Looked it up - "conjugate hyperbola"

So I was thinking about Tissot ellipses again

and pseudo-Riemannian manifolds (where the length of a vector can be zero or negative)

and what the Tissot ellipses would look like

And I think the answer is, it's a hyperbola… plus its conjugate

I've forgotten the precise definition. This sounds a lot like Dupin indicatrices to me.

Your ellipse is only an ellipse at a point of positive Gaussian curvature, right?

It's an ellipse everywhere, if it's a Riemannian manifold.

So it's the metric, not the second fundamental form. OK.

It's a way of visualizing how the chart distorts it.

So it's not just $\|v\|^2 = 1$ (independent of chart)?

Basically, you draw a circle of radius $\epsilon$ around a point in the manifold, and look at the inverse image in the chart, and that's your Tissot ellipse

by circle you mean geodesic circle?

Yeah. Set of points a distance $\epsilon$ away.

So in a normal coordinates chart, it's an actual circle.

As $\epsilon$ goes to zero, the inverse image becomes closer and closer to a true ellipse

@TedShifrin Normal meaning angles are preserved?

no, these are called (geodesic) normal coordinates ...

in the chart, the metric is the identity matrix up to order $2$ at the central point.

I'm not talking about conformality or anything.

Here's what it looks like for the map projection Nat Geo uses

No, nothing to do with Mercator projections.

This is a standard tool in Riemannian geometry, perhaps first envisioned by Gauss.

What is, Dupin?

No, normal coordinates.

Ah, so this?

The exponential map?

Right.

Yeah, the Tissot thingies won't be circles in general for that. (If it's Hyperbolic space, by the way, the ellipses get squashed laterally as you move away from the origin)

(rather than radially, as in the image)

(For the exponential map, anyway. Poincaré's map will have circles.)

At the point that's the center of the coordinate system, the Tissot thingies are indeed circles.

At other points, no.

Ah

Yeah

In the hyperbolic plane, geodesic circles are in fact Euclidean circles (in the usual models), but the center and radius change.

"The usual model" being Poincaré?

upper half-plane or disk, yeah

Yeah

For the Klein-Beltrami model (where geodesics become straight line segments), I haven't worked it out.

So with a pseudo-Riemannian metric, you no longer require $\langle v,v\rangle$ to be positive for $v\ne0$

@TedShifrin That's a nice one

Right.

(What's the word, positive-definite?)

The "inner product" becomes indefinite.

In any case, with pseudo-Riemannian metrics, you can draw the points a distance $\epsilon$ away, but you can also draw the points a distance $-\epsilon$ away

And in some directions you're moving $0$ distance

Sure ... this is where you get time-like and space-like in the Lorentz setting

So if we do the Tissot thing, we'll want to take the inverse image of both the $\epsilon$-circle and the $-\epsilon$-circle. And I think they'll end up being conjugate hyperbolas

Well, sure.

and the assymptotes represent directions in which you move $0$ distance, aka vectors of length $0$

(I dunno if I'm allowed to use the word "length" for pseudo-Riemannian metrics)

Yeah, that's the light-cone in the general case.

Relativity is weird

they called that distance like in the metric case

@AkivaWeinberger and weird is relative :D

Budumtish

In SR and GR, we called that the spacetime interval

Hey guys

Oh, wait, I forgot to take the square-root

By negative distance I really meant negative, uh, $\langle v,v\rangle$

which I guess is imaginary distance?

And then the circles are radius $\epsilon$ and $i\epsilon$? I dunno

If it is < 0 we call that the proper time and the points are timelike separated, if it is > 0 we call that proper distance and the points are spacelike separated, and if it is = 0, well, it is null

Right, yeah

And $c=1$

I was just mentioning imaginary circles (circles of imaginary radius) in here yesterday, DogAteMy.

Ah, were you.

See Pedoe's beautiful book on Geometry.

@TedShifrin To imaginary listeners, maybe?

They weren't very complex, that's for sure.

Ha

circles of imaginary distance is probably very hard to visualise without going 4D

Bit of an unfortunate name, Pedoe

It's just a matter of complexifying.

You know you are doing nothing when you read "Hogwarts: So why aren't the kids “doing it”?" :D

@AkivaWeinberger I've never in my life considered such a thing.

whats the intuition behind defining $\partial_{\overline{z}} = \frac{1}{2}(\partial_{x} + i \partial_{y})$

Dover Publications has very consistent cover art, doesn't it

@JoeShmo: $\partial/\partial z$ gives the holomorphic derivative and $\partial/\partial\bar z$ gives the Cauchy-Riemann operator (i.e., holo iff vanishes).

They're dual to $dz$ and $d\bar z$, so it's all totally natural.

DogAteMy: pretty rudimentary, yeah.

@JoeShmo I would imagine you want to define stuff such that $\partial_z(z)=\partial_{\bar z}(\bar z)=1$ and $\partial_z(\bar z)=\partial_{\bar z}(z)=0$

That's not the right viewpoint, really, DogAteMy. See my comments.

Formally, perhaps you can deduce mine from yours, but meh.

What are $dz$ and $d\bar z$?

1-forms, of course.

Defined how

$dz=d(x+iy)=dx+i\,dy$, etc.

Officially, we're working with complex-valued $1$-forms, of course, not real-valued.

And $d\bar z=dx-idy$?

Yup.

How do you find the "duals"?

so how are said objects dual?

Usual duality of vector fields and differential forms. Dual bases.

ok. im missing something. so $\partial_{z}$ is the total derivative?

If you have a basis for $V$ (resp. $V^*$), you get the dual basis for $V^*$ (resp. $V$) how?

sure

I don't know what total derivative means.

me neither :P

You said it, man.

So $\partial/\partial z,\partial/\partial\bar z$ will be the dual basis to $dz,d\bar z$. Work it out.

ok. so what does $\partial / \partial z$ mean?

ok

Do the little bit of algebra.

In terms of knowing that $\partial/\partial x,\partial/\partial y$ is dual to $dx,dy$, of course.

sure

I'm telling you, in the end, that if you have a holomorphic function $f$, it will satisfy $\partial f/\partial\bar z = 0$ and $\partial f/\partial z = f'(z)$.

alright im working it out

Good :)

I'm trying to remember how this works, and I'm not seeing where the $\frac12$ comes in

I need to review differential forms

OK.

yeah the 1/2 term is mostly what threw me off

Work it out, dammit.

also the fact that there's no negative term anywhere in sight

im doing it!

@AlexFrancisco, Hi Alex. This week I finally had some time to think about your answer posted here: math.stackexchange.com/questions/2931595/… And I don't think I totally get your idea. Of course, those transformations seems to work good, and it's clear that density can be written as some function of distance. But how exactly determining distance helps to pull distance effect from logistic regression output (probability)?

@jakes: I'm not sure I recognize that name from chat (ever).

Found this online:

That's a pity. Tried to move discussion from comments here.

hahaha, DogAteMy

@jakes: I think those are ordinarily private chats.

But wait, there's more

@TedShifrin How can I set up one?

Hey @TedShifrin

hi @CaptainAmerica

I drew fractals today.

I've never set up a room. I think it's easy. DogAteMy, can you advise?

@AkivaWeinberger I$!$ like combinatorics$!$

There's a button here that says "create room" (on mobile you have to click the button on the top-left to get to it, don't know about desktop)

Or actually hold on

https://chat.stackexchange.com/rooms/new

Direct link

thank you

Never knew there was a Stack Exchange site on cooking

"Seasoned Advice"

DogAteMy: I still haven't found the button :(

Oh, if I click on a person's name, I get the option to open a chatroom with said person. That's how it works.

@CaptainAmerica16 If that were so, you'd be busy for a long time!

Fun fact, Hawaii suffers from "vog", or volcanic fog

Like smog but with firemountains

Most of the world is gonna start suffering from fire smog.

Good thing climate change is a lie/hoax.

@CaptainAmerica16 Rip a paper in half, look at the shape of the tear

How come so many people believe in the silly claims that as a mathematician, only 5 people will read the paper you wrote over the next 20 years. I know the number of readers will not be in the millions, but surely people should be able to recognize how absurd that low number would be.

Voila, instant fractal

I've never heard that @TobiasKildetoft

@TedShifrin Hehe funny.

Now I wonder if there is any way to even get a remotely accurate number for the actual amount of readers of a given math paper

@Tobias: I can't quite figure out your comment. Are you suggesting that the average paper has thousands of readers?

@AkivaWeinberger The numbers change most of the time, but usually they are about as low as that (this example was from a terrible answer on Quora)

@TobiasKildetoft Leave an obvious typo, and a money prize for spotting typos

The readership is proportional to your money loss

@AkivaWeinberger I did it :D I can't really see it though.

@TedShifrin No, but probably at least in the decently high double digits (especially within 20 years)

I guess I have no way of knowing, but I wonder if some of my papers have had more than a handful of readers.

@TedShifrin do you have this option only here or on the site as well?

Also depends on what counts as a "reader" of course

Um, let me check, @Jakes. It should be there too.

Hmm, some journals have either a "views" or a "downloads" count, but probably those will only be loosely connected to actual number of readers.

@CaptainAmerica16 I tried ripping a piece of toilet paper the "wrong" way (parallel to the perforation but against the grain)

Pretty sure the irregular shape of the tear is fractal in nature

Not necessarily self-similar, but still infinite detail as you zoom in

(Fractals don't need to be self-similar)

@AkivaWeinberger Yours is a lot more defined. Mine was almost straight down so I couldn't see the pattern. I'm going to try again.

Dunno if toilet paper has more fractal tears than regular paper

Also, maybe rotate 90 degrees and try again? Dunno if paper is

agh, what's the word

"anisotropic"

@TedShifrin yo

heya @Stan

Toilet paper is anisotropic (not isotopic, that is, not the same in all directions). Dunno about regular paper, I meant