Mathematics - 2017-01-16 (page 5 of 5)

Simple

22:00

@sarcopsy yes

sarcopsy

Well then it's 0 when Y is 0, and 1 otherwise, so what are the corresponding probabilities?

Fargle

@Perturbative If you mean "strictly contains an open set containing x", I would call the first one you said the standard convention. If you didn't mean "strictly", they're equivalent, since for every point of an open set there is a basis element about that point contained in the set.

Mike Miller

@PVAL Why is there no Lagrangian sphere?

PVAL-inactive

Because of the argument I gave above

Fargle

@Perturbative: wait, no they're not equivalent unless the neighborhood is already open.

PVAL-inactive

22:04

Lagrangians have neighborhoods isomorphic to the 0-section in the cotangent bundle

Fargle

Come to think of it, I've seen both quite a lot. I'm not sure which one would be the standard convention.

PVAL-inactive

and the 0-section in the cotangent bundle of S^2 has nonzero self intersection

Simple

@sarcopsy when $Y$ is 0, the probability is 0, when $Y>0$, it is $\lambda^y e^{-\lambda}/y!$

sarcopsy

@Simple Er I don't think the probability is 0 at $Y=0$

PVAL-inactive

@MikeMiller Or you nuke it from orbit and use Gromov's theorem that $\Bbb R^4$ has no exact Lagrangians (the primitive 1-form of the symplectic form cannot be exact when restricted to the Lagrangian).

Mike Miller

22:06

I didn't want to nuke.

Simple

@sarcopsy it should be $e^{-\lambda}$

Fargle

@Perturbative: I suppose the difference is that the first convention automatically makes every neighborhood of some point a neighborhood of EVERY point it contains, whereas this need not be true for the second convention (e.g. $[0,1]$ is a nbhd. of $\frac{1}{2}$ but not $0$)

PVAL-inactive

The other argument, (that Lagrangians have standard neighborhoods) is not a nuke and is proved in every beginning book on symplectic geometry.

sarcopsy

@Simple Yup, so whats $P(X=0)$ and $P(X=1)$? That's the mass function of $X$

JessyunBourne

@Astyx I heard back from my prof. It was a typo.

Simple

22:09

@sarcopsy $P(X=0)=e^{-\lambda}$ and $P(X=1)=\lambda^ye^{-\lambda}/y!$

JessyunBourne

I asked it as a question on MSE main.

0

Q: Basis for free Abelian group $A_{n}$ of rank $n$

Let $B = \{ b_{1},b_{2},\cdots , b_{n} \}$ be a basis of the free abelian group $A_{n}$ of rank $n$. If $d_{1}$, $d_{2}$, $\cdots$, $d_{n}$ are $n$ nonzero integers, then I need to prove that the set $\{ d_{1}b_{1},\, d_{2}b_{2}, \cdots , d_{n}b_{n} \}$ is a basis for $A_{n}$ if and only if $d_...

sarcopsy

@Simple No, your formula for $X=1$ is wrong

Simple

@sarcopsy $\lambda e^{-\lambda}$

sarcopsy

@Simple No, that's the probability that $Y=1$, not $X=1$

@Simple When is $X=1$?

PVAL-inactive

@Mike @Ted Eh, I guess there isn't a totally real one either as then the complex structure gives you get a vector bundle isomorphism from $TS^2$ to to the trivial bundle.

I've probably done this exercise before.

Perturbative

22:14

@Fargle, You're right, but I guess it seems kind of odd to allow the use of one or the other definition as a 'sort of convention', when both definitions seem to be fundamentally different

Alessandro Codenotti

I've always used the definition that a nbhd of $x$ is a set containing an open set containing $x$, more often than not we just say "assume all nbdhs in the following argument are open" though since if they're not you can just replace them with this smaller open set

Kaj Hansen

I finally got the opportunity to go on one of my favorite rants on main: math.stackexchange.com/questions/2100766/….

PVAL-inactive

Usually when I say neighborhood I mean the closure of an open set.

Kaj Hansen

Even though it is a thread better-suited for educators

Simple

@sarcopsy when $X=1$, are we summing poisson probability? I am not quite understand indicator function

PVAL-inactive

22:19

@Kaj The students are not taught that rote memorization is easier, it is just a fact that in the way they are tested it just is.

sarcopsy

@Simple An indicator function is 0 when the condition (in this case $Y>0$) is false and 1 when the condition is true. You are right in that you are summing the Poisson, but there's an easier way to calculate it without having to do any sums

Kaj Hansen

@PVAL-inactive, that's not quite what I'm saying. It's more that the rote-memorization path is what's presented by default

I'm near-certain I did not see the derivations of any of that stuff I listed when I was in high school :/

It was always just "here's a formula and it works :D"

PVAL-inactive

These kind of derivations do very little for students unless they are required to produce similar derivations.

That is my experience in teaching them.

Simple

@sarcopsy I don't see the easy way

sarcopsy

@Simple What's $P(Y=0) + P(Y\gt 0)$?

Kaj Hansen

22:26

I still think it can save a lot of effort. For example, for a long time, I didn't have the "distance formula" memorized, and every time I'd draw the two points in the plane and the right triangle with those the endpoints of its hypotenuse and reason it out.

Same thing with the inverse trig function derivatives.

Random Variable

@DanielFischer Is there an example of a function (preferably not overly contrived) that has a Maclaurin series expansion that converges at only a finite non-zero number of points on the boundary of its disc of convergence?

PVAL-inactive

I'm certainly not telling you that these aren't valid or good ways to think about these things.

Kaj Hansen

I memorized them after a while just from having used them so much, but initially I found that to be a huge pain. It saved effort for me.

PVAL-inactive

They are just completely ridiculous to try and give to 99% of the gen. pop.

Semiclassical

Yeah. If I ever memorized what stuff like $\cos(\arcsin x)$ simplifies to, I've forgotten it; I always just do it by drawing the triangle.

PVAL-inactive

22:28

@Semiclassical sure but I am guessing you (and probably everyone else in the chat) knows the derivative for arctan arcsin etc.

Simple

@sarcopsy $\sum_{y=0}^{\infty}\lambda^{y}e^{\lambda}/y!$, how to calculate this sum

Semiclassical

Right, I was just giving another example.

By contrast, some things we use often enough that we do memorize them.

sarcopsy

@Simple Do you know that Poisson variables are only non-negative?

Semiclassical

like, say, the Taylor series for the exponential function.

we know why it's like that, and how to derive it if we needed to.

Kaj Hansen

@PVAL-inactive, I suppose you're right, but I think a lot of that is because most of the gen. pop. has weak fundamentals due to the same stuff happening in earlier courses. A solid understanding of algebraic manipulations and why they work ends up being 90% of one's success in calculus.

Semiclassical

22:29

but at this point I just automatically recognize it as such.

then again, I didn't learn it by setting out to memorize the formula. I've just used it so many times that it's second nature.

Simple

@sarcopsy yes, since we have exponential with it

Kaj Hansen

Granted, I don't have any experience actually teaching simple stuff like arithmetic and "solving for x", etc. So I'm not as acutely aware of the challenges that might be involved with teaching it the way I have in mind.

Semiclassical

I run into it from time-to-time when tutoring intro physics.

Fargle

@Semiclassical Laplace transforms are an example in college mathematics where memorization is stressed, at least in my experience.

Semiclassical

Tables get emphasized a lot, it's true.

PVAL-inactive

22:32

I have had plenty of calculus students that have trouble with stuff like that.

Kaj Hansen

Plus I think not seeing the derivations makes math seem lifeless and sort of contributes to people thinking math is sort of a mysterious witchcraft

Fargle

I can do the derivation itself--it's just an improper integral with some spicy techniques--but often all the properties are summed up in a neat box on the back inside cover.

sarcopsy

@Simple No, I mean that Poisson random variables are meant to be used for $Y\ge 0$ only.

Fargle

@KajHansen A lot of the reason I decided to learn math from an early age is my desire to know why this and why that.

Semiclassical

g(x)=f'(x) -> G(s)=s F(s)-f(0)

Perturbative

22:33

@KajHansen I remember how many of my fellow students used to complain about all the trig derivatives they needed to memorize in a Calc I class, especially before an exam. Sadly they didn't realize that all you needed was the derivative of $sin(x)$, $cos(x)$ and the inverse function theorem to rederive everything

Fargle

It was very unsatisfying to me when I was shown the quadratic formula without a proof in the book, so I asked my tutor to show me and I've never forgotten.

PVAL-inactive

Don't call it "inverse function theorem" please.

Semiclassical

@Fargle The funny thing about completing the square vs. the quadratic formula: If you do grad-level physics, you end up doing quite a bit with Gaussian integrals.

sarcopsy

@Simple If $Y$ takes on only non-negative values, then what is $P(Y=0) + P(Y>0)$?

Perturbative

@PVAL-inactive Why not?

Akiva Weinberger

22:34

I memorized the Mayer–Vietoris theorem. Good luck getting me to rederive that for every exercise.

As a recent example for me.

Semiclassical

And the main trick you do there? Rewrite $ax^2+bx+c=a(x-x_0)^2+d$ and argue that shifting by $x\mapsto x+x_0$ doesn't change the integral at all.

PVAL-inactive

I think the mneomic I learned for the quadratic formula is not PC enough to have remained in the curriculum.

I think I said it to students once and was met with laughter.

Semiclassical

So you end up completing the square a lot in upper-division physics :)

Akiva Weinberger

@KajHansen Do people really teach the distance formula as being separate from the Pythagorean theorem? That's so weird.

Fargle

@Perturbative A buddy of mine and I taught our AP Cal class this, as well as the habit of drawing triangles to simplify everything.

Kaj Hansen

22:36

Our mnemonic was just a song set to pop goes the weasel. What mnemonic are you familiar with that PC comes into question? @PVAL-inactive

Fargle

@AkivaWeinberger They don't necessarily teach it as separate, so much as they don't show the connection. And it can look arcane to someone who's not very experienced with algebra.

Semiclassical

I'm trying to imagine someone devising a mnemonic for Cardano's formula.

Kaj Hansen

Yeah @AkivaWeinberger. Apparently this is a thing in college precalc courses as well, at least where I was an undergrad. I'd show this to all my tutees and they'd be really excited at how much easier it is to think about

Simple

@sarcopsy I am not sure

PVAL-inactive

A negative boy couldn't decide to go to a radical party. The boy was square and missed out on 4 awesome chicks. The party was over at 2 am.

Semiclassical

22:37

Yeah.

oh dear.

Kaj Hansen

LOL

Fargle

@PVAL-inactive I'm not ordinarily a fan of mnemonics, but this one is pretty great.

Semiclassical

A mnemonic for Cardano's formula, I think, would need to either be a rap or a musical number in the vein of "Modern Major General"

Perturbative

@Fargle The students of that class should've bought you a bottle of champagne for saving them from memorizing that big list

sarcopsy

@Simple Well does $Y=0$ and $Y > 0$ cover all the possible values that $Y$ can take on?

Kaj Hansen

22:39

@AkivaWeinberger, yeah, there's gonna be things that ultimately one needs to memorize just because it comes up over and over and the derivation is hard. Gauss-Bonnet for manifolds with boundary comes to mind, e.g. But at the point where the derivations become too difficult to do every time, pretty much everyone is sympathetic to what we're saying.

Simple

@sarcopsy yes

Semiclassical

Hence why we have theorems, really.

Fargle

@Perturbative The students didn't appreciate it as much as the teacher. We ended up doing the entire unit on sequences and series so she could catch the AB class up (our school did things weird and grouped the two into the same room, but gave AB two class periods and us just one).

Semiclassical

Because otherwise we'd have to prove everything over and over again.

sarcopsy

@Simple What is the probability that $Y$ takes on a non-negative number then?

Daniel Fischer

22:41

@RandomVariable I don't think there is an easy example. The easy series either diverge on the whole circle or converge at all but finitely many points, as far as I know.

Akiva Weinberger

When you're given 50 problems on solving quadratics it probably makes sense to have the formula memorized at some point. But

SAWblade

hello all! :)

what's everyone up to?

Simple

@sarcopsy $\lambda^{y}e^{-\lambda}/y!$

Akiva Weinberger

then again, I've been mostly self-taught for all of my math, so I don't really know the other students' struggles

sarcopsy

@Simple No I mean the probability $P(Y\gt 0)$

Simple

22:46

@sarcopsy it is a sum

Fargle

@Semiclassical I think that brings to light the difference between memorization by rote and memorization by understanding. I'm sure you wouldn't call your memory of the properties of injectivity and surjectivity similar to that of someone using, say, PVAL's mnemonic for the quadratic formula.

sarcopsy

@Simple Do you know that for a function to be a probability mass function then the sum of the function over all the all the possible values the random variable can attain is 1?

Socrates

@KajHansen are you familiar with change of basis stuff?

Kaj Hansen

Sure @Socrates, what do you have in mind?

PVAL-inactive

@DanielFischer I think the unabomber worked on something similar. Describing and classifying subsets of the real line which a holomorphic function on the upper half-space can converge at.

sarcopsy

22:49

@Simple i.e. The probability that 'something happens' is 1

Simple

@sarcopsy yes, I do. $P(Y>0) = 1 - P(Y=0)$

sarcopsy

@Simple So you're done right

Socrates

@KajHansen do you read german?

Kaj Hansen

I wish :/

Socrates

ok, just asked wether I have to type the assignments

Ali Caglayan

22:51

Hey @KajHansen

Kaj Hansen

Hey there @Ali

Ali Caglayan

I was asked an interesting question today

Kaj Hansen

What was it @Ali ?

Random Variable

@DanielFischer But one could probably be contrived?

Ali Caglayan

Is it possible to pick a group at random?

Simple

22:51

@sarcopsy Thanks your help. If I want to calculate $E(|Y-X|)$, can I use the linearity

Kaj Hansen

Well, is it possible to pick an element of $\mathbb{R}$ at random @Ali ?

Or an element of $\mathbb{N}$ even ?

Kasmir Khaan

If 5 -letter ``words'' are formed using the letters A, B, C, D, E, F, G, how many such words are possible for each of the following conditions:
(a) No condition is imposed.
7^5

(b) No letter can be repeated in a word.
(c) Each word must begin with the letter A and letters can be repeated.
(d) The letter C must be at the end and letters can be repeated.
(e) The second letter must be a vowel and letters can be repeated.

Socrates

how do you make a subnote before something, such that it doesn't look wierd?

Ali Caglayan

@KajHansen yes that was my thinking

however the full question was as such

Socrates

$=_BA$

Ali Caglayan

22:52

You know what the rank of a group is right?

The question was: What is the expected value for the rank of a random group

sarcopsy

@Simple Not immediately, since $|Y-X|$ isn't linear, so you would split into cases where $Y \ge X$ and $X \gt Y$

Ali Caglayan

If I make the question more well defined by limiting the order up to some n

Kaj Hansen

I think @AliCaglayan. Groups can be decomposed as $\mathbb{Z}^k \oplus \{other stuff}$ for $k \geq 0$, and if we write this so that $k$ is maximal, that's the rank

Ali Caglayan

I wonder what could be said about it

@KajHansen Thats the case for abelian groups

Kasmir Khaan

guys is AAAAA consedered to be a word ?

Ali Caglayan

22:55

In general the rank is the smallest size of the generating set

Kaj Hansen

What is it in general @Ali ?

Ah ok

Ali Caglayan

for non abelian groups

Kasmir Khaan

If 5 -letter ``words'' are formed using the letters A, B, C, D, E, F, G, how many such words are possible for each of the following conditions:

Ali Caglayan

If I knew GAP I might be able to write a program to see the distribution of ranks for groups of order less than some integer

Kaj Hansen

Can someone enlighten me? The difference between the direct sum and direct product of infinitely many sets / algebraic structures differs in that the elements of the direct sum $(a_1, a_2, \cdots)$ are such that $a_k = 0$ for all but finitely many $k$. First, what motivates this distinction? Second, this is very similar to the differences between the product topology and box topology on a product of topological spaces, where the distinction is the same. Is there a connection?

Ali Caglayan

22:56

I have no reason to think it will be nice but something may get out

@KajHansen In the category of abelian groups, direct product/sum is a biproduct

Simple

@sarcopsy I don't see how to calculate $E(|Y-X|)$ when $Y>X$

Kaj Hansen

@Ali, there might be data available on groupprops

Socrates

@KajHansen maybe lets do it in private? This will take a while as I have hit a understanding wall.

Ali Caglayan

@KajHansen also the difference between direct sum and product might have to do with $\hom(A + B, C)=\hom(A, C) \times \hom (B, C)$

Kaj Hansen

@Socrates chat.stackexchange.com/rooms/51908/kajsocrates-chat. ?

Ali Caglayan

23:01

and also $\hom (A, B \times C) = \hom (A, C) \times \hom (A, B)$

Then there are the generalisations of these

Kaj Hansen

That's interesting

Ali Caglayan

which gives you some perspective on why direct products and sums are different

Kaj Hansen

I'm going to be silent for a bit while I help Null

Ali Caglayan

ok

Daminark

Hey everyone, this is something I'm having a bit of trouble on. Now I'm really not sure how to pull this off

So how would one prove that if two curves have the same winding number, they are homotopic?

(In the punctured plane)

I'm trying to see if I can take a curve of winding number $n$ and homotopy it to the curve that travels the circle $n$ times, but I'm not sure quite how to do it

Ted Shifrin

23:08

I doubt you can prove that with what you know, @Daminark.

I don't think you need to prove it to do what you're trying to do.

Daminark

This is a different homework problem

Ted Shifrin

Oh? What is this problem?

Daminark

To prove that two curves in the punctured plane are homotopic if and only if they have the same winding number around the origin.

Ted Shifrin

And of course you know nothing about the fundamental group.

Daminark

Not officially, no. Our professor did say that it was secretly the fundamental group tho

*though

Ted Shifrin

23:12

Can you prove it even for curves that just go around the unit circle (not necessarily $z\mapsto z^n$, of course)?

We really need to lift maps $g\colon [a,b]\to S^1$ to maps $\tilde g\colon [a,b]\to\Bbb R$.

robjohn

@amWhy: sorry, I had to leave. I will be back on my computer in a while. I find the mobile interface hard to deal with.

Ted Shifrin

I don't know a more elementary way of doing it.

Heya @robjohn :)

I agree re mobile interfaces.

@Kaj: I hate to say it, but the difference between product and sum is a categorical one. I.e., if you have maps $f_\alpha\colon Z\to X_\alpha$, then you get a map $f\colon Z\to \prod X_\alpha$, and if you have maps $g_\alpha\colon X_\alpha\to Z$, then you get a map $g\colon \oplus X_\alpha\to Z$. So it's dotted lines in diagrams :D Karim will be so happy to explain it to you :D

PVAL-inactive

Just show that you can homotope to a map onto the unit circle, and then show you can homotope any map to one of constant speed. There's only one of these with a fixed winding number. In the topological category the second part isn't obvious, but it is pretty easy to prove it in the smooth category (if you assume your curves are smooth).

Ted Shifrin

@PVAL: Am I missing something silly? You're saying something wrong.

PVAL-inactive

What am I saying wrong?

Ted Shifrin

23:20

Constant speed isn't the issue. It can backtrack a bunch of times. I don't know how to do this without path lifting.

Mike Miller

Night.

PVAL-inactive

Constant speed is assuming the derivative exists everywhere

Ted Shifrin

How do you prove you can homotop it to a constant-speed one?

It does exist. You're assuming nonzero.

So if I have a general smooth closed path in $S^1$, how do I homotop it to $z^n$ for some $n$?

Night @MikeM.

Daminark

We do have the assumption of smoothness, actually.

Night @Mike

Ted Shifrin

Yeah, smoothness is irrelephant here.

robjohn

23:22

@TedShifrin yeah. Though I do see that there is now a way to reference previous messages!

Ted Shifrin

Well, pings work on mobile the same way, @robjohn.

Oh, reference.

@Daminark: Your professor didn't write an exercise having you prove you could lift as I said, did he?

Akiva Weinberger

@robjohn It's easier than on a PC, in my opinion

Ted Shifrin

Heya, DogAteMy!

Akiva Weinberger

Hello! I should be studying for finals.

Ted Shifrin

I find them equally cumbersome, DogAteMy.

Well, go study :) Bye!

Daminark

23:24

@Ted No, we don't have anything about path lifting at all

robjohn

@AkivaWeinberger is it not just a click on the arrow to the right of the comment?

Akiva Weinberger

...Oh.

I take it back, then.

My excuse is that I go on chat almost exclusively on mobile.

Ted Shifrin

Oh, I didn't know that trick, @robjohn :P

We need a seminar in chat technique.

robjohn

@TedShifrin I encourage the use of links because it helps unwind the many threads that go on here simultaneously.

Ted Shifrin

@Daminark: Differential forms and Stokes's Theorem are great, but they will not product homotopies for you. I expect your professor expects an ad-hoc somewhat pictorial argument.

@robjohn @robjohn: Oh, absitively.

Krijn

23:26

@Ted is in a punny mood

Ted Shifrin

Nah, this is my usual, @Krijn. Just ask my poor long-suffering ex-students.

Mike Miller

That's not a pun.

Daminark

Including path lifting disguised, the other problems that have anything to do with forms are just one about proving exact $\iff$ integral depends on endpoints, and some computations of closedness, pullback and the such

Ted Shifrin

It's more a spoonerism.

robjohn

Mobile chat is much better than when I last tried it.

Daminark

23:27

That's also possible

Krijn

You English speaking persons have such a weird language

Akiva Weinberger

@Daminark Is this for algebraic topology or for complex analysis

PVAL-inactive

@Ted If you assume the map is smooth you can count the number of times it backtracks

Daminark

It's my real analysis class

@Akiva

robjohn

The Chatjax bookmark works or I could not do chat on mobile at all.

Ted Shifrin

23:28

@Daminark: Assuming some sort of "niceness" (which you cannot hope to prove), you can break the curve into pieces at points where it crosses itself and at points where it is tangent to a radius line for the circle. Then you should be able to work with those pieces to do it.

Akiva Weinberger

Oh. So you can't do fancy stuff like integrating around $1/z$ in the complex plane.

Mike Miller

"fancy"?

Ted Shifrin

DogAteMy: He's already integrating $d\theta$.

But that won't produce a homotopy.

Yes, @robjohn: I'm glad I worked out the bookmark trick — cumbersome as it is — from someone else.

Daminark

We don't have all too much machinery to work with, we didn't even reach homotopy in class yet, he sent out an email like "Hey everyone, so one more thing you need to know, homotopy is..."

Ted Shifrin

@PVAL: Yeah, you're basically doing basic intermediate value theorem stuff with the smooth lift.

Professors often do these sorts of things, @Daminark, and don't think through things carefully. Especially professors in courses like yours.

They want to see how resourceful or amazingly creative you can be.

robjohn

23:31

It's not that cumbersome. Or are you talking about installing it with copy and paste. That only needs to be done once. Then I just clivk

Mike Miller

They might do these things and think through them carefully.

Ted Shifrin

Right, @robjohn. I was talking about the original installation.

robjohn

click on the likn

Daminark

But this part of the class is really just differential forms in the (punctured) plane, just that our professor is trying to get us to see how we're able to use the theory we have and pull some stuff off

robjohn

Where did the spell checking go?

Ted Shifrin

23:32

looks under the couch for spell-check

robjohn

The spell-check is in the mail

Ted Shifrin

:)

PVAL-inactive

@Ted My point is assume the map goes to some point x and then backtracks to some y (with the derivative only changing direction once!) . You can easily homotope that piece of the map so it goes directly to y without the derivative changing direction.

robjohn

BBL

Pissed off layman

cya

Ted Shifrin

23:34

@PVAL: You're essentially suggesting my ad-hoc discussion of breaking it into pieces at crossing points and change-of-direction points.

Akiva Weinberger

@TedShifrin Wouldn't it produce the required lift, though?

Ted Shifrin

@AkivaWeinberger Oh, good point.

@Daminark: DogAteMy makes a good point. Integrating $d\theta$ gives you the desired map I was talking about. Then you can do the homotopy up in $\Bbb R$. Ted is a dope

Kaj Hansen

Ted is a dope

3

Ted Shifrin

@PVAL: I didn't understand your totally real bundle isomorphism remark. You would get $TS^2 \oplus TS^2$ trivial, right?

Akiva Weinberger

Is the "a" crossed out? It looks like it but it's hard to tell @KajHansen

Kaj Hansen

23:36

(As the kids these days would say)

Yeah @AkivaWeinberger

Ted Shifrin

Hi, @Kaj. You saw my categorical answer to your question?

Kaj Hansen

I did @TedShifrin, thanks. I saw a post by Pete Clark with further detail as well

Ted Shifrin

Even I admit you need that much category theory :P

Daminark

Ah, aight, I'll try that.

PVAL-inactive

@TedShifrin Well a totally real structure on $S^2$ embedded inside $\Bbb R^4$ would give you an isomorphism between $TS^2$ and $J(TS^2)$ with $TS^2 \cap J(TS^2)=0$ but every $S^2$ inside $\Bbb R^4$ has trivial normal bundle.

So you would get an $S^2$ in $\Bbb R^4$ with normal bundle $TS^2$.

Ted Shifrin

23:38

And certainly the product topology arises as the coarsest (or one of those) topology that makes a map into the product be continuous if all the components are. @Kaj ... I guess the box topology works for maps the other way? Check it out.

Daminark

And lol category theory, I find it really funny that my math professor now is more opposed to category theory than my old physics TA

Krijn

Why are people so against category theory?

Ted Shifrin

@PVAL, I was using $J$ to give me the second summand, which is the normal bundle. So we agree.

@Krijn: I have never liked formalism in mathematics. Just my taste.

Daminark

My physics TA from last year has been teaching us math on Sundays, and he's really heavy with talking about things abstractly, while my math professor is all, yeah it's the dark arts/abstract nonsense

Ted Shifrin

@PVAL: Rather than using topology to say the normal bundle is trivial, I was saying the sum of the two tangent bundles has to be trivial. Which is certainly a contradiction.

Why is your physics TA teaching you on Sundays, @Daminark? And why math?!!

PVAL-inactive

23:41

@Ted You are still using topology to show $TS^2 \oplus \nu$ is trivial though.

Akiva Weinberger

So I guess we can think of it as: $f$ is some group homomorphism we know nothing about, whose domain includes all of the $X_a$. Then the direct sum is everything we know must be in the image of $f$, and two elements of the direct sum are equal only if we can prove they're equal.

Ted Shifrin

I'm using $\Bbb C^2$ has trivial tangent bundle. Nothing much there.

mumbles to DogAteMy that he has finals for which to study

mick

Hi all

Daminark

So unless Schlag is teaching the class, in which case it will be a firestorm, undergrad complex analysis here is really light. At some point last year, my physics TA was like "Yeah, I don't like how you don't get a good course in complex analysis until grad school, you want to meet up on Sundays to do the stuff?"

mick

0

Q: Is $\lim_{n \to \infty} f^{[n]} ( g^{[n]} (z) ) $ analytic?

Let $^{[*]} $ denote composition. Let $f(z) = \sqrt z$ and $g(z) = z^2 + 1$ Define $Q(z)$ as $$ Q(z) = \lim_{n \to \infty} f^{[n]} ( g^{[n]} (z) ) $$ Is $\lim_{n \to \infty} f^{[n]} ( g^{[n]} (z) ) $ analytic for $Re(z) > 11$ ?

complex-analysis limits dynamical-systems radicals

Ted Shifrin

23:43

No, that's not quite right, DogAteMy. This needn't be an embedded direct sum/product.

Akiva Weinberger

What does "embedded" mean here @TedShifrin

Daminark

So we started doing that, I wasn't really following along all too well since I had way less than sufficient background, but it all looked really cool so I kept going.

Over the summer we finished complex, so he was like "Alright guys, now my idea is that we can learn the math to do a rigorous formulation of classical mechanics."

PVAL-inactive

@Ted sorry why is TS^2 \oplus TS^2 non-trivial?

Ted Shifrin

Thinking of the domain as containing all the $X_\alpha$ is misleading, but I suppose it's right.

Daminark

ODEs, forms, calculus of variations, Sturm-Liouville, differential/Riemannian/symplectic geometry

Ted Shifrin

23:45

@PVAL: Can't I do the Euler class of the sum?

PVAL-inactive

No its 0

Ted Shifrin

Oh, no.

PVAL-inactive

I think TS^5 \oplus TS^5=0 for instance.

Ted Shifrin

Right, I need $c_1$. Ooops.

Any time you have an odd factor in a product of spheres, you get trivial tangent bundle.

Akiva Weinberger

@TedShifrin The direct sum need not contain copies of its summands?

Ted Shifrin

23:46

But it's not true for $S^2\times S^2$. Ah, so I need to pull back to that, and then do Euler.

PVAL-inactive

This is a bundle over S^2

OneRaynyDay

Hey guys - here's an interesting question(or at least to me) that I haven't been able to figure out.

PVAL-inactive

not over S^2 \times S^2

Ted Shifrin

@DogAteMy: Yeah, of course it does. But the way you say it makes it feel more like union than direct sum/product.

OneRaynyDay

If you have $n_1 \oplus n_2 \oplus n_3 ... \oplus n_k = 0$ the oplus is a XOR

Ted Shifrin

23:47

@PVAL: Right, I only get it on the diagonal of $S^2\times S^2$. Hmm ... There should be a way.

Akiva Weinberger

@TedShifrin In any case, that was my intuition for the tensor product $\otimes$. I was just adapting it for the direct sum.

OneRaynyDay

And you can subtract any number from $n_{1,...,k}$ (just once, so you can only subtract from 1 number) where the number is from 1 to that number,

Ted Shifrin

I think it's wrong for tensor product, DogAteMy.

OneRaynyDay

Can you prove that it'll never equal 0?

Akiva Weinberger

$f$ would be a bilinear map.

Ted Shifrin

23:48

Oh, for groups it's OK, but not for vector spaces ...

Daminark

And it's still going on up to now. What I've noticed about him in this Sunday class is that he usually presents things in abstract form. So as opposed to using the implicit function theorem or even submanifolds of $\mathbb{R}^n$, he just jumped to saying "Alright, let's say we have a set, we don't want to assume any structure, topological or otherwise" and moves on to talk about charts, atlases, differential structures, all that

Ted Shifrin

@Daminark: That's the usual infatuation of new learners with abstract crap :)

Krijn

@OneRaynyDay I do not understand the question

Ted Shifrin

@PVAL: Interesting. If I had a complex structure, then I could use $c_1$ to make the argument.

OneRaynyDay

@Krijn So basically in words: If you have some numbers and they're xor'd with each other, and you can take away any number from 1 of them, prove that it will never equal 0

Akiva Weinberger

23:49

By "subtract" do you mean actual subtraction, or just getting rid of it from the equation with the opluses @OneRaynyDay

@OneRaynyDay Do $\oplus n_i$ to both sides?

OneRaynyDay

@AkivaWeinberger like from $n1 \oplus n2 \oplus n3... \oplus n_k = 0$

PVAL-inactive

@Ted It has a nonvanishing real section.

OneRaynyDay

to $(n_1-x) \oplus n_2 \oplus n_3... = 0$

Ted Shifrin

DogAteMy: I'm back to saying it doesn't make sense for tensor product. (When you tensor with $0$, you kill everything.)

OneRaynyDay

@AkivaWeinberger But it's not a straight up subtraction - it's a subtraction within 1 of the numbers in the expression

PVAL-inactive

23:51

So its just a matter if the complement bundle is $V+\Bbb R$ with $V$ even Euler class or $V$ odd Euler class.

Akiva Weinberger

@TedShifrin Right. And you can show that $f(x,0)=0$ for all $x$ and all bilinear $f$.

Daminark

@Ted Yeah, and the surprise comes from how it's the reversal of what you usually think of, with mathematicians being more, chase abstraction for its own sake

OneRaynyDay

where x is greater than 1, but the expression of $(n_k-x) \geq 0$

Ted Shifrin

So a copy of each factor doesn't sit inside the tensor product in any obvious way, DogAteMy.

Right, @Daminark. I get it. But this is a UC physics grad student, who probably was a math major as an undergrad. The ones I know were :P

Akiva Weinberger

Ah, OK. My intuition about bilinear $f$s we know nothing about should still work, though @TedShifrin

Daminark

23:53

I vaguely remember that he was more theoretical/mathematical physics

Possibly not unrelevant detail: he did all his schooling in Russia before coming here

Ted Shifrin

@PVAL: So what does a nonvanishing section have to do with $c_1$ of a (complex) rank $2$ bundle?

Akiva Weinberger

Well, it wouldn't be the image of $f$, it would be the vector space generated by its image, actually

Ted Shifrin

Definitely not irrelevant (or irrelephant), @Daminark.

PVAL-inactive

@TedShifrin absolutely nothing.

Ted Shifrin

DogAteMy: The tensor product still doesn't contain copies of the factors.

Akiva Weinberger

23:54

So?

Ted Shifrin

I don't know. You were saying you were getting intuition for direct sum from tensor.

Daminark

Oh the first time you said irrelephant I thought it was just a coincidentally funny typo @Ted

Ted Shifrin

That's why I insisted the second time, @Daminark. Now go do your work :P

Daminark

Yeah, probably should finish this while there's still time

Well, see you around everyone!

Ted Shifrin

Bubye.

Akiva Weinberger

23:56

To spell it out, I'm thinking of the tensor product $V\otimes W$ like this. $f$ is some bilinear map we know nothing about, whose left input is $V$ and whose right input is $W$. The tensor product is @TedShifrin

Ted Shifrin

There's a duality missing in what you're doing, DogAteMy.

But otherwise ...

Akiva Weinberger

the vector space containing its image, where we only consider the things that we know must be in that vector space, and we only consider two things to be equal if they are provably so.

Ted Shifrin

I have no idea what that means.

Akiva Weinberger

(The first thing is just saying it's the vector space generated by its image)

Ted Shifrin

You're still off by duality. Bilinear maps on $V\times W$ leads to $(V\otimes W)^*$.

Akiva Weinberger

23:58

@TedShifrin I mean it's everything of the form $\sum kf({\bf v},{\bf w})$.

@TedShifrin I'm not looking at all $f$s, just one $f$ that's assumed to be as general as possible.

It would turn out that that $f$ would have to be isomorphic to $\otimes$ (that is, $f(v,w)$ corresponds to $v\otimes w$).

Ted Shifrin

I don't have the patience to untangle this right now. So I'm gonna quit and you should go study.

Akiva Weinberger

All right, fine. Bye.

Ted Shifrin

What you're saying isn't right.

Ali Caglayan

hey @TedShifrin

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$Mathematics$ Mathematics