Mathematics - 2014-03-02 (page 2 of 4)

there's a definition: a Sylow System is a collection of Sylow subgroups $S_p$, one for each $p$ dividing $|G|$, such that $S_pS_q=S_qS_p$ for each $p,q$

David Wheeler

They made the observation that e is the "natural base" for exponential functions because the derivative has the "nicest form"....and I was like: uh, how do you differentiate $b^x$ and what's e?

Alexander Gruber

sylow systems exist (and are conjugate) for all solvable groups. so this style of argument, the prime by prime product you're doing, shows up a lot in solvable group theory. you might like it.

@DavidWheeler i don't see a problem.

David Wheeler

3:15 AM

@PedroTamaroff consider an induction argument on the number of prime factors

@AlexanderGruber problem with...?

Mike

@alexander well, that's all group theorh

if a problem wasn't solvable it wouldn't be a theorem :D

Alexander Gruber

@DavidWheeler why should there a subgroup of prime index exist?

@Mike what's that, now?

David Wheeler

@AlexanderGruber Not prime index, prime factors. Two factors would be like $|G| = p^kq^m$

Alexander Gruber

@DavidWheeler sure, but what you're saying is to use $P\cap \prod_{q\ne p} Q=1$, right?

Pedro Tamaroff

@seaturtles So, if the order of $G$ is a prime power we're done. Suppose then it has composite order, say ${\rm ord}\, g=p^a m$ with $m$ coprime to $p$.

sea turtles

3:20 AM

my argument shows that every element is in some product of sylows, not that every element is in the product of any fixed full set of sylows. error was spotted later.

David Wheeler

@AlexanderGruber something along those lines, yeah

Pedro Tamaroff

@seaturtles Oh.

Alexander Gruber

@DavidWheeler there may not be a subgroup of order $\prod_{q\ne p} Q$, though.

for example, take $p=3$, $q_1=2$, $q_2=5$, and $G=A_5$.

David Wheeler

I'm only half-paying attention, have to go to my crappy job soon

Alexander Gruber

i was thinking about that argument, too.

i know how to prove this using big guns but i'm having trouble thinking of an elementary argument.

Pedro Tamaroff

3:23 AM

@AlexanderGruber Check the deleted answer here. Then check the profile description.

Mike

@alsxander I was saying 'solvable group theory = group theory'

because if something wasn't solvable, nobody would have proven it, so it wouldn't be in the theory

Alexander Gruber

@Mike ah i see

Pedro Tamaroff

@Mike Irony is back on.

Alexander Gruber

@Mike what about undecidable problems, like group presentations?

<-- great at parties

@PedroTamaroff what am i lookin for?

Pedro Tamaroff

"My research is concerned with the topology and geometry of nilpotent orbits in complex semisimple Lie algebras."

Alexander Gruber

3:26 AM

that's some heavy stuff. :p

Pedro Tamaroff

But $V_4$ has $4$ cosets in $S_4$.

David Wheeler

i saw a comic which had the line: "so Stokes' thereom says if you take a loop on a surface, you can tell how much swirly you have".

Pedro Tamaroff

It was SMBC.

smbc-comics.com/?id=3278#comic

sea turtles

@PedroTamaroff 24/4=6

Pedro Tamaroff

@seaturtles You didn't see the deleted answer man.

I am being funny.

sea turtles

3:28 AM

oh

Pedro Tamaroff

(Do you think so low of me :'(?)

David Wheeler

My thesis had the preface: you take da loop, and loop dee loop, to get da loop group.

Alexander Gruber

@PedroTamaroff hahaha

he's too deep in infinite group theory

the finiteness has left him

Pedro Tamaroff

Hhehehhe

Mike

paste the answer?

Pedro Tamaroff

3:33 AM

"I assume you mean the last element to be $(12)(34)$. In this case, $H$ has order $4$, so there are $3$ cosets. The first is represented by $e$, the second by $(13)$, and the third by $(24)$. Now, just check that the cosets of these three elements are disjoint, and you are done. "

@Mike Dude the irony is back.

Mike

I'm not at my computer thougg

Pedro Tamaroff

Oh. Are you at your uni?

Mike

yah

Pedro Tamaroff

What's the name of that song.

"Someone's World."

Mike

I had to walk a ways so I could go check a pair of pants.

@Pedro scatman

Pedro Tamaroff

3:36 AM

@Mike Naughty Mike is naughty.

Mike

no, I left my debit card in them :D

Pedro Tamaroff

@Mike Oh.

@Mike Scatman totally has to do with scatological right?

@AlexanderGruber I'm stuck.

Darn it.

Alexander Gruber

@PedroTamaroff hmm ok let's see

we know that every element of $h\in P_1P_2$ can be expressed uniquely as $h=p_1p_2$ for $p_1\in P_1$ and $p_2\in P_2$

can we also say that about $P_1P_2P_3$?

Mike

@PedroTamaroff Scat is a style of music, or something

Pedro Tamaroff

OH.

Mike

3:49 AM

the "beep ba boop ba ba ba boop bop" he does with his mouth

Pedro Tamaroff

@AlexanderGruber Thinking.

Mike

@PedroTamaroff I'm in the zone

Pedro Tamaroff

@AlexanderGruber

If we had one normal Sylow, we would apply the Frattini argument, I think.

Alexander Gruber

@PedroTamaroff i'm not sure if every group has a normal Sylow.

Pedro Tamaroff

Because if $P$ was the normal Sylow, we could write $G=P N_G(P)$, and induct.

@AlexanderGruber Sure.

I'm not sure either.

Mike

3:54 AM

@alexander how's your taste in music

is it.......good

Pedro Tamaroff

Yeah, Alexander be cool.

I think he should enter the zone.

Alexander Gruber

i have truly eclectic taste, @Mike.

Pedro Tamaroff

New problem! Find a group $G$ with no normal Sylow subgroup!

Alexander Gruber

@PedroTamaroff $A_5$

Pedro Tamaroff

@AlexanderGruber Did you get the link?

Alexander Gruber

3:58 AM

@PedroTamaroff yeah i just signed up

what do i do?

Pedro Tamaroff

@AlexanderGruber HAHA. I'm so silly.

@AlexanderGruber Build up a playlist and join the line.

Alexander Gruber

@PedroTamaroff now the question is, what is the smallest

Pedro Tamaroff

@AlexanderGruber LOL

Yes.

Mike

there's also an easy-to-use chat thing on the right

Alexander Gruber

i am attempting to figure out the interface

Pedro Tamaroff

4:04 AM

On the lower left you have alike a checklist violet buitton

Ted Shifrin

Rehi @Alex @Mike @Pedro

Pedro Tamaroff

@TedShifrin Hello there.

Mike

hi ted

what's rational homotopy theory

sea turtles

how is the tensor product of vector bundles defined exactly

Jasper Loy

Hi anon, lol.

sea turtles

4:11 AM

wikipedia just intimates its a vector bundle over the same base space whose fibers are iso to tensor products of the original fibers. which doesn't seem very helpful.

Pedro Tamaroff

@TedShifrin will have fun answering questions now. =D

Ted Shifrin

@seaturtles: Same way tensor prod of vector spaces is.

Mike

my question's not worth answering, i suppose :(

heartbroken here

sea turtles

@TedShifrin if the tensor product is just a vector bundle whose fibers are isomorphic to the tensor product of the fibers of the original two vector bundles, then how does that even depend on the geometry of the original two vector bundles?

Jasper Loy

@Mike Let me give you a hug.

Mike

4:13 AM

not a chance

Jasper Loy

=(

Hey @ethan any news of your applications?

Ted Shifrin

Homotopy theory without torsion ... Sullivan's minimal models are all based on diff forms with rational coefficients. Griffiths, Morgan wrote a Birkhäuser book in the 80s

Ethan

@JasperLoy all coming this month

Mike

oh, so it's not really related to the rationals

Ethan

I should have em all buy the end of march lol

Jasper Loy

4:15 AM

@Ethan Your spelling is terrible, lol.

Ted Shifrin

Rational coefficients!

Mike

Sure, sure.

Here I'm thinking it's some exotic invariant theory for manifolds that are locally $n$ dim rational vector spaces, or something.

Ted Shifrin

@sea: For a line bundle $L$, for example, however $L$ twists, $L^{\otimes k}$ twists $k$ times as much.

Mike

@TedShifrin "We also saw in the above proof that the embedding of $M$ as a submanifold of $\Bbb P^3$ can be achieved by using 3 or 4 linearly independent $q$-forms ($q$=1 except in the few exceptional cases.) We will show in IV.11 that every two meromorphic functions on $M$ are algebraically dependent. Hence we also have that every compact RS of genus $\geq 2$ can be realized as an algebraic submanifold of $\Bbb P^3$.

sounds like Chow's theorem, which you suggested for me for that a while back, is overkill

Ted Shifrin

Yes, Riemann Roch works fine for curves.

Ethan

4:20 AM

@JasperLoy what did i spell wrong

sea turtles

@TedShifrin if $E_1\xrightarrow{\pi_1}B$ and $E_2\xrightarrow{\pi_2}B$ are two vector bundles over $B$ then what is the underlying space $E$ of their tensor product, and what is the projection $E\xrightarrow{\pi}B$?

Ted Shifrin

I don't remember the context of our discussion.

Mike

@TedShifrin That's all I was interested in... surely not every complex manifold is algebraic.

Pedro Tamaroff

@Mike Remember the claim that if $f$ is continuous over the reals and zeros out at $\pm\infty$ then $f$ is uniformly continuous?

Mike

Sure.

Pedro Tamaroff

4:21 AM

I am answering something with that, turns out it is more delicate than I thought.

But I have the proof.

It is nice.

Ted Shifrin

We were talking about analytic submanifolds of $\Bbb P^N$ being algebraic. Of course there are non-projective ones. Hodge Theory gives you necessary topological conditons.

Mike

Well, maybe after next quarter I'll be capable of having a conversation about that.

Ted Shifrin

@sea: You take tensor products of the fibers. You work over open sets where they're both trivial and tensor the respective vector spaces.

sea turtles

so what are E and B?

hmmm

Ted Shifrin

@sea: An important example is the tensor product of the cotangent bundle with itself. Sections are bilinear forms on tangent vectors, e.g., Riemannian metrics.

I can't give you a global description of $E$ As a top space in general. But you usually can't with vector bundles, either.

They're described intrinsically, often, and constructed by gluing together trivializations by transition functions.

sea turtles

4:31 AM

I want a definition or description for which it is obvious that the tensor product's geometry will depend on the geometry of the original two vector bundles. Saying the tensor product is a vector bundle over the same base space whose fibers are tensor products of the original two fibers doesn't seem to imply this.

Ted Shifrin

@Pedro: Like the subtlety of unif continuity on $A$, $B$ implies $A\cup B$ ... Or somethin' ...

Pedro Tamaroff

@TedShifrin Exactly. This one is even more subtle. =)

Mike

that feels like it shouldn't be subtle at all

Ted Shifrin

@sea: I've told you three times. I quit.

sea turtles

saves everything Ted has written in chat for future reference

Mike

4:32 AM

@TedShifrin Earlier today someone wanted to know what orientability was without being able to define a manifold

Ted Shifrin

@Mike: Hint:$\min(\delta_1,\delta_2)$ doesn't work.

Mike

@TedShifrin I'm sure, since it's the obvious thing and you said it's not subtle.

Ted Shifrin

Do it for vector spaces?

Mike

Never.

Ted Shifrin

@Pedro: I don't believe you.

Mike

4:35 AM

One should never trust @Pedro

Pedro Tamaroff

@TedShifrin Well, believe me. =)

Ted Shifrin

doesn't

Pedro Tamaroff

OK, I'll post my answer in a second.

I think it is more subtle.

Mike

What if you're subtly wrong about its subtlety?

Pedro Tamaroff

I might be wrong. I think I am not.

Ted Shifrin

4:38 AM

@Pedro, like me, is rarely subtle!

Mike

@TedShifrin I had a good quesiton for you.

Ted Shifrin

Well? Tennis in the am, so it's almost bedtime.

Mike

I don't remember what it is, though.

:D

Ted Shifrin

You're too young for memory lapses.

Mike

Well, if I'm going to die by 30, I'd better get a head start on these things.

Ted Shifrin

4:40 AM

I hope not ...

Mike

Well, the best artists all go at 27.

Ted Shifrin

My dad made it to 53.

Mike

Most of the best artists.

(I also can't count out Ian Curtis, who died at 23)

Though perhaps I should quit the morbid jokes for now.

Ted Shifrin

Beethoven was young, but not that young.

Yes, you should.

Mike

@TedShifrin After doing some Cech cohomoloy and singular homology I definitely appreciate group cohomology more. Things seem more natural (even though it's far less canonically defined)

Ted Shifrin

4:45 AM

I've never learned it.

Mike

Well, my same statement applies for any of the theories defined by "take a ___ resolution..."

Oh ...

@Pedro

Yes.

so how do i find you guys on here

Mike

4:46 AM

@AlexanderGruber we're... in the same room as you

Pedro Tamaroff

Ah?

Ted Shifrin

Huh? @Alex

Mike

@TedShifrin Though I guess you could do Cech that way with fine resolutions

Alexander Gruber

i mean this plug.dj

Mike

you're definitely in a room with the two of us and only us

Pedro Tamaroff

4:47 AM

@AlexanderGruber I'm Chooose. And Mike is.... Mike.

=)

Mike

pedro is playing joe satriani now

Ted Shifrin

yes, @Mike, quite important in complex geometry, too.

Mike

Joe Satriani is?

:D

Alexander Gruber

lol i'm so confused

Ted Shifrin

Me too.

Alexander Gruber

4:48 AM

@TedShifrin these guys have invited me to listen to music with them using a site that i am not smart enough to figure out how to operate.

Mike

@AlexanderGruber i don't get it. you're in the room with us. i can see you in the background doing nothing

maybe you somehow lost your browser page.

Ted Shifrin

I feel that way frequently @Alex ...

Alexander Gruber

@Mike i am in a room called mollusk, which i created, because i couldn't figure out how to make a playlist or join a room without it.

Mike

i'm so confused.

were you... not... arctic-turn?

Alexander Gruber

@Mike Lol

nope :p

Mike

4:51 AM

who the hell was that

no wonder there was confusion

Pedro Tamaroff

WHAT THE ACTUAL FUCK.

Alexander Gruber

he's still there isn't he?

Pedro Tamaroff

Yeah.

Mike

yes

Ted Shifrin

LOL

Arctic char would be better.

Pedro Tamaroff

4:56 AM

@TedShifrin Here.

It is a bit long winded.

I don't like how it ended up, but I didn't want to waste more time on it.

Mike

@TedShifrin Is there a way of defining orientability for smooth manifolds that doesn't rely on homology?

Pedro Tamaroff

@Mike See the answer up there $\uparrow$.

Mike

You answered my orientability question before I even asked it? :D

Ted Shifrin

Of course, @Mike. Differential forms.

Mike

Are you cheating?

Pedro Tamaroff

4:58 AM

@TedShifrin Did you read the ans? @Mike ?

Mike

@PedroTamaroff I commented on yours

@TedShifrin Is there a way to do it with diffforms that's not cohomology in disguise (i.e., cheating)?

Pedro Tamaroff

@Mike The fuck.

Silly?

Mike

It's so overkill for what the OP wants.

Pedro Tamaroff

It's good for the OP to know!

Ted Shifrin

@Pedro: Did anyone point put to the OP that he wrote continuous, not uniformly continuous? I think you're answering the wrong question because the OP messed up his quantifiers.

Pedro Tamaroff

5:00 AM

@TedShifrin Nah, the OP wanted continuous only.

I raised the stakes.

Ted Shifrin

No cohomology needed, @Mike. Or just define an orientation on $TM$ ... Know what an oriented vector bundle is?

Mike

We talked about this before when we were talking about Grassmanians.

(but the answer is no)

Ted Shifrin

@Pedro: You're doing a yucky proof. My union proof works fine. You need to go down to $\varepsilon/2$.

Pedro Tamaroff

@TedShifrin Oh, you're saying $f$ is uniformly continuous over $[M,+\infty)$. I see now.

Ted Shifrin

Well, that is a misstatement, but ....

$M$ depends on $\varepsilon$.

@Mike: We'll do it some time I'm on the desktop.

Pedro Tamaroff

5:05 AM

Oh, sorry. Then I don't see what you're aiming at.

Mike

@TedShifrin I eagerly anticipate it.

Tonight I'll learn a bit about Riemannian metrics.

Ted Shifrin

@Pedro: Given $\varepsilon$, choose $M$ for $\varepsilon/2$.

Pedro Tamaroff

Just give me the proof.

Well, your proof.

Ted Shifrin

I just did. :)

Pedro Tamaroff

You gave a sketch.

Mike

5:09 AM

@PedroTamaroff A good artist can make a beautiful piece out of a sketch.

I think you're a good artist.

Ted Shifrin

Pick $\delta$ for $\varepsilon/2$ on $[-M,M]$.

Pedro Tamaroff

@TedShifrin OK.

I'm not thinking now. Really. I'm just reading. Keep going. =D

Mike

@PedroTamaroff I disapprove

@AlexanderGruber Are there any two simple groups with the same order?

Ted Shifrin

Now that $\delta$ works everywhere.

Alexander Gruber

@Mike yeah

Mike

5:10 AM

neat

Ted Shifrin

Me too @Mike. I'm going to bed.

Mike

so every finite group by jordan-holder is just a list of positive integers

@TedShifrin How so?

"is" "just"

Try it @Pedro.

5:11 AM

@Mike there are simple groups with the same order but there can't be more than 2

Pedro Tamaroff

Mike

oh

i misread

Pedro Tamaroff

$A_5$ is the only simple group of order $60$!

Alexander Gruber

@Mike but even so the extensions could be different

Ted Shifrin

Triangle inequality!

Pedro Tamaroff

5:12 AM

#factdropping

Alexander Gruber

think $V$ vs $C_4$

Pedro Tamaroff

@TedShifrin I don't see how it works.

Mike

@AlexanderGruber sure, but i was hoping i could codify all fingrps as integers, and not like, pairs of integers

Alexander Gruber

@Mike ah i see

Mike

i don't want to say $(n,1)$ and $(n,2)$ for the first or second simple group of order $n$

Ted Shifrin

5:13 AM

Only group of order 60? Simple, perhaps?

Pedro Tamaroff

You need to use continuity at $M$, @TedShifrin, methinks.

Mike

@TedShifrin Pedro doesn't believe in nonsimple groups.

Ted Shifrin

I leave it for you to think about, @Pedro.

Pedro Tamaroff

I think the proof is wrong.

I really do.

Mike

LOL this question is awesome

Ted Shifrin

5:15 AM

If $x<M<y$ use triangle inequality with $M$!

Mike

Whoa, I didn't read the third paragrap

That question is more serious than I realized

Ted Shifrin

Oh, I was thinking $f\ge 0$. Use $\varepsilon/4$.

Mike

Or the comments...

I take back my LOL. That question is scary. I hope this works out for her.

Sawarnik

@minopret Glad to come to help :)

Pedro Tamaroff

@TedShifrin So you took $|x|\geqslant M$; not $|x|>M$.

Ted Shifrin

5:18 AM

Yes. Sure.

Night, all.

Pedro Tamaroff

Bye.

Anthony

5:32 AM

@PedroTamaroff

You there?

Pedro Tamaroff

YA

Mike

oh no

Anthony

I'm trying to find the normal vector to an ellipsoid, is there any way besides parameterizing the surface?

Mike

it's anthony

Anthony

<3

Actually a prolate spheroid, same thing?

Pedro Tamaroff

5:33 AM

@Anthony I don't think so, no. Just take that effing cross product of derivatives.

Anthony

I see.

Mike

what the fuck is a prolate spheroid

Anthony

Prolate spheroid

A prolate spheroid is a spheroid in which the polar axis is greater than the equatorial diameter. Prolate spheroids are elongated along a line, whereas oblate spheroids are contracted. The prolate spheroid is defined by the equation \mu = c for some arbitrary constant c, in prolate spheroidal coordinates. Properties A prolate spheroid with b > a has surface area S_{\rm prolate}=2\pi a^2\left(1+\frac{b}{ae}\sin^{-1}e\right)\qquad\mbox{where}\qquad e^2=1-\frac{a^2}{b^2} and volume V_{\rm prolate} = \frac{4}{3}\pi a^2 b. The prolate spheroid is generated by rotation about the major ...

Pedro Tamaroff

@Mike Damn names.

They are all ellipses.

Anthony

Wait does it work the same way if I'm in cylindrical coordinates?

mathworld.wolfram.com/NormalVector.html

Mike

5:47 AM

yolo it

Anthony

:(

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$Mathematics$ Mathematics