is it normal that my Lie bracket is always 0 ? In which condition does that happen ? (I approximated my curve locally to the second order using a Taylor expansion and it's derivative to the first order. When I compute the directional derivative of $\dot\gamma$, I always obtain something proportional to $\ddot\gamma$ ....

Hey

hi

What's the problem?

trying to compute the second fundamental form for a curve....

What curve?

If I did thing well, then I don't understand why my Lie bracket is 0....

No idea then :p

only in Calc II

it is a curve defined in a space of symmetric positive definite matrices...

hmm...

I must learn about this

well, sorry, gtg I have a an essay to write...

good luck

@YuriVyatkin is it normal that my Lie bracket is always 0 ? In which condition does that happen ? (I approximated my curve locally to the second order using a Taylor expansion and it's derivative to the first order. When I compute the directional derivative of $\dot\gamma$, I always obtain something proportional to $\ddot\gamma$ ....

In fact, if Y=\dot\gamma is the tangent of my curve and X is any vector, what I get is XY = \ddot\gamma * g(X, Y)/g(X, X) , with $g$ any metric that is used to project vectors on the curve (to build the extension) ... which at the end is obviously symmetrical, and thus cancels the commutator

@WhitAngl Do you know anything about Frobenius theorem? Please take a look at this en.wikipedia.org/wiki/… Later I will try to spend some time on it to understand how it works. Basically, the theorem states that vanishing of the brackets means that one can "integrate" the frame and obtain a submanifold for which it is a coordinate frame.

I don't know this theorem... I'm checking it. But I feel that my calculations would yield a 0 commutator for any metric and any curve, so I feel I'm wrong somewhere

I think I'll post a Math.SE question detailing what I did... it just sounds wierd

(I'm reading the wiki page)

I think I more or less understood the page (at least the formulation for vector fields)... I'm now writing a Math.SE question, this might make more sense :)

question asked! : math.stackexchange.com/questions/263018/… :)

arg, a final question : if X = \dot\gamma, then XX = \ddot \gamma or should I use the extension in that case as well ?

Yeah, it is a useful question. If it is not answered by someone really good I will add a bounty on it for sure ;-) Sorry, I cannot give to it more attention at the moment: things are really falling over my head %)

hehe :) for my Math.SE question, I finally just need someone to check whether what I did is correct : at the end the Lie bracket is not 0, so at lease this is better for me ;)

but do you happen to know if, for X=\dot\gamma, then XX=\ddot\gamma ?

(sorry for the spam - there is noone in the computer science lab that would be able to help me ;) so I'm infinitely grateful! )

I have a feeling that we both are missing something that essentially makes the calculation work.

This is not spam, I guess.

why ? isn't it supposed to be correct ?

Ultimately, when you differentiate something it must be defined in a neighborhood.

I have not give it a good thought. In fact, I am just checking stuff on the comp from time to time when I have a spare minute during my relocation chores

:)

but to differentiate, the neighborhood only need to be define along a 1D curve, right ?

it only needs a 1D neighborhood (?)

so, ultimately, my curve $\gamma$ is the neighborhood I need to compute \dot\gamma \dot\gamma

As I have mentioned, if you differentiate tangential fields in tangential directions, you get the independence of extensions. The second fundamental form is defined as such, but you are using a formula that is a workaround, but the result should be the same.

Sorry again, I can't really concentrate now on the essence of the problem %)

ok, no problem! thanks a lot :)

I even have not read your question yet :-[

It is OK. I am working on my stuff through that.

:)

oh I didn't pay attention, but if I use the formula that I obtain using the extension, I also get $\dot\gamma \dot\gamma = \ddot \gamma$ , so everything turns out ok apparently

I have read quickly your question and noticed that you are a little bit sloppy with the Lie derivatives that appear in your calculations at least in two forms: 1) as the directional derivative of a scalar, and as 1) the Lie bracket of the vector fields.

The Lie derivative does not use any metric at all, so appealing to any sort of a metric does not imply that [X,Y]=0

I did mention it earlier that form the outset you have some coordinates (unrelated to your choice of a frame along the curve). And I hinted albeit not very explicitly that there are closed explicit expressions for both types of the Lie derivative, namely for the directional derivative of a function, and the Lie bracket of two vector fields. These are standard and known very well.

In particular, there is no such a thing as "directional derivative" of a vector field, i.e. UV is meaningless. Your expression with the limit must be a non-tensorial object that depends on all possible choices and lacks naturality. Your constructions really need to be checked carefully.

Your expression for the projection operator look quite obscure for me. I do not understand it.

Reading your question even more I have a feeling that you are missing a point that the Lie bracket defined as [U,V] = UV - VU must be seen as an operator on functions, that is through the action [U,V]f = UVf - VUf = U(Vf) - V(Uf). If you have coordinates of U and V you can find coordinates of vector [U,V].

The expression that I mean age given here ictp-saifr.org/schoolgr/Lecture0Friedman.pdf for instance (and everywhere where people look at coordinate expressions of the Lie derivatives, or just directional derivatives and Lie brackets)

I see that I'm still misunderstanding things... when in the Koszul formula there is g([X,Y], Z) in this case [X,Y] is not an operator, right ?

for my projection, it is simple: we start from \gamma(t) and we add an epsilon times a vector U and we obtain a new point P. I want to find the closest point of P on gamma(t) given a metric h. Let's call this closest point Q. I then assign to P the vector on the curve at Q : this will give me the extension. I thus need to evaluate \gamma^{-1}(Q) which is the parameter of Q, to take the vector V(\gamma^{-1}(Q)). But Q is the projection of P on \gamma : Q = \Pi(\gamma(t)+epsilon*U)

hence the formula I gave.

I managed to talk to my differential geometry professor today! He finally suggested to use a fixed frame rather than having the frame attached to the curve. Doing so I can use the standard formula for the Christoffel symbols for the ambient space, and then pull back the connection to the curve and compute the second fundamental form more easily... I just implemented it and I'm currently testing