Hi, I am here. Will be able to react from time to time for a few hours today %)

thanks a lot! In addition to my questions (I think I'm progressively figuring out how to smoothly extend the vector field), I'm also wondering : if a curve is a constant speed geodesic using one particular frame, does it remain a constant speed geodesic in another frame ? does it remain a geodesic at all ?

As far as I know, there are geodesics and unparametrized geodesics (what Jason DeVito has called almost a geodesic). Constant-speed is an artifact of a special parametrization of a curve. I am not getting at the moment how the parametrization relates to the choice of a frame along the curve. May be you can explain?

Technically, it may be more convenient to you to work with unparametrized geodesics. I heard from Vladimir Matveev that this was explained somewhere in Spivaks volumes. I guess, it is not too hard to obtain such an equation from the standard geodesic equation by substituting a change of variable.

In fact I don't know either. I read the paper informationgeometry.org/Seminar/gauss-Takatsu.pdf that gives a Riemannian metric on the space of gaussian measures. I tried the metric on a simple 1D gaussian and checked along a geodesic whether it is constant speed, and it was not. I sent an email to the author of the paper and he told me that it's just because he uses a different parameterization... which kind of confuses me.

In fact my end goal is still the same : computing $\nabla_{\dot\gamma}\dot\gamma$ projected on the normal bundle of the curve. I thought I had managed until I numerically checked with geodesics and it didn't work. From that, you explained that I was missing a term that I also have difficulties implementing...

So, we have a curve and a frame along a curve, represented in a coordinate patch of the underlying manifold. Therefore we have 1) the coordinates and 2) the frame. Can we use the coordinates to proceed with the commutation coefficients of the frame?

I will leave you for a while, but please look at the following two remarks.

Remark 1. The commutation coefficients aka structure coefficients of a frame characterize this frame. From Frobenius's theorem it follows that if these coefficients vanish the frame turns out to be a coordinate frame for some coordinate system. From here I have a foggy impression that it is possible to compute these coefficients in a way that does not involve a connection.

Remark 2. Technically one can use a connection, of course, provided it is torsion-free, not necessarily the Levi-Civita connection. Unfortunately, I have never got any practical computation that far, so I can only take part in a discussion. This must be known to people who do computational differential geometry (discrete differential geometry).

Please feel free to leave your thoughts here. I will come back from time to time...

ok thanks! I'll think about it. For the moment I tried the extend the vector field outside of the curve by taking the inverse of the projection on the curve, but whatever I try, I get a higher value of the second fundamental form for a non geodesic than for a geodesic (so I don't think this is due to numerical issues!) :s

I didn't know Koszul formula, but looking at it, it seems that I can compute the second fundamental form directly, right ? with X tangent to the curve and Y normal to the curve, I get 2<\nabla_X X, Y> = -Y <X,X> +2 <X, [Y, X]> + <Y, [X, Y]> if I'm correct.... I don't even need to compute the connection! But I need to compute the Lie bracket which isn't great

and requiring that Y is normal to the curve isn't really great either :s

I think I realize I don't understand anything and what I originally did was even more wrong than I thought. To be concrete, the metric is g_V(X,Y) = trace(XVY) for V a symmetric positive definite matrix (the point in the manifold) and X and Y are tangent vectors which are symmetric matrices. When I used to compute the Christoffel symbols, this required, for example, Y g_V(X,X), with X my first basis vector which is the tangent to my curve, [...]

and Y, the derivation, a vector that is not tangent to my curve. What I did is to derive g : lim(g_{V+\epsilon Y}(X,X) - g_{V}(X,X)) = trace(XYX) and that was my coefficient. However I realize that it should be wrong for several reasons : (1) I didn't explicitely do any extension of my vector field outside of the curve ; (2) if Y is normal to the curve, I would expect the derivative to be 0, but this is not what is exhibited by this formula.....!

hello

helloo

hello

@WhitAngl hello

hi!

@Ethan hello!

out of curriousity how old are you?

also what tools do you use in differential geometry?

hi

I am 29, but I am a postdoc in computer science

that means that I took basis classes of differential geometry, but I want to apply that in concrete examples

I'm happy to use whatever tool can solve my current problem which is about computing the curvature of a curve

are you their?

there

yes

why did you study differential geometry

or why do you

is that an entire field of study?

I propose we open another chat for that, then.

how?

whats a postdoc?

@YuriVyatkin: I paste back what I last wrote:

I think I realize I don't understand anything and what I originally did was even more wrong than I thought. To be concrete, the metric is g_V(X,Y) = trace(XVY) for V a symmetric positive definite matrix (the point in the manifold) and X and Y are tangent vectors which are symmetric matrices. When I used to compute the Christoffel symbols, this required, for example, Y g_V(X,X), with X my first basis vector which is the tangent to my curve, [...]

and Y, the derivation, a vector that is not tangent to my curve. What I did is to derive g : lim(g_{V+\epsilon Y}(X,X) - g_{V}(X,X)) = trace(XYX) and that was my coefficient. However I realize that it should be wrong for several reasons : (1) I didn't explicitely do any extension of my vector field outside of the curve ; (2) if Y is normal to the curve, I would expect the derivative to be 0, but this is not what is exhibited by this formula.....!

First of all we need to figure out how to compute the Levi-Civita connection of a metric in any manifold, regardless whether the elements of it are thought as vectors, functions, polynomials or matrices. In the numerical computations such a manifold will be finite-dimensional, that is every time we work with a presentation of this manifold as a piece of R^n with all the structures pulled back to this piece via the chart.

The second fundamental form is then the normal part of this Levi-Civita connection applied to two vectors which are tangent to your submanifold, i.e. the curve in your particular problem.

The Koszul formula is the explicit expression that gives the Levi-Civita connection in terms of the metric and the Lie bracket:

<\nabla_X Y, Z> = 1/2 (X g(Y,Z) + Y g(Z,X) - Z g(X,Y) + g([X,Y],z) - g([Y,Z],X) - g([X,Z],Y))

See the derivation of this formula e.g. en.wikipedia.org/wiki/Levi-Civita_connection or in any standard course of Riemannian geometry

Using this formula we obtain the usual expression for the Christoffel symbols in a coordinate frame, because the brackets of partials vanish. If the frame is arbitrary the brackets produce the "commutation coefficients", and we get the generalized expression for the Christoffekl symbols in a (possibly) non-holonomic frame.

With this in mind there should be a way to pick a concrete example of a manifold and a local frame in an open set of this manifold, and compute the connection coefficients of this frame. This should not involve any metric or connection!

The local frame that we are looking at the moment should be defined locally, that is on an open set, not just along a particular curve. In this situation, using the coordinates (!!!) we should be able to calculate directional derivatives using the standard multivariable calculus formulae.

These coordinates come from the chart in which we represent all the objects that we are dealing with, including the vector fields that form the local frame

Finally, when you endow the curve with a convenient local frame there should be a natural (for you choice) way of extending this frame to a tubular neighborhood of the curve. The theory then must assure that the constructions that you use do not depend on this choice.

Therefore, if all the analytic expressions that are in the foundation of your algorithms are correct, you should get consistent results, but there may also be issues with stability etc.

thanks! could you tell me whether in the expression X g(Y,Z) applied to X=vector not tangent to the curve, and Y=Z=vector tangent to the curve, this should make use of the extension, or this can be computed directly with extending the vector field to a tubular neighborhood ?

I also have difficulties understanding "and compute the connection coefficients of this frame. This should not involve any metric or connection!"... how can we compute connection coefficients without connection (and metric) ? :s

Oops, in the phrase "and compute the connection coefficients of this frame. This should not involve any metric or connection!" the first instance should be read "commutation coefficients". Sorry for the typo.

I am not quite sure about the actual computation that you may perform, but in X g(Y,Z) we need extensions of Y and Z to proceed. A choice of such Y and Z gives a scalar function g(Y,Z) over a neighborhood which can be differentiated in the direction of X.

Initially, you choose a local (non-holonomic) frame, and the commutation coefficient do depend on this choice which is made over a neighborhood. Moving on, you restrict your quantities to the submanifold (the curve). The final result (the second fundamental form) should be independent of choices of extensions.

ok - then I think I'll use an extension which is the inverse of the projection on the curve, using a simpler (euclidean) metric.

The independence of extensions only holds if you differentiate in tangential directions (along your curve).

oh!

Sure, you should use any extension that naturally arises from your construction

so, what extension should I use when I differentiate in a non tangential direction ?

The second fundamental form is by definition the difference of the ambient connection and the intrinsic connection where both connection are taken from tangential vector fields.

The fact that you are using the formula which gives you the normal part of the ambient connection should not contradict the definition

I don't understand.... for example, in the Koszul formula : <\nabla_X Y, Z> = 1/2 (X g(Y,Z) + Y g(Z,X) - Z g(X,Y) + g([X,Y],z) - g([Y,Z],X) - g([X,Z],Y)) , even if I set X and Y to be tangent to my curve, Z is not necessarily tangent to the curve, right?

When you carry on the computations you take ANY extension, but because the theory (calculations!) guarantees that the result will be independent after taking restrictions (projections in a sense!), we should not worry

Correct. The Koszul formula gives you the ambient connection that is able to act in all directions

oh ok... then I can use whatever I want for the extension then ?

Sure. Provided the final results are independent of choices (e.g. the 2nd fund. form)

ok :)

Do I confuse you even more than before? :-)

hehe, no I think I start to get it ;)

I'll try that then... I'll take a simple euclidean projection on the curve and carry out the computation.

(or rather the inverse of the projection)

thanks a lot for all your help!

oh and one last thing :

when I get <\nabla_X Y, Z> for all Z in my basis, I need to solve a linear system to get \nabla_X Y right ?

That would be good. As I told you a while ago, the second fundamental form of a curve is a slippery object because it is ultimately a sort of function with values in the normal bundle of the curve. If I were you I would try to get the clearest understanding of this, and to my mind a generalization of the Frenet formulas is the right way. But, it is up to you :-)

hehe thanks! I'm a bit late on this project to take a different path, but I'll check that out :)

I have not made any concrete computation of this sort, to be honest %) Sure, <\nabla_X Y, Z> implies that to find the connection you need to solve an equation, but why we then obtain the explicit expression for Christoffels? I would use the latter.

I think anyway there is an equation to solve : if I use the expression of the Christoffel symbols that uses the commutation coefficients, I need to solve a similar equation to get these commutation coefficients

If I still don't manage, there is Shing Tung Yau just in the next building... if he can't manage to explain to me, I think nobody can ;) (he is also currently working on more concrete conformal mapping of the brain)

Of course, it may be the way that you are following. I would better ask you how to do this in a concrete situation. In my research I am trying to understand how various invariants of submanifolds arise, and what do they mean.

Wow, I envy you :-) I would ask him some questions too...

cool! You're a PhD student? postdoc?

I can ask questions for you, but I'm unlikely to understand the answer ;) lol

I am finishing my PhD in few months. Manage to make a very little progress (sigh!), but my supervisor believes that we can wrap this into a PhD anyway...

Oh, I was just kidding. I will go to a conference soon, will meet there a lot of people who know everything already ^_^

I guess making significant progress in differential geometry is much harder than in other fields. My differential geometry professor this year (Siu-Cheong Lau) doesn't have many paper but he is good and still prof at harvard

hehe ;) ok!

Yeah. Perelman has not too many papers either, he-he.... He refused to be a prof in Harvard, btw.... And two of IMO guys who I taught geometry study in Harvard now. But I do envy you, of course :-)

you're on the west coast right?

In New Zealand, Auckland University

oh, that's far! New Zealand seems a nice country as well!!

Yes, the country is very good. I enjoy it. We have summer now :-P

hehe, I guess snow will be coming in a couple of days here while I spend my christmas working on a deadline in January :s

OK. I wish you a good luck! It was good to talk to you. Merry Xmas and a Happy New Year ;-)

thanks for all - I think I should manage to compute what I need - and merry Xmas and happy new year too!!