Mathematics - 2017-03-17 (page 4 of 5)

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9:00 PM

We should push to change that @Alessandro

@Alessandro Definitely gonna fix it

We should have all first years do an intro to proofs + categories course

Alessandro Codenotti

@Daminark nah, I'm fine thanks

What I'd really like is a proper set theory course for undergrads

Hey guys another short question for complex analysis. I just read about a theorem that if you have a connected set with a single hole, f holomorphic and $γ_1,γ_2$ closed curves around the hole then the integral of those curves is the same. I just read about it in german literature but I'm not satisfied yet, any suggestions or can you confirm this?

I don't know any set theory

@Felix.C Have you learned residue thm yet?

9:01 PM

No

also, is $f$ just holomorphic on that set? or everywhere holomorphic

Anyways, let's just dive in

@Felix.C Homotopic curves integrate to equal values for holomorphic functions. This is true.

What would be the difference?

It would make a huge difference

in the latter case, we would have both integrals to be 0

Hi guys

9:03 PM

in the former, that would not be true

Question: Let $\sigma\in S_n$ be a permutation. We define the permutation matrix $P(\sigma)$ as the $n\times n$ matrix:
$$
P(\sigma)_{ij}=
\begin{cases}
1&i=\sigma(j)\\
0&i\neq\sigma(j)
\end{cases}
$$
We see that $P(\sigma)$ permutes the columns of the identify matrix according to permutation $\sigma$. The mapping $P\colon S_n\to GL_n(N,\mathbb R)\colon\sigma\mapsto P(\sigma)$ is an injective groupsmorphism.

Also, I suppose $\gamma_i$ have the same winding numbers around the hole.

@MeowMix They would still be equal.

Assume we have $\sigma_1=(13)(2)$ and $\sigma_2=(1)(23)$. It follows that $\sigma_1\cdot\sigma_2=(132)$. However, $P(\sigma_1)\cdot P(\sigma_2)$ gives a permuted identity matrix (columns are permuted) according to $\sigma=(123)$.

@BalarkaSen I meant "both integrals being 0" would not be true

So it doesn’t hold that $P(\sigma_1\cdot\sigma_2)=P(\sigma_1)P(\sigma_2)$. What actually is true in this example, is that we have $P(\sigma_2\cdot \sigma_1)=P(\sigma_1)\cdot P(\sigma_2)$, i.o.w., $f(ab)=f(b)(a)\neq f(a)f(b)$. Am I misinterpreting something here, because I would say that $f$ does not represent a homomorphism...

9:04 PM

Question didn't say that.

My question did. :P

Alessandro Codenotti

I should start complex analysis next week, not looking forward to it

Because he asked the difference between everywhere holomorphic and holomorphic just on that set

Alessandro Codenotti

Especially because the course will repace the AT one

Hi @ShaVuklia @Alessandro

Alessandro Codenotti

9:05 PM

Hi @Astyx

Hi @Astyx

:)

@Felix.C Anyways, I propose the following contour

whips out MS paint

@AlessandroCodenotti Complex analysis can be great.

Think of it as 1-dimensional complex differential forms theory :P

Let $\gamma$ be the contour around that hole, okay?

The nice thing about complex analysis is that it's not so pathological like real analysis

9:06 PM

And let $\gamma_1$ be any contour contained in our connected set

@MeowMix Ah sure, yes if it's holomorphic globally I agree then it would be 0 since we can find can find $F$ such that $F'=f$ but I disagree with your second statement

@Felix.C which statement would that be?

Alessandro Codenotti

Yeah, I've been told comples analysis is beautiful, but I would have liked to see more AT

@BalarkaSen aha, now it's all clear!

Once you know a function is differentiable, you just have an arsenal of ridiculously overpowered theorems that are just nice and shouldn't be true

But they are

Anyway, I propose that the integral around $\gamma$ is equal to the integral around $\gamma_1$

Do you see why?

9:07 PM

lol

@Meow Mix "in the former, that would not be true"

Akiva Weinberger

Is "$\forall\epsilon_1,\epsilon_2,\exists N:\forall n>N,(a_n<\epsilon_1\land a_n<\epsilon_2)$" ever more useful than the usual version (with one epsilon)?

@Alessandro Well, you can always study AT by your own. Take it up as a project.

Akiva Weinberger

This was we don't need to worry about mins and maxes in the proofs

Automatically analytic, $f(z)dz$ is a closed form, etc

Actually so there's this theorem called Liouville's theorem

Which says that a bounded function is constant

9:08 PM

@Felix.C Because it wouldn't. The integral doesn't have to be 0 there

You can use it to give a clean proof of the fundamental theorem of algebra

@Daminark Cauchy's theorem then becomes the Green's theorem.

If there are singularities inside of the hole

Alessandro Codenotti

Sure, I think I'll keep working on Hatcher's book. The notes from my professor also cover up to singular homology and Mayer-Vietoris, but there's no time in this course

@Felix.C If you have a curve with winding number > 0 around a pole of a holomorphic function, your integral can never be 0.

9:10 PM

@ShaVuklia The operation for matrices is multiplication, while the operation for permutations is composition

The idea is that if a polynomial $P$ has no zeroes, you know that $|P(z)| \ge \mu > 0$

Anyways, @Felix.C Let's look at a contour

@Astyx Yes, I am aware of that

Then what is the problem ?

But $f(a\circ b)$ should then equal $f(a)\cdot f(b)$

9:10 PM

The proof I saw was that if there is a hole and we have $γ_1,γ_2$ around that whole then we can use $γ_3,γ_4$ to connect them and to show that the integrals of $γ_1,γ_2$ are equal but I'm not sure whether I understood everthing right

So then you can look at $\frac{1}{P(z)}$, and note that it is bounded above by $\frac{1}{\mu}$, and below by $0$. It's bounded and entire, hence it's constant

Please excuse the bad drawing.

user image

@Astyx Well, if you look at my example, it seems that we have $f(a\circ b)=f(b)\cdot f(a)$

Unless I'm interpreting something wrong

something with the order

I'm going to "tape" two contours together by the edge

@MeowMix It's not entirely trivial to show that integral around $\gamma_1$ is the same as $\gamma$.

You need the Green's theorem/Cauchy's theorem.

Akiva Weinberger

9:12 PM

Once you have Cauchy's theorem, though

which says that it's zero on nullhomotopic/nullhomologous paths

we have $a=(1)(23)$, $b=(13)(2)$. Therefore $b\circ a=(123)$

however, if you multiply the matrices, you seem to get $(132)$ @Astyx

@BalarkaSen I disagree respectfully. :P

Right. And you can perturb $\gamma_1 - \gamma$ to be null.

user image

Akiva Weinberger

Well, it's already nullhomologous even without putting in any extra lines. The extra lines make it nullhomotopic.

9:13 PM

Consider this path $\gamma_2$

@MeowMix You disagree that you need Cauchy's theorem to do it? Then you're wrong.

@BalarkaSen I meant green's theorem

Akiva Weinberger

"It" being $\gamma_1-\gamma_2$, I guess

Cauchy's theorem is Green's theorem

@ShaVuklia $$(P(\sigma_1) \cdot P(\sigma_2))_{i,j} = \sum_{k=0}^n P(\sigma_1)_{i,k}P(\sigma_2)_{k,j} = \sum_{k=0}^n \delta_{\sigma_1(k),i}\delta_{\sigma_2(j),k} = \delta_{\sigma_1(\sigma_{2}(j)), i}$$

9:14 PM

Green's theorem = Cauchy's theorem in complex analysis

Oh yeah... Sorry

Like, you use Cauchy Riemann equations and then Green's theorem to derive Cauchy

@Astyx oh okay, I'll need a minute to read my way through that:P

Anyways, @Felix.C What I'm going to do here is add the contour integrals around $\gamma_2$ and $\gamma$, and so we'll get the contour integral around $\gamma_1$

do you see why that is true?

@MeowMix the hole is inside $γ$ right? Now you use that the path outside the hole give the integral = 0 Since those are simply connected domains right? Then you follow that $γ,γ'$ are the same?

9:15 PM

@AkivaWeinberger Yeah, true. I guess you can't directly show nullhomologous chaps integrate to zero without perturbing them to be nullhomotopic though.

@Felix.C Exactly

And thus, all of the contour integrals inside of our set are equal

It should be 0, not 1 by the way

@MeowMix There's a little subtlety. The path $\gamma$ you have drawn doesn't produce a simple closed curve (aka contour). You need two paths joining the contours, like a tube.

And then make the tube arbitrarily thin to see that the integrals are same.

Lol, maybe I should do algebraic topology this spring

@BalarkaSen You're right.

9:17 PM

@MeowMix Ok thank you really a lot for taking the time!! Really appreciate it. Since this is the beginning of my second semester at university I'm sometimes uncertain to trust myself ;)

But yeah, what you said is the idea.

Akiva Weinberger

@BalarkaSen I'm not sure what you mean

@Astyx ah yea, I didn't even notice at first:P

@Felix.C No problem. A lot of people feel the same way

Akiva Weinberger

Triangulate the thing it's a boundary of

Nullhomologous specifically means it's the sum of boundaries of lots of triangles. Each of those go to zero.

9:18 PM

@AkivaWeinberger That's peturbing it to be sum of nullhomotopic things.

Akiva Weinberger

Oh

Do you agree with the calculations ? @ShaVuklia

Akiva Weinberger

Yeah, OK then

@Astyx I'm figuring out the last equation

Up until there I agree

Akiva Weinberger

@MeowMix Here's an interesting contour:

9:19 PM

Woah

Akiva Weinberger

Say those two points are poles. Then that contour isn't nullhomotopic; you can't homotope it to a point

Does the purple/black represent which direction its turning?

But integral around it is still zero

Akiva Weinberger

Can you show that the integral along that contour must be zero anyway?

I don't know what the colors are for. The arrows indicate direction.

Alessandro Codenotti

Hi @Ted

9:20 PM

Hi @Ted

Salut @Ted

Hi @Alessandro, @Balarka, DogAteMy, Zach, @Astyx.

Hey @Ted!

Oh, and Demonark.

@Akiva This corresponds to aba-1b-1 in the wedge of two circles, doesn't it?

9:21 PM

@AlessandroCodenotti Jinx you owe me a sods /s

Hi@Ted

@Astyx I'm not sure I agree actually

Hi @ShaV :)

Akiva Weinberger

Yeah

Hi

Hi @Ted

oh wait

one more sec

9:21 PM

So it's the simplest example of such a contour I guess

sorry it's my first time working with the Kronecker delta :P

@Alessandro: Working on my reservations/plans for Europe. Crazy complicated.

Haha take your time @ShaV

@AkivaWeinberger And, the red/green are the only singularities?

Akiva Weinberger

Yeah

9:22 PM

Sounds like Zach needs to high-tail it to chapter 8 of my book.

So the holomorphic function is like $\frac{1}{(z-1)(z-2)}$

Alessandro Codenotti

@TedShifrin too much bureaucracy?

I was just thinking $1/(z^2-1)$

Well if there are two of them, then perhaps they're more like doubularities

works

Akiva Weinberger

9:23 PM

Same thing, essentially

LOL, no, @Alessandro. Just trying to coordinate with two sets of friends, and figure out lots of lodging and logistics ... car, train, bus, ferry, plane ...

puts Demonark on warning

Akiva Weinberger

@MeowMix Suppose only one of the two highlighted points were singularities.

What happens to the curve?

Alessandro Codenotti

isn't chapter 8 the differential forms one?

Come on, it was an isolated pun

Now, @Ted, you can't just let the kid jump into Ch 8! Remember how much we fought when I wanted to do that without the prereqs? :P

9:24 PM

@AkivaWeinberger Well you can't pull it apart because the singularities are in the way

@Astyx OK, I finally get it :P

Demonark, but essential all the same.

Alessandro Codenotti

@TedShifrin Aha, I see, of course I suppose those 2 groups are in different countries?

Akiva Weinberger

@MeowMix Would that still be true if there were no green singularity in the way?

or if there were no red singularity?

@Alessandro: One for Paris. The other for Nice->Italy->Croatia.

9:25 PM

@AkivaWeinberger No, you would be able to homotope it to like a figure 8

@ShaV Cool ! And the last term is in fact $P(\sigma_1\circ \sigma_2)_{i,j}$

@Alessandro Yes, that one chapter of notorious fame.

Akiva Weinberger

@Balarka: You, Zach, and DogAteMy are driving me to an early grave.

Akiva Weinberger

Why

9:25 PM

you are right indeed @Astyx

@Ted Today on my geometry test, I had this weird right triangle thing to prove and since I had some time left, I included a proof w/ linear algebra as a joke :P

Beautiful

0

Q: Group of units of direct sum of rings is isomorphic to direct sum of the groups of units

Let $R_{1}$, $R_{2}$, $\cdots$, $R_{m}$ be rings with identity. I need to prove that the following group isomorphism holds: $U(R_{1} \oplus R_{2} \oplus \cdots \oplus R_{n}) \simeq U(R_{1}) \oplus U(R_{2}) \oplus \cdots \oplus U(R_{n})$. I surmise that induction is going to be necessary he...

abstract-algebra ring-theory group-isomorphism direct-sum

alright, I am truly convinced now. However, I do still wonder where I was making a mistake in my example @Astyx

So my guess is you messed up matrix product when computing those :p

9:26 PM

Was it correct, Zach?

@AkivaWeinberger Sorry, what do you mean by why? :P

yea probably!

I will go through it again

Akiva Weinberger

I guess that makes sense, since $u\cdot v=0$ is equivalent to the Pythagorean theorem

My favorite proofs are those which basically consist of nuking a potato from orbit

5

Thanks a lot for this proof :) This was an excellent exercise for me, actually!

9:26 PM

For instance you took $i=\sigma(j)$ instead of the opposite

Akiva Weinberger

@MeowMix Why are we driving Ted to an early grave

@TedShifrin Well, yeah. But I asked her to grade my normal proof, since she'd think I was crazy otherwise

At least I think. It was a rather easy question.

oh really? let me see

She might not know the fancier stuff, Zach. But if you tell me, I will grade it.

Hahaha I'll try to recall it and write it out

9:28 PM

busted

Last quarter in analysis, we had proven that if a function's derivative was in $L^p$, the function itself was Hölder continuous with exponent $\frac{1}{p}$, and I remember using this fact in order to prove that $C^1$ functions were Lipschitz continuous in a compact set

@AkivaWeinberger Why weren't we earlier?

Alessandro Codenotti

@Daminark I suppose you already know how to show that $2^{1/n}$ is irrational for $n\ge 3$ by a simple application of Fermat-Wiles? (Which is way too weak to show that $\sqrt{2}$ is irrational unluckily)

See I've seen that but it may be circular

Akiva Weinberger

@Daminark $\sqrt[3]2$ is irrational. Proof: if not, we'd have $\sqrt[3]2=\frac ab$, or $b^3+b^3=a^3$, in direct violation of Fermat's Last Theorem. Unfortunately, this method is not strong enough to deduce the irrationality of $\sqrt2$.

9:29 PM

My favorite proof of irrationality of sqrt(2) is Gauss lemma

LOL @silly DogAteMy proof.

Alessandro Codenotti

I'm afraid there's more than one thing named Gauss lemma. The one about the gcd of the coefficents of a polynomial?

Akiva Weinberger

Oh, no, Alessandro beat me to it

it says if something is irreducible over the integers it's also irreducible over the rationals

rip @Akiva

9:30 PM

x^2 - 2 is easily irreducible over the integers

Akiva Weinberger

How do we prove that?

Alessandro Codenotti

@BalarkaSen yeah, the one I'm thinking about is used to prove this

it's basically rational root theorem

I think Eisenstein is a little easier than Fermat, eh, @Balarka?

Oh, yeah, duh ... quadratic.

Right.

9:31 PM

I've just been in touch with the publisher and they're finally going to do a corrected printing with all the mistakes corrected.

What would a corrected printing be without all the mistakes corrected?

braces for smack

That's nice, for multi @Ted?

Akiva Weinberger

Redundancy is neither useless nor pointless, Zach

@MeowMix "we fixed the typos you sent us but we added some more along the way"

Yup, Demonark. I've been asking for 10+ years.

@Balarka: That happened to me with my linear algebra book the first edition.

9:33 PM

@MeowMix you usually have a different name, don't you?

@Balarka It's like "Where's Waldo" but for retired math professors

Alessandro Codenotti

@BalarkaSen kinda like debugging a computer program, you always introduce more bugs

@Ted lol whoops. What did they do?

Akiva Weinberger

Oh, speaking of, @Ted,

you have a somewhat humorous typo on page 227.

@Ted Is it still going to be expensive?

9:34 PM

God, it seems like that publisher is such a headache

Alessandro Codenotti

@AkivaWeinberger the error correcting codes part of the abstract algebra course convinced me exactly of that

@MeowMix Don't quite get that reference

@Astyx I don't see where I make a mistake though;
$$
\begin{pmatrix}
0&0&1\\
0&1&0\\
1&0&0
\end{pmatrix}
\cdot
\begin{pmatrix}
1&0&0\\
0&0&1\\
0&1&0
\end{pmatrix}=
\begin{pmatrix}
0&1&0\\
0&0&1\\
1&0&0
\end{pmatrix}.
$$
This matrix corresponds with $\sigma=(123)$, while we need $\sigma=(132)$.

Always expensive, unfortunately.

Oh, oh, DogAteMy.

I remember some time ago you mentioned that they had jacked the price way higher than you wanted, and yet they're still dragging their feet on things like this...

Akiva Weinberger

9:35 PM

@BalarkaSen Maybe you know it as "Where's Wally"?

What are the permutations @ShaV ?

Nope

That's why I never published the diff geo book, Demonark. I'm fed up with it all.

DogAteMy: You gonna tell me?

Left matrix: $(321)$, right matrix: $(132)$

wait wait wait

:)

9:36 PM

@BalarkaSen It's a series of books which consist of images, and within those images is a carefully placed "Waldo" that you have to locate

First matrix is (1 3)

correct

Ah

Akiva Weinberger

@TedShifrin Theorem 5.5.4 has a LaTeX mixup

and the second is $(2\ 3)$

9:36 PM

Yup

Example:

user image

Oh, yikes. That must have gotten messed up when I fiddled with something else. Thanks, DogAteMy.

And last (321) = (13)(23)

@Akiva what's the typo?

Akiva Weinberger

It's a book full of double-page spreads like that. You need to find a specific character (named Waldo, or Wally in the U.K.)

9:37 PM

I found Waldo!

A half-dead mosquito is crawling on my computer screen. Would my guilty conscience allow me to squash it to death?

Put the critter out of his misery

Or let him die in pain and suffering

Either one is fine by me

there is my mistake then, because I interpret this matrix as $(123)$

instead of $(132)$

Maybe I'm gonna write a book called "Prove Waldo's theorem"

An extra backlash, DogAteMy. Dunno how that happened. Anything more substantive?

9:39 PM

the first column of the identity matrix is the second column now, no?

@MeowMix Interesting aspect. Let's be a goth and disembowel the mosquito in a ritual murder

so 1 or mapped to 2, I would say

Akiva Weinberger

No, don't think so @TedShifrin

And the second column of the identity matrix is the third column now, therefore 2 is mapped to 3

hence, $(123)$

Oh, haha @Akiva

"smvect"

9:40 PM

If you call A that matrix, what is A(1,0,0) ?

Akiva Weinberger

I didn't even know that command existed

$\smvect ab$

Wait, no

@Meow Is Waldo even on this picture ?

maybe it's defined in his preamble

Sorry, I don't understand what you mean by $A(1,0,0)$?

oh wait

DogAteMy: I have a whole file of my own macros.

Akiva Weinberger

9:40 PM

@Astyx Yeah (I found him, too)

@Astyx Do you think I'm a psychopath?

multiplying the matrix by vector (1,0,0)?

We'll all answer that one, Zach.

That I'd give you a picture without Waldo?

@MeowMix Too bad

9:41 PM

Just to waste your time?

actually that'd be kind of funny

Akiva Weinberger

He's actually on the other side of the screen

I found his brother (or cousin or whatever), but no sign of him

if that's what you meant, it's the first columns of A @Astyx

@ShaVuklia Yes, and that's also the product of A with the collumn matrix (1,0,0) (which is the first vector in the canonical basis of $\Bbb R^3$)

@Meow You never know

That's how I calculated it, I didn't interpret it any other way actually

but how does that knowledge help me?:p

Akiva Weinberger

9:43 PM

Huh, yeah, there's both Waldo and Wenda

Finally found him

what does he look like

Akiva Weinberger

https://en.wikipedia.org/wiki/Where's_Wally?

He has a striped red and white hat

and glasses

Well if you note $e_1$, $e_2$ and $e_3$ the canonical basis of $\Bbb R^3$, you'll find that $Ae_i = e_{\sigma(i)}$

9:45 PM

@AkivaWeinberger Oh I saw Wenda too

oh, nerdish

Here $Ae_1 = e_3$, $Ae_3 = e_2$, $Ae_2 = e_1$

So $\sigma = (321)$

Like 5 years ago, when I was doing my Bachelors, there was a law student named Waldo at my uni

@BalarkaSen Interesting, when I posed your circle question to my friend, he just gave an explicit solution

Who... ended up missing

And everyone thought it was a joke when they were told that "Waldo was missing" >.<

9:47 PM

@MeowMix That's what I did

when I first got it, I mean

Wow, my dad just used two expletives towards my cats

And not your average expletives either. Like words that are considered traditionally offensive.

You're next, Zach.

Akiva Weinberger

smbc-comics.com/comic/2014-07-04

spotting waldo is harder than it looks. i'll just listen to some music

Is Waldo a common name ? In french it was translated as Charlie, which not that rare

9:48 PM

@Astyx It's kind of an odd name to have

If you're born any time past the 2000s

Charlie Hebdo?

Akiva Weinberger

Also: smbc-comics.com/index.php?id=20

(the only era I know)

For instance @Ted

It was no more common in my day, Zach.

Akiva Weinberger

9:49 PM

Apparently in Israel they translated it to Efi, which I've never heard

@Akiva Do you know them all off by heart in order to find them so quickly ?

@ShaV Did you see what I wrote up there ^ ?

Akiva Weinberger

@Astyx No

Oh, short for Efraim. I guess that makes sense

@Astyx Ah right, I had no idea I could look at it that way. However, my book states that the columns of the identity matrix are permuted according to the permutation. Why can't I just look at how those columns are permuted? It seems to me that $(321)$ corresponds with
$$
\begin{pmatrix}
0&0&1\\
1&0&0\\
0&1&0
\end{pmatrix}
$$

wait

:l

Something's wrong.

Here the second collumn is left untouched @ShaV

9:52 PM

sorry

Still wrong.

And what you're saying is completely equivalent to what I said (figure out why as an exercise)

Unless you're writing your permutations backward.

Well we know that the third columns should be in the second column, right?

because $3\mapsto 2$

$e_1$ goes to $e_3$, so the first column should be ...

9:53 PM

so that's why my last column is in the second place

It's like moving all the collumns right by one (except the last, which is sent to the first)

$e_2$ goes to $e_1$, so the second column should be ...

Scratch what I said (quite) some time ago about algebraic topology, it interferes with group algorithms

well, if $e_1$ goes to $e_3$, then the first columns of the identity matrix should become the third column in the new matrix?

that is how I see it

So the matrix should be $\begin{bmatrix} 0&1&0\\0&0&1\\1&0&0\end{bmatrix}$.

9:55 PM

ohhhh wait

I think I'm confusing rows and columns here

somehow

Remember that the $j$th column tells you where $e_j$ is mapped.

Think of your elements $1$, $2$, $3$ as the standard basis $e_1,e_2,e_3$

You're not helpful, DogAteMy.

Akiva Weinberger

Sorry

okay I think I'm getting my mistake

I had naively assumed something

9:56 PM

It's not a group morphism anyway is it @Akiva ?

I will slowly reread what you've all said, thanks a lot for the help!

LOL, it's an anti-morphism, @Astyx.

@BalarkaSen It actually looks like any pair of irrationals works here

you want linearly independent

For the circle problem you mentioned

9:58 PM

Feel free to ask if you're still unsure afterwards @ShaV

Actually, that's not true. I take it back

@Ted When will you be around Paris again ? I'll write it somewhere

Wait.

I guess some brain food will replenish my thoughts. I'll think about it then. Bye math.SE

along with the adjective I added that works

Greenleaf is a pretty great band

@Astyx: June 5-11.

Akiva Weinberger

9:59 PM

@Astyx You need to reverse the multiplication also, I guess

Have a nice few minutes without me

/s

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