A geodesic on a Riemannian manifold $(M,g)$ with the Levi-Civita connection $\nabla$ is defined as a curve $\gamma(t)$ such that $ \nabla_{\gamma'(t)}\gamma'(t) \equiv0$.

Can we show that generally, the second fundamental form of a geodesic as a one-dimensional submanifold of $ (M,g) $ is $ 0 $?

  • $\begingroup$ Hello @Bach. As F.T. pointed out, I have misread this question. His answer is correct. Feel free to mark his answer as the correct answer. $\endgroup$ – Ernie060 Sep 17 at 15:06

EDIT: I have misread and misinterpret the OP's question. The answer of F.T. is correct. This answer explains that the second fundamental form of a submanifold along a geodesic of that submanifold does not vanish in general.

No, in general the second fundamental form along a geodesic does not vanish.

As an example consider a (regular) curve $\alpha$ in a surface $S$ in $\mathbb{R}^3$. Let $T$ be the unit tangent vector and $N$ the unit normal of the surface along the curve. Define $V = N \times T$; this is a vector normal to the curve but tangent to the surface. (This frame $T$, $V$, $N$ is called the Darboux frame).

Then $$ T' = k_g V + k_n N, $$ where $k_g$ is the geodesic curvature and $k_n$ is the normal curvature of $\alpha$.

The curve $\alpha$ is a geodesic iff $k_g =0$. However, the part $k_n N$ is not necessarily zero. This part vanishes iff $k_n=0$, i.e. iff the curve is an asymptotic curve. Note that in the case $k_g=k_n=0$, then $T'=0$, so then the curve is a straight line.

If you consider a geodesic on a surface with positive Gauss curvature, then the second fundamental form along the geodesic never vanishes, since the normal curvature cannot be zero in any direction. This follows from the formula $k_n = k_1 \cos \theta + k_2 \sin \theta$, where $k_n$ is the normal curvature in the direction $\cos \theta e_1 + \sin \theta e_2$ and $e_1$, $e_2$ are the two principal direction with respective principal curvatures $k_1$, $k_2$.


I believe that the answer of Ernie060 is quite confusing as he is considering a curve in a surface in a 3-manifold. He is actually claiming that the geodesics of the surface are not geodesics of $\mathbb{R}^3$ which is clearly true for a generic surface. For those who are interested in minimal submanifolds, I would like to point out that these phenomena are very common. For instance, the Clifford torus is a minimal submanifold of the sphere but not of the euclidean space. (See DoCarmo Riemannian Geometry)

The answer to your question is actually yes.

Assume we have a general submanifold $\Sigma$ of $(M,g)$. The second fundamental form is defined as: $$ A(X,Y)=(\nabla_X Y)^N $$ with $X,Y \in T_p\Sigma$ and $\nabla$ Levi-Civita connection of M. If we consider $\Sigma$ one dimensional, (as $\gamma'(t)$ form a basis of $T_{\gamma (t)} \Sigma$ for all t) the second fundamental form is identically equal to zero by definition of geodesic.

  • $\begingroup$ Hello F.T., it is only now that I see that I have completely misread the OP's question . Next time, feel free to leave a comment. $\endgroup$ – Ernie060 Sep 17 at 15:05
  • $\begingroup$ Dear Ernie060, Unfortunately, I don't have enough reputation to comment under your answer. Next time, if I am allowed to, I will comment. $\endgroup$ – F.T. Sep 19 at 7:07
  • $\begingroup$ @ F.T. Ah, no problem, I understand! $\endgroup$ – Ernie060 Sep 19 at 7:46

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