\top \beta_p$ has full rank as well.

MatheinBoulomenos

@ÍgjøgnumMeg oh btw, there's a workshop on Iwasawa theory one week before the uni starts

Semiclassical

I'm trying to show that $\beta^\top \alpha$ is nonsingular.

MatheinBoulomenos

in Heidelberg

ÍgjøgnumMeg

@Mathein ooo! I'll be in Heidelberg on the 28th of September

Semiclassical

9:45 PM

blah, $R'$ should be $R_p:=N-R$

MatheinBoulomenos

@ÍgjøgnumMeg I'll email you the announcement as a pdf

ÍgjøgnumMeg

Cool thanks :)

MatheinBoulomenos

it's by Greenberg himself

Semiclassical

I guess the better statement is that $\beta^\top\alpha$ is also full rank.

ÍgjøgnumMeg

damn

Ultradark

9:47 PM

@Semiclassical I changed my question to sum over the twin primes. This converges right?

Semiclassical

What?

Ultradark

$$ S=\sum_{n=1}^\infty \ln(p_n^2)K_1(\ln(p_n^2)). $$ Here $p_n$ is the nth twin prime and $K_1$ is the modified bessel function of the second kind.

So summing over $p=3,5,5,7,11,13,17,19,...$

Semiclassical

well, it depends. If there's finitely many twin primes, then that sum truncates eventually and so converges trivially.

Ultradark

okay

Semiclassical

If there's infinitely many twin primes, however, then you're in the same boat as before: $p_n\to \infty$, so $\ln(p_n^2)K_1(\ln(p_n^2))\to\infty$

ÍgjøgnumMeg

9:49 PM

@Mathein oh so it's twice a week!

Semiclassical

If $p_n$ is any infinite sequence that grows without bound, you have no prayer of getting an infinite series

s.harp

$K_1(z)$ is asymptoitcally proportional to $1/z$, so that product doesnt go to infintiy but the sum will not converge

Semiclassical

oh, you're right. I was thinking K_1 was the one that blew up at infinity

Ultradark

@s.harp for the primes or the twin primes?

s.harp

actually nvm im not sure about what bessel function we are talking about

@Ultradark for any unbounded sequence

Semiclassical

9:53 PM

hm, actually, the large-n asymptotics for K_1(z) are e^(-z)/sqrt(z)

ÍgjøgnumMeg

@Mathein cheers for the heads up, very cool

Semiclassical

so that might actually be dying fast enough for the sum (for primes or for twin primes) to exist

I wouldn't expect to get anything computable out of that---bessel functions have f* all to do with primes as far as I know

(in general bessel functions don't give a damn about nice values)

MatheinBoulomenos

@ÍgjøgnumMeg it's quite possible that you wouldn't have got to know about it, I'm not sure if it's publicly on the internet, since my advisor just sent me an email with the pdf and not a link

ÍgjøgnumMeg

@Mathein thanks for that then! :D

MatheinBoulomenos

you're welcome :)

Ultradark

10:00 PM

bessel functions are related to primes @Semiclassical?

MatheinBoulomenos

@Ultradark bessel functions are definitely studied in analytic number theory, but I don't know any details

Secret

$e^{\frac{b}{c}}=n, \frac{b}{c}=\ln(n), b =\ln(n^c), e^b=n^c$

$\ln(n^c)=m, n^c=e^m$

$\ln (n)= \frac{m}{c}$

MatheinBoulomenos

@ÍgjøgnumMeg let me know when you have time, then we can meet up and I can show you the city and the uni and talk about the different profs etc.

ÍgjøgnumMeg

@Mathein cool! I'll be in Heidelberg on the 28th of September, setting up my flat (if I find one by then... lol) and then I'm in Bonn on the 1st and 2nd of October, but after that I'm just cramming complex analysis lol

Ultradark

Can I come @MatheinBoulomenos?

MatheinBoulomenos

10:10 PM

@Ultradark uhm.. I don't really know you, sorry. Do you mean to study at Heidelberg? you need some solid German knowledge for that

Daminark

How's it going everybody?

ÍgjøgnumMeg

Hey @Daminark !

MatheinBoulomenos

Hi @Daminark!

Pretty well, thanks. And yourself?

Daminark

Doing aight, thanks! Next week is TA training and the week after classes finally start for the fall!

ÍgjøgnumMeg

@Daminark nice :) What are ya taking/TAing

MatheinBoulomenos

10:14 PM

@Daminark @ÍgjøgnumMeg oh btw, my prof confirmed that I can work under him over next summer as a "research assistant in computional arithmetic geometry" or as a friend of mine put it, "MAGMA bimbo"

ÍgjøgnumMeg

that's cool af

MAGMA bimbo is a fine title

MatheinBoulomenos

yeah not sure which one of those to put in the CV

Daminark

Each time you send it change it

ÍgjøgnumMeg

MAGMA hoe

Daminark

I'm gonna be TAing (computational) Calc 2

Semiclassical

10:16 PM

@Ultradark no, they don’t (as far as I know). That’s what I meant

ÍgjøgnumMeg

Cool :)

Alessandro Codenotti

@ÍgjøgnumMeg that's unlucky, I'll be in Bonn from the 3rd

Daminark

As for what to take, probably AG, rep theory, and something else

ÍgjøgnumMeg

@Alessandro sad :(

MatheinBoulomenos

@Daminark TAing + AG + rep theory sounds really nice

ÍgjøgnumMeg

10:17 PM

@Daminark nice

ANT, Mod. Forms, and a proseminar

lol

debating whether or not I should take higher analysis but idk

Daminark

I might take/audit one of the intro classes (algebra, AT, analysis) since my knowledge of the stuff is a bit iffy, also might do a topics in combinatorics ("Crystal bases"), or analytic NT

ÍgjøgnumMeg

Sounds cool!

MatheinBoulomenos

higher analysis is measure theory + some other stuff

the other stuff depends a bit on the prof

ÍgjøgnumMeg

you think it's worth doing that? I know so little analysis that it might be good lol

idk I should make a decision at some point

Daminark

What's your proseminar on?

mick

10:22 PM

4

Q: Proof (or hints towards proof) for asymptotic shape of orbit $0 \to 1 \to b \to b^b \to \cdots$ with certain class of $b$?

In a paper by Baker & Rippon (1983) the property of being convergent or divergent for iterated exponentials $z_{h+1} \to b^{z_h}$ with $b$ complex and $z_0=1, z_1=b, z_2=b^b, \cdots$ for classes of the bases $b$ have been established - a problem which has been considered here in MSE a couple of t...

ÍgjøgnumMeg

@Daminark point-set topology, with a bit of AT at the end

and my talk is on covering spaces lol

MatheinBoulomenos

idk, if you think you want to do some analytic stuff

I'm going to do intro Japanese, adic spaces, moduli problems in AG, rigid analytic geometry, linear algebraic groups and I'm going to TA intro abstract algebra

mick

0

Q: What Julia set has shape 8?

We know the Julia set for $f(z) = z^2 $ is a unit circle. It looks like a zero. I wonder what analytic function has a Julia set that is just the shape of an 8 , and nothing else ?

dynamical-systems topological-groups complex-dynamics

MatheinBoulomenos

maybe I'll drop something if it gets too much

ÍgjøgnumMeg

@Mathein oh I'm going to take Russian too!

MatheinBoulomenos

10:24 PM

nice

make sure to register for the language course asap, I already registered for Japanese

ÍgjøgnumMeg

adic spaces looks like it'd be interesting

but I'd die lol

oh okay, cheers

MatheinBoulomenos

if you haven't seen any AG, it's going to be pretty tough

idk if you have access to log into the lsf yet

ÍgjøgnumMeg

yeah lol, I'll have AG in 3rd semester I guess

MatheinBoulomenos

you can just register for the language courses in the lsf

ÍgjøgnumMeg

yeah I have no uni ID

Daminark

10:25 PM

Oh point-set, rip

ÍgjøgnumMeg

Indeed.. but I have to do it at some point hahaha

MatheinBoulomenos

at least you also do some AT in the end

Daminark

This is true

ÍgjøgnumMeg

yeee

MatheinBoulomenos

fundamental group and covering spaces, that's some beautiful stuff

Daminark

10:27 PM

I still have to learn that properly tbh, I basically know a list of true facts in AT which I desperately bash together when it's time to solve problems

ÍgjøgnumMeg

lol

is the classification of riemann surfaces an AT problem?

MatheinBoulomenos

no

but AT definitely plays a role

ÍgjøgnumMeg

fair, I seem to remember a guy who was doing AT for his undergrad dissertation talking about riemann surfaces

ah fair

MatheinBoulomenos

Riemann surfaces means you need to involve some complex analysis, that's beyond pure AT

Konformist Liberal

Hello! ncatlab.org/nlab/show/Introduction+to+Topology+--+2 I don't get the first sentence of the proof of Lemma 2.6. Don't we need that the open neighborhood is in the image?

MatheinBoulomenos

10:32 PM

Riemann surfaces are surfaces with extra structure, that's a "finer" classification question than just topologically. For example, the unit disk and the complex plane are homeomorphic, but not isomorphic as Riemann surfaces (by Liouville's theorem)

Daminark

Was it just classification of compact Riemann surfaces or of Riemann surfaces in general?

MatheinBoulomenos

hey @Ted

Ted Shifrin

Hey @Mathein and Demonark.

ÍgjøgnumMeg

Liouville's theorem is "bounded entire functions are constant" ye?

MatheinBoulomenos

yes

Ted Shifrin

10:34 PM

Ye

ÍgjøgnumMeg

ye

hey @Ted

Ted Shifrin

Hey @ÍgjøgnumMeg

Daminark

Ye

MatheinBoulomenos

Riemann surfaces in general I think reduces to uniformization + understanding subgroups of the automorphism groups of the three simply connected Riemann surfaces

which seems hard enough

Ted Shifrin

Well, you still need to understand $3g-3$, @Mathein.

MatheinBoulomenos

10:38 PM

$3g-3$?

The dimension of the moduli space for genus $g$ curves when $g>1$.

MatheinBoulomenos

ah, I see

Daminark

Yo Eric

Érico Melo Silva

10:57 PM

yo how’s life @Daminark

ÍgjøgnumMeg

@Eric are you part of an fb group called actually good math problems

Daminark

Going p well, I've got orientation (mainly TA training) next week, after that it's classes

Érico Melo Silva

@ÍgjøgnumMeg not that ik of lol

@Daminark sick, r u in madison already

Daminark

Yup

Already did quals actually

Érico Melo Silva

how do u like the town

Daminark

11:10 PM

It seems decent at first glance, though I haven't been around that much

Érico Melo Silva

cool cool

Ted Shifrin

Heya mr Eric

It's a cool town, Demonark, but just wait for all the snow :P

Daminark

"a cool town" heh

Érico Melo Silva

hi @Ted

ALannister

11:48 PM

0

Q: Gauss-Newton Method not converging for my function

I need to solve the optimization problem $$ \min_{x\in \mathbb{R}^{3}}f(x) $$ where the function $f$ is defined as follows: $$ f(x_{1}, x_{2}, x_{3}) = \frac{1}{2}\left[\left(2x_{1}-x_{2}x_{3}-1 \right)^{2}+\left(1-x_{1}+x_{2}-e^{x_{1}-x_{3}}\right)^{2}+\left( -x_{1}-2x_{2}+3x_{3}\right)^{2} \ri...

matrices optimization eigenvalues-eigenvectors matlab jacobian

oh by the way, can somebody please give me a link to the robjohn thing where I can get my latex to render properly here? This is a new laptop.

Transcript for

$Mathematics$ Mathematics