hmm. maybe I should try that with $((x - h)^2 + (y-k)^2)^2$ instead

though it looks to have the same problem, as above, hrm

I can get it easily enough if I go to rational functions instead, but I'd rather something polynomial :/

I do seriously wonder if you can ever get even a local minimum if a section just below and a section just above is like three disjoint circles.

yeah.

I can't convince myself either way.

Ok, maybe some other point can be a minimum. Like even in your example a saddle point is inevitable.

right.

(similarly in the 120 degrees thing a monkey saddle or sth like that will be inevitable. I don't know what the model of a generic critical point is where the Hessian vanishes but the 3rd order tensor doesn't)

Alright, I am really chickening out now because I am procrastinating on work

lol

this is technically research work, but i'm a bit half-awake for it right now

Got it! @BalarkaSen

$f(x,y)=(1 + 2 x) (1 - 2 x + x^2 - 3 y^2)$

good to hear

...except that's upside down

oh well. $(1+2x)(3y^2-(x-1)^2)$

eh, whatever

good work

wait...something is wrong with that.

that's got f=0 for all x=-1/2

that's not good

aagghhh it's a monkey saddle

really?

well, probably not. but close enough:

It darn sure looks like a monkey saddle. Is that really a minimum?

I mean, pictorially, of course not.

Hi @TedShifrin

Hi @Balarka

@MikeM: How did you figure out studiosus?

think so. this is what I get if I zoom in:

i changed the mesh style

so yeah, it's got that dip in the middle

@BalarkaSen You'd need a normal form for (symmetric) cubic forms.

Right, @TedShifrin

@TedShifrin He changed his name.

@TedShifrin I'd be curious if you know a simple example of what I have in mind. one that actually works :/

I know what I did wrong

Also, I was looking at an old question of mine and Moishe had a comment that I knew was studiosus.

Aha .... I'm not sure we have any better idea who he is now, @MikeM :P

aaanyways, gotta go for now

I don't know what you're talking about, @Semiclassic, but let me know sometime.

and then the two comments after that

I guess we can think about Morse theory a little bit, @Balarka, although not with degenerate critical points.

trying to figure out whether one can do that with polynomial $z=f(x,y)$

i thought i'd come up with an example, but I did it wrong :/

I don't see circles ...

good evening everyone

@TedShifrin Moishe Cohen has an academic website etc.

@TedShifrin IIRC the cubic version of Morse theory is known as Cerf theory but I have no idea if this is the kind of thing it deals with

Oh, really, @MikeM ... actually in Korea?

@BalarkaSen That's the reductionist catchphrase I've given you.

@TedShifrin No idea. Seems like he's in Israel.

The name would suggest Israel, but I thought I clicked and saw Far East somewhere in his profile. Let me look.

I figured.

faculty.vassar.edu/mcohen

His MSE profile says South Korea

His Vassar webiste says Vassar. :P

Maybe different persons?

Visiting from Israel, yeah.

So the Seoul is a lie.

That would seem exceedingly unlikely, @Balarka.

He knows a number of people I know, actually, but Ph.D. from LSU. Interesting.

Well, glad to know who he is :P

From his MSE contributions I can't call him anything other than brilliant. Good luck to him and his career.

Interesting that his CV shows that's he's in Electrical Engineering at the Technion.

Yes, good to have people like him around for the rest of us mortals :)

My guess is he's now going on the job market and would like to be able to cite his MSE contributions.

indiana.edu/~jfdavis/notes/duh.pdf

I should write a PDF like this. Of course, I forget the things that I usually forget.

heh, cool article

Right. If I wrote such a handout, I'd have forgotten the results.

@Balarka: Any un/interesting diff geo to discuss?

@TedShifrin: I have a line of curvature on a surface. Is there a way I can detect it? I know that if I compute the IInd fundamental form of the unit tangent vector with itself, I get the principal curvature at each point. But what if I just know the curvature of the surface, and not $k_1$ and $k_2$ individually?

principle = noun, principal = adjective (or sometimes noun)

oops darn

You can only discern principal curvatures by looking at the normal slices of max/min normal curvature. There's lots of ways you can solve $k_1k_2 = K$. :)

As Mike likes to say, Gaussian curvature is intrinsic and principal curvatures clearly depend on the embedding and are extrinsic.

ah, fair enough

That makes sense.

If the surface has more symmetry to it — like a surface of revolution — then of course we know the principal directions. And if it's a minimal surface, we might know something ...

Some cool exercises coming up in the next sections, too, @Balarka.

In the case of a surface of revolution, does it suffice to see the obvious curves (circles of revolution, and the longitudes) are orthogonal?

No, that + justifying why one of those families must be lines of curvature ... suffices.

You can look at the Gauss map explicitly and see why those directions are eigenvectors.

I just got an email from someone I asked a research question to that was (no subject). I was terrified it was going to be "Everything you said was wrong."

I usually demonstrated that in lecture geometrically. I don't think I wrote it out that way in the notes.

You should be thrilled that the person replied, @MikeM.

No, no, I asked them in person. I was afraid they thought on it. :P

It was actually an email organizing a carpool. It always is.

Ohhhh ... so no response to your question.

no, I got that last time as an "I think so". (I think so too.) I was afraid it had turned into a "God no."

haha

I am a grad student, hence perpetually afraid.

Get over it.

@TedShifrin Right. It's easy to justify why the parallel curves are lines of curvature, I think.

Easier to justify the meridians (profile curves), but you should think through both.

By parallel I mean the longitudes (not the circles). :P Sorry about that.

I also asked you to think geometrically about why the rulings and helices are asymptotic curves on the helicoid.

The circles are parallels. Because they're all parallel.

Right, right, I misspoke.

@TedShifrin The normal of the helices stays normal to the helicoid.

But the parallels are easy, too, because the Gauss map along them is a constant vector plus something normal to the circles. So it's easy to see how it changes.

No, the principal normal of the helices is along the rulings!

I messed up the TNB frame. Right.

It's tangent to the surface

Cool.

Hey, I was wondering if anyone might be able to shed any light on a PDE problem I have? math.stackexchange.com/questions/1968789/…

(Disregard if this isn't the place to post it)

How do you know $\lambda=(2n)^2$? @wowdavers

I haven't sat down and separated variables to work it out.

Bad form, by the way, to write a summation formula in terms of $n$ and not have $n$ everywhere in the formula.

And where is $k$ in your solution?

@TedShifrin Ah, I see, so you're saying the normal to the surface along the circles project to a constant vector on the plane orthocomplement to the normal of the circle.

I'm saying the component along the axis of revolution is constant, @Balarka.

By symmetry :)

@TedShifrin I've worked it out - I'm pretty positive that $\lambda=(2n)^2$ - substitution of $\lambda$ into the equation gives me the weird non-integer n.

That agrees with what I said. Nice!

What about $k$, @wowdavers?

Yeah, sorry about that - should have substituted the n's.

$k$ is a constant

But you need $k$ in your solution!

@Balarka Do you know the Riemannian statement of the uniformization theorem? Have I talked to you about this?

It should be in the u(x,t) function, which is the solution. u(x,0) is an initial condition (exponential goes to 0).

*I mean exponential goes to 1

@MikeMiller No, I don't. I think you did a little bit, but I forgot because I didn't understand.

$k$ needs to appear in your time exponentials, doesn't it?

Yeah, $B_n*e^{−λkt}...$?

Let $\Sigma$ be a smooth surface without boundary. Do you understand why there's a canonical equivalence between oriented conformal classes of Riemannian metrics, and complex structures on $\Sigma$?

Oh, sorry, you have it in there. I missed it. My humble apologies.

No need to apologize

@MikeMiller Not immediately, at least rigorously. Probably because holomorphic maps are conformal.

One direction is clear (complex structure ---> oriented conf. classes of Riemannian metrics). The other direction is probably not so obvious to me

OK, so I agree that boundary conditions give us $2n$ as the frequency for the sin wave. The problem, @wowdavers, is that their initial condition is not compatible with the boundary condition. So there is a contradiction in the problem.

@Ted

@TedShifrin I realized the error and asked my prof - and apparently you can still solve it. Something to do with trigonometric identities?

Suppose $M$ is given an orientation and a Riemannian metric $g$. Then one can define a complex structure $J$ by $J(v)$ = "the unique vector such that $\{v, Jv\}$ is an oriented orthonormal basis of $T_x M$".

It is straightforward to verify that $J$ depends on the conformal class of $g$; i.e. $e^fg$ gives the same $J$ as $g$.

Oh, I knew that but I forgot (you told me). I should write an article like that too.

@wowdavers: But $\pi\sin(3\pi/2) = 0$ fails to hold, so that's garbage.

Can you have a look at my answer here?

Yes, there are identities like $\sin 3x = \sin 2x \cos x + \sin x \cos 2x$, but so what?

Is there any work around to this, or is it impossible to solve?

I know it physically doesn't make sense either - infinite heat flux is implied.

The initial conditions and boundary conditions are in contradiction with one another.

So I take it that you can't solve the problem at all?

@BalarkaSen Going backwards, given $J$, you can define the orientation and also, given any vector, what the unique "oriented orthonormal vector to it" is. That suffices to define a conformal structure in dimension 2.

I agree.

@TedShifrin

@wowdavers: Ask your teacher how $\sin (3\pi/2)$ is equal to $0$?

The uniformization theorem says that the universal cover of a Riemann surface is isometric to one of the standard three. Because complete simply connected surfaces with constant curvature are one of those very same standard three (and their metrics give rise to the usual complex structures), the Riemannian uniformization theorem says "Every metric on a surface without boundary is conformal to a complete metric with constant curvature."

So for what values of $t$ do the boundary conditions hold? We'll have to have a very discontinuous solution if you say $t>0$ but not $t=0$.

This probably doesn't help you understand the geometry of embedded surfaces, since I don't think there's any good way to think of conformal changes in that setting. :P

I think I found an error with a memorandum.

What you just typed there isn't a contradiction, @wowdavers. What our problem has is a contradiction.

Doesn't $E = G$ and $F = 0$ give necessary and sufficient conditions on $\Bbb{I}$ to be conformally equivalent to the plane?

Oh, I guess it is a contradiction if $u(0,x)=1$ for all $x$ including $x=0$ and $x=L$.

The solutions of the heat equation are continuous, so I don't know what they have in mind. You'd better talk to your professor, @wowdavers.

I have no idea what that means.

Yes, to the plane in the usual metric, @Balarka.

Remember, I never learned anything about this.

Right. I worked out that exercise.

The metric tensor has to be a scalar multiple of the identity, @MikeM.

@TedShifrin Oh, OK.

@MikeMiller Interesting reformulation.

@BalarkaSen Well, one can write down a formula for what happens when you change the metric. If I use $s$ for "curvature" (what else would $s$ be? :D), I have something like $s(e^fg) = e^{-f}(s(g) - \Delta f)$.

I'd like someone to double-check my answer.

I see. I wonder if I can do that by hand using what I have learnt of surfaces.

$s(e^f g) = e^{-f}(s(g) + \Delta f)$.

Do you know GB yet?

Nope.

I pretend to know the statement but no, I don't know it yet.

OK.

Hey everyone, just quick question. Is it possible to define a compact set as the double union of open subsets of a metric space?

@ahorn: I didn't have time to read your answer carefully. Did you explain carefully why one has to check that $t=0$ and $t=1$ cannot occur?

I understand why the problem is flawed, but I have input saying that 1. you can solve, 2. that you can't - and the prof is saying the former.

How does the solution go from $-3$ to $0$ instantaneously as $t$ increases? @wowdavers

So go talk to the prof, @wowdavers.

Well, I can say that if you're trying to prove uniformization for a closed surface $\Sigma$, you're trying to solve the equation $\text{sgn}(\chi(\Sigma)) e^f - \Delta f = s(g)$.

I last taught this material in 1986 :P

So we have "reduced" the uniformization problem to solving PDEs, which people are OK at.

what's a double union?

@Perturbative No, that would not be correct

Interesting!

@MikeMiller Clearly not I.

Heya @Alessandro

hi @Tobias

@TedShifrin Hi

Hi @Ted

I shall try to work it out, thanks @TedShifrin

If $\Sigma = T^2$, this is solving a linear PDE involving the Laplacian - $\Delta f = -s(g)$. That this is solvable is now well known.

Flat torus or usual torus, @MikeM?

Any torus!

If it's the flat torus we already solved it.

When $\chi(\Sigma) < 0$, this is a bit harder, since it's no longer linear (that nasty $e^f$). But there are still methods.

Oh, I wasn't actually following .... but ok.

@MikeMiller cute

@TedShifrin If it's the flat torus one could of course take $f=0$.

there's probably some boundary conditions involved too?

@TedShifrin I didn't explain that because $t\neq 0$ and $t\neq 1$ is given in the question. I'm going to chat.stackexchange.com/rooms/43593/constructive-feedback

There's no boundary. I said closed surface.

Gotcha

When the surface is noncompact, it's harder. Basically you pick a compact exhaustion $\Sigma_n \subset \Sigma$, solve that equation with the boundary condition $f_n|_{\partial \Sigma_n} = -n$, and then prove that that the pointwise limit $\lim f_n$ exists and is smooth. The boundary condition is what forces the conformal metric to be complete.

Could be that I want it to be equal to $n$. Hell if I remember.

@Alessandro, e.g $\cup_{i=1}^n\left(\cup_{\alpha=1}^{\beta}G_{\alpha}\right)_i$

@BalarkaSen You don't know scalar curvature in higher dimensions, but I told you the definition. So one natural generalization of the uniformization question is "On a closed manifold, is there a conformal metric of constant scalar curvature?"

@Perturbative how is that different from just a union?

Mhm

Uhm I'm missing something, isn't $\bigcup\limits_{\alpha=1}^{\beta}G_\alpha$ a single set?

@TobiasKildetoft, I assumed it would be, convention wise, just like how people refer to 'double summation' instead of just 'summation'

That's significantly harder in higher dimensions and was finished in 1984. It's called the "Yamabe problem". I have no damn clue how it was done.

@Perturbative Anyway, how would this relate to being compact at all?

@MikeMiller Does that answer it in the positive?

yeah

surprising

From a PDE perspective it lies at exactly the boundary between the kind of equation where you can solve it easily and where you can't solve it. So it lasted a long time.

So it all boils down to analysis at the end? (I'm curious how to get the PDE though. There's no natural correspondence w/ Riemannian metrics anymore (of course not!))

Oh, sure there is. Scalar curvature still has a formula for what happens when you change the metric. Let me find it.

@Perturbative can't every open set be written like that? And in Hausdorff spaces (so also in metric spaces) the compacts are all closed

Ah, interesting.

@TobiasKildetoft I'm trying to find a simple expression for the definition of a compact set. Rudin defines "$K$ as a compact set of every open cover of $K$ contains a finite subcover. So if $\{G_{\alpha}\}$ is an open cover of $K$, then there are finitely many indices $\alpha_1, ... , \alpha_n$ such that $K \subset G_{\alpha_{1}} \cup .. \cup \ G_{\alpha_{n}}$"

That's the standard definition, @Perturbative.

You may start with a very uncountable open cover, even, but still a finite number will suffice for a given compact subset.

The actual PDE is different in higher dimensions. You can write it as $$\psi^{n+2/n-2}s(\psi^{4/n-2}g) = 4\frac{n-1}{n-2} \Delta \psi + s(g)\psi.$$ Note that here $\psi > 0$ and that the Laplacian $\Delta$ depends on the metric $g$. (The Laplacian also depended on the metric before.)

@TedShifrin, is there a simpler way to express $K$ though? Instead of having of having to say "finitely many indices" couldn't we just represent it as a union from $i$ to $n$?

yikes

that looks bad

I think the solution was quite complicated and involved more geometric ideas than "eh just solve it".

But I don't know much about that stuff. There's also an associated flow (like Ricci flow). I'd be interested to know if that's applicable in all dimensions as a proof yet.

No, @Perturbative, because you have no idea which open sets (i.e., which indices) will be the ones.

@Alessandro Written in the way I wrote it?

@TedShifrin Ahh, okay

A lot of good theorems in geometry are ultimately proved by delicate PDE approaches. Calabi's conjecture is another famous one.

The smarmiest phrasing is "There are Calabi-Yau manifolds".

heh.

You don't want to hear this crap, and Ted doesn't want you to hear it, so I'll stop.

I don't know what those are though

oh, I do want to hear. But I am probably not in a position to appreciate more than what you said.

Mumble mumble chern class mumble mumble Ricci tensor mumble mumble.

I'll bug you and Ted when I am in such a position though, for sure!

Learn what holonomy is and we'll talk.

excellent, that's in the next chapter in Ted's notes.

But I am in like section 2 of this chapter so that'll take time :P

You still have GB to learn first.

No, GB comes right after holonomy.

oh

The boundary term for GB is really asking about holonomy. He'll see.

Sure. I see why you'd do that.

I am the wrong person to teach geometry anyway. My taste is all wrong.

@Balarka: You really should read section 3.3, too, and see how elegant differential forms are for all the stuff.

@TedShifrin, question for you: if I write $\Bbb{I}_p(v, v)$, I say "inner product $v$ comma $v$" or a variation of that aloud and in some contexts say "metric". what do I call $\Bbb{II}_p(v, v)$, then?

Some day you should teach the undergrad curves/surfaces course and learn it, @MikeM.

second fundamental form on $v$?

Or $v,v$ if you want.

@Ted: True. But I mean even abstract geometry.

I don't mind confusing quadratic form and symmetric bilinear form.

eh, too much to say :)

Then keep quiet @Balarka :D

@perturbative yes

"2 of v comma v"

Now I'm wishing I'd gotten some of my undergrads to video my diff geo lectures the last semester, but no one volunteered :(

I like classification theorems, automorphism groups, moduli spaces, and things of a topological flavor.

@Alessandro Ted pointed out above that it can't be represented the way I wrote it

I spent a lot of my career teaching things that were not my research specialty, @MikeM. Depends on what you're interested in doing ...

I see MM has results on when free product of groups acting on S^1 also acts on S^1, @MikeM.

@Perturbative the compact sets can't be represented this way

constant curvature existence questions... but some of the more classical and beautiful stuff isn't my thing. I never liked homogeneous spaces.

You're just a damn topological analyst, @MikeM, not a geometer.

I was proud when I could recently give a colleague an easy classification of 2- and 3-manifolds with transitive isometry group.

@Alessandro, my bad, I throught you were referring to the compact sets, my apologies

no problem, no need to apology

If I take a topology $\tau$ on a space $X$ and order its basis by inclusion there won't in general be a minimum but only some minimal basis, is that correct?

I remember not so long ago when @Alessandro was a baby mathematician who didn't know much ... how far you've come :P

@Ted I'm going to make that phrase catch on!

Is there a quick way to factorising quadratics for x^2 + bx + c currently im listing all the factor pairs and checking them, its time consuming and tedious, wondered if there are some neat tricks to factorising faster?

I don't think there's necessarily that, @Alessandro.

Suppose you take the rational balls at rational points in $\Bbb R^2$ (or intervals in $\Bbb R$). What's your minimal basis?

I like geometry too, but my tools aren't built to really tackle Riemannian questions, other than questions about psc metrics on 3-manifolds.

I still know nothing @Ted! I learnt what a topological space is some 3 weeks ago to be honest

@WDUK: Think about factors of $c$?

@MikeM: You'll recall that I am farther from Riemannian geometry than you.

yeah im doing that currently then finding which one is the sum that equals b

(Or $-b$, yes.)

That's the way to do it.

so writing down all the factor pairs is the only real way to doing it ?

Well, or thinking through which ones can plausibly have a chance of working, yes.

@Ted That's fair. I also don't know shit about complex geometry.

By the time @Danu's done, you will :P

@WDUK That's just using Vieta's formulas.

@Anubis: Way toooo fancy for a beginner.

Maybe if I'm established I'll try to break into su3, g2, spin7.

never heard of it before lol

It's what you're doing. Sum of the roots is $-b$, the product is $c$.

I think I owe Danu an explanation of Engel structures, since his advisor just posted a paper about them.

oh so thats the name for it, cool !

I have this growing sense of guilt when I google the first and second fundamental forms of surfaces which I have already computed before but forgot

It's not a bad idea to keep a little notebook of examples. :)

I have google open for most conversations.

Hm, I see @Ted, if I have a basis which is a subset of the set of rational balls at rational points I can just eliminate one of them and get smaller one

@Balarka: I finally wrote myself a solutions manual for the diff geo book because I got tired of repeatedly working out the same problems every year I taught the course. (And I let students do a bunch of different problems, so it was getting tedious. Of course, it also enabled me to improve the text and reorganize the problems somewhat.)

That idea sounds good. But I am also not very organized, so I am not sure how it'll turn out. I'll give it a go tomorrow morning.

@Alessandro: Yeah, but good luck finding a "smallest." :)

@TedShifrin Hah.

Nice.

but if I keep iterating this construction the infimum need not to be a basis so I can't reach a minimal element

Right.

What I should really do is start keeping a notebook of accessible geometry and topology problems.

That's pretty typical for most topological spaces, @Alessandro.

makes sense now, I don't know why I expected some minimal basis earlier

just optimism probably :P

I think it would make for a discrete topology in the end, @Alessandro.

At least in the Hausdorff case.