@robjohn Back from the park?

hey all, quick question regarding math.stackexchange.com/questions/1055942/…

I get that I should probably rearrange to form a simultaneous equation, then substitute and solve for one term then the other, but I can't quite work out how

@teadawg1337 yep

normally it'd be easy, but none of the individual terms seem conducive to being used in a simultaneous equation

@robjohn How was it?

I'm rusty at this :(

@SwapnilTripathi OK

@PedroTamaroff I guess the correct way to do it would be this : $H$ is normal in $G$, so $gHg^{-1} = H$ for all $g \in G$. Thus, $gPg^{-1}$ is a subgroup of $H$. In particular it's a Sylow $p$ subgroup of $H$ by counting order. Then by Sylow's 2nd theorem, there is some $h \in H$ such that $hPh^{-1} = gPg^{-1}$. Thus, $(h^{-1}g)P(h^{-1}g)^{-1} = P$. Thus, $h^{-1}g \in N_G(P)$. Thus, $g \in H \cdot N_G(P)$

If G acts on a set X, and H is a subgroup which also acts transitively on it, then G=HStab(x) for any x in X.

(same argument)

@PedroTamaroff Interesting. What happens if $G = Gal(L/K)$ acts on the field $L$ by galois automorphisms and some subgroup $H$ of $G$ acts transitively on $L$?

@anon Jinx

what do you mean by acts transitively on L?

surely not the normal meaning of the word...

I dunno. H acts on L by Galois auts of course.

Hmm.

I guess fixed point freely on the basis of L as a K-vector space?

"the" basis?

I know, I know, it doesn't make sense.

I also take it you're speaking of infinite Galois extensions?

nah

well then, are you familiar with the normal basis theorem?

by that you mean simple extension theorem?

there cannot be any basis for L/K on which a proper subgroup of G(L/K) acts transitively

oh.

hmm.

let me think.

@BalarkaSen no, not the primitive element theorem, the normal basis theorem

What is $\Pi(S^n)$, where $\Pi(...)$ is the fundamental groupoid?

@Alyosha $\pi_1$?

It's trivial if n > 1

Oh groupoid

groupoid, not group

It's still trivial for n > 1

@BalarkaSen i.gyazo.com/319e367b1f50230124248e13772d2eac.png.

You mean it's the 1 object, 1 morphism category?

What object? What category? It's the trivial groupoid.

The 1 object in the cat of groupoids, yes, if you mean that.

Alyosha is asking what you mean by trivial groupoid

Yes.

since Pi(S^n) certainly has an infinite number of objects

@anon But all of them can be homotoped to e_x_0 no?

That was my confusion.

Oh no basepoint

Yes.

Darn it. I never worked with groupoids.

So it's nontrivial.

nLab has a convoluted definition of trivial groupoid.

well, any two paths between chosen points a and b are homotopic I think, so

@BalarkaSen I think the answer is almost 'the trivial groupoid'.

you can just define the trivial groupoid to be the groupoids of the isomorphism types of $\Pi(\Bbb R)$ and be done :P

It seems the definition of the trivial groupoid is a cat with objects in $X\times X$, with $\text{Mor}(x,y)$ having precisely one morphism for each $x,y\in X$.

@Alyosha if that's true, then this is a trivial groupoid

But $S^2 \ne I \times I$, surely?

actually I don't understand your $X\times X$ thing. what's $X$?

ncatlab.org/nlab/show/groupoid

i dont even understand the definition

See example $5$ at the bottom.

@robjohn I'm one step away from getting the closed form of $$\sum_{n=1}^{\infty} \frac{H_{4n-3}}{(4n-3)^3}$$

@Alyosha where did it say the objects were in $X\times X$?

presumably $X$ is the object set, and $X^2$ collects all of the morphisms together

Oh right, so $X=S^2$ I believe.

See example $4$.

It's the groupoid $\{0, 1\}^2$

It is a setoid (I don't know what that is, maybe the discrete cat on $X$).

it defines setoid pretty clearly

basically, you have an object set X, and exactly one morphism between any two objects x and y, which you can denote (x,y), in which case X times X collects all of the morphisms together

$\Pi_1(\Bbb R)$ just consists of infinitely many equiv classes of points, non?

That's what $\Pi_1(S^n)$ consists of for $n > 1$

we're treating groupoids as categories

then I'm out

in that sense, what does "consists of infinitely many equiv classes of points" mean

i can't possibly abstract nonsense everything

also, for the fundamental groupoid, you are modding paths by homotopy, distinct points remain distinct

Sure, but then doesn't $\Pi_1(\Bbb R)$ then consist of $[e_{x_i}]$s for all $x_i \in \Bbb R$?

such abstract nonsense

the objects of $\Pi_1(\Bbb R)$ are just real numbers. between any two real numbers there is a unique morphism.

@anon that's what i mean

objects of $\Pi_1(S^2)$ then should just consist of points in $S^2$

@MikeMiller what $\Pi_1(S^n)$ is.

@Alyosha the definition of $X\times X$ is the collection of pairs $(x,y)$, yes. but they are just using $X\times X$ to refer to the groupoid.

X \times X is just the base set of the morphisms, i think.

@MikeMiller @anon Is there a name for $\Pi(S^n)$? As far as I can see it is (for $n \ge 2$) the cat with objects elements of $S^n$ and with the hom classes having cardinality $1$ (and for $n=1$ having countably infinite cardinality).

Why isn't $\Pi(S^n)$ a perfectly good name?

LOL

the trivial groupoid on $S^n$

It's just groupoid isomorphic to $\Pi_1(\Bbb R)$ if i am not wrong

eww, only by virtue of R and S^n having the same cardinality. again eww.

well yeah.

why are you ewwing?

it's precisely the way one should identify the trivial objects in a category

because those isomorphisms are ugly. there is no reason to be thinking about them.

OK, thanks.

I may return with more, similarly naive, questions.

glad to be unhelpful :)

if it's so much ugly even to identify where the identify the trivial objects in the category of groupoids mapped by the functor $X \to \Pi(X)$, then I care about $\Pi$ as much as I care about a table.

throws $\Pi$ out of sight

Pi(X) is the collection of all homotopy classes of paths between points. You can concatenate paths if the endpoint of the first is the startpoint of the other. Don't let the formal abstraction intimidate you.

I know what Pi is

I just don't care about a groupoid

It's just so confus

as you wish

I'm alright about thinking of garbled up topological spaces but NOT alright about thinking of garbled up topological invariants.

@anon speaking of which, have you seen my approach of studying profinite groups by solenoids and topologically garbled up continuums?

hrm? no. sounds interesting, although if your previous sentence's use of the word "garbled" is any premonition, maybe I should be wary...

how does it go?

" math.stackexchange.com/questions/1055692/cyclotomic-polynomial/…; @BalarkaSen

@anon Consider a bunch of $S^1$s. Let the maps be $S^1 \to S^1$ as $x \to x^p$.

This forms an inverse system.

lets see, solenoids are inverse limits of S^1 in the cat of topological spaces right?

Hello @anon :)

@anon Yes.

@SwapnilTripathi hello

So I suspected a connection with $\Bbb Z_p$

it's because the cayley graphs of $\Bbb Z/p^i\Bbb Z$ are really circles with $p^i$ points identified, which coarsely looks like a $S^1$

so the inverse limit of these cayley graphs of 'Z/p^i's, in the category of geodesic metric spaces, looks like the solenoid

turns out that there is a connection : it's the fiber product of R and Z_p.

explicitly, $\lim \, S^1$ is isomorphic to $\Bbb R \times \mathbf{Z}_p/\Bbb Z$

@BalarkaSen what do you mean by the p^i points identified?

interesting

@anon wrong use of words. i mean a circle with p^i-th roots of unities marked. (the nodes of the graph)

yeah

now, Question : what's the fundamental group of the solenoid $X$?

turns out X is not path connected

but, $X$ can be realized as the fiber product of Z_p and S^1

looping once around S^1 takes the lifted path (intuitively) to one leaf of product to another

So another question : can there be a notion of monodromies for these spaces?

if so, is the monodromy $\Bbb Z_p$?

I am having trouble visualizing $\varprojlim S^1$ as $\Bbb R\times{\Bbb Z}_p/\Bbb Z$ (I assume that is a product in the cat of top spaces). What are the projection maps to the $i$th copy of $S^1$ in the system?

it's not that, @anon. it's a fiber bundle over $S^1$ but it's not a trivial one.

@anon $X$ is the inverse limit ... -> R/p^2Z --> R/pZ--> R/Z

@BalarkaSen yes

@MikeMiller ah

Now there is an embedded copy of Z/p^iZ inside each copy of R/p^iZ

@MikeMiller it is R \times Z_p/Z

@BalarkaSen yes

@anon So by uniqueness of inverse limit, there is a morphism $\mathbf{Z}_p \to X$

yes

On the other hand, those R/p^i s are sitting inside R

so there is a morphism $\Bbb R \to X$

R/p^iZ sits inside R?

@anon I mean there is a natural morphism

I sit inside my bed.

@BalarkaSen there are morphisms R->R/p^iZ you mean?

right

so yeah, R->X by universal property I guess

so there is a morphism $\Bbb R \to X$ by uniquenes of inverse limits

so take the pullback of $\Bbb R \to X \leftarrow \mathbf{Z}_p$

@anon HERRO STRANGE HOOMAN.

herro

Wonders if he'll ever know the initial of anon's name.

That is $\Bbb R \times_{\Bbb Z} \mathbf{Z}_p$, i.e., $\Bbb R \times \mathbf{Z}_p/\Bbb Z$

@PedroTamaroff I know anon's first name.

It's Jasper.

Everyone knows.

@WillHunting I want to learn it from him.

Or her.

he's definitely a guy

@PedroTamaroff I also know the first letter of his last name.

@anon I told him that.

@anon So there is a morphism $\Bbb R \times \mathbf{Z}_p/\Bbb Z \to X$. I haven't proved but I am totally sure this is an isomorphism

@BalarkaSen Then don't act like you have. Tsk tsk tsk...

@PedroTamaroff well you're always here to blow my (unbranched) covers.

It's 24 days to the new year.

@pedro I am very troubled by the problems of this world.

However, @anon, I am interested to make the monodromy idea work.

@WillHunting Yes, me too.

Did you know lipstick is made with fish?

delicious

It'd be nice if we had something like $\pi_1(X/\mathbf{Z}_p) \cong \mathbf{Z}_p$-ish.

@PedroTamaroff Next year, I will try to solve some problems of my country. I hope that in doing so, I do not get arrested, jailed or fined.

@BalarkaSen The definition of grupoid is pretty simple, however. It is a category where every arrow is invertible.

@PedroTamaroff sheesh

again with the general nonsense

@WillHunting If you don't get arrested, jailed or fined, maybe the change is small.

@PedroTamaroff I don't mind a warning letter from the police though. That is OK for me.

But don't go around killing people or something.

@anon The current idea is to use Cech methods.

Cech fundamental groups, Cech lifting or something of that sort. I am trying to formalize them to make us of them in the solenoid.

Bah... another downvote.

I'll use this method to study Gal(\bar Q/Q) and thus rule over the whole world [evil laughter]

@pedro I will go to sleep. See you in my dreams.

@BalarkaSen Rule over this one too :-) [evil laughter] $$H_1+\frac{H_5}{5^3}+\frac{H_9}{9^3}+\frac{H_{13}}{13^3}+\cdots$$

I have already told you that I am not interested in any way whatsoever about series/integrals/limits @Chris'ssis

@BalarkaSen Did you say that? I don't remember exactly the terms.

@Chris'ssis That's easy.

@Chris'ssis chat.stackexchange.com/transcript/36?m=18843069#18843069

@PedroTamaroff To be honest with you, not sure if anyone on MSE can do it (there might be some exceptions though :D).

@BalarkaSen Yeah, you said some.

Since you are in MSE, that means you cannot do it either ;)

@Chris'ssis It's trivial.

@PedroTamaroff You never know until you try it.

@Chris'ssis I just did. I solved it. In my head. While watching a movie.

LEL

I didn't even need pen and paper.

@PedroTamaroff lol, OK

srsly just integrate an appropriate function over an appropriate countour

but you must do it in real methods @Pedro

@PedroTamaroff show me the whole solution. I let you think of it till the end of the year. @BalarkaSen might help you, he says he's an expert in complex analysis.

I @Chris'ssis?

I never said I am an expert in anything.

@Chris'ssis Its just a series Chris. It is not relevant.

@BalarkaSen Don't be shy. We know you're an expert.

Pedro knows waay more analysis and algebra than me.

@PedroTamaroff It's not just a series. You might say it's just bla bla about almost anything you meet in mathematics.

@Chris'ssis Not really.

I said it is not relevant. That's my point here.

Why are you guys even trying to get into a debate? She gave me a challenge problem, I said I don't want it, and done.

Let's not make a mess of this chatroom guys seriously.

@PedroTamaroff Relevant for what?

@Chris'ssis For anything.

I'm outta here

@PedroTamaroff Well, that doesn't make sense.

Other than flippin' your meat to others about how you can compute it.

@Pedro flipping meet only gives you a join.

:18946434

Los Angeles Police Department chopper? Seriously?

@Chris'ssis That's also trivial.

@PedroTamaroff "It is amusing since, "essentially", the sum is the energy of a photon gas with discrete wave numbers. I guess @Fabian answer exploits this fact. – Felix Marin"

@BalarkaSen I'm sorry if you don't know what "the helicopter" is.

@PedroTamaroff I don't see why it's relevant :P

@PedroTamaroff Here is a thing you find for the first time from me: I use integrals and series for tackling a very famous math problem.

@Chris'ssis OK?

@BalarkaSen Exactly.

confus

@PedroTamaroff You do mathematics, I mainly attend an art.

@Chris'ssis "...tackling a very famous math problem."

What is this famous problem?

@Chris'ssis Everyone that does art calls it art. That's meaningless.

@PedroTamaroff I'll tell you more if one day I'll succeed.

@Chris'ssis Are you afraid you wont?

Are you trying the Riemann Hypothesis?

>

That's the only famous open problem I know of that has an interpretation using integral and series

@Pedro RH can be interpreted using integrals though, to our dismay.

:P

@BalarkaSen I am not an anti-integrals guy.

I don't know why you have this impression.

Integral (?) theory is fantastic.

integration theory?

oh?

But there's a big difference between the theory of integrals and computing stuff.

yeah, sure

so you should be an anti-closed form guy

What Pedro is trying to probably say that integration theory has problems much more beyond evaluating integrals, whether numerical or closed form

Calculus is more than computing closed forms, sure

But RH can be interpreted in the here-is-an-integral-with-this-conjectured-closed-form way, @Pedro

Who here is a pro at trig substitution?

@Chris'ssis I got that "Inside Interesting Integral" book as a pdf online :D

@Hippalectryon Do you like it?

@Chris'ssis Hello, i heard through the grapevine that ur a pro at integration

@Chris'ssis No idea I haven't started reading it yet :) but the $\displaystyle\int\dfrac{d\text{ cabin}}{\text{cabin}}$ was funny

@EricLawson Hi :-)

@Hippalectryon hehe, it's good to be read. These days I read Ramanujan more (trying to develop some complex series).

:)

@EricLawson That's fine.

@Chris'ssis Just keep me informed :)

int (1/2) sec^3(x) du

@EricLawson Why don't you try the integration by parts?

@Chris'ssis yea maybe i could do that

$$\int_{0}^{\infty}\frac{(1-12t^2)}{(1+4t^2)^3}\int_{1/2}^{\infty}\log|\zeta(\si‌​gma +it)| d\sigma dt$$ $$=\frac{\pi(3-\gamma)}{32}$$

@Pedro that is equivalent to RH, for example

@EricLawson My way is very fast and clever at the same time. Just think a bit of what I told you.

@EricLawson Put '$' between the latex chunks :)

@EricLawson Use the integration by parts as shown below $$\int x' \sqrt{1+4x^2} dx$$ Write down what you get.

@EricLawson Perfect. Now note that you may write things as follows

$$\int \sqrt{1+4x^2} dx = x \sqrt{4x^2 + 1} - \int \frac{(1+4x^2)-1}{\sqrt{1+4x^2}} dx$$

@EricLawson split the last integral and you get one like the integral in the left side but with the changed sign. Use this!

haha that should be so obvious to me

@EricLawson :-)))

@Chris'ssis Oh, did you write down the proof for $\arctan^3(\sin(x))$ ?

@Hippalectryon No, I attended other kind of stuff. There is much to write there (not hard).

indeed

An annoying amount of tedious work.

@Hippalectryon Wait ...

@Hippalectryon I also know the cubic version.

And the generalization ?

@Hippalectryon It cannot be done for some reasons (at least not with my actual knowledge). I need to study some specific series for getting there.

right?

How do you keep track of all those you have already done ? How do you make sure you aren't solving something you have already solved ? @Chris'ssis

@Hippalectryon I keep in mind everything.

@EricLawson it's not just $\sin(x)$ but $\sinh(x)$. Yeah, sure, you can do it in more ways.

@Chris'ssis I have no idea where sinh(x) is coming from

From Romania :3

@EricLawson Can you explain in details how you got that integral above? $1/2\int \sin^2(u) du$

@EricLawson you have $$\int \sqrt{1+4x^2} dx$$

This is integration by parts

(Note that $\ln x$ is positive iff $x>1$, which explains why I wrote stuff like $\ln\frac43$ rather than $\ln\frac13$.)

Uh, yes. Fixing...

Fixed.

Also, I'm sorry... I actually know the answer to this. This is more like a challenge from me to you...

@EricLawson You have some mistakes above ... Note that $\sin^2(x)+\cos^2(x)=1$

derp

okay

@EricLawson and $\displaystyle \operatorname{cosh}^2(x)-\operatorname{sinh}^2(x)=1$

Hint (for my thing): $e^x\ge x+1$ for all $x$, with equality iff $x=0$. (Graph it to see why this is. It's only true for $e$, by the way—"$2^x\ge x+1$ for all $x$" is false.)

Ah okay

Thats why its sinh

@EricLawson Indeed.

Thank you, i think i get it now

@EricLawson Now you have 2 solutions.

so its legit same steps just with sinh

@anon Have you worked through Matsumura before? I've come upon some seemingly contradictory statements.

nope

darn

worth a shot

@Hippalectryon Mangoldt.

@PedroTamaroff Yeah thanks :) just when you posted that it came back to my mind

@Hippalectryon You'd be amazed to see the form of the cubic version ...

@Chris'ssis Amaze me :)

@Hippalectryon Not now. :-)

Ok

Hi. Does the set of all integers satisfy the Peano axioms? I thought they did since there's a bijection from $Z$ to $N$, but $Z$ has no infimum, so I don't believe it satisfies Axiom 7. Is my reasoning correct?

@Ashwin yes i looked at it, seems a bit obscure, but i can see the substitutions work out beautifully, i have gotta do some homework atm i will study it right after I am done

@Ashwin It was worth asking anon about it is what I meant. The book is very meticulous and usually correct, so I suspect my reasoning is faulty rather than the book's claims

@MikeMiller What is it about?

I am quite confident that Matsumura does not have it in for me.

@Pedro Grade, projective dimension, ext.

@Ashwin I read that there exist several sets that satisfy the Peano axioms.

@MikeMiller 3deep5me

how stupid, I misread his definition of something

@MikeMiller something: anything, or anything else. <- That one?

yeah, that one

@Hippalectryon

@MaryStar: Decide in $\Bbb R^d$ for which $r$ the function $|x|^r$ is integrable (and which not). Then you can use that to answer your question by taking various multiples of $r$.

@Hippalectryon ^^^ (in one line - I know, it's hard to imagine such a thing)

@teadawg: I have long ago removed most posted solutions from the web. I may still have some on my office computer. Which ones have you worked?