Mathematics - 2017-06-29 (page 9 of 9)

Semiclassical

20:00

So there's some finite probability you never return, hence transient

GFauxPas

But if p=q, 0 is recurrent

Which is intuitive

Semiclassical

Same for all integers, in fact

GFauxPas

Sounds plausible

Semiclassical

Every integer is recurrent if the chances are equal and transient otherwise

GFauxPas

Ill buy it ;)

Semiclassical

20:02

Yeah

Secret

29

Q: If I publish under a pseudonym, can I still take credit for my work?

Is it possible to publish under a nom de plume (pen name, pseudonym)? ... and still take credit for the work? I may have a chance to publish with the professor I am working with soon and if I am on the list of authors I would like to use a nom de plume, but I would also like to use the publica...

So...

I have this plan in my far far future:

GFauxPas

We proved the $1-|p-q|$ i think but im not intetested in the proof now

Semiclassical

Yeah.

Tedious way is to write down the probability of returning with n steps

GFauxPas

So that would be ... an infinitely large stochastic matrix :o

Semiclassical

And then take the limit

Secret

20:05

Since it is currently scheduled that I will be catching up on abstract algebra and other topics related to the integral project in the future, and suppose I (or we if I had some other collaborators) got lucky and did find something worthwhile, obviously the first step to get acceptance in the mainstream is to publish it. However

Danu

Right, so a multiplicative sequence $K(1+x_1+x_2+\dots)=1+K_1(x_1)+K_2(x_1,x_2)+\dots$ is supposed to be determined by a power series via the relation $K(1+t)=f(t)$

Semiclassical

Sure, but the matrix itself will be simple

GFauxPas

We only did finite matrices in class

Semiclassical

It'll be p on the first subdiagonal and q on the first superdiagonal

Danu

In a mult. seq. you are allowed to plug in all things of the form $1+a_1+a_2+\dots\in \cal A^{\Pi}_1$ where $\mathcal A$ is a graded, commutative $R$-algebra with unit, $\cal A^\Pi$ are the formal power series in it and $\cal A^\Pi_1$ are those that start with $1$

So if you consider the special case $\mathcal A=R[t]$ where $t$ has degree one

Then plugging in $1+t$ makes sense

Mike Miller

20:08

ok

Danu

So what you get is $1+K_1(t)+K_2(t,0)+K_3(t,0,0)+\dots$

so it is kind of surprising at least to me that you can recover all of the multiplicative sequence (recall that multiplicative means that $K(ab)=K(a)K(b)$ for $a,b\in \mathcal A^\Pi_1$)

GFauxPas

And 0 everywhere else?

Danu

So now on to the way how to recover it

Mike Miller

sorry remind me are the K_i symmetric?

Danu

No

Mike Miller

20:09

No of course not my bad

Danu

The $K_i$ are homogeneous of degree $i$

Secret

there's a catch. (Unless I can find collaborators that are from the maths community), that will mean I will be literally trying to publish all by myself onto some suitable journal. While yes at such far ahead in the future I will definitely have an institution email address to refer to, since I am a chemist and never have actual pure maths training, if I post under my real name and got rejected, that might unecessarily affect the reputation of my institution.
What should be the best course of action? By that time, I might be 40 or 50 something, should I take formal maths degree in that fiel…

Chris Nguyen

Guys help on number 23 plz

Mike Miller

yes

Chris Nguyen

mualphatheta.org/useruploads/files/…

help on 23

Danu

20:09

if the elements $x_1,\dots,x_n$ has degrees $1,\dots,n$

Mike Miller

yeah I got it

Danu

OK

So how to recover it: Say we want to recover $K_j(x_1,\dots,x_j)$

Mike Miller

I see why you think this is the splitting principle now. Please go on

Danu

Take any $n\geq j$ and consider the $n$-fold product $f(t_1)\cdots f(t_n)$. It is symmetric in the indeterminates by construction, hence can be expressed in terms of polynomials of elementary symmetric polynomials. Arrange the expression order-by-order (the order is given by the sum of the exponents; all $t_i$ have order one). You obtain something like $f(t_1)\cdots f(t_n)=1+P_1(\sigma_1)+P_2(\sigma_1,\sigma_2)+\dots$ for polynomials $P_j$ which contain all the order-$j$ terms. In fact $K_j=P_j$

The independence of the choice of $n$ follows from some elementary symmetric polynomials shit, as does uniqueness of the $K_j$ recovered in this way

Mike Miller

I suspect there's a way to translate this entirely into bundles and line bundle language, yeah

GFauxPas

20:13

Yes ofc 0 everywhere else because the needle cant skip squares

:)

Danu

And they do satisfy the relation $K(1+t)=f(t)$ since $$K(1+t)=1+K_1(t)+K_2(t,0)+K_3(t,0,0)+\dots=1+K_1(\sigma_1(t))+K_2(\sigma_1(t),\‌sigma_2(t,0))+\dots=f(t)f(0)\cdots f(0)=f(t)$$

My chatjax is giving me an error on that $\sigma_2$ but it's correctly typed...

dafuq

Mike Miller

The algebraic component of the splitting principle is that the map BU(1)^n -> BU(n) is injective on cohomology, and I think what you're saying here is something like that the map from a product of a bunch of A's to A[[t]], or something like this. I don't quite understand the algebra well enough

Danu

So what exaclty is going on here... I'm so weirded out on this

GFauxPas

@Semiclassical if p=q the needle will visit EVERY number at least once almost surely?

Balarka Sen

@Danu It's a character limit thing

Danu

20:15

The problem is mostly that this symmetric polynomial stuff is essentially black magic to me

@BalarkaSen Oh...

Mike Miller

But i think you want to do it st the nth level for each n and get an injection of some relevant algebras

Danu

@MikeMiller I understand the splitting principle purely in terms of projectivized bundles and the Leray-Hirsch theorem.

Semiclassical

@GFauxPas Good question. I think so?

Mike Miller

The way you interpret the splitting principle is that char class formulas are true if determined on line bundles, and here you're saying formulas are true if they're true on line bundles

Krijn

@Mike I have finished The Crying of Lot 49 for quite some time now, still don' t get it

Danu

20:17

So is that really what I'm saying though? I don't quite understand how this $K(1+t)$ corresponds to taking a line bundle @MikeMiller

Mike Miller

you're saying the same thing algebraically, with a graded algebra replacing the graded "algebra" of vector bundles

Danu

I think this is the crucial ingredient that I'm missing

Mike Miller

Using the $t_i$ that are all degree 1

Corresponds to line bundles

Danu

In some way... but I don't understand exactly how

GFauxPas

Id think so, because each state is markov

Mike Miller

20:18

Line bundles are degree 1?

Semiclassical

Right.

Danu

Yeah, of course, I got that far :P

GFauxPas

No matter where i am, the probability of returning to that state ignores how i got there

Danu

But it's not so clear to me what EXACTLY K is supposed to model

Mike Miller

And you're saying things are determined by their values on monomials in degree 1 elements

Danu

20:19

it's not just "give me a vector bundle and I give you a class"

it's more like "give me some class and I give you a different class related to it"

so I don't see what exactly "line bundle" translates to

Semiclassical

It depends on this being 1D, though. I think it's false if you work in 3D?

Chris Nguyen

How do you post a link in latex

Mike Miller

I don't understand the question. I doubt I can help though.

Danu

@MikeMiller Maybe we can be very concrete

Chris Nguyen

I copied and pasted a link but the whole things is not in blue

GFauxPas

20:20

Well.were on a strip

Danu

Consider the multiplicative sequence given by formal inversion of a power series with leading term 1

$K(a)=a^{-1}$. What that does is take a Chern class, for instance, and give you the Chern class of a bundle such that the sum is trivial

GFauxPas

So 1d

Mike Miller

I do not know multiplicative sequences in a concrete helpful way

Which is why I can't help you

Semiclassical

Sure

GFauxPas

And whyd you skip.2d

Danu

20:20

kk

GFauxPas

?

Danu

Know anyone here who does?

Mike Miller

sorry

Ted would be the only one...

Danu

Oh btw, I got my rejection letter from Bonn's PhD program today :sadface:

(but I already knew that was coming since I'd talked to the guy I wanted to work with there and he'd said no)

Mike Miller

I'm sorry :(

GFauxPas

20:21

:(

Danu

Still sad

Andrew Thompson

Wait, @Danu, didn't you just enter your msc?

Semiclassical

From Wikipedia's page on random walks: "Will the person ever get back to the original starting point of the walk? This is the 2-dimensional equivalent of the level crossing problem discussed above. It turns out that the person almost surely will in a 2-dimensional random walk, but for 3 dimensions or higher, the probability of returning to the origin decreases as the number of dimensions increases. In 3 dimensions, the probability decreases to roughly 34%."

Andrew Thompson

Or am I getting old?

Danu

@AndrewThompson No.

This is the end of my third year

GFauxPas

20:23

I was doing 1d case

Andrew Thompson

Oh. IIRC you started with physics and changed to math, right?

Danu

But I'm sure that I still give the impression of being an undergrad :D

Yes, I switched about a year ago. Though I did some courses before that

Semiclassical

Sure, but that's why I skipped to 3D

Andrew Thompson

Ahh, then it makes sense. Which other places are you applying to?

Danu

Several others in Germany

Semiclassical

20:24

It's the first dimensionality for which 0 isn't recurrent when the probabilities are equal

Andrew Thompson

Good. Bonn is very highly competitive, I imagine.

Danu

I guess it's the most prestigious place in Germany.

GFauxPas

Whats intetesting in the p=q case is that the rv "how many steps until you come back" has infinite mean

Andrew Thompson

What was your thesis on?

Danu

It's not done yet, but it's on complex geometry (but rather the differential-geometric side of it)

Andrew Thompson

20:29

Like a proper ex-physicist :)

Danu

Maybe? I'd say it's lacking in the differential-equations side of things.

For a physicist, that is.

Andrew Thompson

I wouldn't know. My DG-friends are all afraid that the world will slowly turn them into physicists.

Danu

Really? Lol

They must have no idea about what physicists do

Sha Vuklia

@Ted uhm, if $(aN)(bN)$ can be considered as $\{ahbh'\mid h,h'\in N\}$, then it is easy to show

Andrew Thompson

They don't (and neither do I.)

Danu

20:31

Unless they're into this global analysis kind of stuff. That's pretty close to physics (all calculations in local coordinates...)

Mike Miller

I have no idea what physicists do.

Secret

Any sufficiently mathematicians who have published paper experience before here?

Mike Miller

Just ask your question. Nobody's going to volunteer themselves in response to that.

Secret

I already asked my question at least 20 something posts above, (but I will repost it if necessary)

Astyx

@Akiva It's still 70 + 21 though

Also hi/rehi chat

Secret

20:37

REPOSTED:
I am pondering about a plan in the far far future:
Since it is currently scheduled that I will be catching up on abstract algebra and other topics related to the integral project in the future, and suppose I (or we if I had some other collaborators) got lucky and did find something worthwhile, obviously the first step to get acceptance in the mainstream is to publish it. However there's a catch. (Unless I can find collaborators that are from the maths community), that will mean I will be literally trying to publish all by myself onto some suitable journal. While yes at such far ah…

Krijn

@Danu Lol' d

Zee

@Secret some journals have anonymous submittions

You get peer reviewed incognito I mean

Secret

hmm ok, didn't knew that

Danu

@Krijn :)))

Zee

@Secret am no mathematician but my advice is to focus on getting concrete results and all else is trivial in some sense, this isn't philosophy after all

Danu

20:42

...or should I say :((((( hahaha

@Zee define "concrete results", else this is philosophy after all

Zee

Something that a mathematcian can use in his research

Krijn

@Danu I have the same feeling

After 5 years of studying mathematics, I realised that I have almost no knowledge of mathematics

Just the trivial thingies

Secret

Hey @Ted, have a very academia related question for you (see above)

Ted Shifrin

Hi @Danu

Secret

@Zee I think for me, the hardest thing is not getting the results (positive or no goes) but how to present them coherently and that the proofs make sense

Faust7

20:47

morning @TedShifrin

Ted Shifrin

@Secret: I can't really give solid advice. It really depends whether the "something" is of interest and worth publishing. Most refereeing in math is not done blind, which means that a referee would know your name and your (lack of) academic position, but that doesn't mean that it wouldn't be accepted with the right content and aim at academic journal.

Hi @Faust7.

Faust7

@Krijn i know how u feel

Danu

Welcome back, Ted

Ted Shifrin

Thanks, Danu.

Secret

I see, I will keep that in mind, thanks guys

Danu

20:49

I hope the trip back wasn't too bad

...cause I GOT SOME QUESTIONS FOR YOU

Ted Shifrin

I'm still pretty brain-dead after the 24-hour day of travel.

Danu

It's mostly philosophy

and about some of your favorite topics

Krijn

Nice philosophy or eww philosophy

Astyx

Philosophy is best served when brain-dead

Zee

lol

Ted Shifrin

20:50

@Astyx: When the server is brain-dead or when the recipient is?

Danu

So I'm looking to give an explanation of both the Chern-Weil set-up for characteristic classes and the one through multiplicative sequences

Astyx

Both, for optimal results

Danu

Both give ways to define characteristic classes, but in a kind of different way

Ted Shifrin

To me, Chern-Weil is geometric and explicit. The other stuff is formal algebra.

Danu

I just discussed this with Mike a bit

Exactly, the multiplicative stuff is pretty algebraic

but it's still somehow really close to the other things

Ted Shifrin

20:52

Well, it's all pretty algebraic (invariant theory underlies Chern-Weil). But I stress "formal."

Danu

1) Where are the invariant polynomials in the multiplicative sequence approach? Are they hidden somewhere? The invariance of the polynomials seems crucial in the Chern-Weil approach

Sergey Dylda

Hello! Anyone familliar with model theory? Need some help with notions.

Ted Shifrin

I suppose they're buried in the (invariant) cohomology of the classifying spaces.

Danu

Do you somehow always construct things to correspond to invariant polynomials in the mult. seq. approach? Where is this hidden?

I don't really think about classifying spaces for any of this, I must admit

Ted Shifrin

I haven't thought about multiplicative sequence stuff since grad school.

I no longer have Hirzebruch's book to refer to, either.

Danu

20:54

2) What exactly is the correspondence between the splitting principle and the way one reconstructs a mult. seq. from its characteristic power series?

I just explained the latter process to Mike in case you wanna take a look

49 mins ago, by Danu

Right, so a multiplicative sequence $K(1+x_1+x_2+\dots)=1+K_1(x_1)+K_2(x_1,x_2)+\dots$ is supposed to be determined by a power series via the relation $K(1+t)=f(t)$

But I can't quite pin down what's going on in the background; it might be hidden in the black magic that is elementary symmetric polynomials

Ted Shifrin

Yeah, the orthogonal and/or unitary invariant theory is buried in those polynomials, I'm pretty sure.

Danu

So did Hirzebruch come up with both of these things at exactly the same time?

I mean the mult. seq.'s and the splitting principle

It sounds to me like they have to somehow be exactly the same thing, viewed in two ways

Do you know the history?

Ted Shifrin

Not any more, I don't.

Danu

Are there any books where this is recorded, maybe? Any sources?

Ted Shifrin

Does Hirzebruch not give some history in the preface of his book?

Spivak may have some in Volume 5, too.

Danu

21:00

Maybe... I haven't looked. Thanks

Krijn

I could really use an algebraic geometer at the moment

Ted Shifrin

@Sha: I don't know if you settled your question or not. But here's a very brief argument for what you wanted. From $(aH)(bH) = (ab)H$ we multiply by $a^{-1}$ and get $HbH=bH$, so $b^{-1}HbH = H$, which says that $b^{-1}Hb = H$.

Krijn

Although, maybe I should be able to solve it myself

Probably not though

Ted Shifrin

@Krijn: I suspect your question will be too algebraic for me.

Krijn

@TedShifrin Probably just a tiny bit

You know that if you have an effective divisor $D$ of degree > 2g on a Riemann surface, then it is base point free

So by Riemann Roch you can find a map with fiber equal to $D$ (above say 0)

Ted Shifrin

21:04

Oh, this is the kind of stuff I used to love. Go on.

Krijn

But a Riemann surface is just a nice curve over $\mathbb{C}$

Danu

Sounds nice to me, too :D

Krijn

And I want to do the same for a curve over $\mathbb{F}_q$

Danu

EWWWWW

hahahaha

Ted Shifrin

LOL ... oy.

Krijn

21:04

But then the usual argument doesn't hold anymore :(

Stupid vector space over finite fields are stupid

Danu

Is it me, or is Hirzebruch's original proof of the signature theorem 10x nicer than the "corollary of Atiyah-Singer" approach? I have started hating spin bundles :(

Balarka Sen

I absolutely love how Krijn kept you two interested long enough to deliver the punchline/final question.

Ted Shifrin

Yes, @Balarka, you'd think he had practiced that skill.

Why, @Krijn, are vector spaces stupid in the finite case?

Krijn

@TedShifrin Because subspaces of lower dimension can still cover it

Ted Shifrin

Hmm ... Can you expound/expand on that a bit?

Krijn

21:08

As $D$ is base point free, you get that $L(D)$ has dim $L(D - P)$ + 1 for every point $P$ in the divisor

Ted Shifrin

Yes.

Krijn

So in the $\mathbb{C}$ case, this is enough to say, well, there must be some $f$ with fiber $D$

Not so much in the $\mathbb{F}_q$ case

Ted Shifrin

So I'm trying to think through this argument.

Daminark

Hey @Ted! Long time no see!

Ted Shifrin

Heya Demonark.

Daminark

21:12

Also hey @Balarka, @Danu, @Krijn

Krijn

Heya

Daminark

How's it going?

Balarka Sen

Hi @Daminark

Daminark

@Balarka Peter May's lecture is gonna start soon :P

Ted Shifrin

So what'd you learn in the month of June, Demonark?

Balarka Sen

21:13

Oh right. @Ted: Peter May's algebraic topology lectures which Daminark is attending defines $H^n(X; G) = [X, K(G, n)]$.

Let us wait for reaction.

Ted Shifrin

LOL

I'm getting ready to write off Demonark as a hopeless case, anyhow.

Daminark

Some number theory, in particular it's been working up to Dirichlet. There's the main atop which has been more difftop-y, though we did talk about cobordism very briefly which is new.

Then there's Peter May's atop, which as Balarka said, is doing things in a very specific way

Krijn

Yay for number theory

Ted Shifrin

Yes, the formal, algebraic world view. I'm acquainted with lunatics who think like that.

Balarka Sen

I am picking up number theory from Daminark

Daminark

21:16

He's never even taught like this before, it's an experiment to see if starting with cohomology and Eilenberg-Steenrod axioms work

Balarka Sen

Though I haven't caught up with Fourier analysis over finite groups yet. I will, this weekend.

Krijn

If anybody wants to talk some number theory; I need the distraction for a moment

Balarka Sen

@Ted: They prove $H^n(\Sigma X; G) = H^{n+1}(X; G)$ from $K(G, n) = \Omega K(G, n + 1)$.

Daminark

Supposedly this is the modern way of thinking about it shrug. Amusing nonetheless

Ted Shifrin

A bunch of guys at UGA do combinatorial number theory stuff with Fourier analysis, @Balarka. Some cool stuff by Magyar, Lyall, and their students.

Daminark

21:17

@Krijn we've got a number theory study group where we're working through Weil's NT for beginners

Ted Shifrin

Demonark: I once had a review of an NSF proposal tell me I was doing old-fashioned Italian-style algebraic geometry. They meant that as a slap. I find it mostly a compliment. Suffice to say, they really had no idea what we were doing.

Balarka Sen

Oh, cool stuff.

Daminark

@BalarkaSen I'll try to explain it better this weekend now that I actually got completely straight

Krijn

Daminark

@Ted lmao, some people I guess are not fond of classical stuff

Balarka Sen

21:20

I actually adore the identity; my slogan for this is "cohomology is dual to homotopy". I am put slightly lopsided that May is taking this as a definition. It is amusing.

Danu

@BalarkaSen Hurr durr cohomology is dual to homology :D

Balarka Sen

No, that's the catch. That's the standard duality.

H^n(X; G) = [X, K(G, n)] is the thing I am talking about.

Danu

I know, I was joking

Balarka Sen

Ah, Ok

Danu

Also because I don't know what that second thing means because I don't know homotopy theory

Daminark

21:21

I think I kinda remember Peter May expressing disMay at homology being introduced via chains or whatever. He sees coomology and E-S as being more natural, and generalizing to other cohomology theories, which is what he wants to get into. Also mentioned something about K-theory

Ted Shifrin

Yeah, people like me think about cohomology from a totally different perspective.

Balarka Sen

Yeah K-theory is an example of a generalized cohomology theory.

Simply Beautiful Art

Let $p_n$ be the $n$th prime. Can the following limit be computed?
$$\lim_{N\to\infty}\dfrac{\sum\limits_{3<p_n<N}(p_n+1\mod3)}{2\sum\limits_{3<p_n<N}(p_n-1\mod3)}$$where $0\le a\mod b<b$.

Daminark

His tentative approach to homology is that cohomology is dual to something, therefore homology exists

Balarka Sen

I think it's represented by, what, BSO?

Ted Shifrin

21:22

@Hippa: Salut.

Hippalectryon

@TedShifrin o/

Ted Shifrin

First time I've seen you since our lunch.

Hippalectryon

I was here some hours ago

Balarka Sen

@Danu Based homotopy classes of maps from X to K(G, n), where K(G, n) means space (unique upto homotopy equivalence - this is a fact) with pi_n = G and pi_i = 0 for every other i.

Ted Shifrin

But I wasn't?

Hippalectryon

21:23

I mean, we were both here a couple of times before, but I didn't speak up because I was busy

Simply Beautiful Art

Basically, it translates into "are more prime numbers equal to $2$ or $1$ mod 3?"

Ted Shifrin

Oh.

Hippalectryon

So, how are you ?

Akiva Weinberger

@SimplyBeautifulArt So, twice the number of $1~\rm mod~3$ ones divided by the number of $2~\rm mod~3$ ones?

Ted Shifrin

Finally made it home the night before last after a (literal) 24-hour travel day.

Croatia is stunning.

Akiva Weinberger

21:24

Oh, there's a $2$ on the bottom.

Simply Beautiful Art

Yeah

Akiva Weinberger

I'm sure the calculation exists.

Daminark

Nice

Simply Beautiful Art

Me neither, but checking up to $N=100$, it looks like 1

Akiva Weinberger

It probably is, I think

One

Daminark

21:28

Oh @Balarka apparently the probability talks were actually nifty, Araske did go to those and gave me a rundown, Lawler's style is apparently very good

Krijn

I have known things like this, @Simply, but I forgot it I think

Balarka Sen

@Daminark You should take all the REU + bootcamp

literally everything

Simply Beautiful Art

Oh really? Interesting

Krijn

Or at least closely related stuff

I'm going to google a bit

Simply Beautiful Art

:| Well, I honestly wouldn't know if the limit even exists. :-)

Daminark

21:30

He sorta introduces a problem and builds theory out of it, which I appreciate. Probability that a random permutation has a fixed point, random walks on the lattice, paradoxes, etc

Balarka Sen

Cool stuff

Danu

@TedShifrin @MikeMiller It seems that section 11 of Lawson & Michelsohn is close to what I'm looking for... I think.

Krijn

@Simply, I think the ratio is 1

Because both sets have Dirichlet density 1/2

Although that density is not the same as yours

Faust7

BLARG

Simply Beautiful Art

@Faust7 ?

Faust7

21:35

@SimplyBeautifulArt ?

Simply Beautiful Art

._.

Faust7

^^ hows ur day going

Krijn

Faust7

anyone remeber how to solve an ode by seperation of variables?

Simply Beautiful Art

Hm, okay @Krijn

Akiva Weinberger

21:38

What's Dirichlet density? The density of them compared to primes in general? @Krijn

…What's $m$?

Krijn

@AkivaWeinberger Almost

$m$ should be $N$

Akiva Weinberger

Oh, OK.

Krijn

It's quite a long thing to TeX, but wiki gives it quickly

Akiva Weinberger

So they both have Dirichlet density $\frac12$.

Krijn

Yes

But I know there's a better argument, and I'm still looking for that

Akiva Weinberger

21:42

Oh, OK, Wiki says that if they have a natural density, they have a Dirichlet density equal to it.

So, if Semi's limit exists, it will be $\frac{1/2}{1/2}=1$.

Wiki suggests that they do have a natural density as well

Krijn

No, Wiki solves it?

Akiva Weinberger

So it should equal $1$, then.

Krijn

If a subset of primes A has a natural density, given by the limit of

(number of elements of A less than N)/(number of primes less than N)

then it also has a Dirichlet density, and the two densities are the same.

The two densities are the same

Akiva Weinberger

@Krijn If it has a natural density, though.

But it seems to suggest that it has a natural density a little further down.

Krijn

Oh, sorry yes

Akiva Weinberger

21:45

> For example, in proving Dirichlet's theorem on arithmetic progressions, it is easy to show that the Dirichlet density of primes in an arithmetic progression $a + nb$ (for $a$, $b$ coprime) has Dirichlet density $1/\phi(b)$, which is enough to show that there are an infinite number of such primes, but harder to show that this is the natural density.

Krijn

This is clearly one of the cases of "in practice"

Where clearly means "Otherwise I have no clue"

Akiva Weinberger

> In practice, if some "naturally occurring" set of primes has a Dirichlet density, then it also has a natural density, but it is possible to find artificial counterexamples: for example, the set of primes whose first decimal digit is $1$ has no natural density, but has Dirichlet density $\log(2)/\log(10)$.

Krijn

According to MO: See Lang's Algebraic Number Theory, Ch. VIII.4 and XV

Sooooo, of to the gen lib rus ex

Akiva Weinberger

The point is, the answer to @Semi's question is $1$.

Krijn

Yes, but we are in some nice algebraic number theory now

So let's forget that question

Akiva Weinberger

21:51

Analytic number theory

Krijn

Hmm, this is really close to both sides though

I like this about number theory, almost all of these easy-to-ask question are incredibly hard to answer and often need a lot of (beautiful) theory to be answered

dms.umontreal.ca/~andrew/PDF/PrimeRace.pdf sums it up quite nicely

Semiclassical

How did I show up in this? :)

Krijn

@Akiva confused you and Simple, I think

Semiclassical

Ahhh

Krijn

Lol, the first value where there are more primes 1 mod 3 than 2 mod 3

608,981,813,029.

Akiva Weinberger

22:05

I do mean @Simply. Sorry

Simply Beautiful Art

:P

Sha Vuklia

@Ted oh wow, okay, I had no idea we were allowed to show the elements like that. I will have to look at it carefully tomorrow. but thanks!

Also, guys: Let $G$ be a group and $N\subset G$ a normal subgroup. My book says that if $G/N$ is abelian, then $[G,G]\subset N$. (And the other way around). The proof is quite simple, but I’m a little bit confused because we all know that $N=\{e\}$ is also normal, but that would imply that $[G,G]=\{e\}$. Should they have said that $N\neq\{e\}$?

$[G,G]$ is the commutator subgroup btw

Krijn

No, only if $G/\{e\}$ is abelian

Which implies $[G,G]$ is trivial of course

Sha Vuklia

oh right

I forgot about the factor group

thanks

Ted Shifrin

22:24

@Sha: It's probably best to work with elements rather than sets, but that should give you the idea.

Faust7

23:28

i got a stupid question

Chris Nguyen

hi

wow my question got downvoted because I didn't post a question but a link

so dumb

Faust7

the area of a disc is $ \pi r^2 $ when i thinking of the volume of a sphere i want to think $ \int_{-r}^{r} \pi r^2 dr $ where is my missing 2?

follow by the question wth is it that im integrating

actually i think i know what im integrating two cones of height r

but why .>

Josh Wonser

What missing 2? Integrating from r to -r gives you everything

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$Mathematics$ Mathematics