@TedShifrin hum so if all orbit are periodic it means that T does not depend on $x$?

@JeSuis: No, if all orbits are periodic, the period certainly will depend on which orbit you're on. Consider the vector field $(-y,x)$.

@TedShifrin You mean in the sense that I can always bring $x_0$ to $x_0'$ by a translation of the torus?

Oh wait. That one has constant period. But if I do $(-y,x)/(x^2+y^2)^\mu$, then ...

@Danu: By additivity of integrals, you're translating in the vector space, then modding out to get the torus. So that's a translation of the torus.

@TedShifrin Sure, okay. Anyways

I am reading an exercise when it says that if one has a periodic orbit then all orbit are periodic, what am I supposed to prove ? not sure how can I formalize the question @TedShifrin

@JeSuis: Oh, assuming the vector field is linear?

now , yes

OK, then it's true.

Now, the properties are (i) it is holomorphic (I can undersatnd the proof: localize and then do it "by hand" in $\Bbb C^n$) (ii) it induces a bijection $\bigwedge_0\operatorname{Alb}(X)\cong H^0(\operatorname{Alb}(X),\Omega_{\operatorname{Alb}(X)})\cong H^0(X,\Omega_X)$ (which I don't understand)

is that so clear for you? :p @TedShifrin

Certainly not true for most vector fields that have the occasional closed orbit.

@PVAL Everythinf works.

(iii) it is functorial for holomorphic maps---this is not so bad.

I mostly don't understand the second because the proof says "also proven by a local argument" and that's just not enough for me.

Yes, @JeSuis.

What does $\Lambda_0$ mean, @Danu?

@TedShifrin I don't know. I'm assuming he typo'd and it hsould be $\bigwedge^0$??

@TedShifrin formalizing it, what am I supposed to prove ? I am a bit lost there

No, actually I just don't know.

Well, find the notation and tell me, @Danu :)

Or maybe let's just think about the latter isomorphism and ignore the first.

@TedShifrin Yeah, perhaps that's just notation he introduces. I was mostly interested in the second.

I'm fairly sure though that the $\bigwedge_0$ never occurs before.

@JeSuis: For the linear case, you should be able to write down an explicit solution using matrix exponentials.

@Danu: How do you get holomorphic differentials on a torus?

Right, @JeSuis.

When does such a thing have a closed orbit?

what do you mean by closed orbit?

What do you mean by a periodic orbit? :P

@TedShifrin I don't really know. I think I've heard at some point or other that you get them from holomorphic forms on $\Bbb C^n$, and now that I think about it you need something periodic (doubly so?).

Hello could soneone hekp me with the limit as x goes to 1 from the negative of (arccos^2(x) )/(1-x)

ahhh, ok closed'='periodic, the flow is $\phi_t=x(t)$ right

Right, @Danu.

Do we know dimensions of cohomology, @Danu?

So you know what the orbits of $x'=Ax$ looks like depending on the nature of the eigenvalues of $A$, @JeSuis?

@TedShifrin Of a torus? Sure... $n$ choose $k$ in dimension $k$ for an $n$-dimensional (real) torus

Because the CW boundary maps are all zero

We're doing $H^0(\Omega^1)$, right?

Yeah

So what's the dimension?

Dimension of Alb? Depends on $X$, I'd say?

@TedShifrin I know that the solutions depends on eigenvalues of $A$, yes

Hello anyone?

Sure, @Danu. I'm asking for $\dim H^0(\Omega^1)$ ...

LOL, @user379685. There's a lot of traffic in here.

What is $\lim_{x\to 1^-}\dfrac{\arccos(x)}{1-x}$?

I cant read this im on mobile sorry

The square of arccos divided by 1-x

lol @TedShifrin you're too good of a person :P

I removed the square, @user379685. What is that limit?

Why $1$, @Danu? Say $\dim H^0(X,\Omega^1_X) = g$. What's the dimension of Alb?

@TedShifrin Yeah, I just deleted this. Sorry.

Infinity

@TedShifrin formellement c'est quoi les orbites de $x'=Ax$?

No, @user379685. Hint: You should recognize this as the definition of a derivative.

anyone here know eye doctor terminology?

Ummm

@JeSuis: Les trajets ... the flow-lines.

@TedShifrin Well since Alb is $H^0(X,\Omega_X)^*$ quotiented by a lattice, its dimension is then just $g$.

OK, so tell me the answer to my question.

Im lost :c

Write down the definition of the derivative of arccos at $x=1$ (computed from below), @user379685.

@TedShifrin: $M$ is my surface, $u$ and $v$-curves are lines of curvatures. Since $k_1 \neq 0$ is constant, and principal curvature is precisely the curvature of the line of curvature $u$-curve (WLOG) is of constant curvature. If it's torsion-free, I can invoke classification of curves and get done.

And @user379685, don't forget to tell me what arccos(1) is.

@TedShifrin Damn, thanks. So it should be half of the dimension of $H^1$ on Alb, so $g/2$.

@TedShifrin en fonction des valeurs propres on a des trajets différents

We didnt had derivatvies yet

Ted is getting freaking overloaded here :P

No, @Danu. Guess again.

@TedShifrin What are the doubly ruled complex surfaces? Are there only two?

sheesh

Then I have no idea how you're supposed to do it, @user379685. Do you find limits by plugging numbers in a calculator?

@TedShifrin Wait, what? Didn't you ask for the dimension of $H^{1,0}(\text{Alb})$? Isn't that half of the dimension of $H^1(\text{Alb})$, which is $g$ choose $1=g$?

@MikeM: two?

No by doing some manipulations

Right, @Balarka. You have to also explain why the normal curvature agrees with the curvature of the curve.

Square of arcsin x divided by 1-x can be done without derivatives

@TedShifrin Really, only one?

I guess I believe you.

Nevermind, I do believe you.

Well, @user379685, that's not of the 0/0 form.

@MikeMiller What's the best NV DLC?

What did you like about the game most?

It's challenging having 4 mathematical conversations.

Changing arcsin to pi/2 - arccos makes this computable

I applaud your courageous effort, @TedShifrin!

Nvm

@MikeMiller I liked the weird missions. Like in the bunker with the plants, the ghouls in the rocket factory.

@TedShifrin I was going to said: Bravo à Ted for this simultaneous conversations

I see, @user379685. So you substitute u = arccos(x) and rewrite the problem. Duh.

I didn't think of that at first.

Dead Money is a story-driven and mostly railroaded quest. You cannot back out once you go in. Honest Hearts has a barebone quest but is great for exploring and really nice to look at. (I've actually been to Zion and it's fucking beautiful.) Old World Blues is funny and about weird science shit. The combat is kinda bad though and you'd better get used to having to hit robots with axes or whatever.

Fucking bullet sink.

@user379685: It's actually a clever problem.

Maybe it's actually 5 conversations. :D

@MikeMiller Actually been? Does not everyone get to go?

Hi

@TedShifrin Literally everyone in the room was asking your help when Mike joined in :P

I don't like it much but it might be up your alley. And Lonesome Road is more like Dead Money, but it literally only has one other character. If you like shooting shit and hearing him talk now and then, it's nice. I liked it, but most people don't.

claps

@0celo7 Zion national park.

I resign @Balarka. But I'm glad you're thinking about those questions.

@MikeMiller I know, I talked to the guy in the cave about it.

Howdy @dsillman ... out of school already?

Oh. I mean, I've been there with my own skin and bones.

Oh, you mean IRL.

Good night all and thanks @TedShifrin

yup a short day

Bonne nuit, @JeSuis :)

need a 6th question @Ted :P? Good evning by the way

LOL, hi, @Alessandro

@user379685: It works out rather nicely. :)

So did I misunderstand your question, @Ted? Btw, feel free to tell me to come back later

(or not at all)

Your dimension isn't right, @Danu. Remember what dimensional real torus we have.

Could you elaborate ted?

@TedShifrin Herpaderp, again. So $g$ it is.

@user379685: Did you see my line saying you should put u=arccos(x) and rewrite everything?

OK, @Danu. Now finish answering your own question :P

U squared divided by 1 - cos u

Right, @user379685. Now find the limit as u->0.

@Danu: I wonder if I can get an MSE prize for this :P

@TedShifrin Sorry, missed this message. Well, if $S_{\alpha(t)} \alpha'(t) = k_1 \alpha'(t)$ then $S_{\alpha(t)} \alpha'(t) \cdot \alpha'(t) = k_1$ (modulo arclength parametrization of $\alpha$) so $\Bbb{II}(\alpha', \alpha') = k_1$. And that's precisely normal curvature.

I told Lozansky I should start charging you and him :P

Its still 0 over 0

@TedShifrin You really should! If you cared about reputation points I'd give you a bounty!

Right, @user379685, but you should know how to do this one. Standard manipulation.

Hmm

@Balarka: Sure. But why does the curve have constant (regular) curvature?

So I have equality of dimensions. Great. But how do I get inclusion in one direction? The torus here is not directly a quotient of $X$ or anything like that, so I don't see how.

Oh, actually...

It's the fundamental theorem of calculus, @Danu.

?!

What's the derivative of $\int_{x_0}^x \omega$?

Wow ... it's boring in here all of a sudden. Takes a nap

@MikeMiller I started Sierra Madre

but one of those things destroyed me after I got God

Oh god its so clever

@TedShifrin: Meusnier's, I think. The tangent to the curve is the principal direction. The principal normal and the curve normal are both perpendicular to that, so the angle between them is $0$.

Huh? @Balarka

@TedShifrin $\omega$? Man I'm feeling real weird about this---I have no idea how this even connects to anything

@user379685: You should have had that problem (maybe upside-down) before.

Any $1$-form (but in our case it'll be holomorphic), @Danu.

I didnt

My prof has all sorts of hard problems

Weird, @user379685. (1-cos x)/x^2 is a totally standard problem, maybe even in the book.

@0celo7 I think that gets a lot more fun once you get inside.

@user379685: I think it's a cool problem. I just didn't think about how you should be doing it.

But I like that one a lot, especially if you explore to find all the secret stashes.

I love exploring in NV.

Thank you ted bye guys

Bubye :)

@TedShifrin If I just write stuff in coordinates then I wanna say e.g. $\omega=fd x$ and the derivative of $\int_{x_0}^x\omega$ is $f$

@TedShifrin: Did I say something horrible again?

But we're doing a path integral in $n$-space, @Danu.

@Balarka: Yes. Why do $\mathbf n$ and $\mathbf N$ agree along this curve?

@TedShifrin Okay, so then I'll just sum over coordinates and have $\sum f_i dx_i$ and end up with $\sum f_i$, no?

@Danu: smack

We're talking about the derivative of a function of $n$ variables. This will be again a $1$-form.

You mean exterior derivative?

Sure ... Or gradient or whatever.

@TedShifrin Oh duh, I did say something horrible.

Hmm.

@Balarka: Interestingly, some (presumably) undergraduate answered a diff geo question and completely misunderstood how $\mathbf n$ and $\mathbf N$ can be different on curves. I had to politely point out he was misinformed :P

heh

Ah, my paycheck just arrived. :)

of course, curves inside planes can have totally messed up geometry, although the torsion is zero.

Hence you need to know that actual curvature agrees with normal curvature.

Right. Hrm.

So I agree that you can apply Meusnier once you know something about the two normals :)

BTW, you can hate this question, but I still love it :P

Does someone know a book or another reference in which the equivalence between $n$-dimensional Hausdorff measure and the Lebesgue measure in $\mathbb{R}^n$ is shown?

Hmm, for sure it's in Federer's million-page tome on Geometric Measure Theory. :P

lol

Maybe in Frank Morgan's little introductory book on GMT.

I no longer have these books, so I can't look.

Aw. I was going to plan a trip from Dec 1 - 6 but I have to give the last topology talk of the quarter.

Put the talk on video, @MikeM.

And answer all possible questions while you're at it.

I guess I could try.

It's not the 4-color talk. It's a talk about uncountable R^4s.

Ohhh ... Freedman stuff.

Mostly Taubes, actually. Doesn't need much input from Freedman.

So @TedShifrin Just working in local coordinates I don't get anything that I think is enlightening... Denoting the antiderivative (w.r.t $x_i$) of $f_i$ by $F_i$, I have $d\int_{x_0}^x \sum f_i dx_i=\sum_i d F_i= \sum_{i,j} \frac{\partial F_i}{x_j}dx_j$ where the diagonal part is the original form, but also off-diagonal stuff might appear I guess.

I'm thoroughly confused.

no, no, @Danu. You're confuzled.

The answer is that you get $\omega$ back again.

Think about computing the partial with respect to $x_i$.

OK, @Ted, curve normal is normal to both the principal directions, isn't it?

You take a path that moves just in the $x_i$ direction from $x$.

thanks @Ted, there are both in my uni's library so I'll check them out, in any case it seems that I should look in the "geometric measure theory" section, I wasn't sure where to look

When I originally said "$\omega$?", that's actually what I was saying. But I figured it was wrong by your respondse :P

@Balarka: Look at a general surface of revolution.

Oh, @Danu, I'm sorry. I thought you were saying you didn't understand.

Oh right.

@Alessandro: I don't know that many books that seriously discuss Hausdorff measure.

It's just a curiosity, the professor stated this equivalence without proof so I'd like to see one

@TedShifrin Hey Ted

If you know the definition of Hausdorff k-dimensional measure, @Alessandro, I think for the case of n, you should be able to see it pretty directly from inner measure/outer measure. You're just using balls specifically instead of arbitrary open sets.

Lozansky, you came in after the craziness. Six different people were asking me questions simultaneously. My brain is dead now.

I am confused about something I shouldn't be. Guess I'll sleep today and start fighting again tomorrow.

@Balarka: I believe you need to use more stuff in this problem. You have more info from Lemma 3.3, etc.

But go sleep!

I thought I didn't need 3.3 for the curvature part.

If I do, oh well!

I was trying to not use it :P

That's where you're going to use it!

There's actually differential geometry going on here.

@TedShifrin That's alright, I'll just try and wrestle with this Möbius transformation myself :P

I see. Interesting.

@Lozansky: I'm still waiting for payment on my previous bill :P

Not a bad problem after all.

thinks about smacking Balarka

Hm, it makes sense intuitively, maybe I'll try writing a proper proof by myself before looking it up @Ted

lol

Cool @Alessandro