@MatheinBoulomenos how to show that any cyclic group is abelian?

each element is $g^k$

@LeakyNun it's obvious QED

lol

do you have a nuke using cohomology / category theory / etc @MatheinBoulomenos

a+b = b+a

(applied to the exponents)

I don't know what a cyclic group is so I'm just going to say nothing and read

reading is good :P

@LeakyNun not sure if this is nuked enough, but if $G=C_n$, $\Bbb C[G]=\Bbb C[x]/(x^n-1) \cong \Bbb C \times \dots \times \Bbb C$ (using the factorization of $x^n-1$ over $\Bbb C$) which implies that all irreps are one-dimensional which implies that the number of conjugacy classes equals the number of elements in $G$ as a corollary of sum-of-squares and number of irreps=number of conjugacy classes

which in turn implies that $G$ is abelian

aha

nice

staahp, staaahp, it's already dead

r.i.p.

finite abelian fourier transform is actually interesting @Semiclassical

This is what I was Talking about!

@Semiclassical beating a dead horse?

@MatheinBoulomenos ooops! i didn't see this message

Anybody know how the vector field on that sphere is defined?

@MatheinBoulomenos are you still around?

yeah

great :P

so can we try to solve the question using their hint?

$F=\{ x^2,y^2 \}?$

not sure, I thought you wanted to ask about $Q_{2^n}$

yea

Oh wait, im talking about the $C_n$

two dim. rep.s

real ones

I'm not sure how to do that with the hint, as you said, the real Jordan form is not useful for computing powers

alright

so i will use your way. so i need to fully understand it :P

so we get that $\phi : C_n \to GL_2(\Bbb R)$ is orthogonal for eahc $g\in C_n$

for a cyclic group, you only need to know what a generator does

so $\phi(\sigma) = A$

$\sigma$ a generator

$A\in O(2)$

how can we proceed?

$A^n = I$

as you already said $A$ is either a reflection or a rotation. reflections are easy, as they are diagonizable over $\Bbb R$, you have either $\mathrm{diag}(1,1)$, $\mathrm{diag}(1,-1)$ or $\mathrm{diag}(-1,-1)$

and if it's a rotation, then by the condition $A^n=I$, it's necessarily a rotation by $k \cdot 2\pi/n$ for some $k \in \{1, \dots, n-1\}$

@MatheinBoulomenos I've looked into the book, but it appears that the last proposition of chapter 4 is 4.4.3., I could not find a 4.4.4.

@s.harp it's on page 106 in my book

The old, classic proof by nonexisting reference :P

@MatheinBoulomenos so that's enough?

@user123 not quite

you also need to work out when those matrices are similar

so basically a rotation for $\theta=k2 \pi /n$ is similar to a rotation for $\theta=2\pi - k2\pi/n$ (rotating in the opposite direction)

@mathein I guess this means there is a second edition, I'll see if i can find that

@s.harp yes, I do have a second edition

@MatheinBoulomenos alright thanks..

@user123 to prove what I said about these similarities, you can e.g. compute the complex eigenvalues (for one direction)

i will try to do it, and i will ask tomorrow the teacher what he meant by the jordan form, maybe we missed another maybe simpler way

there is also another way if you know about group algebras

basically all you need to do is to factorize the polynomial $x^n-1$ over $\Bbb R$ if you do

Here's a question which I feel like I should know how to answer

@AlessandroCodenotti congrats!

I've got polynomials $f_1(x),f_2(x),g_1(y),g_2(y),h(x,y)$

If I pick out two of them which contain $x$ and compute the resolvent, I'll get conditions on the polynomial coefficients for the two polynomials to share a common root x

If I compute the resolvent of h(x,y) and f_1(x) in particular, I'll get some function R(y)

blah, I don't know what I'm on about

demonic @Alessandro: Of course, I am not surprised.

Hi @TedShifrin

Hi @students, whoever you be.

So, here's something (possibly) interesting. I haven't decided for sure yet.

Let $M=\begin{pmatrix} 1 & u & x & y \\ u & 1 & z & 1\\ x & z & 1 & v \\ y & w & v & 1\end{pmatrix}$. The principal 3-by-3 minors are then given by $f(u,x,z),f(u,y,w), f(v,x,y), f(v,z,w)$ where $f(x,y,z)=1-x^2-y^2-z^2+2x y z$. Let $H(u,v,x,y,z,w)=\det M$.

Let $\text{res}_x(P(x),Q(x))$ be the resultant of polynomials $P(x),Q(x)$ with respect to $x$.

to simplify notation, I'll drop the dependence of $H$ on $x,y,z,w$. It's still there

What I'm finding then: $\text{res}_v(\text{res}_u(H(u,v),f(u,x,w)),f(v,x,y)) = (1-x^2)^4 P(x,y,z,w)$

where $P$ is some annoying polynomial

what's slightly interesting is this: $\text{res}_v(\text{res}_u (H(u,v),f(u,y,w)),f(v,x,y)) = (1-y^2)^4 P(x,y,z,w)$

Polynomials are suppose to be our friends :P

and similarly if I do f(u,x,z), f(v,z,w) or f(u,y,w), f(v,z,w)

different factors out front depending on what variable is common to both polynomials

but same P(x,y,z,w)

@Ultradark that is quite easy to cook up.

what does the chaos of a wave mean

generally

rehi demonic @Alessandro

Rehi

Rather late for you!

Yeah I was actually going to sleep

LOL, so your avatar entered without you?

I checked to see if there was some interesting math going on

@AlessandroCodenotti what sort of math do you do?

oh, none of that!

More interesting than sleeping :P

rehi @Semiclassic

Set theory and model theory are the things I'm most interested in. I also like topology

It is very disconcerting to me that someone trying to learn about Lebesgue measure can't correctly negate a definition of continuity. Sigh.

howdy @loch @Dair

Is the following true: If $G$ is the internal semidirect product of $N \unlhd G$ and $Q \le G$, and $\phi_1 : N' \to N$ and $\phi_2 : Q' \to Q$ are isomorphisms, then there is some $\theta : Q' \to \aut N$ such that $G \cong N' \rtimes Q'$

@Ted Howdy

Hi @TedShifrin

Wow, it's like a subway just pulled into Times Square here.

I think I was able to prove it, but the proof so super long and tedious. I just want to make sure I didn't make any mistakes along the way.

Yeah, too tired to do any more math/CS for today,,,

I really wish I knew a good hyperreal analysis book...

You mean $\theta$ maps to $\text{Aut}(N')$?

@TedShifrin "that set is the complement of a $G_\delta$ so it is even Borel" is not the intended answer I guess

I honestly don't like the Keisler one...

@TedShifrin yeah, sorry.

Robinsons's Nonstandard Analysis is the standard one, @Dair (pun intended).

It's like 70 dollars -_-

LOL, @Alessandro: Well, it's nice to understand the set of points of discontinuity. :)

I guess compared to bio text books that's nothing...

There are some Russian online "bookstores" where you can "purchase" a very cheap "ebook copy"

I still wish it were like in the 10 dollar range. Rip no job...

Not sure every book is worth buying, TBH.

I wish I knew a good but cheap :P non-standard analysis book.

Can't you "pretend" and borrow books from SDSU?

Most good books aren't cheap.

Alain Robert's Nonstandard analysis book is <$15 through dover.

I know more bad books than good books.

@TedShifrin I'm not really at sdsu anymore...

Some will comment mine are amongst those.

Same with Martin Davis's.

@TedShifrin ur books are p solid

also, idk what sdsu's library is like...

But don't you have "connections," Dair?

You off the plane already, @Eric?

I have alumni status at UCSD because of the Berkeley alumni thing...

Aha.

@TedShifrin im in the south bay for like a week before i go back to chi

I didn't know they did that.

Ohhhh ... @Eric ... quarter break back there?

@anakhro I'll check that out...

All sorts of fun eating to be had ...

indeed spring break woot woot

yeah it's dope but parking is supposedly no bueno at UCSD...

@Dair definitely download from libgen or something like that before purchasing

I can testify to that, @Dair. I've done it several times.

Though at $15 you can't really go wrong...