Mathematics - 2017-03-31 (page 5 of 5)

Ted Shifrin

21:00

@Vrouvrou: dis-moi exactement le problème.

Meow Mix

> In short, it's what I call mythmatics. Here are some examples:
>
> 1. There is an infinite set.
> 2. Non-terminating radix representation can be used to represent any "real number".
> 3. There are irrational numbers.
> 4. An infinite sum is possible.
> 5. 1/3 = 0.333...
> 6. 1 = 0.999...
> 7. The integral is an infinite sum.
> 8. Numbers can be derived using sets.
>
> These are fallacies one finds in pseudo mathematics.

^ John Gabriel quote?

Ted Shifrin

No, Zach, and I have no interest.

Balarka Sen

@PVAL-inactive lol

Daminark

Oh lord @Zach what in the world is... GAH, I'm just gonna get back to my manifolds pset

Meow Mix

Anyways, I ought to go read section five

Semiclassical

21:01

Life's too short for pseudomath.

Ted Shifrin

I've posted on Facebook that arithmetic must be an advanced science, because all the Republicans are now denying arithmetic just as they deny science.

Demonark, every time you say pset it's funny, because I haven't heard "problem set" since MIT days. No one at UGA called homework that.

Meow Mix

I've never identified myself as republican or democrat, simply because neither fully express my views.

Pset makes me think "Percocet"

Balarka Sen

poset

Ted Shifrin

We're not going to devolve into politics. But I had to say that given all the current news.

Vrouvrou

j'ai une distance $d'=\ln(1+d_1)$ ou $d_1(X,Y)= |x_1-y_1|+|x_2-y_2|$, on demande de trouver la boule fermé de centre O et de rayon 1 par rapport à la distance d' , j'arrive à $B_{d'}(O,1)=\{(x_1,x_2)\in \mathbb{R}^2, |x_1|+|x_2|\leq\exp{(1)}-1\}$ est ce que je peux continuer plus loin ? @TedShifrin

Daminark

21:03

@Ted A number of professors say homework, but students say pset more

PVAL-inactive

@Ted At least the US will soon have the greatest military ever finally.

Ted Shifrin

Oh, ok, @Vrouvrou. D'abord, si c'est une boule fermée, il faudra $\le$ et non $<$, n'est-ce pas?

Meow Mix

I cannot believe these headphones are not working.

Vrouvrou

désolé erreur de frappe @TedShifrin

Ted Shifrin

OK, ça me semble correcte, enfin.

@PVAL-inactive What a g'damn relief.

Demonark: It's just funny, cuz I don't think anyone at Berkeley said it, either. Just MIT, to my recollection.

Danu

21:07

So I have this bundle $\Bbb CP^1\to Z\to \Bbb M=G_2/SO(4)$. Fact: $H^*(G_2/SO(4);\Bbb Z)=\Bbb Z[g_4^M]/(g_4^M)^3+\Bbb Z_2[u_3]/u_3^3=0$, i.e. infinite cyclic in degrees 0,4,8; 2-torsion in degrees 3,6. Fact: $H^*(Z;\Bbb Z)=\Bbb Z[g_2,g_4,g_6,g_8,g_{10}]/(g_2^2=3 g_4,g_2g_4=2g_6,g_4^2=2g_8)$, i.e. free in even degrees, with the given relations between powers of generators.

What I'm looking to do is find $p_2(Z)$ to get a hold of $c_3(Z)$. To figure out the Pontryagin class, I use $TZ=\pi^*M\oplus T\pi$, where $T\pi$ is the vertical tangent bundle (of rank 2). Fact: $c_1(T\pi)=g_2$, i.e. $p_…

(I can also send this text in a pdf if you'd prefer @Ted)

Ted Shifrin

This is looking like more than I'll have stamina to go through carefully, @Danu. I'm actually still quite sick :( But send it to me.

Danu

@MikeMiller If you're bored, but definitely don't feel obliged ^

Ted Shifrin

Mike has to work on his own stuff.

Vrouvrou

@TedShifrin donc je ne peux pas allez plus loin ?

Ted Shifrin

Pas autant que je sache, @Vrouvrou. On peut faire un dessin.

Meow Mix

21:12

I just realized that "The mitochondria is the powerhouse of the cell" would be grammatically incorrect

Ted Shifrin

You have a plural subject and a singular verb.

Meow Mix

Yeah.

Ted Shifrin

Drives me nuts when people say/write "The data shows that ..."

Danu

Hahaha

Vrouvrou

ok je vois merci @TedShifrin

Ted Shifrin

21:14

OK, @Vrouvrou.

PVAL-inactive

I thought one of the pronounciations of data was singular

and one was plural.

Ted Shifrin

Media versus mediums

I know nothing about data.

Vrouvrou

c'est un losange n'est ce pas de centre (0.0) et de diamètre 2(e-1) @TedShifrin

Meow Mix

"Whomst"

Danu

@TedShifrin sent.

Ted Shifrin

21:16

c'est plutôt un carré, je crois, @Vrouvrou.

Danu

I hope you get well soon

I am finding myself having to read papers in French (on the classification of manifolds with nearly Kaehler structures)!

Meow Mix

I speak fluent spanglish

Ted Shifrin

So it seems, @Danu, that saying $p_2(Z) = \pi^*p_2(M)$ is making the mistake I made of saying that $p_1$ of the fiber can't appear because it has rank 2.

I'm fine with reading in French, thank you :P

German is a bit harder for me, but I've done that, too.

Meow Mix

Funny thing, I convinced, in fifth grade, my class that "Spaingland" was a country which speaks "Spanglish"

Danu

@TedShifrin Like I said; it gives the correct result (and the other thing doesn't...).

Ted Shifrin

21:17

Zach, you're experienced at being obnoxious.

Alucard

:D

Ted Shifrin

@Danu: I wonder if all the "facts" are correct, or if there might be some alternative fact or misapplication.

Alucard

Hello and bye folks

Vrouvrou

@TedShifrin c'est un carré penché alors lol $|x_1|+|x_2|\leq (e-1)$

Danu

@TedShifrin Ask me about any of them and I can prove them or give you the precise line in Salamon's paper that proves it.

Ted Shifrin

21:19

C'est un vrai carré, Vrouvrou!!

Meow Mix

@Ted do you have any cool complex analysis problems I would be able to solve? I already figured out Akiva's wacky contour.

Danu

That is, with exception of the cohomology of $G_2/SO(4)$. But I can quote that from other papers, and corroborate it with my partial calculations which almost determine it.

Ted Shifrin

Stay focused, Zach. You're jumping way far ahead again.

Danu

^

Vrouvrou

Pourquoi, j'ai enlever la valeur absolue et je trouve un losange @TedShifrin

Danu

21:20

Orrrr become a physicist and do all the cool shit but don't really know what you're talking about :D

7

Balarka Sen

@MeowMix Akiva's contour thing was not complex analysis. It was actually a winding number problem; baby algebraic topology, sort of.

Ted Shifrin

A line of slope $1$ and a line of slope $-1$ are orthogonal, right, room?

Balarka Sen

As in, there was not any actual analysis in it.

Danu

@TedShifrin I want to say yes :P

Ted Shifrin

Me too.

Danu

21:21

But I'm afraid it's a trap

Admiral Ackbar - "It's A Trap!"

►

Balarka Sen

Sure, $y = x$ and $y = -x$

Ted Shifrin

@Vrouvrou is arguing that $|x|+|y|\le C$ is a diamond and not a square.

No, @Danu, your computations are evidently a trap.

Danu

I really really checked them well---I enlisted three fellow students, and a professor.

Dattier

Hello

Balarka Sen

@TedShifrin I think it's a diamond too.

Vrouvrou

21:22

@TedShifrin can i do it en ligne ?

Balarka Sen

Square with the diagonals placed on the x and y-coordinates.

Danu

But it's probably something stupid. The only reason why I'm strongly doubting you and Mike is because (i) my supervisor said something different (ii) his thing gives the right result (iii) I vaguely remember agreeing with him, despite being aware of the direct sum decomposition; but that's probably useless haha.

Ted Shifrin

A diamond is a rhombus that is not a square. If the angles are all $\pi/2$, isn't that a square??

Danu

@TedShifrin Maybe he just means a 45 degree rotated square

Balarka Sen

I never saw that definition of diamond. It's a square upto rotation; it just looks like a diamond.

Ted Shifrin

21:23

@Danu: But that's why I asked you to verify in Hirzebruch and Milnor, and not trust me.

Who says a square is no longer a square if it's rotated?

Vrouvrou

@TedShifrin vous me parlé a moi ?

Ted Shifrin

WTF?

Balarka Sen

@TedShifrin I am not claiming it's not a square. I'm claiming it looks like a diamond :P

Ted Shifrin

smacks Balarka rigorously

Meow Mix

@Ted Wow, section 5 was pretty short

Ted Shifrin

21:24

@Vrouvrou: Je m'en fous. :)

Meow Mix

hopes he doesn't make a joke comparing the section to my attention span

Vrouvrou

?????

Balarka Sen

@Ted Fixed.

Danu

@TedShifrin Well, I verified what there is to verify. Which is not much. But yeah.

Ted Shifrin

But isn't that what you're saying Mike and I said must be wrong?

Vrouvrou

21:25

@TedShifrin pourquoi cette réponse , je ne vous ai rien dit

Ted Shifrin

J'ai dit que c'est un carré. C'est un carré. C'est tout.

Si on fait virer un carré, c'est tout de même un carré.

Balarka Sen

@Vrouvrou It's just not a square in the usual sense. It's diagonals are along the coordinate axes. Ted is right.

Danu

@TedShifrin Well, I don't really think any of you explained my initial question. Mike just started repeating that it doesn't matter. I'm wondering if you can't find an example of a complex vector bundle that doesn't split as complex but does split as real. Maybe the Chern numbers don't come out the same way.

Vrouvrou

ok ne soyez pas méchant avec moi merci

Danu

Don't be offended @Vrouvrou

Mathematicians get stern when they think you're wrong :P

Vrouvrou

21:28

je remarque juste que le point (e-1, e-1) n’appartient pas l'ensemble

Ted Shifrin

Et vous insistez tout le temps que je réponde à vos questions. Il ne faut pas me les poser.

JeSuis

Hi!

Ted Shifrin

Ce point là n'a rien à faire avec ce carré.

Bonjour, @JeSuis.

Meow Mix

Is it bad if I don't like the white filling in oreos?

Ted Shifrin

I hate oreos altogether.

JeSuis

21:29

@TedShifrin how are you

Vrouvrou

pour un carré décliné c'est un losange c'est tout , enfin merci pour votre aide @TedShifrin et désolé pour le dérangement

Ted Shifrin

Sick, @JeSuis, thanks for asking :)

And losing patience.

Balarka Sen

I like bourbon biscuits.

Ted Shifrin

Oh good, @Balarka is turning into a drunk.

Balarka Sen

Not the actual bourbon

PVAL-inactive

21:30

I like bourbon ice.

Balarka Sen

Just the biscuit :)

SteamyRoot

I like bourbon

Ted Shifrin

@Danu: I certainly know examples that don't split holomorphically, but otherwise I don't think that's possible. But in our case, can't you tensor the real splitting with $\Bbb C$ to get the bundle you're talking about?

@Balarka: But there's bourbon in the biscuit, presumably.

I actually hate bourbon. But that's a different story.

PVAL-inactive

how

JeSuis

@TedShifrin arf, bon rétablissement. Fortunately I have no questions today :P

PVAL-inactive

21:32

its delicious

Ted Shifrin

I do not like sweet liquor, @PVAL.

Unless it's a brandy type thing, and then sweet still bugs me mostly.

Good for you, @JeSuis.

Danu

Bourbon is like shit whisky

Balarka Sen

@TedShifrin I certainly hope not.

JeSuis

lol @Danu

Ted Shifrin

Why the name, Balarka?

Danu

21:33

(not meant to be taken entirely seriously)

PVAL-inactive

@Danu You're drinking the wrong bourbon then.

Ted Shifrin

Oh, @Balarka is right.

Danu

@PVAL-inactive I don't know... I checked around quite a bit before I got some. Name some that you think are good... Maybe I know 'em

PVAL-inactive

Maker's is reasonably cheap and delicious.

Ted Shifrin

@Danu: You can respond to my math question at your leisure. I'm out of here for now.

Balarka Sen

21:34

Google says from the European royal House of Bourbon.

Vrouvrou

have a good night all bay

Ted Shifrin

Just saying, @PVAL, I have 15 kinds of gin but I hate bourbon.

Bonne nuit, @Vrouvrou.

Danu

@TedShifrin Yeah, I know. I see the reasoning.

Ted Shifrin

If you're doing it non-holomorphically, @Danu, you can use a hermitian metric, just as in the real case, right?

JeSuis

In class today we discuss about the fact that if we choose randomly two matrices in $GL_n{\Bbb{C}}$ the probability that the generated group is a free groups is 1.

Danu

21:35

@TedShifrin I mean

SteamyRoot

Here's a fun activity to do with your biscuits: youtube.com/watch?v=T1QD3uLa3IY

Ted Shifrin

@JeSuis: Yup, probability $1$ they won't commute.

Danu

In my case, one of the bundles I'm using to split is not a complex bundle

So I cannot split it as complex bundles.

Ted Shifrin

Oh, right.

Danu

So that doesn't work

JeSuis

21:35

@TedShifrin it's a wonderful result.

Danu

This is why I'm suspicious

Ted Shifrin

But if one of them is, then you get a splitting with an orthogonal complement.

But not your intended question.

Danu

I'm hoping/suspecting that maybe something goes wrong

Maybe you get something like a real isomorphic bundle which has no complex structure at all?

So no formula for Chern numbers

Ted Shifrin

I still don't see what goes wrong with tensoring the real splitting.

JeSuis

it was about Zariski Topology, hyperbolic geometry.

Danu

21:37

Me neither, to be honest.

Ted Shifrin

Wow, that much, @JeSuis? Oh, I see ... Zariski topology, of course. But I guess I don't see why hyperbolic geometry is needed.

Danu

I'm jsut certain it cannot come out right this way.

Ted Shifrin

Well, that's sure a good puzzle, @Danu. You can explain the answer to me over drinks in June :)

Danu

But by now I'm convinced enough that it's not just me being super silly and so I will not be too ashamed to ask Kotschick next time I see him in a week or so...

Balarka Sen

@SteamyRoot I have a better activity to do with my biscuits

Danu

21:38

After that, I'm going on holidays (?!)

Balarka Sen

Eat 'em

Danu

@PVAL-inactive ZRight, I heard of that. Do you know Knob Creek?

JeSuis

@TedShifrin It was to exhibit a free group.

Ted Shifrin

Sure, @JeSuis, but if $A$ and $B$ do not commute, they generate a free group.

What else is there to say?

Balarka Sen

@TedShifrin Really? What if A and B satisfy some other relator?

PVAL-inactive

21:39

@Danu I haven't drank enough to say much about it.

Ted Shifrin

Oh, good point.

Good thing I'm not an algebraist. :P

Danu

I think the internet told me that KC was supposed to be a pretty good bourbon. Honestly, it was just super boring and plain. Have you tried some whiskeys before (good ones, either smokey (Ardbeg, Laphroaig, Lagavulin etc.) or other ones like maybe Talisker or Bunnahabhain)? I find those so incredibly much more interesting.

Ted Shifrin

I'm not a big whisky fan at all, so I'm not paying detention to this. Bye for now, all.

Balarka Sen

You can probably pick infinite order matrices with prob 1. It's not obvious to me you can pick free of all relators

Danu

The bourbon flavor just seems to be a solid grainy/sweet maybe vanilla-ish thing. Not much to it, as far as I could tell.

@TedShifrin Detention?! lol. Bye Ted! Get well soon!

Balarka Sen

21:41

Bye.

PVAL-inactive

@Danu yah its mainly vanilla and sweet.

s.harp

I just drink scotch

JeSuis

@TedShifrin If I am not mistaken the argument was to say that relation form a closed set in Zariski topology and so zero measure, and to exhibit a free group tt was necessary to look for it in the half plan of Poincaré.

PVAL-inactive

in other words delicious things

unlike wood

smoke

s.harp

the ones you mention I like a lot (espc laphroaig)

JeSuis

21:43

@TedShifrin Bye, bonne journée. :)

@BalarkaSen did you see this result before?

Balarka Sen

I didn't.

But it doesn't surprise me too much, for probably unrelated reasons.

PVAL-inactive

I don't think bourbon has much in the way of "complex tastes"

Balarka Sen

I was just talking about a biscuit, people! :)

JeSuis

Ha, I send an email to my professor to get more details on this. I was amazed.

Danu

Exactly... The lack of complexity. Gets real boring!

@PVAL-inactive I don't know man... If you go beyond casual drinking in a bar every now and then... Those typical flavors get hella boring! Gotta get some depth going :D

(needless to say, I'm a huge fan of whiskey)

PVAL-inactive

21:51

Delicious doesn't get boring for me :/

Danu

Oh well. I respect your choices, wrong as they may be ;-)

But really though, to each his own I guess.

PVAL-inactive

Having hard liquor when I'm out is usually pretty rare for me as I don't like being robbed.

Danu

Yeah, I understand. I drink most of my whiskey at home---the price difference is huge. It's pretty affordable this way.

PVAL-inactive

I think the last whiskey I bought out was an $11 dollar drink that I figured would be around a triple. It ended up being a single.

Danu

That sucks :\

Meow Mix

22:13

Ted's back in town?

Daminark

So let's say you have a function $h:\mathbb{R}^{n+1}\to \mathbb{R}$ for which $0$ is a regular value, and $M = h^{-1}(0)$ is compact. Now, consider the map $F(x) = \frac{\nabla h(x)}{|\nabla h(x)|}$. Is it true that every point in $S^n$ is a regular value of this?

Scratch that

Balarka Sen

22:29

just think about a hypersurface :)

Daminark

Hypersurface is a manifold with codim 1, right?

Balarka Sen

Yeah.

I see you wanted compact. That can be done too; make a sphere really flat on the top so $F$ is constant there

Daminark

Yeah, the idea is to prove that the map is surjective

Now, I got that the preimage of any regular value is finite

So if everything in the image of $F$ is a regular value, then I know that $F(M)$ is infinite, and thus has a limit point in $S^n$

Balarka Sen

But that's not true because of the examples I gave, right?

$dF = 0$ if a bit of the surface is really flat.

Daminark

I'm still processing it, I'm slow with visualization

OK I can kinda see it

Balarka Sen

22:35

You can also make $dF$ kill a single dimension instead of all. Say, look at the torus embedded in the standard way (like a donut lying on a table). $F$ is constant along the "topmost" circle of the donut, so $dF$ kills a dimension.

($F$ is basically the unit normal along the surface)

Daminark

Merp

Well Sard's might help?

Balarka Sen

That says there is a regular value. What are you trying to prove?

Anyway the expert has arrived

Daminark

I'm trying to prove surjectivity, and I thought that Sard's says that the critical values form a set of measure zero?

Ted Shifrin

@Balarka: Isn't $F$ a scalar function?

Balarka Sen

@TedShifrin Scalar functions where?

Ted Shifrin

22:39

Oh, never mind.

I didn't scroll far enough.

Balarka Sen

It's the unit normal along the hypersurface, right?

Ted Shifrin

Yup.

Balarka Sen

Gauss map. Your favorite :)

Ted Shifrin

Indeed. :)

Titles of so many of my papers ...

Balarka Sen

@Daminark I don't know how Sard's might help with this.

Cool exercise, BTW.

Ted Shifrin

22:41

Wait. So what's the exercise?

Balarka Sen

He wants to prove the Gauss map is surjective.

PVAL-inactive

@Ted @Mike Do you know a reference for the fact that there is a unique (up to isometry) complete simply-connected hyperbolic space?

Ted Shifrin

For a compact hypersurface?

Daminark

Yeah

Ted Shifrin

@PVAL: There's a unique complete simply connected space of any constant curvature.

Reference ...

Balarka Sen

22:42

It's interesting to understand what happens if it's not surjective, @Daminark.

Ted Shifrin

Probably Joe Wolf's book on spaces of constant curvature has it, @PVAL. Let me check DoCarmo or Kobayashi/Nomizu.

For nonpositive curvature, there's of course Cartan-Hadamard.

Yeah, K-N prove (last theorem in Volume I) that any two simply connected, complete Riemannian manifolds of constant curvature $k$ are isometric to one another.

PVAL-inactive

@TedShifrin Does Cartan-Hadamard give isometry?

Ted Shifrin

Yup, @PVAL.

You get the exponential map from a fixed tangent space with an obvious metric as an isometry onto.

Balarka Sen

@Ted Do you see why the (2, 3, 7) triangle group acting on the hyperbolic disk has a fixed point on $S^1_\infty$?

Ted Shifrin

I don't really even know what that means, @Balarka. I should, but I don't.

@PVAL: It's also Theorem 4.1 in Chapter 8 in DoCarmo.

Balarka Sen

22:49

Group of orientation-preserving isometries of the hyperbolic plane generated by reflection along sides of the hyperbolic triangles with angles $\pi/2, \pi/3, \pi/7$

Ted Shifrin

So we need to decide if that is hyperbolic, parabolic, or elliptic, @Balarka?

Balarka Sen

Now you've lost me in turn :)

What do those mean?

PVAL-inactive

@Ted I don't have any of those books on me atm so I'm probably gonna try and do it as an exercise (gasp)\

Ted Shifrin

It's pretty deep, @PVAL. K-N do it by using Cartan's equivalence problem format.

Cartan-Hadamard is a nontrivial theorem, too. But you can quote that no problem for your case.

A hyperbolic isometry is elliptic if it has a single fixed point (think rotation), parabolic if it has one fixed point at infinity, and hyperbolic if it has two fixed points at infinity. @Balarka [I'm copying this out of an exercise in my diff geo notes.]

PVAL-inactive

This is for a talk that is probably getting quite survey-ish.

So its good for me to know the idea of how to prove everything.

Balarka Sen

22:52

@Ted Ah. So yeah, parabolic I guess.

Ted Shifrin

Oh, good grief, @PVAL. Well, I wouldn't do any of this in a seminar talk.

Can you figure out a LFT matrix representation for the isometries, @Balarka? If so, parabolic is $|\text{trace}| = 2$.

There's probably a good geometric way of thinking about this, but I've never thought about it.

PVAL-inactive

@Ted We are having a seminar for the Nielsen-Thurston classification, and we just proved the classification.

My hope is to explain as much of the 3-d motivation as I can.

Balarka Sen

I could try to if I knew what an LFT matrix is (that trace fact is nice). I am doing this for a topology problem.

Ted Shifrin

@PVAL: The uniqueness of space forms (up to isometry) is a standard well-known fact, but OK.

linear fractional transformation, silly, @Balarka :P

Balarka Sen

Oh.

Ted Shifrin

22:55

The isometry group is $SL(2,\Bbb R)$, after all.

in the upper half-plane model.

Balarka Sen

Sure, sure. Just couldn't decode that acronym.

Ted Shifrin

Sure, my fault.

So my exercise 15 in section 3.2 is to classify by trace and then to explain what the three types look like geometrically.

Balarka Sen

That's a very cute classification of parabolic isometries.

Ted Shifrin

I gave the hint to think about action on horocycles.

So maybe you should do the exercise and explain it to me. I've forgotten :)

Balarka Sen

But does that tell me if all of those fix a specific point? Parabolic would seem to say every isometry fixes a single point individually.

<--- knows no hyp. geom.

Ted Shifrin

22:58

Oh, well, good point.

Perhaps if you look at the generators (not the reflections, but their products), you can see they have a common eigenvector.

So these pairwise products would have to commute, then.

This kind of stuff is surely in Joe Wolf's book, too.

Which I gave away ... I hate having no library to speak of.

PVAL-inactive

@Ted Most of Thurston's constructions put explicit complete metrics locally isometric to H^3 on 3-manifolds, in ways that make it completely opaque to me to see them as a quotient of H^3. I knew that fact but really didn't know why (and dont think my audience will either) it was true.

Ted Shifrin

What do those particular angles tell you, @Balarka? Why are they so important?

Balarka Sen

@TedShifrin Bookmarked.

Ted Shifrin

Right, so Thurston is certainly calling upon this classification result, @PVAL.

Balarka Sen

@TedShifrin I have no idea. This appears as the fundamental group of a homology sphere (which admittedly also 2,3,5 does) and using that fact about fixing point at infinity on the circle, I can apparently construct - by a theorem of Thurston - a C^0 foliation which cannot be made C^1.

I think you don't have that for (2, 3, 5).

That it's a homology sphere group is essential for the construction.

Ted Shifrin

23:03

Interesting, Balarka. I mean we could take an explicit model for the triangle (one side on a vertical ray, the others semicircles we just construct), and then try to write down the matrices. I know I could do all that in an hour or so.

Danu

o/

Balarka Sen

Yeah, but that's a bit uninspiring. I thought there'd be a visually evident way to see this.

Danu

Did you see the correction I sent you, @Ted? I typo'd a factor (and then the error propagated)

Ted Shifrin

Yeah, @Danu, thanks. I'm still puzzled by our conundrum.

Me too, @Balarka. As I say, Joe Wolf's book of spaces on constant curvature discusses various crystallographic groups carefully, so I bet it's in there.

Danu

I made some big progress in structuring my thinking on the overall to-do list :-)

Ted Shifrin

23:05

And it's surely in Thurston or someone like that somewhere.

Danu

I need a lot of stuff by Wolf

Also Gray

Ted Shifrin

Ah, my old compatriots.

Danu

No Grey Wolves just yet...

But yeah

3-symmetric spaces, isotropy irreducible homogeneous spaces, nearly Kaehler structures...

Daminark

Somehow I'm beginning to wonder if regular/critical values are the way to go about things here

Ted Shifrin

Yup ... Gray did a bunch of stuff like my thesis with volume of tubes formulas, too.

Demonark, the proof that pops into my mind immediately uses stuff you don't yet know. What set-up did he give you?

Daminark

23:08

So $h:\mathbb{R}^{n+1}\to\mathbb{R}$ and $0$ is a regular value

Balarka Sen

@TedShifrin Never heard of that book. Interesting.

Daminark

Now, $M = h^{-1}(0)$ is compact

Ted Shifrin

Oh, it might be helpful, @Balarka.

Balarka Sen

@Ted: I think you an cook up a proof a la Hilbert.

Ted Shifrin

Huh? Proof of what?

Daminark

23:09

Then let $F(x) = \frac{\nabla h(x)}{|\nabla h(x)|}$, show that $F$ is surjective onto $S^n$

Balarka Sen

Like his proof of compact constant negatively curved surfaces in R^3.

Ted Shifrin

That's a super hard PDE (or complex analysis) theorem, basically, @Balarka.

Demonark: What machinery do you have? Absolutely nothing?

PVAL-inactive

@Balarka This should be something as simple as dividing S^1 by the limits of your geodesics coming from the sides of your triangle

Balarka Sen

Err? I mean the proof that there isn't any embedded surface in R^3 of constant negative Gaussian curvature. Am I confusing names?

The proof with the planes

Ted Shifrin

No, that's a hard theorem, @Balarka. You're thinking of something like Fary-Milnor for curves or some convexity result?

PVAL-inactive

23:12

The reason it doesn't work for the (2,3,5) case is that no such triangle exists.

Daminark

We don't really have much machinery

Ted Shifrin

I know. :(

So if it weren't surjective, you'd have a value that was regular by default (i.e., not in the image). So that would tell you degree 0, which would lead to issues by later stuff. But ...

Balarka Sen

@Ted It is a hard theorem to prove any embedded surface in R^3 has a point of strictly positive Gaussian curvature? It's in your notes.

Ted Shifrin

That has nothing to do with Hilbert, @Balarka.

Balarka Sen

Huh.

PVAL-inactive

23:14

I saw a proof of that in a 1-line MO question recently.

Ted Shifrin

There's no compact surface of nowhere-positive curvature.

Yeah, @PVAL, the geometric idea is 1 line.

Balarka Sen

Yes, I mentioned compact above; was weary of saying it again and again

Ted Shifrin

Nothing to do with Hilbert, I repeat.

Balarka Sen

I think you want that proof idea here

@TedShifrin Got it. Was misremembering then

Ted Shifrin

Nothing to do with constant negative curvature (and necessarily noncompact).

That uses actual differential geometry ideas, and I don't see how it proves the Gauss map is surjective.

Semiclassical

23:16

hi chat

Daminark

We've also proven on this homework that if the derivative a smooth map $f$ from one manifold to another is always injective, then you can find local charts to get $\psi^{-1} \circ f \circ \phi(x_1,\ldots,x_m) = (x_1,\ldots,x_m,0,\ldots,0)$

Ted Shifrin

But maybe it's a Morse theory idea. Take a direction that is not hit by the Gauss map, and consider that height function.

Daminark

We also know that preimages of regular values are submanifolds

Balarka Sen

For any vector you can take a plane with normal in that direction and pull it closer to the surface so it's tangent to it, @Ted

Ted Shifrin

Right, Demonark, and there is a companion for surjective — indeed, for constant rank, although I don't usually do that in the undergrad course, either.

Oh, I see. It's just what I said. Consider the height function and it has a maximum somewhere. Now let's shut up and let Demonark think.

You have the same idea, @Balarka, but you have to be a bit careful, yes.

Daminark

23:18

@Ted In analysis we did the constant rank theorem for mappings between open sets, so I imagine in general we find charts and then use that to get it, right?

Ted Shifrin

Yes, Demonark, precisely.

Balarka Sen

Yeah, I leave Daminark to fiddle with stuff

Daminark

Alright, thanks!

Ted Shifrin

I'll have to give Demonark my favorite question soon: Proving that a smooth retract of a manifold must be a smooth submanifold :P

OK, I'm off to watch Federer.

Daminark

Lol, the GMT Federer? And see you!

skill patrol

23:19

Enjoy

Balarka Sen

@Daminark I was about to say!

Sheesh.

Daminark

(That's the only Federer I know of)

PVAL-inactive

Where's @Daminark's question?

Balarka Sen

It's the tennis Federer I suppose

Daminark

Makes sense

And @PVAL it's some time back, basically that given a compact hypersurface $M$ (or at least one which is the preimage of a regular value of smooth function, which I'd guess is exhaustive but I'm not sure), its Gauss map is surjective

And based on this

Meow Mix

23:25

Hi

Daminark

So given $v\in S^n$, basically take its tangent space and push it far off, so that $M$ is between it and the origin, and then pull it toward the surface.

Balarka Sen

Yeah.

Daminark

The distance between the manifold and our affine space should be achieved by some point, and at the first point of intersection, this should be tangent to $M$ as well?

The picture suggests that

PVAL-inactive

@Daminark So @Ted knows the answer to this but wants you to think about it?

Daminark

So it seems

I'm more thinking out loud at this point

PVAL-inactive

23:36

Alright I won't give it away then.

It took me a few minutes.

Balarka Sen

I think Daminark's picture is good

If he can make this into a proof we're happy

PVAL-inactive

Tell me if you want a hint

Daminark

Would it be fair to kinda shift around the manifold a bit? Maybe do it such that the desired point is now a scalar multiple of $v$?

Somehow, the fact that a manifold is situated in a particular location feels like it shouldn't really matter

Like if we rotate the whole picture somewhat maybe that'll help

Simple

23:52

In my ML textbook, a Lipschitz continuous function is a function $f$ whose rate of change is bounded by a Lipschitz constant $L$ such that $|f(x)-f(y)|\leq L\Vert x-y\Vert_2$ for all $x,y$

We should have $\Vert f(x)-f(y)\Vert_2\leq L\Vert x-y\Vert_2$ instead $| f(x)-f(y)|\leq L\Vert x-y\Vert_2$

Do I miss something?

Semiclassical

$f:\mathbb{R}^n\to\mathbb{R}$? @Simple

Simple

yes

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