Mathematics - 2016-05-18 (page 1 of 3)

Balarka Sen

04:32

@MikeMiller Thanks!

user116211

Is there any difference between $\sum_{ij}$ and $\sum_i\sum_j\;?$

SNB77

05:00

Hi

What is the meaning of a surface being "closed"

Balarka Sen

It means it doesn't have a "boundary".

SNB77

In my textbook, it says "If the surface is closed, then the boundary curve is empty and the surface integral is 0"

closed like a sphere is closed?

Balarka Sen

Yeah, sphere is closed.

SNB77

but the surface integral through a sphere is not always 0

Balarka Sen

A surface integral on a sphere is always $0$, by Stokes' theorem.

Why do you think it isn't?

SNB77

05:03

but

because like in Divergence theorem, surface integral over sphere is not 0

Balarka Sen

Can you give me an example?

I am losing context here, surely you mean you are integrating a gradient of something on the sphere?

SNB77

wait actually, I think that only happens is F = curl(A) for some A

Balarka Sen

Right that's how I interpreted it.

Your book says that, right? Because I agree it's not true that all surface integrals over the sphere is 0.

SNB77

yep, i just didn't read the sentence before about it

Balarka Sen

OK, great.

SNB77

05:10

suppose F = del(f) for some f, so it has a vector potential, and is conservative. Then does that still apply?

the thing about the surface integral over a sphere being = 0

Balarka Sen

You're integrating F over the sphere?

SNB77

yeah

Balarka Sen

Um, no, it need not be true, I don't think.

SNB77

so just because F = del(f) for some f, it does not necessarily follow that F=curl(A) for some A

Balarka Sen

No, definitely not.

SNB77

05:17

ok great. Thanks!

Balarka Sen

06:07

Hi, @Huy.

Huy

morning @BalarkaSen

Balarka Sen

what's up?

Huy

@BalarkaSen: about to teach, just revising stuff for graduation exams

you?

Balarka Sen

not much, about to start doing math again.

Huy

good, have you been cooking in the meantime?

Balarka Sen

06:18

yeah, i can boil water now

Huy

very good. soon you'll be ready to mix water and syrup.

Mike Miller

@Huy can you teach me the quasi isometry proof

Huy

@MikeMiller: maybe later, the question I asked last week was related to the proof. I need to work on a few parts for a bit before I'm convinced

@MikeMiller: the proof in F&M contains a lot of statements without proof or reference, and some intuitively make sense and some don't, but most aren't that easy to prove (at least for me)

Balarka Sen

06:39

@Huy what would mixing water and syrup accomplish though?

Mike Miller

fair enough

Balarka Sen

09:52

Hi @Kari

Kari

Hey, @BalarkaSen!

How's it going?

Balarka Sen

It's ok. Not doing much math this week.

How about you?

Henry

Hi!

Can someone help me with this question? math.stackexchange.com/questions/1790178/find-the-polynomial-px

Its about polynomials

Secret

10:10

https://www.youtube.com/watch?v=gpM_rnQBCr0
I wonder how can one construct a proof that the symmetric way that this ad shows will ensure it took the least number of steps to completely exchange all business cards for n people?

SteamyRoot

11:02

I don't suppose anyone has CARAT installed?

user 1618033

11:16

@robjohn Cool news! I developed a new master theorem that will appear in my second book. I think a powerful master theorem (one piece) is enough for a single book.

(stuff unseen before in no environment because it wasn't invented - my estimate is that a master theorem can be used to easily write 5-10 books (300-400 pages) each one)

I love them.

Theoretically, you can use them indefinitely, I mean you can write infinitely many books, but after a while the stuff you get looks too ugly for a potential reader.

But, yes, theoretically you can write infinitely many books using each one. Limitless potential!

user 1618033

11:38

@Semiclassical btw, almost forgot to ask you if you slayed that limit I gave you some days ago, the easier version.

Preparing for a trip. Back later if possible.

Henry

11:53

Hi!
Can someone help me with this question? http://math.stackexchange.com/questions/1790284/finding-the-prime-factors

Its about number theory (related to prime factorization)

Tobias Kildetoft

@Henry Do you know Euler's totient funtion? (it is not strictly needed here, but it will make things a bit simpler)

felipa

12:14

Does anyone see a simple proof for math.stackexchange.com/questions/1790290/… . I feel it is meant to be obvious

but sadly it's not obvious to me

Juan Fran

sry I never really thought seriously about probability theory

felipa

@JuanFran no problem

I think it's going to be very simple for someone :)

Balarka Sen

Hello again @JuanFran.

felipa

12:37

what is the operation called on two matrices A and B where you take the inner product of the concatenation of the columns of A and the concatenation of the columns of B?

Akiva Weinberger

12:51

@felipa $A^TB$?

felipa

@AkivaWeinberger thanks!!

It's the Hilbert Schmidt inner product apparently

Clarinetist

Is anyone here familiar with maximum likelihood estimation?

Huy

you should be

don't you do that to get an estimator for the mean, variance, etc. ?

Balarka Sen

13:06

I will hopefully take statistics in my school this year.

Huy

I don't think you'll like it

Balarka Sen

well, hopefully I'll prove you wrong. :)

Huy

:)

Juan Fran

13:19

hi @BalarkaSen sorry I was AFK

Balarka Sen

No worries.

Akiva Weinberger

@felipa Well, the trace of that, probably

(Sum of diagonal elements)

Ali Caglayan

13:53

Hello @BalarkaSen

How are things

Balarka Sen

Hi @AliCaglayan

@AliCaglayan It's alright, still learning math. How about you?

Ali Caglayan

@BalarkaSen Yes. Studying for exams

Then university

Balarka Sen

Good luck with the exams.

Ali Caglayan

Thank you very much

@BalarkaSen Perhaps you would be interested in this

Have you ever seen diracs belt?

Balarka Sen

Yes.

Ali Caglayan

13:59

The demonstation that Su(2) double covers So(3)

Plate trick

In mathematics and physics, the plate trick, also known as Dirac's string trick, the belt trick, Balinese cup trick, is any of several demonstrations of the mathematical theorem that SU(2) (which double-covers SO(3)) is simply connected. To say that SU(2) double-covers SO(3) essentially means that the unit quaternions represent the group of rotations twice over. == The trick == One way of doing the trick is to rest a small plate flat on the palm, then perform two rotations of one's hand while keeping the plate upright, ending in the original position. The hand makes one rotation passing over it...

Balarka Sen

I have seen it.

Ali Caglayan

I stumbled upon it yesterday I found it very interesting

Balarka Sen

Although more classically it's used as a demonstration of fundamental group of RP^2 being Z/2.

Ali Caglayan

I guess its rare to get a demonstation in algebra

Balarka Sen

That SU(2) double covers SO(3) is not really an algebra fact.

Henry

14:02

@TobiasKildetoft Thanks! I've solved it.

Ali Caglayan

@BalarkaSen I read about coming from Lie algebras

Here is a small problem for anyone interested. Given a matrix that can be written as $\left [\gcd(i, j)\right]_{ij}$ component wise, what is it's determinant?

Balarka Sen

I mean SU(2) is the group of unit quaternions, which is the 3-sphere. That double covers RP^3, which is precisely what SO(3) is.

Ali Caglayan

@BalarkaSen That is a simple way to think about it

Balarka Sen

I suppose this double cover preserves the group structure though, so perhaps there is more to it than just the topology.

anon

I submit that Sp(1) is the group of unit quaternions and SU(2) is the group of 2x2 special unitary matrices.

Balarka Sen

14:06

Well, they are isomorphic, aren't they?

anon

Yep.

But so is the multiplicative group of positive reals and the additive group of all reals, e.g.

One doesn't use the notation for one to refer to the other. Pet peeve of mine when everyone calls SU(2) unit quaternions.

Balarka Sen

@Ali Why would you want to think about it in a complicated way? The simpler the better. But I am talking about the picture, I do understand that it might be more than just a map between the topological spaces.

@anon Fair enough.

Why is thinking of the unit quaternions as Sp(1) more preferable than SU(2)? Just curious, I do not know these groups too well.

anon

Well, two distinct questions here. One is of notation. From the outset, Sp(1) and SU(2) refer to different things. They happen to be isomorphic. So one could in principle identify them. Which is the second question: under what circumstances is it a good idea to identify them? For the purpose of having the double cover of SO(3), I think it's a lot easier and more direct to just think about Sp(1) than SU(2).

A lot more people know about SU(2) than quaternions though which is I suspect part of the reason they do that.

Balarka Sen

"For the purpose of ... I think it's easier and more direct to just think about Sp(1) than SU(2)" I was asking why you think that, actually.

anon

One can prove Sp(1) double covers SO(3) using basic geometrical properties of the cross product. How would you do it for SU(2)?

(Also, do you know why they're isomorphic?)

@AliCaglayan Sp(1) or SU(2) is the universal covering group of SO(3). u.c.g. can be constructed explicitly as the group of endpoint-preserving-homotopy-classes of paths emanating from the identity. a path in SO(3) from the identity is essentially a continuous family of rotations, which is actually more akin to our intuitive mental picture of what a rotation is than an actual rotation. (a single rotation has only a before and after, and no in between.)

In turns that for paths in SO(3), rotating 360 degrees around an axis is distinct from not rotating it, but 720 degrees isn't. If one affixes an object to an unmoving environment using, say, belts, one can interpret the twisting of the belt as encoding a memory of how the object was rotated, and hence the trick.

Balarka Sen

14:18

I don't know how to construct a map SU(2) --> SO(3) directly. I usually just show SU(2) is isomorphic to the group of unit quaternions (which is easy, just write down the matrices for 1, i, j, k) and construct the map by making the unit quaternions act on $\Bbb R^3$ by rotations. I think the last bit requires some work.

anon

You conjugate span(i,j,k) by unit quaternions to get rotations. :-) I've written about the geometry in a number of answers under arctic tern.

Henry

Can anyone suggest me some book to study analytic NT?

Balarka Sen

@anon I'll have a look at those answers.

Thanks.

Henry

Something to solve questions like these brilliant.org/problems/exploring-divisor-function-4

And these brilliant.org/problems/good-bye-brilliant

robjohn

@Henry I have used Borevich and Shafarevich

Henry

14:21

Wow! Looks Good. Thanks a lot!! @robjohn

Does it cover Dirchlet Convolution?

@robjohn

user 1618033

Nice question from a middle school test

People from a city speak either French or English. If 64% speak French and 58% English, how many speak both languages?

I received tons of wrong answers (unfortunately).

Balarka Sen

@robjohn Whoa Shafarevich wrote an analytic number theory book.

That's pretty cool.

robjohn

@Henry It doesn't do much with convolution,

Henry

Oh.

robjohn

That could be in either an algebraic or analytic NT book

Henry

14:30

Still a good book though

robjohn

it is

Henry

No Problem, I'll find it elsewhere :)

I've been told that convolution is analytic NT

Also, does inclusion - exclusion come in NT? I thought it was combinatorics.

Balarka Sen

Combinatorial flavor of number theory, yes.

Mostly used in sieve theory.

Henry

Someone used Inclusion Exclusion to solve this NT problem brilliant.org/problems/exploring-divisor-function-4

Can someone tell me how inclusion exclusion is used here?

user 1618033

@Henry @BalarkaSen is our expert in NT problems

Balarka Sen

14:38

I do not know number theory.

In particular, I won't be able to help.

@AliCaglayan Oh, I just remembered. There's a cute "application" of the belt trick which can be used to demonstrate a constructive proof of the sphere eversion theorem. You might want to look into that, there's a youtube video of it somewhere. It's incredibly cute.

Not to mention surprising.

Ali Caglayan

@BalarkaSen What is it called?

Balarka Sen

Eh, maybe it's not quite the belt trick but it looked a lot like it. In any case, it's worth watching.

@Ali "How to Turn a Sphere Inside Out".

Ali Caglayan

@BalarkaSen Oh yeah I have watched that.

Balarka Sen

Ah, ok.

Ali Caglayan

@BalarkaSen Yeah it was an argument with turning numbers

Balarka Sen

14:47

Um, no, winding numbers can be used to show why it fails for the circle.

There is no homotopy-through-immersions of the identity and the antipodal map from the circle to R^2. Because these have different winding numbers.

Henry

14:59

I have posted an elementary set theory question on MSE, I'm having a difficulty proving it. Can anyone please help? math.stackexchange.com/questions/1790497/show-that-aa-geq-2n-1

Obliv

15:12

can someone tell me how to solve this $\frac{d^2x}{dt^2} = -(\frac{k}{m})x$

Kenny Lau

this is simple harmonic motion

Obliv

yep

Kenny Lau

the answer is $x=A\sin\left(\sqrt{\frac km}t-\theta_0\right)$

Obliv

how do you get that solution

Kenny Lau

well, firstly, do you know that $\frac{d^2x}{dt^2}=v\frac{dv}{dx}$?

robjohn

15:14

@user1618033: good morning

Obliv

yes

Kenny Lau

what do you get when you use that substitution?

Obliv

if you integrate with respect to $dx$ on the right side you get $-\frac{kx^2}{2m}$

user 1618033

@robjohn Hi (here is 18:15 :D)

Obliv

which isn't sinusoidal

Kenny Lau

15:15

but the left hand side becomes $\frac12v^2$

you're missing a constant also

Obliv

so $\frac{v^2}{2} = -\frac{kx^2}{2m} + C$

how is that turned into a sine/cosine function?

Kenny Lau

Express that, with $v$ being subject

Obliv

what do you mean subject?

Kenny Lau

Well, $v=...$

Obliv

oh

$v = \sqrt{-\frac{kx^2}{m}+C}$

Kenny Lau

15:18

now, note that $v=\frac{dx}{dt}$

Obliv

$\int dx = \int {\sqrt{-\frac{kx^2}{m}+C}~dt}$

like so?

Kenny Lau

not so fast

take the reciprocal of both sides

Obliv

$\frac{dt}{dx} = \sqrt{-\frac{m}{kx^2} + C}$

Kenny Lau

and then you can go ahead

remember that you try to group terms with the same variable

so, $x$ and $dx$ on one side, $t$ and $dt$ on the other side

Obliv

oh I see, then solve for $x$ at the very end

Kenny Lau

15:21

yep

oh, you forgot to take the reciprocal of the right hand side

oh, you took it wrongly

Obliv

oh

Kenny Lau

remember there's a $C$ there

Obliv

I thought $\frac{1}{C} = C$

like it's a constant regardless

Kenny Lau

but $\frac1{x+C}$ is not $\frac1x+C$

user 1618033

@robjohn I got some fantastic result today, but I had to work for it a pretty long time.

Obliv

15:24

oh

Kenny Lau

that would be an integral of the form $\int\frac{dx}{\sqrt{a^2-x^2}}$

which means trigonometric substitution (or Euler-substitution)

Obliv

I guess in this case it would be useful to keep the $C$ as $\frac{1}{C}$

Kenny Lau

not really

robjohn

@user1618033 That's great! I am home today, but I have a lot of things to take care of here.

Kenny Lau

$\frac{dx}{dt}=\sqrt{C-\frac{kx^2}m}$

$\frac{dt}{dx}=\frac1{\sqrt{C-\frac{kx^2}m}}$

robjohn

15:25

@user1618033 I will try to take a look at the problem you sent

Obliv

oh .. lol

user 1618033

@robjohn you refer to the limit I sent to you? If, yes then that is great. One more thing, try also the variant without the zeta expresson in front, it is very nice.

Obliv

@kennyL Isn't there another way to represent this harmonic motion? I feel like there could be a more general periodic function that one could construct

user 1618033

The limit is $0$, but this is less important. For the version I sent to you, the limit is $1$. Mathematica goes crazy after a while and gives wrong results.

Kenny Lau

@Obliv using the formula above?

13 mins ago, by Kenny Lau

the answer is $x=A\sin\left(\sqrt{\frac km}t-\theta_0\right)$

Henry

15:27

@user1618033 Can you post the question here, I like limits.

Obliv

instead of $\sin$ couldn't we use $f(x+P) = f(x)$? i read this is a general definition for periodic functions here en.wikipedia.org/wiki/Periodic_function

user 1618033

@Henry That one not really now (I have some plans with it).

@robjohn it's nice to notice some things there for the simpler version without zeta in front.

Henry

@user1618033 Oh ok. No problem :)

Kenny Lau

@Obliv I believe it must be $\sin$, because that's what solving that integral gives

user 1618033

@Henry :D

Agawa001

15:31

hmm for a respectfully notable room as mathematics, that is too much few chatters to be in

or maybe it is summer !

hence holidays

user 1618033

@robjohn btw, someone contacted me these days showing some interest for buying my book project. The thing is that I wouldn't accept any offer ever, no matter what. This is not a business.

It was unexpected though.

Henry

@Agawa001 Maybe you can help me. Can you tell how is Inclusion Exclusion used here?

brilliant.org/problems/exploring-divisor-function-4

Or you @user1618033

user 1618033

@Henry I'm not into these problems for a long while, although I admit they are cool. When I return to this stuff, I'll let you know.

Agawa001

@Henry if noone from the members rowed right there to the margin could help you, how would u expect from me to do me poor math thing ?

user 1618033

@Henry Maybe you like this inequality

If $P$ degree$>2$ and $P'$ has all real roots, then

$$\frac{\displaystyle \int_a^b \frac{1}{P'(x)}\ dx}{\displaystyle \int_a^b \frac{1}{P''(x)}\ dx}>\frac{P'(b)-P'(a)}{P(b)-P(a)}$$

Agawa001

15:39

look there is some heavy-reputed members as robjohn who are more dignified, just ask dot ask to ask @Henry

Noah Harris

Can anyone help me find the explicit formula for the recursive series $(k+1)a_{k+1} = 2a_{k-1}$

Henry

@user1618033 Do you know a Latex rendering website?

Kenny Lau

@Henry Read the description of this chatroom

@NoahHarris all the even terms are skipped?

also, what is the initial term?

Henry

Haha! Thanks @KennyLau

Noah Harris

$a_0 = 1$

Kenny Lau

15:42

So all the odd terms are skipped?

Henry

Oh, but I don't want to install, I want online Latex renderer...

Noah Harris

Yeah, they should be I think

They're zero anyway

Kenny Lau

@Henry You don't need to install

just a bookmark

it's a javascript code

Obliv

@kennyL I see why. The $\sin$ function is a specific periodic function. If the equation was not $F = -kx$ and instead something that produces a periodic function like the first image here en.wikipedia.org/wiki/Periodic_function then sin wouldn't be the solution to such an integral

Kenny Lau

so $a_2=\frac{2}{2}=1$

Obliv

15:44

I don't know of any equations that would produce such a periodic function and I can't imagine how getting the solution would be though

Agawa001

@KennyLau you became regular here? see how this place is useful but it is just summer holidays, so you r ought to wait another 2 or 3 months

Kenny Lau

$a_4=\frac{2a_2}{4}=\frac12$

$a_{k+2}=\frac{2a_k}{k+2}$

$a_k=\frac2ka_{k-2}$

$a_2=\frac22$

$a_4=\frac24\frac22$

$a_6=\frac26\frac24\frac22$

@NoahHarris Can you see it now?

Henry

@KennyLau That's magic!

Latex has now appeared!

Kenny Lau

nice

Henry

Thanks.

@user1618033 I have noted your question. Seems interesting. I'll solve it in home later.

gtg, bye everyone.

user 1618033

15:50

@Henry You'll try you mean. :-)

Noah Harris

uh I see $\frac{2^{n/2}}{n!!}$

user 1618033

Some tutoring sesion here. Later.

Kenny Lau

yes @NoahHarris

Noah Harris

yikes.

Kenny Lau

single dollar sign to make it in-line

Noah Harris

15:50

Yeah

Henry

"Mean", funny. I'll first try Mean value theorem :D @user1618033

Noah Harris

Well, that's not nothing. But don't think that's what I was supposed to get

Kenny Lau

well

I can make it better

Noah Harris

For reference this is what I'm trying to do: math.stackexchange.com/questions/1789074/…

Kenny Lau

sure, I can do that

$n!!=2\times4\times\cdots\times n=2^{n/2}(n/2)!$

Balarka Sen

15:57

@MikeMiller I'm trying to get back to math. Where should I start? Guillemin-Pollack chapter 1? If there's anything specific you'd want me to do before that, I'd have a try too.

Noah Harris

Oooh. That looks helpful, thanks @KennyLau

Kenny Lau

@NoahHarris bonus question: how do you convert $n!!$ where $n$ is odd?

Balarka Sen

I want to start by reading some manifold theory as we planned to do.

Mike Miller

@anon I'm not sure I agree it's that much more direct. you have an action of $SU(2)$ on $\Bbb C^2$; projectivize to get a linear action on $S^2$; extend over rays to get an element of $SO(3)$. I guess you could complain that there's work to be done in seeing that the thing you get in the end is an element of $SO(3)$

@BalarkaSen I gave you those linear algebra problems, but there's not really any serious need to do them - they tie in to some geometry, but geometry we're not ready for yet. G&P ch1 is not a bad idea. You know ch3, so you might as well do 1-3 and the exercises.

I think you already know what's in ch1 but it's harmless to go through again anyway.

In fact I suspect you're already aware of quite a lot of it. There's no harm in not doing stuff you already know how to do.

Balarka Sen

Alright, great, I'll start on GP 1-3 for now and occasionally think about those lin. alg. problem you gave. I'll ping you with the solutions of (what I think are) hard and interesting exercises from GP if you want too, if that's ok to you?

Mike Miller

16:05

Yup

There are some fundamentals worth learning, but once we're done with those we can figure out where you want to go with all this

Balarka Sen

@MikeMiller I guess, but I suppose I need to do the exercises to get comfortable with it. I also usually try to learn things by heart: e.g., I can probably prove all the theorems that are in Hatcher, except perhaps Poincare duality, so I'd want to give it a read.

Well, Hatcher minus appendices and chapter 4

@MikeMiller Alright. I should also note that I probably won't be able to get things done super-fast as in what happened with differential forms from Ted's book, since I have not-so-pressing school work (school hasn't really started and won't start anytime soon, but I want to get ahead a bit).

I hope that's ok. I promise that I won't take a decade to learn it though.

Mike Miller

I am skeptical. But sure.

Balarka Sen

I mean, in contrast to what happened with multivariable calc.

Mike Miller

Multivariable was fast?

Balarka Sen

Well, apparently I took too much time to learn it than was needed. I am just saying even though I go slow on G-P, I promise that won't happen.

Mike Miller

16:16

Oh, ok.

Balarka Sen

16:38

Hi @TedShifrin!

Ted Shifrin

hi @Balarka

Did you finish all the Stokes's Theorem stuff?

Balarka Sen

You haven't been here for quite a long time!

@TedShifrin Yeah.

Ted Shifrin

I've been back once or twice. About to disappear again for 2+ weeks.

Balarka Sen

Aw. Are you on a trip again?

Mike Miller

here*

Ted Shifrin

16:40

About to be.

@MikeM: If I stopped to edit everyone's spelling and grammar, I'd never read any math.

Balarka Sen

Ah. Happy journey.

Ted Shifrin

@MikeM: It's pretty clear I'm the one who got the quotient topology wrong. Sloppy mental miscue on my part.

Mike Miller

I already don't do any math, so no harm.

@TedShifrin Well... I agreed with you. :)

Ted Shifrin

Clearly points are closed in the leaf space. Now I'm trying to decide what the correct example (if any) might be where there can be dense points.

Anyhow, makes me more pleased with my charges. And glad I can be irresponsible and retired :)

Mike Miller

Maybe you were thinking of torus foliations.

Balarka Sen

16:45

@TedShifrin Speaking of Stokes' theorem, interpretation of Stokes' theorem (at least the 3d version) using flux (i.e., using the Hodge dual of the forms) is much more clear, I agree now.

Ted Shifrin

You mean solid torus, @MikeM? Somehow I doubt it.

Mike Miller

No, @Ted, irrational lines.

I agree the Euclidean case is probably unsalvageable.

Balarka Sen

Namely, integrating (summing up) the divergence ("things flowing out of small nbhds around points") you get the flux along the boundary surface ("amount of thing which flows out of the surface").

Ted Shifrin

Oh, surely points aren't closed in that case.

Right, @Balarka: That's the physicists' "proof" that I was suggesting to you for Green's Theorem (but you can do it for work just as for flux).

Mike Miller

yes, in that case the leaf space is uncountable indiscrete

Balarka Sen

16:48

Ah, I see.

Ted Shifrin

I gather I've missed more contentiousness from our favorite chatter who keeps changing names.

Did you do the exercise 8.5.22, @Balarka? That's actually a very important concept for modern geometry.

And you saw 8.7.7, 8.7.9, 8.7.12?

Balarka Sen

The standard version of the Green's theorem might be a bit unintuitive as stated, but thinking of curl as "the amount of thing some small ball rotates around it's axis when placed at a point under the action of the vector field", it becomes "small swirlies add up to big swirly on the boundary" which should be intuitive because one can think of swirls in opposite direction cancels out like this:

Ted Shifrin

Sure, of course.

Mike Miller

there's probably some nasty foliation of higher-dim Euclidean space, since I'm pretty sure your Haefliger-Reeb theorem comes from Poincare-Bendixson theory, which doesn't really generalize

Ted Shifrin

Right, @MikeM. I actually am not sure I've ever thought about this.

And all my books that might be interesting to look at for this are long given away.

Mike Miller

16:53

it's not in candel-conlon

in hunting for it, by the way, found the charming result: "if $M$ is compact and has finite fundamental group, it has no real analytic codim 1 foliation"

Ted Shifrin

Hmm ... Does that come from lamination theory or something?

Balarka Sen

@TedShifrin Yep, but I didn't do 8.7.12. I liked Milnor's proof of hairy balls.

Ted Shifrin

Pedro recently ran into that Milnor proof somewhere and posted it on his blog, I think.

You had inquired about 8.7.12 earlier. That's why I mentioned it.

What about 8.5.22 and calibrations, @Balarka?

Obliv

guys if I'm to prove that if $x$ is an element of the group $G$, then $\{x^n~|~n\in \mathbb{Z}\}$ is a subgroup of $G$, how am I sure that $G$ is closed under exponents?

Ted Shifrin

What do you mean by "closed under exponents," @Obliv?

Balarka Sen

16:57

@TedShifrin Hmm, I looked, that I didn't do. Is that the proof of the minimal surface equation using forms you mentioned earlier?

Mike Miller

@Ted: It's reduction to Poincare-Bendixson (pulling back the foliation to a singular foliation on a well-chosen disc) and then some holonomy analysis; the contradiction comes because they get that the holonomy is (germinally) trivial on one side of the curve but not the other

which is nonsense if you're real analytic

Ted Shifrin

Yes @Balarka

Obliv

like that $x^n \in G$ @tedS

Ted Shifrin

Ah, very cool, @MikeM.

Are you asking why all the elements $x^n$ actually belong to the whole group $G$, @Obliv? I do not understand.

Balarka Sen

Oh cool then I have to do it! I noted it down.

Mike Miller

16:58

you have calibrations in your book? very cool...

Obliv

I'm asking that based on the information given is it a valid assumption that $x^n \in G$?

Ted Shifrin

It's a powerful argument, @Balarka. They appear in an exercise only, @MikeM.

Mike Miller

seems hard to do a whole lot with them at this point though, but I guess that's the minimal surface equation you mentioned

Ted Shifrin

Yes, @Obliv, because $G$ is closed under multiplication and inverse to start with.

You need to show that $H=\{x^n\}$ is non-empty and closed under multiplication and inverse.

Obliv

are all groups closed under multiplication?

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