So the derivative of the unit tangent vector of a curve is normal to the curve. If that curve is contained in a surface, and if the derivative of the unit tangent vector is normal to the surface as well, then the curve is a geodesic.

Given any two points on a surface, the shortest path between them that lies completely on the surface is a geodesic.

The converse is not true, however

Yeah 'cause you've got great circles that go the wrong way 'round

Not so great now, are we

Arcs of the great circle, yes

You can fix this by saying geodesics are critical points of the arclength functional

So I haven't read chapter 3, which is presumably where he proves that the geodesics are the shortest paths.

I think so. I think he does a hands-on argument

You could read my answer, alternatively :)

Here

I'm curious to see how the chapter goes, because while I intuitively believe that it's plausible that something whose normal vector is normal to the surface is a geodesic, I'm not sure I would know how to prove it.

Or how to convince myself of it, even.

Thanks, I'mma read that

I wonder if there is a neat way to plot the surface of the value of a functional. One issue is the axes have to be "path valued" and it is not clear how to do that

There is none, it's a function on an infinite dimensional space

(He consistently writes "geodetic"?)

Yeah must be a typo

(Oh, apparently that's a word)

(but unrelated)

@AkivaWeinberger I think the key idea is to prove it locally

If you have a small ball $B_\epsilon(0) \subset T_p M$, say $B = \exp B_{\epsilon}(0)$ be the "geodesic $\epsilon$-ball" around $p$

Guess so, cause it is simply too large. But suppose we only want to numerically approximate it by putting a mesh in the domain, then I think one of 3B1B's video provide some insights on how to construct such a GUI:

Take a point $q \in \partial B$. You want to prove that the unique geodesic $\gamma$ joining $p$ and $q$ has length $d(p, q)$.

To be clear, what you're writing as $g(v,v)$ is what do Carmo writes as $\langle v,v\rangle$, right? Specifically, it's the inner product on the tangent space that $v$ lives in.

@BalarkaSen Arright

Yep

> Theorem: A path $\gamma$ in $\Omega_{p, q}M$ is a critical point of the energy functional (i.e., $d\mathcal{E}(\gamma)(X) = 0$ for all $X$ in $T_\gamma \Omega_{p, q}M$) if and only if $\gamma$ is a geodesic of $M$.

So this means that, if you've got a geodesic, wiggling it a little won't change its energy much? (And if it's a local maximum, wiggling it a little will only decrease its energy.)

Wait, why do we have $[\partial_t, \partial_s] = 0$? @BalarkaSen

Well, it means, if you wiggle the geodesic $\gamma$ to get a 1-parameter family $\gamma_s$ of geodesics (fixing the endpoints) then the infinitisimal rate of change of energy $\mathcal{E}(\gamma_s)$ is zero at $s = 0$

@AkivaWeinberger Coordinate vector fields commute

Ah, I see

$\alpha$ (the variation) is a chart of $M$ around $\gamma$

With $\alpha_*(\partial/\partial t) = \partial_t$ and $\alpha_*(\partial/\partial s) = \partial_s$

(cont.) The GUI will probably look something like this (finger can swipe points around)

As $[\partial/\partial s, \partial/\partial t] = 0$ in $(-\epsilon, \epsilon) \times [0, 1]$, $[\partial_s, \partial_t] = 0$ in $M$

where the numerical version, the points on the mesh are joined by splines

Integration by parts?

Wait hold on don't tell me

:) OK

So $g(\nabla_{\gamma'}X,\gamma')+g(X,\nabla_{\gamma'}\gamma')$ is the derivative of something

Yes :)

Remember that $\nabla$ was our metric connection

So I guess it's the derivative of $g(X,\gamma')$ in the direction of $\gamma'$

or, rather,

the derivative of $g(X,\gamma')$ with respect to $t$

Correct.

So if you integrate that over $\gamma$, you get $g(X(\gamma(1)), \gamma'(1)) - g(X(\gamma(0)), \gamma'(0))$

However, $g(X,\gamma')|_{t=0}^{t=1}$ is zero since $X(\gamma(0))=X(\gamma(1))=0$

Yup

so the integrals of $g(\nabla_{\gamma'}X,\gamma')$ and $g(X,\nabla_{\gamma'}\gamma')$ are negatives of each other

Arright, got it

So the theorem is clear, right?

Yeah

Now what's the relation between energy and arclength?

Wait, hold on

Say we have any curve in the plane between $(0,0)$ and $(1,1)$ that only ever goes to the right and up. Won't its energy be constant?

Because of Pythagoras

That's not a smooth path though

But yes

I don't mean staircasing, I mean like, say, the parabola $y=x^2$ only moves up and to the right between those two points

Or wait

Oh.

Does the arclength parameter mess this up

...Hm, it shouldn't

I don't see what you mean by energy is constant

...or, wait

@BalarkaSen Like, I was thinking that should have the same energy as the straight-line path

Oh no absolutely not

Right, I think I see it now actually

If we were integrating over $dx+dy$ then yes, but we're integrating over $dx^2+dy^2$

if you get what I mean

Not super coherently, but yes, the arclength parameter butts in for sure.

There's a relation between the energy and the arclength

Which brings back to my question ;)

Right, if it were $\gamma(t)=(t,t^2)$ then the energy would be the same as a straight line, but that's not unit speed

Yep

Good catch. Holy shit you can read all of my answer now jesus christ

I talk about this issue somewhere below

Why is the $H_1$ norm defined like that

It looks weird

Well it comes from an inner product is the point

You shouldn't read that specific paragraph because it's not really relevant for you

Let me tell you the relation, btw. If $\gamma$ is arclength parametrized, $\mathcal{L}(\gamma)^2 = \mathcal{E}(\gamma)$. (In general, $\mathcal{L}(\gamma)^2 \leq \mathcal{E}(\gamma)$ by Cauchy-Schwarz)

Where $\mathcal{L}$ is the arclength functional

I guess the point is that you're changing the metric to $\tilde{g}(X, Y) = \int_\gamma g(X, Y) + \int_\gamma g(\nabla_{\gamma'} X, \nabla_{\gamma'} Y)$?

Yep

@BalarkaSen ...Why?

If $\gamma$ is arclength parametrized $g(\gamma', \gamma') = 1$ isn't it.

Ohh

Wait but then we'd have $\mathcal L(\gamma)=\mathcal E(\gamma)$, no?

Ah, er, you're right.

Mental typo on my part.

Wait no

No I think I was right.

> Passing to the energy functional $\mathcal{E}$ implicitly mods out by the gauge action as $\Omega_{p, q} M /\text{Diff}_+(I)$ can be identified with the subspace $\mathcal{A} \subset \Omega_{p, q} M $ of arclength parametrized curves between $p$ and $q$ in $M$, where the two functionals $\mathcal{E}$ and $\mathcal{L}$ agree.

I see

If I arclength parameterize $\gamma$, the domain of $\gamma$ isn't $[0, 1]$ anymore

(I mean, I don't quite know what a gauge action is, but clearly in this context it's the reparametrization)

It's $[0, \mathcal{L}(\gamma)]$

It's $\int_\gamma$ anyway though

Oh wait

I pullback to get an integral over an interval with the $t$ variable

$\mathcal E(\gamma)$ becomes $\int_0^1L^2~{\rm d}t$, yeah

@AkivaWeinberger Right, gauge action in physics is when you have a group $\mathcal{G}$ action on your coordinate space $\mathcal{M}$ (in general an infinite dimensional space) under which a certain set of differential equations, say, are invariant

It's pretty much a group action over a huge ass space

Mike does gauge theory over a space of connections, for example

(Don't ask me about that though)

I have seen a paper describing general relativity as a gauge theory over the configuration space

So the above quote isn't strictly speaking true, then, since $\mathcal E$ and $\mathcal L$ don't agree over the arclength parametrized curves

It was a 56-dimensional fiber bundle

@AkivaWeinberger Yeah, I meant $\mathcal{E}$ and $\mathcal{L}^2$. I should edit that in

But then again, minimizing $\mathcal L$ is the same as minimizing $\mathcal L^2$, which agrees with $\mathcal E$ over constant-speed parametrized curves

Yup

(or whatever you'd call them)

So you have a proof that arclength-minimizing things are geodesics anyway

Using this indirect way with the energy

u can use the ~jet bundle~ to show such a thing

Right so I think I get everything except for why you changed the metric to $\tilde g$

The new metric makes $\widetilde{\Omega_{p,q} M}$ into a Hilbert manifold

What's a Hilbert manifold

Manifold locally like a Hilbert space

Infinite dimensional manifold modeled on Hilbert spaces

I.e., charts are maps to open subsets of Hilbert spaces

Why isn't $\Omega_{p,q}M$ a Hilbert manifold

Well, smooth paths can converge to bad objects

Ah, so it's not complete?

Yup

And so the Solobev norm measures how close a smooth path is to a "bad object"?

What's an example of a "bad object"?

A non-smooth curve

I see

Like the limit of the sigmoid curve to a step function

@AkivaWeinberger Well, there are multiple uses of having the Sobolev norm. Firstly, you can enlarge your space to a larger space complete under the Sobolev norm. Secondly, you retain the notion of differentiability in some way, because you can have "weak derivatives"

Oh, so a Vee shape (like the graph of the absolute value function) is allowed?

Worse things than that are allowed

$H^1$ paths are absolutely continuous with tangent field in $L^2$

Let's just work with Euclidean space. Elements of $H^1([0, 1] ; \Bbb R^n)$ are paths $f : [0, 1] \to \Bbb R^n$ such that $f$ is absolutely continuous and $f' \in L^2([0, 1]; \Bbb R^n)$.

Where $f'$ is the weak derivative

That is, $\int_0^1 f g' = \int_0^1 f' g$ for all smooth functions $g$ on $[0, 1]$

What's an example of something that's not in $H^1$?

A discontinuous path.

Maybe you want a better example lol

What's an example of something continuous that's not in $H^1$?

The Cantor function I think

sniped

Mhm

It doesn't even have a weak derivative

Would something like $(x,x\sin(1/x))$ not be in it as well?

> Unconfirmed proposition: If anything that serves as a counterexample, it is often begin with the word "cantor"

@AkivaWeinberger I'm doubting that.

Oh, that has infinite arclength I think

(Not sure)

You should ask about the analytical meaning of Sobolev spaces to 0celo7

I know infinitisimally little about it

Only that you can have elliptic regularity with it

That's why I used it in the answer, really. If $\gamma$ is a geodesic in this new $\widetilde{\Omega_{p, q}}$, it's actually smooth

Elliptic regularity meaning?

It's the principle that if a function in an appropriate Sobolev spaces satisfies an elliptic PDE, then that function is smooth. For example if $f \in H^1([0, 1]; \Bbb R^n)$ is solution to an ODE, then $f$ is smooth.

That's why H^1-geodesics are smooth. They satisfy the ODE $\nabla_{\gamma'} \gamma' = 0$

Arright

(Ellipticity is a condition on the coefficients of the PDE that 0celo7 told me but I forgot)

In any case, the only relevant bit to me is the part saying how, when you minimize $\mathcal E$, you end up with geodes

ics

You end up with cool rocks, yes

For example the Cauchy Riemann equation is an elliptic PDE I think

That's why solutions to it are automatically analytic

Hm

@AkivaWeinberger yup

I should say that this is not the standard proof of arclength minimzer $\implies$ geodesic

I can guess

lol

So I am trying to remember the standard pf

So take the geodesic ball $B = \exp B_\epsilon(0)$ in $M$ around $p$ like I said before

There's a theorem in doCarmo I believe which says the metric $g$ on $M$, restricted to $B$, can be decomposed as $g_{(r, \theta)} = g_r + g_\theta$ where $(r, \theta)$ is the "polar coordinate" on $B$ indicating the point of distance $r$ from $p$ and lying on the geodesic emanating from $p$ with tangent vector $\theta$ (abuse of notation)

Where $g_r$ is the radial part of the metric

And $g_\theta$ the metric on the sphere at radius $r$ in $B$ (need not be standard metric on $S^{n-1}$ in any sense)

My notation is all fucked up

Er, the metric is just defined on the tangent space, though, isn't it? Or do you mean like a legit metric in the sense of metric spaces where people use actual fucking metrics because words have meanings, dammit

lmao but no I mean the Riemannian metric

For every point $x$ on $B$, the tangent space $T_x M$ decomposes as $T_x M = T^1_x M \oplus T^2_x M$

Where $T^1$ is the 1-dimensional subspace of $T_x M$ tangent to the geodesic joining $p$ and $x$, at $x$

Oh OK I guess

Well except for $T_pM$ itself

And $T^2$ is the orthogonal subspace tangent to $\exp(\partial B_\epsilon(0))$.

Rightrightrigh, I meant $B - p$

This is literally a geodesic polar coordinate near $p$

In particular this means geodesics emanating from $p$ in $M$ are all orthogonal to the geodesic sphere $\exp(\partial B_\epsilon(0))$.

And then since an inner product on $V\oplus W$ is the sum of an inner product on $V$ and an inner product on $W$...

Yup

Right so now what

I feel like this should prove the geodesic joining $p$ and and a point $x$ on the geodesic sphere $\partial B$ is the path of least length joining $p$ and $x$.

Because it's orthogonal to $\partial B$

Oh I think I see then

Any path can be made shorter by projecting it onto the geodesic

'cause, uh, theorems

Err

It depends on how exactly the $g_r$ and $g_\theta$ change as $x$ moves I guess

Actually $g_r$ is just $dr^2$

Hi,I have a problem with a car moving in a circle and I have the wheelbase and SteeringAngle and I have to find the radius of the circle and that would be WheelBase/sin(Math.toRadians(SteeringAngle) but what should I do if the steeringAngle is negative ?

The usual metric on $\Bbb R$.

Oh yeah

It's $g_\theta$, the spherical part, that's bad

Oh so I see

$g = g_r + g_\theta \geq g_r$

@O.Rares What's a wheelbase?

distantace between tires of the car that moves in a circle

Oh

So $g(\gamma'(t), \gamma'(t)) \geq g_r(\gamma'(t), \gamma'(t)) = |\gamma'(t)|^2$

And a steering angle of 0 corresponds to going in a straight line?

(Aka a circle of infinite radius)

I guess if it's negative, it's the same radius as if it were positive but it'd be turning in the opposite direction

no if it's negative the circle is smaller

hello

someone know the "deformation lemma" ?

@O.Rares Does a steering angle of 0 correspond to going in a straight line?

@BalarkaSen Ya

yes

Then it being at, say, -30 degrees should be the same as it being at 30 degrees, except going in the opposite direction

in the image the angle is 30 degrees

If 30 degrees is turning clockwise then -30 degrees should be turning counterclockwise

but at the same radius

hm let me try

So just take the absolute value of your expression is what I'm saying basically

So like, for any path $\gamma$ between $p$ and $x$, $\int_0^s \sqrt{g(\gamma'(t), \gamma'(t)} dt \geq \int_0^s |\gamma'(t)|dt$ which is greater than the Riemannian distance between $\gamma(s)$ and $\gamma(0)$ because that's $\text{inf}_\gamma \int_0^s |\gamma'(t)|$

Where $\gamma(0) = p$ and $\gamma(s) = x$

And $\gamma$ stays in $B$

for some reason it may be correct but I didn't got the correct answer, I need to get x and y of the new position if I have the wheelbase the angle and the distance that that car made and for 1.75 3.14 -23.00 I got 1.06 2.89

There's a less confusing way to parse this. $B - p$ is diffeomorphic to $(0, \epsilon] \times S^{n-1}$ and the metric on $g$ under this diffeomorphism becomes $(Euc_{\Bbb R}, g_{S^{n-1}})$ where $Euc_{\Bbb R}$ is the Euclidean metric on $\Bbb R$

Take the negative of the x coordinate @O.Rares

still wrong

Ugh my notation is becoming too badass for me to write coherently

@AkivaW You were right previously, we're projecting

To the first coordinate $(0, \epsilon)$

And under this projection the metric only decreases

So minimize on the standard metric on $(0, \epsilon)$, which is just the straightline path from $0$ to $\epsilon$, and then pulling that back by the diffeomorphism we get the unique geodesic from $p$ to $x$ in $B$

Hello, guys. Can every riemannian manifold $M$ of dimension 2 (a "surface") be isometricaly immersed into $\Bbb R^3$? If not, what are the conditions on $M$ which make this possible?

No

Take the projective plane as a counterexample

I think that might be true if it's orientable?

Wait, isometrically embedded

Not even true

Even asking immersion (not embedding) it is not possible...?

Hmm

The flat torus isn't isometric in R3

there's no flat torus in $\mathbb R^3$

Oh, indeed

This is true if you have low regularity

The flat torus can be $C^1$-isometrically embedded in R^3

what does it look like

Very corrugated

disgusting

that one?

Yes

@BalarkaSen Has this result a name?

it's part of Nash embedding theorem

It's apparently called the Hévéa Torus

@BalarkaSen Thanks! (nice pics haha)

it's very wiggly

Wiggling is a crucial part of h-principles

important image

@AndersonFelipeViveiros The approximate intuition is that when you have a $C^1$ isometric embedding, curvature stops making sense

Because curvature requires at least two derivatives

@BalarkaSen wouldn't it have a distributional curvature

Even if it does it's probably not useful enough to have a restriction :P

By the way is there any website that lets us graph multivariable functions?

wolfram alpha?

@AndersonFelipeViveiros In any case, more concretely, by Hilbert's theorem(?) a compact, smooth surface in $\Bbb R^3$ always has a point of positive curvature. So, for example, there are metrics of constant curvature -1 on surface of genus $g$ for $g \geq 2$, which cannot thus be embedded in $\Bbb R^3$. (Similar for the flat torus)

@BalarkaSen are you familiar with filters?

@BalarkaSen What about the hyperbolic plane?

I don't understand how corrugation can pack the 4th dimension into 3 dimensions

is it not embeddable

in particular, I'm thinking about why it is true that a set S is compact iff every filter containing S meets every neighbourhood of some a in S

oh you said compact

nvm

@BalarkaSen But can it be immersed?

because the flat torus, despite being a 3-manifold, has a 4 dimensional extent

@Slereah Nope. Hilbert's theorem also says there are no embedded complete smooth surfaces in R^3 of constant negative Gaussian curvature.

so how does corrugations get around that and pack away that extra degrees of freedom

@BalarkaSen whaaat

what about the revolution of the hyperbole, though

Pseudosphere? Not complete!

Where's the singularity?

I think the fact I stated before is not called Hilbert's theorem. I forget the names. This is the real Hilbert's theorem

is it at infinity on the hyperbole branch?

The pseudosphere isn't the surface of revolution of the hyperbola

It's the surface of revolution of the tractrix

It's surface of revolution of the tractrix

Sniped

Oh did you actually mean hyperbola

The surface of revolution of that doesn't have a constant negative Gaussian curvature

hm

Right but you said it has positive curvature somewhere

That's for compact surfaces

I said two things

Ah OK.

For the pseudosphere, it goes off asymptotically to the z = 0 plane

That's why it's not complete

what a ridiculous shape

Well, that's the double

Which is not a smooth surface

Yeah if you want it to be smooth you have to cut it in half and then you lose completeness

What about the elliptic hyperboloid

@LeakyNun Nope, I don't know about filters. I think Akiva does

If I take the octagon from here and draw it on that shape, I'll end up with a partially constructed double torus of constant curvature.

From there I just need to glue the appropriate edges.

@Slereah Not constant curvature

Somehow.

@BalarkaSen oh it has to be constant

For Hilbert's theorem, yeah

Well, I know about ultrafilters, which are a special case

The flat torus, without reference to any embedding space, is inherently a 3 dimensional object?

@AndersonFelipeViveiros Hm, I think it's still true for immersed surfaces (in the compact version I said I mean)

It's a two dimensional object

It's $\Bbb R^2/\Bbb Z^2$ (the latter being the group)

ah right, I had it solid

(i.e. thinking of D x D and then take one of its two congruent cells (A duocylinder is consists of two flat tori bonded together if I recall) instead of thinking of S x S)

Solid torus is $D \times S$

right so flat torus is $S \times S$

well all toruses

tori

the interpersonal.SE questions are blowing up in my HNQ feeds lately

maybe I should ask a question there about how to strike a conversation with a girl about blowing up seven points on the plane or something of the sort

if only we still had hats

Hi,

The forum art of problem solving is not open, for all (for reader) ?

Any one have a coumpt here,you can open this link : artofproblemsolving.com/community/c506306h1568869_fibo222018

Huh

excuse my english, @Secret do you understand ?

It seems the Casimir effect is too faint to violate the ANEC in any realistic setting

sad

not open for me either

Oops

Wrong chat

nvm

> You have requested a top secret page for which you have not received security clearance. Or maybe you have clearance, but you are not logged in. Click here to log in.

@Secret is it not a joke ?

I don't think I will joke on something serious

@Secret : what's that mean ?

I work for a secret project, and I don't know that ?

Do you usually logged in when you use that forum?

A some one of this site have this message :

You have requested a top secret page for which you have not received security clearance. Sorry about that. Perhaps if you do 10 push-ups, I will let you in.

the question is not secret the answer yes

Calculate : $F(2^{2^{2018}}) \mod (2^{89}-1)$, $F$ the sequence of Fibonacie

bye

Sorry, the last question

what about this link : forums.futura-sciences.com/science-ludique-science-samusant/…

can you open that

and find this question in french

thanks

yeah, opens fine

thanks

Hi guys, very silly question. The following function is trivially integrable in $(-\infty,x]$ right?

$$u(t) = \begin{cases} 1 & t \geq 0 \\ 0 & \text{otherwise} \end{cases}$$

If this is true the sequence $$g_n(t) = \frac{1}{1+e^{-nt}} \leq u(t)$$

and therefore by the dominated convergence theorem we have

$$\lim_{n\to+\infty}\int_{-\infty}^x g_n(t) dt = \int_{-\infty}^{x} \left(\lim_{n\to+\infty} g_n(t)\right) dt = \int_{-\infty}^{x} u(t) dt = x$$

sorry I meant $$\int_{-\infty}^{x} u(t) dt = \max(0,x)$$

geometrically, the heviside step function integral is the area under the graph, which is indeed $\max (0,x)$

algebraically, I don't know how to prove that.

@AkivaWeinberger But is it turning 30 degrees F or 30 degrees C?

btw, someone must have banned me from interacting with pseudohuman, cause all commands did nothing in sandbox

oops lol

and btw, she is no longer in sandbox, someone must have decomissioned her

lmao

tbf we did fuck around too much with it

ALL FOR SCIENCE THO YOLO

(That's, ahem, a highly star-worthy message, as a Rick and Morty enthusiast can immediately tell)