Mathematics - 2018-12-20 (page 2 of 2)

ÍgjøgnumMeg

20:00

lol

Mike Miller

20:12

I am ok. I don't have time to think about your question, unfortunately.

Arrow

@MikeMiller too bad. Ah well, have a nice day then!

Semiclassical

@MatheinBoulomenos variance is a word, but I'm not sure it's what you intend here

(variability, maybe?)

MatheinBoulomenos

@Semiclassical I mean the difference between covariance and contravariance

Semiclassical

ahhh

jrh

Quick question, is there any operation represented by a capital (kind of cursive looking) N? E.g., N(x + y + z)? I saw this in a paper; earlier on in the paper N is the number of samples but the equations are written in a different font so I'm not 100% sure. The paper is this arxiv.org/abs/1509.06113 and the symbol near the beginning of page 3 if that's easier.

At this point I'm thinking it's just saying to multiply by the number of samples; when I try to copy/paste the character from my PDF reader it comes back as standard old ASCII N.

Leaky Nun

20:24

@MatheinBoulomenos wtf

the symbols are links to their definitions

I'm impressed

quid

@jrh it likely means a normal distribution, note that there are two parameters separated by a comma.

But I am not sure. In LaTeX and MathJax one would get the symbol via \mathcal{N}

Vrouvrou

hello

any help here:

1

Q: Prove that $\max(\max_{t\in[0,1]} |f(t)|,\max_{t\in[0,1]} |f'(t)|)\leq \frac12 ||f'||_{L^2([0,1])}$

i want to prove this: $$\max(\max_{t\in[0,1]} |f(t)|,\max_{t\in[0,1]} |f'(t)|)\leq \frac12 ||f'||_{L^2([0,1])}$$ where $f\in C^1([0,1])$ and $f(0)=f(1)=0$ What I do is : (using Cauchy Schwartz inequality) $$ |f(t)|=|\int_0^t f'(t) dt|\leq \int_0^t |f'(t)| dt\leq (\int_0^1 |f'(t)|^2 dt)^{\frac12...

MatheinBoulomenos

@LeakyNun wow, I never noticed that

jrh

@quid Thanks for the tip!

quid

Indeed.

You are welcome.

Leaky Nun

20:33

@MatheinBoulomenos do you need ordinals and such to prove the characterisation of finitely generated?

MatheinBoulomenos

@LeakyNun it's the same as the proof that Noetherian in terms of ACC is equivalent to all submodules f.g.

it needs countable choice, but no ordinals

Leaky Nun

I considered that

I failed to see the sameness

MatheinBoulomenos

hmm right there's a difference in one direction

you probably need Zorn

Ted Shifrin

hi @Mathein

MatheinBoulomenos

Hi @Ted

If $M$ is not f.g. then let $(M_i)_{i \in I}$ be the collection of all f.g. submodules. Choose a maximal subset $J$ of $I$ that is totally ordered by exclusion (exists by Zorn), then $\bigcup_{j \in J} M_j = M$ or else we contradict maximality. so $(M_j)_{j \in J}$ is a chain like we wanted

I find that easier to write down that the ordinal stuff

but it still uses Zorn

Leaky Nun

20:45

exclusion?

Daminark

Hey guys

Ted Shifrin

Hi slothful Demonark.

MatheinBoulomenos

hi @Daminark

Alessandro Codenotti

Hi everyone

Demonark is no longer demonic? @Ted

Ted Shifrin

hi demonic @Alessandro

MatheinBoulomenos

20:58

hi @Alessandro

Ted Shifrin

It's built in to his name, @Alessandro.

CaptainAmerica16

Hi all.

Ted Shifrin

heya @CaptainAmerica

CaptainAmerica16

It is so cold where I am right now.

Ted Shifrin

How cold is so cold?

CaptainAmerica16

21:00

I don't know how cold it is in the house, but outside is 37 F

Ted Shifrin

Hmm, that's not very cold. Try below 0 a few days.

Or at least below freezing. This is December, almost January, after all.

CaptainAmerica16

I just don't like winter weather. I don't think we've had temperatures below freezing since like 2016.

Ted Shifrin

Well, that's highly unusual. Aren't you in NJ or something?

CaptainAmerica16

Like 2 hours from. Center of PA

Ted Shifrin

Oh, my sister's near Redding. They get tons of snow and cold all the time.

oops, Reading.

Leaky Nun

21:04

lol

English

Ted Shifrin

English is absurd.

CaptainAmerica16

We've rarely gotten snow either

lol

Daminark

At least I'm too slothful to be evil

CaptainAmerica16

I hope it snows on Christmas

Ted Shifrin

Let's not get carried away, Demonark.

CaptainAmerica16

21:08

I think I'm giving up on the concavity problem too. I think I'm missing something really simple, but I just can't figure it out.

Ted Shifrin

You are very close.

The hint is that it will not be differentiable at the integers.

CaptainAmerica16

:thonk:

The first thing that came to mind are sharp corners.

I'll try to draw it out

Leaky Nun

power sets are ridiculous

it lets you make impredicative objects

CaptainAmerica16

Wait, it a piecewise function...maybe there's a hole at every integer?

Fargle

@LeakyNun Agreed. Even for finite sets---iterating the power set on a singleton set 64 times, we end up past 2^63 - 1, which is clearly the largest integer.

Leaky Nun

21:15

lol

Eli

So I came here, out of desperation I might add, to ask for help in solving a problem. But now that I'm here I was wondering 'isn't that what everyone or most would do' , so I really don't know if I can ask in the first place. Maybe this is just a place for math magicians to relax and have number fetishes, but maybe they like to help as well... What is this place? :P

CaptainAmerica16

Just ask; don't ask to ask

per the room description

Ted Shifrin

@CaptainAmerica: No, no, remember the two functions agree at all the integers?

Leaky Nun

@Eli well I would say this place is for mathematics at or beyond undergraduate level

CaptainAmerica16

fudge

Leaky Nun

21:17

@Eli and people might answer your questions if they want to; nobody is obliged to answer any questions

MatheinBoulomenos

I think $\Bbb{N} \cup 2^{\Bbb N} \cup 2^{2^{\Bbb N}} \cup \dots$ is a very reasonable object

Ted Shifrin

ignores @Mathein

Leaky Nun

@MatheinBoulomenos I think $2^\Bbb N$ is already not very reasonable

we can't even tell its size

MatheinBoulomenos

@Ted what did I do?

Ted Shifrin

@LeakyNun Huh?

Fargle

21:18

@LeakyNun It's Pretty Big.

Ted Shifrin

hi @Fargle

Fargle

Heya @Ted.

Ted Shifrin

How do I not know it's $\aleph_1$?

Eli

Alright, thanks, here it goes: Basically I have this sequence {1,2,3,4,7,5,11,8,16,12,22,6,29,17,9,37,46,23,10,30,56,67,13,38,18,79,92,14,47,1‌06,57,121,19,24,68,31,80,15,25,93,20,9,32,107,...,n } and I need a function for it. I've already solved a couple, this is the last one, after I know that function, I will conquer all! muhahah. (full: math.stackexchange.com/q/3045860/333955)

Leaky Nun

you mean how do you know it's not $\aleph_1$?

Ted Shifrin

21:20

Isn't $\aleph_1$ the cardinality of the reals?

Leaky Nun

that's what continuum hypothesis states

Ted Shifrin

@Eli: I don't think you're going to get any answers to that from anyone here.

Alessandro Codenotti

No, that's the continuum hyp.... Damn snipers

Ted Shifrin

I don't play these games of not assuming CH or AC. Go play by yourselves.

Semiclassical

my usual tactic when I see a long sequence is to plug it into OEIS

Eli

21:21

@TedShifrin why not?

Semiclassical

alas, that seems to come up empty here

Leaky Nun

@Ted I don't think CH is widely accepted

Eli

@Semiclassical already tried that :/

Alessandro Codenotti

ZFC proves $|\Bbb R|=2^{\aleph_0}$ but it's consistent that $2^{\aleph_0}$ is way bigger than $\aleph_1$. It could be every $\aleph_n$, $n\geq 1$ or even bigger than $\aleph_\omega$

(it cannot be $\aleph_\omega$ though)

Leaky Nun

do we need C?

Ted Shifrin

21:23

@Leaky: In the mathematical circles I traveled in, no one cared about this.

Leaky Nun

exactly

but I think they use AC a lot

Alessandro Codenotti

Yes, without C I don't think you can even say that $|\Bbb R|$ is an aleph

Semiclassical

I don't think there's anything particularly offensive about this kind of question, but there's nothing I find especially interesting in it

Leaky Nun

I mean, ZF proves $|\Bbb R| = 2^{\aleph_0}$

Alessandro Codenotti

Oh, yes, that's true

Eli

21:24

Are there like steps or methods of finding or converting a sequence into a function f(x) = something ? For the more complicated one that is.

Ted Shifrin

@Leaky: We certainly don't shy away from proofs that use AC or Zorn, no.

Leaky Nun

@Ted I think we're saying the same thing

Ted Shifrin

I have never found this stuff interesting. Shrug.

Leaky Nun

I think we all know that

Semiclassical

I find it interesting that such questions can be asked, but not much further

MatheinBoulomenos

21:25

@Eli a function is not necessarily a closed formula and a finite number of points never determines a function

Ted Shifrin

Anyhow, all I said was that I didn't think Mathein's infinite union of sets was reasonable.

Leaky Nun

wait what?

Eli

@MatheinBoulomenos well it's based on a pattern, so there must one...

Ted Shifrin

And you said $2^{\Bbb N}$ wasn't reasonable (which surprised me).

Eli

Out of curiosity, why doesn't my chat display the $.. equations?

Leaky Nun

21:27

predicativists reject the power set axiom :)

Semiclassical

it's not enabled by default

use the latex in chat link in the upper right

Ted Shifrin

puts Leaky on ignore now

Semiclassical

I'm trying to figure out a good label for people who reject the "rejecting axioms is interesting" axiom

Alessandro Codenotti

But anyway while there is no doubt about whether AC should be accepted or not for usual mathematics, there isn't such a clear consensus on CH. Probably because it's mostly irrelevant

CaptainAmerica16

@LeakyNun whats a predicativist

Leaky Nun

21:28

@TedShifrin hey that's religion discrimination! :P

Eli

@Semiclassical thanks :)

Fargle

As someone who admittedly hasn't studied logic or set theory deeply, I haven't used CH literally even once.

Leaky Nun

@CaptainAmerica16 there's a link on the word

Eli

@MatheinBoulomenos to follow up, maybe you suggest something concerning the following: math.stackexchange.com/q/3045860/333955

@MatheinBoulomenos maybe you could*

Leaky Nun

@Semiclassical hey euclid definitely didn't believe in the axiom of infinity

Alessandro Codenotti

21:30

@Fargle yeah exactly, unlike AC, CH doesn't really come up often in ordinary math

CaptainAmerica16

Eh, i guess i get the gist

Eli

@LeakyNun Infinity doesn't exist unless we make it exist? That sounds weird...

Leaky Nun

that's maths

CaptainAmerica16

What's it called when you keep drawing the same picture over and over again, but expect different results each time?

Leaky Nun

has anyone here encountered any anti-foundationist?

$Q := \{Q\}$

Eli

21:34

@CaptainAmerica16 Determination...

CaptainAmerica16

@Eli I was thinking more along the lines of insanity - but that works too.

MatheinBoulomenos

what happens if you drop extensionality? you might end up with many different empty sets or something like that

Alessandro Codenotti

Foundation is also kinda useless in ordinary math

Leaky Nun

@MatheinBoulomenos oh no, that's madness

or I suppose we can live with multisets

Alessandro Codenotti

But it breaks a lot of the nice structure of $V$ if you drop it

MatheinBoulomenos

21:37

if you drop extensionality you could just replace all "equal" with "contain the same elements" and end up with something equivalent

since actual equality becomes intractable

Daminark

@MatheinBoulomenos a bunny will cry

MatheinBoulomenos

good point

CaptainAmerica16

Eli

I'm facing a problem, I can set a bounty in 2 hours, but I'm hesitant, either 150 rep, but I lose chat privileges or 100 rep and keep chat privileges... hm....

CaptainAmerica16

@Ted I tried to make the points as noticeably sharp as possible.

Semiclassical

21:48

So $f(x)=n^2(n+1-x)+(n+1)^2(x-n)$ for $n<x<n+1$?

Leaky Nun

@Semiclassical I think you're extrapolating too much

Semiclassical

feh

Leaky Nun

wait what's the difference between interpolation and extrapolation?

Fargle

@LeakyNun Interpolation is "between" your data, extrapolation isn't

Using 1992 and 1994 data, you can interpolate 1993 data, or extrapolate 1995 data

user525966

@AlessandroCodenotti One question is what value am I allowed to use for variables?

I can't tell what's a valid way to label variables

Leaky Nun

21:58

@Fargle interesting

Alessandro Codenotti

Variables are usually labeled $x_1,x_2,\dotsc$. Formally the language should have countably many fixed symbols for the variables, but often you use other symbols as well because it's more comfortable (x,y,z etc.)

user525966

@AlessandroCodenotti I mean in an actual proof

Like for example how would you prove user21820's example ￢∀x∈S (P(x)) → ∃x∈S (￢P(x))

Alessandro Codenotti

I suppose you're working in a system with some inference rules? You need to decide which to apply, that takes practice

user525966

what's the usual system most people use?

Alessandro Codenotti

22:20

First order logic with what I think it's called an Hilbert style deduction system. But user21820 like Fitch's style and I know nothing about that

user525966

how would you do that one hilbert style?

Leaky Nun

@MatheinBoulomenos noch hier?

ich versteh nicht lemma 1.5.10

MatheinBoulomenos

@LeakyNun $\mathrm{Hom}_{\Bbb Z}(\Bbb{Z}[G],A)$ and $\mathrm{CoInd}^G(A^\circ)$ have different $G$-actions

Leaky Nun

then how are they isomorphic :o

MatheinBoulomenos

see the proof: you don't take the identity as an isomorphism

this is a surprising lemma because it tells you that the $G$-module structure on $\mathrm{Hom}_{\Bbb Z}(\Bbb{Z}[G],A)$ is independent from the $G$-action on $A$

Leaky Nun

22:32

for example?

(do you just do maths abstractly like this...)

MatheinBoulomenos

for example take $G=\Bbb Z/2\Bbb Z$ and $A=\Bbb Z$ and let the generator of $G$ act by $x \mapsto -x$. Then $\mathrm{CoInd}^G(A^\circ)=\mathrm{Hom}_{\Bbb Z}(\Bbb{Z}[G],A^\circ)$ is just $\Bbb Z^2$ where the generator $G$ switches the two copies. $\mathrm{Hom}_{\Bbb Z}(\Bbb{Z}[G],A)$ is also $\Bbb Z^2$ but the generator of $G$ acts by $(x,y) \mapsto (-y,-x)$. The lemma tells you those two $G$-actions are isomorphic

maths researcher

hello guys

i have a question

I want to prove that the set of integers is a closed set

so what I did was that I said

Leaky Nun

hot take: ${}_NA/I_GA$ should really be called $\hat{H}^{-0}(G,A)$ while $A^G/N_GA$ is $\hat{H}^{+0}(G,A)$ @MatheinBoulomenos

MatheinBoulomenos

nah, Tate cohomology is fine as it is

for example, it's two-periodic for cyclic groups

Leaky Nun

what's it with shifting by 1

maths researcher

22:40

assume the sequence $x_n$ converges to x and x is not an integer then setting epsilon epsilon to be 1/2 implies that for some n>N $x_n$ $\in$ B(x,1/2) which is a contradiction

is my proof correct?

Leaky Nun

@MatheinBoulomenos that's... interesting

MatheinBoulomenos

you asked for an example :P

maths researcher

@LeakyNun, @MatheinBoulomenos is my proof correct?

may you please help?

B is the ball by the way centered at 1/2

i mean centered at x

with radius 1/2

anyone, please?

MatheinBoulomenos

@LeakyNun the lemma is used in the proof of dimension shifting, which is pretty important

Leaky Nun

what is dimension shifting?

MatheinBoulomenos

22:50

for every $G$-module $A$, there's a $G$-module $A^*$ such that $H^{n+1}(G,A)=H^n(G,A^*)$ and another $G$-module $A_*$ with the corresponding property for homology

and on Tate cohomology for $G$ finite you get $\hat{H}^{n+1}(G,A)=\hat{H}^n(G,A^*)$ and $\hat{H}^{n-1}(G,A)=\hat{H}^n(G,A_*)$ for all $n \in \Bbb Z$

Leaky Nun

that...

is quite interesting

MatheinBoulomenos

that's a pretty useful proof technique. Suppose you want to show a statement about $\hat{H}^n(G,A)$ for all $n$ and all $G$-modules $A$ ($G$-finite), then it suffices to show this for one $n$ by dimension shifting (often you take $n=0$ or $n=-1$)

Leaky Nun

@MatheinBoulomenos what's a goal that I can aim for?

MatheinBoulomenos

for example $|G|$ annihilates $\hat{H}^n(G,A)$ because $\hat{H}^0(G,A)=A^G/N_GA$ and for $a \in A^G$ one has $|G|a=N_Ga$

Leaky Nun

I mean, when I studied Galois theory, it was for quintic polynomials; when I studied homology, it was for dimension invariance

now it feels like just wandering around without any goal

MatheinBoulomenos

22:55

you can do it for local class field theory

Leaky Nun

that I can aim for

so group cohomology is for LCFT?

MatheinBoulomenos

that's one application, yes

GCFT can also be proved by group cohomology, but it's harder

for LCFT see Serre "Local Fields" or math.ucla.edu/~sharifi/algnum.pdf

Leaky Nun

are there... simpler applications?

MatheinBoulomenos

Kummer theory?

not sure if that counts, you can do that without group cohomology as well

Leaky Nun

oh well I guess that's reasonable

is there a higher level proof of hilbert 90 though

apart from manually computing the cohomology

MatheinBoulomenos

22:59

Hilbert 90 in a more general form is related to Galois descent

Leaky Nun

oh right there is non-abelian group cohomology

@MatheinBoulomenos which is a special case of faithfully flat descent right

MatheinBoulomenos

yeah

Leaky Nun

see, I can imitate words with no meaning

@MatheinBoulomenos so a galois map is faithfully flat?

MatheinBoulomenos

yes, $K \to L$ is faithfully flat

Leaky Nun

is every field extension faithfully flat?

or only separable ones?

MatheinBoulomenos

23:08

every non-zero $K$-algebra is faithfully flat, I think the Galois condition just gives you a nice description of the descent as invariants

Leaky Nun

oh

oh right of course it is, they're all free

MatheinBoulomenos

Here's what Serre writes about "forms" (this is a bit informal and vague. Let $K/k$ be a field extension and $X$ be an "object" defined over $k$, we say that an object $Y$ defined over $k$ is a $K/k$-form of $X$ if $Y$ becomes isomorphic to $X$ when the ground field is extended to $K$.
The idea is that if $K/k$ is Galois, then the set of $K/k$-forms of $X$ is up to $k$-isomorphism is classified by $H^1(\mathrm{Gal}(K/k),A(K))$ where $A(K)$ is the set of $K$-automorphisms of the scalar extension of $X$ to $K$

Leaky Nun

interesting

MatheinBoulomenos

so when you take $X=k^n$, then you get that $H^1(\mathrm{Gal}(K/k),\mathrm{GL}_n(K))$ classifies $k$-vector spaces $V$ such that $L \otimes_k k^n \cong L \otimes_k V$

but thes are all isomorphic to $k^n$, so $H^1(\mathrm{Gal}(K/k),\mathrm{GL}_n(K))=1$ which is Hilbert 90

Leaky Nun

but then how do you define cohomology?

MatheinBoulomenos

23:16

$H^1$ with non-abelian coefficients is defined explicitly with cocycles and coboundaries

Leaky Nun

that's sad

MatheinBoulomenos

not sure if higher non-abelian cohomology groups exist

haven't seen them so far

Leaky Nun

very sad

MatheinBoulomenos

also note that $H^1(G,A)$ won't be a group when $A$ is non-abelian, just a pointed set

there's also singular $H^1(X;G)$ for non-abelian $G$ with a similarly explicit definition

Leaky Nun

but, I mean, don't I know enough group cohomology for Kummer theory already

or maybe should I aim for artin-schreier theory

MatheinBoulomenos

23:19

there's a generalized Kummer theory which covers both simultaneously in Neukirch

Leaky Nun

of course there is

MatheinBoulomenos

Neukirch ANT IV.3.3

Leaky Nun

dank

MatheinBoulomenos

but yeah undestanding how Hilbert 90 implies Kummer theory and Artin-Schreier-theory is a good first goal for group cohomology

Leaky Nun

I doubt they shift dimension though

MatheinBoulomenos

23:23

that's true

another goal might be Schur-Zassenhaus

Leaky Nun

no butterflies? :(

MatheinBoulomenos

this follows from the application I gave above of dimension shifting that $|G|$ kills $\hat{H}^n(G,A)$

Leaky Nun

@MatheinBoulomenos cool

Forever Mozart

today I went deep

MatheinBoulomenos

and also involves a concrete interpretation of $H^2(G,A)$ and $H^1(G,A)$ in terms of group extensions

Forever Mozart

23:25

down the rabbit hole

and came up empty

no proof

misery

math is hard

so that's been my day so far

Leaky Nun

@MatheinBoulomenos wo kann ich es finden?

MatheinBoulomenos

Rotman homological algebra 9.1 has a detailed proof that $H^2(G,A)$ classifies extensions

Leaky Nun

du weiss viele bueche

kennst*

MatheinBoulomenos

but Rotman refers to some other books for the full proof of Schur-Zassenhaus

he only proves the case for an abelian normal subgroup

the reduction to that case can be found in the references that Rotman gives or in math.uconn.edu/~kconrad/blurbs/grouptheory/schurzass.pdf

Leaky Nun

it just had to be kconrad lol

Balarka Sen

23:40

You can just write down the proof that $H^2(G, A)$ classifies extensions

It's not hard

MatheinBoulomenos

yeah, true

the cocycle condition translates to associativity in the group that you construct

Balarka Sen

The relevant cocycle turns out to be the obvious measure of failure of a set-level section to be a homomorphism

It's pretty cool.

Leaky Nun

oh, Schur-Zassenhaus has nothing to do with Artin-Schreier

MatheinBoulomenos

no, just another cool application of group cohomology

Leaky Nun

what was group cohomology created for?

Balarka Sen

23:48

stop asking stupid questions and go actually study something

group cohomology would demystify if you spend some time thinking about it

Leaky Nun

fair enough

MatheinBoulomenos

I'm not sure about the history. Special cases were known before algebraic topology or homological algebra existed, they were used in group theory. At some point in the 20th century number theorists realized that group cohomology is really useful for their stuff. I don't know much about the time inbetween

Leaky Nun

Contact

For playing the game Contact, where one person tries to "defend"...

MatheinBoulomenos

oh I played that irl before, that's fun

Balarka Sen

there's a short history of cohomology theory here

section 2.5

MatheinBoulomenos

23:52

@Balarka thanks

Balarka Sen

honestly it feels like if you had one word to describe 20th century mathematics, "cohomology theory" is the best you'd get

that shit pops up everywhere

Daminark

Did someone say cohomology?

I have been summoned

@BalarkaSen what about 21st, if you're familiar?

Balarka Sen

i dunno m8

higher category theory :3

Leaky Nun

thanks @BalarkaSen

Balarka Sen

You should take that description of 20th century mathematics very facetiously though

Leaky Nun

23:55

I mean thanks for your link, of course

Contact

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$Mathematics$ Mathematics