Mathematics - 2017-04-03 (page 2 of 3)

5:26 PM

@MaryStar yep.

@BAYMAX nah. My knowledge of Chua's circuit is limited to the Wikipedia page

@TedShifrin just checked, and there's nothing evident in the pile at the front office

So as far as I can tell, the graders should have gotten the right number of papers

Akiva Weinberger

5:44 PM

@BalarkaSen How about this: Give the boundary a uniform density and a gravitational pull. Each point will be pulled towards the boundary; at $t=\infty$, assuming no point ever feels zero gravitational pull, each point will end up at one of the points on the boundary.

what are you trying to do?

Akiva Weinberger

It shouldn't be too hard to show that no point ever feels zero gravitational pull, or too show that this is continuous at infinity.

This fails for, say, deformation retracting a disk onto its boundary, since the point at the center feels zero force.

Ok.

Akiva Weinberger

@MikeMiller It's this deformation retract where it's intuitively obvious that it exists, but kinda tedious (and inelegant) to actually describe one explicitly

I think your earlier description worked and wasn't too tedious.

5:57 PM

@arctictern To see where Bott Periodicity on KO comes from, first introduce $\mathcal F_k$, the space of Cl_k-linear Fredholm operators on a Hilbert Cl_k-module. Then the index defines a Cl_k-module, and it is not too hard to see that this provides an equivalence of pi_0 with the appropriate degree KO-groups.

Actually, just as you can prove (as Atiyah and Singer do) that $\mathcal F_0 \simeq \Bbb Z \times BO$, and that $\mathcal F_{k+1} \simeq \Omega \mathcal F_k$ and that $\mathcal F_k$ is a classifying space for Cl_k K-theory.

Because Cl_k K-theory is the same as Cl_{k+8} K-theory by the periodicity theorem for Clifford algebras, you get Bott periodicity

We finally have a math major in the chat!

@Major_Math

Hi Demonark, @MikeM

How's it going @Ted?

Next question.

What did you learn in diff top today?

:/

6:14 PM

Didn't have it today, difftop is a Tuesday/Thursday class for me

Ohhhh ...

I guess it was for me too when I took it back in 1972 :P

Haha, yeah

I find that I kinda prefer MWF classes

Hello folk

Hi chat

Like especially when tired, my attention span is going to have to be pushed to the limit in order to stay focused properly for an hour and a half

Hey @Eric and @Semi!

6:22 PM

I liked TR for undergrad diff geo, because some days get a bit involved and I don't want to stop in the middle of a big proof. But otherwise I pretty much always taught undergrad classes MWF.

Hi @Eric

how is everybody's day going

Pretty well, we started Banach-Mazur which was very amusing, and also very powerful, as we came to realize

How're you doing with grad geometry, @Eric?

so far it's been mostly review, we've defined affine connections, proved that the metric fixes the Levi-Civita connection and a bunch of review of some diff top stuff

next class is geodesics and stuff

I'm guessing he's doing everything in terms of $\nabla$ as opposed to diff forms

6:25 PM

in my reading course with neves we've been doing willmore surface stuff and conformal stuff

yeah he's using do Carmo

I figured ...

So you're trying to get to the proof of the Willmore conjecture?

(in the reading class)

and so far he's sticking pretty closely to it so no differential forms are on the horizon

yeah we're reaching for a bunch of special cases first and we'll get to the proof if we have time

right now i have to read a paper by robert bryant on willmore surfaces

I highly endorse anything written by Robert.

I think I read that many years ago, actually.

yeah glancing at it it looks like he'll be using moving frames

gets excited

Of course :)

I don't think that paper uses the method of equivalence, though, but I might be misremembering.

6:30 PM

anyway i have many computations i must do

so i shall be off

Keep me posted :P

farewell chat folk

I will @Ted

Bubye.

See you @Eric!

Lol I'm curious about this moving frames stuff

You can see a rudimentary exposition for surfaces in section 3.3 of my notes.

I keep thinking I should be ambitious and type up my graduate course notes, but it hardly seems worth all the trouble. It would be a major project to do a good job.

6:33 PM

Right, yeah we'll be doing some stuff with that when we get to your book

Yeah typing up notes is pretty tough, especially when it's all in one go

@BalarkaSen I need your help with something so I am calculating the $\pi_n(SO_3)$ for $n \geq 3$

Well, I'm reasonably experienced. My three published textbooks were all typeset using all my LaTeX stuff. But the worst part for the grad geo stuff (a whole year course) would be drawing figures.

Hi, Karim!

I proved that $SO_3$ is homeomorphic to $RP^3$

hi @TedShifrin

I know a few people who are typing up daily notes for classes, and in difftop the guy who's doing it is actually incorporating a lot of nice diagrams

how are you

6:35 PM

Hey @Adeek!

Do you know about the long exact homotopy sequence for a fiber bundle?

yeah @TedShifrin

that is what I was thinking about

@Daminark hi

@TedShifrin so we have the following projection map

OK ... So have you considered $S^1\to SO(3) \to S^2$?

$p : S^3 \rightarrow S^3/ \{-+1\}$

this will be a covering map

Use \pm

6:36 PM

so we get the following long exact sequence

Well, my way you need to know the homotopy groups of $S^2$, and your way you need to know them for $S^3$. Neither is trivial to know.

$\pi_{r + 1}(S^3) \rightarrow \pi_{r + 1}(SO_3) \rightarrow \pi_r(\pm 1) \rightarrow \pi_r(S^3) \rightarrow \pi_r(SO_3) \rightarrow \pi_{r - 1}(\pm 1) \rightarrow $

@TedShifrin we can use the fact that we know homotopy groups of $S^3$

Oh, you know them for $S^3$? That's nontrivial. Then it's easy.

is it true that $\pi_r(\pm 1) = 0$ ?

Except for $r=0$, sure.

PVAL-inactive

6:38 PM

What's the 145673th homotopy group of $S^3$?

Way to go, @PVAL :P

okay awesome so we have aside from $r = 0$ we have isomorphism between $\pi_r(S^3) \rightarrow \pi_r(SO_3)$

I don't care about that. It's more important to see why pi_r(S^0) is obvious.

Be careful.

Well, I'll leave the discussion to the actual topologists in the room :P

Alessandro Codenotti

Hi chat

6:40 PM

okay :D @MikeMiller @PVAL-inactive you agree so far since we have $\pi_r(\pm 1) = 0$ for $r \neq 0$ we have isomorphism $\pi_r(S^3)$ and $\pi_r(SO_3)$ ?

You also don't need the homotopy LES to see covering space has the same pi_* as the base but sure

Hey @Alessandro!

Be careful, Karim. What about $r=1$?

Hi @Alessandro.

oh right

PVAL-inactive

@BalarkaSen They're both a direct consequence of lifting property of fiber bundles.

6:41 PM

@TedShifrin For $r = 1$ they are isomorphic...?

les is slightly more tedious

Oh you mean why it follows from LES

yeah

@PVAL Yeah, but...

Um, @Balarka, really?

6:42 PM

Sorry take that back

You or me?!!

In any case I hope it's clear why pi_3(S^0) is trivial. Otherwise I'll step away.

I take it back. I misunderstood the question. Ofc they are not isomorphic for $\pi_1$.

PVAL-inactive

I think the lifting property of fiber bundles is probably the most important theorem in algebraic topology.

Anyone wanna argue that point?

I've thought about how to teach a course where that comes in almost instantly.

Hey @Astyx!

6:44 PM

Hi chat

oh right @TedShifrin yeah for r = 1. We have $\pi_0(\pm +1) = Z_2$ as it is the number of path connected component

What's up ?

Not too much, how about you?

Started revisions for my exams

So far so good

Nice, good luck!

6:47 PM

okay cool I got it @BalarkaSen and @TedShifrin

@Adeek Geometrically, the point is for S^n with n > 1, maps from S^n to base lifts to maps from S^n to universal cover. That is why you get an inverse to the map $\pi_n(\tilde{X})\to \pi_n(X)$ by lifting giving the isomorphism.

it is trivial as you say

right @BalarkaSen neat

@BalarkaSen why is $\pi_r(\{\pm 1\}) = 0$ for $r \neq 0$ ?

I thought you just said you understand that?

What does $\pi_r(\pm 1)$ mean?

Hi @Ted, feeling better yet ?

yeah

7:00 PM

Not really, @Astyx, but thanks for asking. ... Probably going to see my doctor this afternoon.

@BalarkaSen $\pi_r(\pm 1) = [(S^r,1) \rightarrow (\pm 1,1)]$

@TedShifrin are you sick ?

If $r > 0$, what is a map $S^r \to S^0$?

Yup. Respiratory/sinus thing going around everywhere ... :(

There are not many such maps.

@TedShifrin I hope your better soon.

7:02 PM

Sorry to hear it :/

Thanks, Karim.

Darn, still? Well, wish you a speedy recovery

@TedShifrin Not good, get well. Drink a lot of warm water.

@BalarkaSen well if $r \geq 0$ then we must have $f(1) = 1$ but $f(-)$ could be anything why is it

@Adeek No it can't be anything.

7:03 PM

That doesn't help with bacterial infection :D

why ?

What does $1$ mean in the domain, Karim?

That's what I am asking you

oh yeah it is continous

righttt

never mind yeah it must be constant yeah

@Ted Probably helps with the sinus though.

7:05 PM

yeah okay

@Adeek Why must it be constant?

because $\{ \pm 1\}$ is discrete

So? Continuity and codomain being discrete isn't the point.

Well, only one quarter of the point.

yeah so maps continous maps are constant so we must have that there is only one class which constant path

constant map from $S^r$ to $S^0$

not path

You still haven't said why it's constant. Not all continuous maps to discrete spaces are constant.

7:07 PM

well because we have that $S^3$ is path connected

it is locally constant

sorry

Just connected. For $r > 0$, $S^r$ is connected. That is the point.

$S^r$ is path connected

yeah

@TedShifrin Did exercise 13 again.

Was less hard than it felt back then :)

LOL, it really wasn't that hard except for understanding how you're done.

Yeah, you just want to rotate the frame so that $\omega_{12}' = \omega_{12} + d\theta$. And you know $\omega_{12} = df$ locally, so choose the angle carefully so that...

7:13 PM

Oh @Ted from glancing through Milnor/GP plus looking at our syllabus, this week we're scheduled to do homotopy and Brouwer's fixed point theorem

I don't know what this has to do with integrability though.

Prob gonna have vector fields as well, since we were supposed to do that last week but instead did regular/critical values

How're you going to prove Brouwer?

My favorite is through transversality, which I don't think you covered yet?

Ah, sure, it's just a regular value argument. Baby transversality.

$d^2=0$ is the baby case of integrability for Frobenius, @Balarka.

Yeah, regular values and Stone-Weierstrass

7:18 PM

@Daminark Smooth approximation, yes.

And yeah we're not doing transversality until basically halfway through the quarter

@TedShifrin $d$ being the exterior derivative? How?

Oh, sure, nevermind.

Frobenius is a serious generalization of that.

Yeah it literally says kernels of the k-forms which are differential graded ideals of $\Omega^*(M)$ are precisely the integrable subbundles.

Well, the kernels of them.

@Ted Does that mean if $K = 0$ then $\omega_{12}$ generates a foliation on my surface?

No, you have to work on the frame bundle to get something globally well-defined.

7:31 PM

Got it.

Fawad

> If $r$ and $s$ are roots of the quadratic equation $ax^2+bx+c=0$, then $$ax^2+bx+c=a(x-r)(x-s)$$

Those both $a$ are same? (Just making sure)

Hi chat

Lunch time. I'm outta here.

See you @Ted!

And yeah @Fawad

Enjoy lunch

Bon appétit @Ted

Yes @Fawad

7:44 PM

Suppose we have a differential equation y''+P(x)y'+Q(x)y=0.

We set y(x)=u(x)v(x) to transform the differential equation into the form u''+q(x)u=0.

Is u(x) a solution of y''+P(x)y'+Q(x)y=0 ?

Fawad

@Astyx that $a$ needs to be an integer?

Alessandro Codenotti

Someone want to talk about a probability exercise?

8:05 PM

@Fawad not necessarilly

@Alessandro what is it ?

Alessandro Codenotti

I pick a point $p_1$ uniformly in $[0,1]$, then pick a second one uniformly from the longer between $[0,p_1]$ and $[p_1,1]$. I'm interested in studying the random variable $X$ representing the value of the leftmost point

Sha Vuklia

He guys, I need to show if the following limit exists,
$$
f(x,y)=\frac{x^3-xy^3}{x^2+y^2},
$$
for $\vec x\to 0$. I tried out the following:
$$
\frac{x^3-xy^3}{x^2+y^2}\leq\frac{x^3-xy^3}{x^2}=1-\frac{y^3}{x^2},
$$
but obviously I'm stuk here. A similar approach by omitting $y^2$ gives the same problem. What could I do next?

8:20 PM

Hello @Astyx !! Do you maybe have an idea about my question above?

No @MaryStar

v needs to be a solution

IIRC

@Astyx Why does v have to be a solution?

Compute y' and y'' and substitute them in the equation

Helios

Hello everyone!

Would anyone be as kind as to help me with an algebra exercise I have not been able to do after spending hours on it?

@BalarkaSen still around ?

8:35 PM

Yes

could you help me with this part ? I want to get the kernel

@Alessandro I get something like $P(X \in [t, t+dt]) = \begin{cases}{{4\over 3}(3/2-t)dt \text{ if } t\in[0,1/2]\\ {4\over 3}(1-t)dt \text{ otherwise}}\end{cases}$

nvm @BalarkaSen I got it

Helios

I posted the exercise at: math.stackexchange.com/questions/2215895/…

I was told to have a look at a factor group, but I have not been able to solve it

@BalarkaSen I want to determine all quotient morphism from $SO_n$ to G?

with discrete kernel

8:43 PM

What have you tried?

so we know that the fundmental group of $SO_n$ must be $Z_2$

and we know all such maps must map fundmental group injectively onto G

Also, what is $G$?

Do you want to classify discrete normal subgroups of SO(n) or what?

we want to find all such G

@Adeek Huh?

we want to find all G for which the map $p : SO_n \rightarrow G$ has discrete kernel this map is surjective as well

8:46 PM

That's equivalent to classifying discrete normal subgroups of SO(n).

why ?

You tell me!

@ShaVuklia Do not create blow-ups by deleting part of the denominator. Terrible idea! What does it mean to say $\vec x\to 0$? It means $r=\|\vec x\| = \sqrt{x^2+y^2}\to 0$. So estimate things in terms of $r$. [You'll see examples done like this problem in my lectures, too.]

can i classify it using the subgroups of the fundamental group instead?

the fundamental group of $SO(n)$ is $\mathbb{Z}_2$

for $n\ge 3$

I don't know how you want to do that by looking at fundamental group. I don't care. This can be done by hand.

8:50 PM

what do you mean @BalarkaSen ?

done by hand

This is a group theory problem, not an algebraic topology one. See my normal subgroup comment above.

Hi @Semi

Hi chat

@Helios: You need to specify exactly what you know and what you don't when you post a question like your group theory question. Normal subgroups and quotient groups (that's another way of saying factor groups) are standard after a few weeks in a group theory course.

Rehi @Semiclassic

Hi @ted

8:54 PM

@TedShifrin It's pretty neat that $\omega_{12}$ is exactly $\nabla_{(-)} e_1 \cdot e_2$. I get the connection with holonomy now (integrating $\omega_{12}$ along loops give the holonomy, and it's obvious from that what the connection with $K$ is modulo Stokes')

$\omega_{12}(v)$ equals that, @Balarka. :)

@ted right now the status of the missing papers seems to be: "ask again later"

Fixed.

Sorry, @Semiclassic. This sucks. And it's not fair to the students in limbo.

Agreed.

8:55 PM

@Balarka: I personally would just write it as $\nabla e_1\cdot e_2$.

Good idea.

I did check the pile at the front office and confirmed that there weren't any student papers mixed into that.

So whatever happened occurred after my job as proctor concluded.

Has no one asked the particular TA to double-check that he didn't leave some papers somewhere?

@Alessandro I might have made a silly mistake, I'll check again later

i presume they have.

I'd have asked them myself at today's TA meeting, but...

8:58 PM

It strikes me that 3.3. is essentially all of your notes on surfaces summarized: first and second fundamental forms are built in, Theorema Egregium is trivial, Gauss Bonnet and holonomy-curvature relation is literally Stokes' theorem, and all

Basically, yes, Balarka, and lots of the exercises are revisiting old exercises with much nicer techniques.

True.

Well, when the prof says they won't be there at the TA meeting there's really no incentive for TAs to actually show up

Obviously, it's missing a lot of the conceptual stuff on the shape operator, lines of curvature, asymptotic curves, ... and theorems proved back in Chapter 2.

And indeed no one did

9:00 PM

The professor really is dropping the ball, @Semiclassic. Perhaps there was a true family emergency or something for this.

Right, agreed.

Yeah, that's fair.

The other factor in this, though, is that this is an emeritus prof who is being phased into retirement. And while there are certainly profs who remain motivated right up to retirement, they're not the only sort

@Ted: What would you suggest for me if I were to say that I wanted to learn some Riemannian geometry (which I do)?

Well, @Semiclassic, there are plenty of undedicated ones who are quite young, for that matter. ... Emeritus would mean already retired. Are you saying he has retired and is teaching after that?

My suggestion; learn what Berry curvature is and then teach it to me.

I'm not sure, if I'm honest. I should be more careful.

9:04 PM

Well, Balarka, the standard text is doCarmo (big, not little). If I had brainwashed you to do things with moving frames and forms, then that might not be the best choice.

Ok, apparently not yet retired

The 5 volumes of Spivak are a bit much.

There are some French texts that I do not know super well, but that are well respected, Balarka.

(translated, of course)

For reference: en.m.wikipedia.org/wiki/Berry_connection_and_curvature

Hmm. I indeed have heard of do Carmo. What are those French texts?

Berger & Gostiaux is one, but there's another.

9:07 PM

@Semiclassical 2hard4me

sigh

Gallot, Hulin, Lafontaine is another, but I don't know it at all.

Ah, I have heard of the latter.

I am partial (although there are no exercises) to Chern, Chen, Lam Lectures on Differential Geometry — obviously plenty of forms in that.

Of course, as you know, I can provide you with a lot of exercises. :P

Whoa, that's a lot of recommendations. For clarification, I think I want to learn more intrinsic (Riemannian) than extrinsic geometry right now.

9:12 PM

Well, even for intrinsic, there's extrinsic (studying submanifolds of your ambient manifold). Like geodesics, totally geodesic submanifolds, etc.

Ah, right, fair enough.

The answer here /55641) gives it as: "[Barry Simon] proved that the Berry phase is the holonomy of a (connection of) the Hermitian line bundle given by

Oh blah

Doing this on my phone

Bah, I am too slow.

@TedShifrin Chern-Chen-Lam seems to have a lot on Finsler geometry. No idea what it is.

Oh well. Point being, I suspect I should know those phrases

One chapter, not a lot. It's a topic that Chern and then Bryant have tried to bring to the foreground. It's basically geometry of a norm, rather than geometry of an inner product.

9:18 PM

Ah, alright

@Astyx But why is then v a solution but not u?

Interestingly, many of the foundational theorems (like about geodesics and completeness) still hold, @Balarka.

Strange!

You can't have stuff like a natural connection geometry on them, I guess? Can't talk about angles.

Consider the free loop space of a Riemannian manifold. It has an energy functional, whose critical points are the geodesics. This space has an O(2)-symmetry. If you do the same thing in Finsler geometry, you only have the S^1 symmetry. No reflections of geodesics.

There is an interesting question. Let M be a closed Riemannian manifold. Does it have infinitely many prime closed geodesics (prime meaning it's not just a geodesic you traveled n times)?

@Balarka: Sure, you still have connections. Just not a metric connection. But there are plenty of connections in life that don't come from metrics.

9:22 PM

It's know for any metric on S^2, any metric on a manifold whose $\pi_1$ has infinitely many conjugacy classes, and any manifold whose free loop space has sufficiently complicated cohomology.

@Ted That's what I meant by "natural" I guess. There's probably going to be a lot of connections compatible with the Finsler structure.

But Katok gave an example of a Finsler manifold (I think a Finsler structure on $S^2$) that has exactly three prime closed geodesics.

@MikeMiller Oh, interesting.

So somehow the extra $O(2)$ symmetry is crucial to the question

I haven't worked enough with Finsler to know enough about it. ...

9:24 PM

That's all I know. It's like Riemannian geometry without a time reversal symmetry a lot of the time.

But, @Balarka, doing projective differential geometry, there's projective connections that have nothing to do with Riemannian or Hermitian structure.

@MikeMiller That's a fun intuitive description.

@Ted: Ah. What's a projective connection?

It's basically what happens when you try to differentiate in projective space (or a projective bundle) by lifting locally to affine space.

a connection preserving a projecti've structure 🙃

But there are normalizations akin to the normalizations you get with the Levi-Civita connection for a Riemannian structure.

bids Balarka good un-sleep time

9:29 PM

Ah alright.

Yikes, it's 3 AM already and I wanted to listen to a talk on foliations

Q: What is an equivalent condition to $\sigma f = f\sigma \ \forall \sigma \in S_X,$ where $f: X \to X$?

I invented something

1

Let $f: X \to X$ be a map of sets and $\text{Sym}_X$ be the group of permutations of $X$. Then if $f$ commutes with every permutation $\sigma \in \text{Sym}_X$ what else can we say about $f$? We can say that if $|X| \gt 1$, then $f$ is not constant since if: $X= \{a,b,c\dots\}$ and $f(X) = c$....

So it seems obvious, right?

So asking if it's already been studied

I too invented something. I call it fast array iteration leaping, or FAIL for short

? and d are either zero or involve @ symbols.

F a = a + 1

F a # .... # b # d = F a # .... # b

F a # b # .... # c = F[a] a # b - 1 # .... # c

F[1] a # .... # b = F a # .... # b

F^[k] a # .... # b = F^[k-1] (F a # .... # b ) # .... # b

F a # ? # .... # ? # 1@0 # b # .... # c = F a # ? # .... # ? # 0 # .... # 0 # a # 1@0 # b-1 # .... # c, with a amount of '# 0's

F a # ? # .... # ? # d@0 # b # .... # c = F a # ? # .... # ? # O(d)@a # .... # O(d)@a # a # d@0 # b - 1 # .... # c, with a amount of '# O(d)@a's

If you apply the above rules to something like F 5 # 5@5@5@5 # 5, you'll end up with an extremely insanely large number. I don't think you'd be able to easily grasp F 5 # 5 even, for that it is larger than 10^10^10^... a million powers of ten.

Hello again everyone!

@Daminark Howdy

How's it going @Simply?

9:42 PM

Large to put it simply :P

Haha

@SimplyBeautifulArt please explain more about what you're doing. In human language pls

mixed with math

googology.wikia.com/wiki/User_blog:Simply_Beautiful_Art/…

@FruitfulApproach Sure, I'll explain it here

We start with the simplest case:
F a = a + 1
F 3 = 4

No, surround with $ $

Oh, okay, sure

9:53 PM

$ dollar signs

Then use the chatjax startChatJax bookmark google it

then LaTeX shows up here

Now we add one #:
$Fa\# b = F[a]a\#b-1$

Yeah, I know

What is $F, \#$

?

Defined above.

18 mins ago, by Simply Beautiful Art

? and d are either zero or involve @ symbols.

F a = a + 1

F a # .... # b # d = F a # .... # b

F a # b # .... # c = F[a] a # b - 1 # .... # c

F[1] a # .... # b = F a # .... # b

F^[k] a # .... # b = F^[k-1] (F a # .... # b ) # .... # b

F a # ? # .... # ? # 1@0 # b # .... # c = F a # ? # .... # ? # 0 # .... # 0 # a # 1@0 # b-1 # .... # c, with a amount of '# 0's

F a # ? # .... # ? # d@0 # b # .... # c = F a # ? # .... # ? # O(d)@a # .... # O(d)@a # a # d@0 # b - 1 # .... # c, with a amount of '# O(d)@a's

And so we have
$F3\#2=F[3]3\#1$

I can't get to above

Then just listen along

You good so far?

9:55 PM

Please make a concise definition like: A group is a set together with a binary operation that satisfies these 4 axioms:

etc