A curve $\alpha$ on a riemannian manifold $(M,g,\nabla)$ is a geodesic if $\nabla_TT=0$, where $T$ is the tangent vector field. A generalization of this geodesic equation suggests that $\nabla_TT=\rho T$ where $\rho$ is a real valued function. To get such curves one can define a new symmetric non-metric connection $$\bar \nabla_XY=\nabla_XY+u(X)Y+u(Y)X$$ where $u$ is a 1-form. In this case the geodesics of the new connection satisfies $$\nabla_TT=-2u(T)T$$ If we could determine a 1-form $u$ such that $-2u(T)=\rho$, then the geodesics of the new connection are those curves that satisfy the general geodesic equation of the old connection. I have derived the curvature tensor of the new connection in terms of the old curvature tensor, the 2-form $du(X,Y)$ and $S(X,Y)=X(u(Y))-u(\nabla_XY)-u(X)u(Y)$. I try to get simple examples of such curves. Now I asked the following.
My question is
(1) Are there any hints to determine the 1-form using the only given condition
(2) How can I determine the importance of these curves(minimizing distance, ...).

Alex and Benoit said that $\rho$ should be zero and so my trial to fix the problem failed. But we have two important issues
(1) A new symmetric non-metric connection(I noticed that Pandey, Tripathe and Agashe defined and studied many semi-symmetric non-metric connections but I didn't realize the importance of introducing these new connections to a manifold equipped with the Levi-Civita connection) Is it important to study such new connections.
(2) General geodesics: I solved the problem for general geodesics of the plane with constant $\rho$ without using the one form, they are still straight lines with some strange properties(given initial position and initial velocity i got two straight different straight lines(a geodesic and a general geodesic) passing through the same point with the same velocity). Are there any hints to fix the problem of solving the general geodesic equation for arbitrary manifold.

Thanks in advance.


migrated from mathoverflow.net Feb 4 '14 at 13:24

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  • $\begingroup$ \#1: any $1$-form has a kernel, so you cannot find one unless $\rho=0$. \#2: are you sure that any generalization is useful and worth considering? $\endgroup$ – Alex Degtyarev Feb 4 '14 at 10:09
  • $\begingroup$ For #1, do you mean that $\rho=0$ over $M$ or a specific domain. For #2, this is my question a help to determine their importance $\endgroup$ – Semsem Feb 4 '14 at 10:20
  • 5
    $\begingroup$ The geodesic equation is to be interpreted as acceleration=0, it thus defines geodesics that have constant speed. The equation you propose means the acceleration of your curve is tangent to the curve, therefore it corresponds to geodesics with arbitrary parametrization (and $\rho$ is exactly the scalar acceleration). This is hardly new, and more fitted to Math.SE. $\endgroup$ – Benoît Kloeckner Feb 4 '14 at 10:29
  • $\begingroup$ This is what we know thanks to geodesics. Now, we have certain group of curves that satisfies this equation and i want to determine their importance. $\endgroup$ – Semsem Feb 4 '14 at 10:56
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    $\begingroup$ I mean that, if you want $u(T)=\rho$ for any vector $T$, then $\rho=0$. And, as Benoît Kloeckner suggests, what you are trying to introduce is a reparameterization of geodesics. $\endgroup$ – Alex Degtyarev Feb 4 '14 at 11:08

I hope the following can clarify what is discussed in the comment.

Let $\rho $ be a function on a Riemannian manifold $M$ and $\gamma:[0,1]\to M$ be a geodesic. Let $T(t)$ be the unit tangent vector. Let $U = aT$ for some positive function $a :[0,1] \to \mathbb R$. Then $\nabla_U U = \rho U $ can be written as

$\rho(\gamma(t)) a(t)T(t) = a\nabla_T (aT) = a(t) \dot a(t)T \Rightarrow \dot a(t) = \rho (\gamma(t))\Rightarrow a(t) = \int^t_0 \rho(\gamma(s))ds + a(0)$

Thus, if you consider the parametrization $c: [0,d]\to [0,1]$ such that $\dot c(s) = a(c(s))$, then the curve $\alpha = \gamma \circ c$ satisfies

$\dot \alpha (s) = a(c(s)) T(c(s)) = U(c(s))$.

Thus $\alpha$ satisfies $\nabla_{\dot \alpha} \dot \alpha = \rho \dot \alpha$. Note that $\alpha$ is geometrically the same as $\gamma$. So it makes sense that you got straight line in your calculation. That is, you got a geodesic anyway, but with some other parametrizations.

  • $\begingroup$ Your answer is really notable. I know that "being geodesic" is not invariant under re-parametrization. So, you proved that any geodesic could be re-parametrized to satisfy the general geodesic equation. Finally, only affine parametrization preserves geodesics. Thank you for appreciated endeavor. $\endgroup$ – Semsem Feb 5 '14 at 11:54
  • $\begingroup$ Your simple idea lead me to understanding some new about general geodesics, Thanks again, confidence upvote $\endgroup$ – Semsem Feb 7 '14 at 23:54

The curves you are describing are exactly reparametrizations of geodesics. This extends without problems to general linear connections on $TM$ (usually assumed to be torsion-free). One calls to such linear connections projectively equivalent if they have the same geodesics up to parametrization, and it turns out that this is equivalent to being related as $\hat\nabla_XY=\nabla_XY+u(X)Y+u(Y)X$ for a one-form $u$. (To find $u$ explictly in a situation as described in your question, one just has to solve an ODE.) The whole topic was studied classically as "projective equivalence of affine connecitons" and "projective differential geometry". This is because projective space as a homogeneous space of the group of projective transformations can be viewed as the homogeneous model of a smooth manifold endowed with a projective equivalence class of connections. There has been quite a lot of renewed interest in these topics (and their relations to Riemannian geometry) during the last years, since the are closely related to conformal geometry and, more generally, parabolic geometries.


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