Mathematics - 2016-10-24 (page 1 of 5)

Physics Guy

00:00

Ever thought of taking a course on basic topology ?

@user3502615 I'm studying the mathematical aspects of string theory.

Hello, Mr. Shifrin

Adeek

@TedShifrin can I discuss with you something quick ?

Ted Shifrin

Hi @PhysicsGuy

You just did, Karim :P

Adeek

haha

So I just want to know if something is correct.

Suppose p is a prime and G is a group such that $|G| = p(p + 1)$. Prove that G has normal subgroup of order p or a normal subgroup of order p + 1.

so using counting argument I showed that there are p elements outside of all p-sylows.

user3502615

@PhysicsGuy

Adeek

i.e Let P be a sylow p-subgroup of G. Suppose it is not normal. Using that assumption I extracted that there must be p elements outside of all p-sylows.

Ted Shifrin

00:07

Um, assuming not just 1, then you account for $1+(p+1)(p-1) = p^2$ elements. Agreed.

Adeek

yeah

user3502615

call me crazy but ive never been interested in physics

Ted Shifrin

I love physics, @user3502615 ... just don't know enough.

Physics Guy

@user3502615 I'm not calling you crazy

Some people don't like physics. that's just normal.

user3502615

@Adeek by "counting argument" do you mean principle of induction?

Adeek

00:08

@user3502615 no

counting argument as @TedShifrin has shown above.

Ted Shifrin

You need to use Sylow, though, @user3502615.

Adeek

Then Let A be the set of elements that don't lie in any of the p-subgroup of G. We showed that $|A| = p$. Let q be any prime dividing p + 1. Then A must contain some element of g of order q ? Is that true ?

Ted Shifrin

Agh, Karim. Don't make this complicated. Make it simple.

Adeek

what would you suggest?

Physics Guy

Ok, I leave, bye ;)

Ted Shifrin

00:10

Bubye, @PhysicsGuy.

Hmm, I was being stooopid, Karim.

Adeek

hm I mean if I show that A must contain some element g of order q. Then everything follows smoothly

Ted Shifrin

But for any prime $q|(p+1)$, Sylow tells you there can be only $1$ subgroup.

Adeek

but why is that the case (smoothly not in the case of infinitely differentiable haha)

Ted Shifrin

Haha.

Oh, not if $p+1=q(q+1)$. Hmm.

user3502615

@Adeek

sorry

Adeek

00:12

yeah by cauchy theorem we know that such an element exist but why can it be in A ?

Ted Shifrin

Because $q\nmid p$.

Adeek

ohh I see

Ted Shifrin

I bet there's a good group actions argument that's less cumbersome.

Adeek

yeah

is it true that $N_G(P) = P$ ?

Ted Shifrin

What's $P$?

Adeek

00:15

P is the sylow p subgroup of G.

Axoren

@BalarkaSen That's rather annoying. Is there a finite number for which that many points will correspond to a unique parabola? Thanks for giving specific counterexamples.

Ted Shifrin

What do you know about the normalizer and the number of conjugate subgroups?

Hi @Axoren: Love it when you're annoyed :)

Axoren

Most people do, it's a curse.

Ted Shifrin

I'm sure more people try to annoy me than try to annoy you :P

Axoren

You just know more people, likely.

Adeek

00:16

the number of the conjugates of the sylow P subgroup is given by the index of the normalizer.

Axoren

Rather, more people know you.

Ted Shifrin

Right, Karim.

So I win by default, @Axoren.

Adeek

so we have $[G : N_G(P)] = p + 1$

Ted Shifrin

Karim: Can you argue that the identity together with the other $p$ elements not of order $p$ must form a subgroup?

Adeek

oh ok yeah I agree

Ted Shifrin

00:19

Yes, right. So what do you infer?

Adeek

yeah $N_G(P) = P$

Axoren

@TedShifrin At the very least, you're top billing.

Ted Shifrin

"Most hated," you mean, Axoren?

Adeek

so if we have g of order q it can't be inside of $N_G(P)$

Axoren

First name on the sign.

Ted Shifrin

00:19

A lot of good that does.

Adeek

let x be a generator of P. Then all powers $x^i$ are generators of P. Then none of the $x^i$ centralizes g.

Ted Shifrin

Who's $g$?

Adeek

otherwise we would get $g \in N_G(P)$.

g is an element of A of order q.

Ted Shifrin

Oh, an, not the.

Adeek

yeah

Therefore we have $\{g,xgx^{-1},...,x^{p - 1}gx^{-(p - 1)}\}$

are all distinict

all of them have order q.

Then, each of them must lie in A, so A must be that set above.

Ted Shifrin

00:22

Well, $A$ has way too many elements.

So you're arguing that anything in a $p$-Sylow subgroup commutes with the guys left out?

Adeek

So every element of A has order q therefore q must be the only prime dividing p + 1, so p + 1 is a power of q, and so $A \cup \{e\}$ must be a unique sylow q-subgroup of G.

Ted Shifrin

Nonsense, Karim

Take $|G|=5\cdot 6$.

Adeek

A has p elements @TedShifrin

Ted Shifrin

Actually, groups of order 30 are sorta interesting, as I recall. It's been 5 years or so since I last taught algebra.

Adeek

wow

Ted Shifrin

00:24

$12$ is also interesting. Both are of the form $p(p+1)$.

So try actual examples to test out your reasoning.

Adeek

hm what you don't like about my argument ?

oke

Ted Shifrin

$6$ is not a prime power.

Adeek

I proved that A which as a set has cardinality p. I proved that every element of A has order q.

Ted Shifrin

That's clearly wrong, Karim.

Take $p=5$, $p+1=6$.

Adeek

oh

Ted Shifrin

00:27

Uh huh. :D

Adeek

yeah that works @TedShifrin

Ted Shifrin

What works?

Adeek

I think my argument works as well

in this case

Jessy Cat

Can anyone in here help me show that the map provided in the answer to this question is surjective? math.stackexchange.com/questions/1972519/…

Ted Shifrin

Karim: Seriously? $6$ is neither a power of $2$ nor a power of $3$, so you'd better fix your argument.

Adeek

00:31

oh I see

wait a sec

Ted Shifrin

@Jessy: You're making life too difficult. If I take any nonzero rational number $a$, what is $\langle a\rangle$?

(Hint: The same is true in any field.)

Jessy Cat

The cyclic group generated by it.

Ted Shifrin

Oh, I was doing rings. What are you doing?

Jessy Cat

groups

Adeek

yes I see my issues @TedShifrin

Ted Shifrin

00:33

You need to clarify that in the question (or did I miss it?).

Adeek

yeah

Jessy Cat

Yeah, I specifically mention groups in the question, @TedShifrin

Ted Shifrin

OK, @Jessy, so let's look at a concrete example. (OK, sorry.) $a=1$ and $b=7/9$.

What's the isomorphism?

Adeek

hm yeah

p + 1 is a power of q or it could be a multiple of q with something else.

Jessy Cat

Honestly @TedShifrin I don't know. Is that supposed to be somehting I'm supposed to be able to tell just by being given two numbers?

Ted Shifrin

00:34

Yup. :)

Jessy Cat

How??

Adeek

but then why must it be unique ?

Ted Shifrin

Hint: Whatever gets killed in the first quotient group has to get mapped to something that gets killed in the second quotient group.

Adeek

hm

Jessy Cat

By killed, what do you mean?

Ted Shifrin

00:35

Become 0 in the quotient.

Jessy Cat

You mean 1

this is an additive group

Ted Shifrin

Yes, and ...

what gets killed in the second?

Jessy Cat

Well, 1 in the first group is already dead

Adeek

maybe I should ask it as a question

Ted Shifrin

Karim: This is in all sorts of books. Dummit and Foote surely do this.

You were on the right track. I'm claiming that everything in $A$ together with the identity must be a subgroup that commutes with the $p$-group.

Jessy Cat

00:36

In order to turn $b$ into 0, it needs to be added to its reciprocal

i mean it's negative

Not its reciprocal.

Adeek

I see

Ted Shifrin

No. Addition of a number is not a group homomorphism!!

$0$ has to map to $0$.

Jessy Cat

Usually

Ted Shifrin

ALWAYS.

(FOR additive groups)

Jessy Cat

I know. sarcasm.

Axoren

00:38

At least always.

Ted Shifrin

I never can read sarcasm in chat.

Jessy Cat

But how does that show me that f: x->x/a is surjective?

Ted Shifrin

Unless it's mine.

Huh?

Axoren

Poe's a jerk for inventing Poe's Law.

Jessy Cat

That's what I need to show in order to understand the answer someone posted to my question.

Ted Shifrin

00:39

Stick with my example. What homomorphism are you going to take $\Bbb Q/\langle 1\rangle \to \Bbb Q/\langle 7/9\rangle$?

Jessy Cat

I don't know

Ted Shifrin

Review what I said.

Jessy Cat

something that maps 0 to 0

that's what I understood from what you said.

Ted Shifrin

When I talked about killing things.

Jessy Cat

okay, f: x -> 9/7x?

Ted Shifrin

00:40

Yes, the equivalence class of 0 must map to the equivalence class of 0. Indeed.

Why 9/7?

Jessy Cat

x->-7/9x?

Ted Shifrin

Stop and think, dammit.

Don't just guess.

Jessy Cat

I'm not just guessing. I am a very visual person and I have a very hard time understanding something unless I see it.

I don't really understand what you mean by killing things!

Ted Shifrin

Well, you're wrapping an interval of rational numbers around and making a circle, identifying 0 and 1 in the first creature. Similarly, identifying 0 and 7/9 in the second creature.

When you look at $G/H$, you set things in $H$ equal to the identity element in the new world. That kills them all off.

Jessy Cat

What do you mean "identifying 0 and 1" and "identifying 0 and 7/9"

Ted Shifrin

00:42

I mean identifying. They become the same in the quotient.

Doesn't your teacher explain anything?

Jessy Cat

No, apparently not.

This isn't helping.

I'm sorry I asked.

Ted Shifrin

OK ... I won't address any more of your questions.

Jessy Cat

Well, it would help if you didn't get so mean about it.

Ted Shifrin

I'm not being mean. I am being frustrated, yes. But it's better that I just ignore your questions.

Jessy Cat

"think, dammit" and "doesn't your teacher explain anything" aren't helpful.

Ted Shifrin

00:44

Are you paying my salary here? :D

Jessy Cat

I wish I were.

Ted Shifrin

I agree I'm not infinitely patient today. I will let you get help from others and I'll not deal with it.

0celo7

Hi ted.

Jessy

Ted Shifrin

Hi 0celo.

0celo7

Just finished typing my review guide for analysis

@TedShifrin Also, I think $C^1$ and strongly differentiable are equivalent.

Strongly differentiable at $x$ implies $df(x)$ is continuous, and vice-versa.

So they're equivalent.

Ted Shifrin

00:56

Well, @0celo, I realized (and I think I commented to you) that there's a question of where.

0celo7

Of where?

Ted Shifrin

Strongly differentiable at $p$ is $C^1$ at $p$ plus knowing the function is differentiable near $p$.

If you work on an open set, then, yes, they're equivalent.

0celo7

@TedShifrin Yeah.

The proof isn't trivial though (imo)

Ted Shifrin

I still don't think I would ever teach this stuff, but it's a moot point now :)

No, it's not.

0celo7

As with many things, one needs to cleverly use the mean value inequality

Ted Shifrin

00:58

If only a small segment of advanced researchers ever use this, I don't see the point of teaching it.

0celo7

the $||f(x)-f(y)||\le M||x-y||$ thing

Ted Shifrin

Oh, sure, mean value inequality gets used a hundred times in my multivariable math book.

Well, I haven't counted, so maybe that's an exaggeration.

Balarka Sen

@Axoren 5

Ted Shifrin

Oh oh ... Balarka didn't sleep again.

Balarka Sen

any conic is determined uniquely by 5 points

@TedShifrin I did actually, this time!

Ted Shifrin

00:59

Indeed ...

Any cubic is determined by 9. :)

0celo7

what time is it in your part of the world?

Balarka Sen

Yup

Jessy Cat

Too late.

Balarka Sen

about 6:30

Jessy Cat

8:59, and I want to go to bed somet4ime tonight.

Ted Shifrin

00:59

I can never get used to time zones being off by a half hour (mod 1 hour).

Balarka Sen

Daylight savings something something?

Axoren

@BalarkaSen Thanks. Is there a succinct argument for why that's true?

Ted Shifrin

In the US daylight savings shifts by one hour.

@Balarka: Do you want me to answer so you can get ready for school?

Yes @Axoren :P

Balarka Sen

@Axoren Totally! Think of a generic conic as a polynomial of 6 terms, with 6 coefficients each term.

Ted Shifrin

Huh?

Six terms, actually.

Balarka Sen

01:02

Fixed.

Ted Shifrin

6 coefficients TOTAL, yes :)

Axoren

Oh, that makes sense.

Ted Shifrin

And passing through a point is a linear condition on those coefficients, @Axoren.

Balarka Sen

I meant one coefficient for each term.

Ted Shifrin

But the equation can be multiplied through by a nonzero constant, so that takes away 1 degree of freedom.

Balarka Sen

01:03

Yah. So you're going to boil down to intersecting 5 hyperplanes in a 5-space

That's a unique point, generically.

Ted Shifrin

Only if you projectivize.

Five hyperplanes in 6-dimensional vector space.

You should have let me take over :P

Balarka Sen

I'm shoving that compactification under the rug.

Ted Shifrin

I should just leave. :0

0celo7

wait, no

Balarka Sen

Nah, go on. I'd leave soon anyway

0celo7

01:05

I have a question about the definition of $C^2$

Ted Shifrin

LOL. Have a good day at school, @Balarka.

Axoren

When you start talking about projective spaces, that's where I know very little about how the geometries work.

And I start thinking less geometrically than I normally do.

Balarka Sen

Hence the shoving under the rug. But Ted's mad about that, so...

Ted Shifrin

That's why I left it out. You get 5 hyperplanes in $\Bbb R^6$, @Axoren, all independent, so they intersect in a $1$-dimensional subspace.

That $1$-dimensional subspace corresponds to $c(a_0,\dots,a_5)\in\Bbb R^6$.

Hardly "mad," @Balarka.

Axoren

When you say subspace. Which type of space?

Linear or projective?

Ted Shifrin

01:07

Linear.

Axoren

Huh. Okay.

Ted Shifrin

I said $\Bbb R^6$ :)

A line in $\Bbb R^6$.

Axoren

I don't immediately see why it's a line in $\mathbb R^6$

Ted Shifrin

So your conic has the equation $a_0x^2+a_1xy+a_2y^2+a_3x+a_4y+a_5=0$, or any constant multiple thereof.

Independent equations each reduce the dimension by $1$.

Jessy Cat

i wish i were dead.

Axoren

01:08

Oh, the coefficients.

That's clearer, actually.

0celo7

@TedShifrin A function $f:U\to\Bbb R^n$ is differentiable at $x\in U$ iff each $f_i:U\to\Bbb R$ is differentiable, right?

Ted Shifrin

Yuppers @0celo.

Mike Miller

You want $\to \Bbb R$.

Balarka Sen

@TedShifrin Just joking. Also, it's not school today, just have to go to a physics tuition I signed up for, but now regret for it.

0celo7

@MikeMiller thanks.

Ted Shifrin

01:09

tuition ≠ tutorial, @Balarka :P

0celo7

@BalarkaSen Tuition?

Balarka Sen

grr

Ted Shifrin

You ok with it @Axoren? Lots of pretty arguments like this in baby algebraic geometry.

I've taught Balarka both to smack and to grrr. My life is complete.

Axoren

It makes sense. It's just weird to think about it like that at first.

Balarka Sen

It's pretty weird, yes

Ted Shifrin

01:11

Well, you're learning about projective duality without even knowing it :)

Balarka Sen

But you just learnt about the moduli space of conics

Ted Shifrin

See, @Axoren, now you know soooo much :P

Mike Miller

@JessyCat If you mean this, there are plenty of people and organizations you can talk to, and should. If you don't, you shouldn't joke or insinuate.

Axoren

Is this somehow a group action? Because $\{a_i\}$ is from $R^6$, but then you have this other bit $x^2, y^2, xy, x, y, 1$. This seems like a linear transformation with two different fields.

Balarka Sen

OK, gotta go.

Ted Shifrin

01:14

Bye, @Balarka.

No, @Axoren. We're just looking at the vector space of (non-homogeneous) quadratic functions. It's a 6-dimensional vector space.

But when we set one = 0, constant multiples don't matter. That's why God invented projective space.

Axoren

Right, there's an equivalence between anything on lines intersecting the origin.

Ted Shifrin

There is a group action of $\Bbb R^\times$ (nonzero reals under multiplication) on the vector space.

That's what you just said, actually.

Axoren

Yeah, I was wondering if that's what it was.

I might have said it backwards

But it's either one of those groups acting on the other.

Ted Shifrin

I would just say the points of each $1$-dimensional subspace are all identified.

Huh? Only one group so far. :)

Axoren

Alright, that's where I'm missing it.

Ted Shifrin

01:17

Group acting on vector space.

Axoren

I was under the assumption that all vector spaces were groups.

0celo7

@TedShifrin Ok, it basically amounts to showing $df(x)[h]=(df_1(x)[h],\dotsc, df_n(x)[h])$

Ted Shifrin

But you don't want that. As an additive group, a vector space needs a bazillion generators.

Axoren

I see. We avoid calling it a group to avoid justifying it as a group. The fact that it's a vector space is enough for group action. The fact that it's a set is enough for a group action.

Ted Shifrin

Sure, @0celo, but that is clear. Just like you can always break down a linear map $T\colon V\to\Bbb R^m$ to $m$ linear functionals.

Not justifying, @Axoren. That's too little structure.

0celo7

01:19

@TedShifrin yeah but you have to show that each component is $df_i(x)$

Ted Shifrin

Not difficult, 0celo.

0celo7

I didn't say it was

Just checking

Axoren

When defining a group action on a vector space, is it generally easier? I haven't really done it enough to know if the structure has some streamlining benefit.

Ted Shifrin

Easier than what?

Axoren

I've only really approached the concept in the barebones setting of $G \times A \to A$ where $A$ is just a set.

Ted Shifrin

01:20

That's fine.

I was just objecting to your thinking of the set as a group. :)

Axoren

What does the additional structure of a vector space $A$ give in that context?

Ted Shifrin

It means multiplication by nonzero scalars makes sense :)

Axoren

Wait.

Mike Miller

It means that the map $A \to A$, given by $v \mapsto gv$, is linear for all $g$.

Ted Shifrin

Do we really need that, @MikeM?

Axoren

01:22

Oh jeez, that's huge, actually.

Mike Miller

@TedShifrin Huh? Yeah?

Ted Shifrin

I just need to know that multiplication by a scalar sends a line to itself.

I mean, ultimately, it matters, of course, when I want to define mappings on projective space, etc.

Mike Miller

A group action on an object should preserve the structure of that object...

Axoren

Which means it preserves linearity.

That's funky.

Ted Shifrin

Well, ok, but Axoren was just talking about the vector space as a set. I don't disagree with your wanting more structure.

Mike Miller

01:23

@TedShifrin He asked what the addition structure of a vector space gives... It gives two things.

Ted Shifrin

Did he ask that? All I care about is scalar multiplication. :P

I was objecting to his thinking of the vector space as an additive group.

Mike Miller

OK, presumably I didn't pay enough attention to the conversation.

Sorry for butting in.

Ted Shifrin

I do that a lot. Nah. Butt in entirely.

Axoren

Additional structure, not addition structure.

But your chiming in was still helpful.

Ted Shifrin

@Axoren: Agreed. But originally you wanted to think of the vector space as an additive group, remember?

Mike Miller

01:25

Typo.

Ted Shifrin

That's when I objected.

Axoren

Yeah.

Ted Shifrin

Anyhow ... I think you understand.

Axoren

I need to learn function spaces better.

It's one of the topics of my Monday Wednesday classes

Supposedly, we're going to be talking about Hilbert spaces, and might touch upon space-filling curves.

Ted Shifrin

Nice topic.

I don't see space-filling curves connected to Hilbert spaces, though. Maybe the Hilbert cube, though.

0celo7

01:29

@TedShifrin what's gonna be on a first semester analysis midterm?

Ted Shifrin

Don't ask me, 0celo. Ask your professor.

Axoren

@0celo7 Analysis

Ted Shifrin

I've already said I'd never teach such a course the same way.

Good answer, @Axoren.

0celo7

@Axoren well, or topology

Axoren

Also, spelling.

If you get your name wrong, the wrong person gets the grade and you end up with a 0.

0celo7

01:31

He probably knows my handwriting

Axoren

That's good. One less subject to study.

:P

Ted Shifrin

LOL

Has he told you whether it's all proofs, some examples/counterexamples, etc.? Such courses often ask for examples or counterexamples.

Mike Miller

Surely you know all this crap anyway.

0celo7

@MikeMiller If he asks me to prove $\ell^p$ is Banach I'll be in trouble

Knowing things and knowing the proofs are different

Ted Shifrin

That's not an unreasonable proof if there's time.

0celo7

01:33

although I did work that one out for extra credit on a homework

@TedShifrin not if you need to prove the Minkowski inequality too

Ted Shifrin

But if it was only extra credit homework and not done in class, that's not a reasonable test question. ... No, I would accept basic facts.

But something that appeared only on extra credit should not show up on an exam.

Mike Miller

Minkowski and Holder.

I don't think Minkowski is absurd to prove.

Ted Shifrin

There's only an hour and a lot of stuff has been covered ...

0celo7

> Included: all the handouts posted, including HW solutions (be sure to check you have the latest posted version)
Sections in Fleming's text: 2.4 to 2.10, 3.1, 3.3, 4.3, 4.4.
You may be asked for definitions and statments of theorems (in addition to short proofs.)

Mike Miller

Not that it's important to be able to do tedious manipulations.

0celo7

01:36

dropbox.com/s/wxfvpiz9383bj22/m447%20study%20guide.pdf?dl=0

this is basically what we've done so far

I typed this up while reviewing

Ted Shifrin

Fleming. So he's teaching differential forms and Lebesgue integration?

0celo7

yes, in the second semester

Ted Shifrin

That course was my one B as an undergraduate. I've always found that ironic :P

0celo7

idk why he uses Fleming

all the Banach space stuff is not in there

nor is Arzela-Ascoli, Stone-Weierstrass

Ted Shifrin

I like the book, but it's very idiosyncratic. Well, he's off the deep end. Fleming is plenty hard enough. ... That's all single-variable, Fleming is multivariable analysis.

0celo7

01:39

He's off the deep end?

Ted Shifrin

I've already said that's my opinion several times.

0celo7

Single variable? Check page 3 of the notes

We did all of those in great generality

Ted Shifrin

I mean they standardly appear (e.g., in Rudin) in a single-variable analysis course. They can be put in more general settings, yes.

0celo7

we did not do them in single variable analysis

single variable analysis was very slow, but I really liked it

Ted Shifrin

So your prof presumably picked Fleming for the multivariable stuff, but even then he's doing more subtle stuff.

0celo7

01:43

Yeah, we're following Fleming more closely since we started derivatives

but the strong differentiability is completely out of the prof's head I think

Ted Shifrin

Nah, there are notes out there.

0celo7

Doesn't mean he's using them

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$Mathematics$ Mathematics