Knowing almost nothing about differential geometry, I have to understand this statement from the book that i am reading (page 115):
Recall that a geodesic on a Riemannian manifold $M$ is a path $\gamma(t)$, which minimizes the distance
$$L(\gamma) = \sqrt{\int_{t_1}^{t_2} |\dot{g}(t)|^2 dt}$$
among all curves $g:[t_1,t_2]\rightarrow M$ constrained by the boundary conditions $g(t_1)=\gamma(t_1), g(t_2)=\gamma(t_2)$, and this minimization property should hold true, whenever $t_2$ is close enough to $t_1$. It is equivalent to stating, that the accelleration of the curve $\gamma$, viewed from the tangent space to the manifold vanishes identically.
What does that last sentence mean? My Interpretation is that $\langle \ddot{\gamma}(t), T_{\gamma(t)}M \rangle = 0$, i.e. that the second time derivative of the curve is orthogonal to the tangent space. But intuitively, I don't see, why that should be an equivalent statement to the minimization property. I would rather expect that any curve on $M$ has its acceleration in the span of the tangent space of $M$.
Best Answer
The definition of geodesics is not exactly what you stated. A curve is a geodesic if it minimizes locally the distance between points, not globally. But this is not so important.
In the case where $M$ is a submanifold of $\mathbb{R}^n$ with the induced metric, that is, $M \subset \mathbb{R}^n$ and if $u,v \in T_pM$, that is $g_p(u,v) = \langle u,v\rangle$ the usual scalar product, then a smooth curve $\gamma : I \mapsto M$ is a geodesic if and only if its acceleration is orthogonal to $M$, that is, for all $t\in I$, $\gamma''(t) \perp T_{\gamma(t)}M$.
If $(M,g)$ is an abstract riemannian manifold, there is no ambiant space to state that the acceleration of a curve is orthogonal to $M$. In fact, there is nothing that can define the acceleration on an abstract manifold! Even if smoothness is well-defined, second derivative of a function is not well-defined as it is closely coordinate-dependant. So there is no possibility to talk about high order differentiation.
The key is that if $M$ is a riemannian manifold, there exists an intrinsic object call the Levi-Civita connection or covariant derivative, denoted by $\nabla$ or $D$ usually, that allows you to talk about acceleration. It is an algebraic objet that allows you to derive vector fields in the direction of vectors fields. In case $M\subset \mathbb{R}^n$ with the induced metric, the Levi-Civita connection is the orthogonal projection onto the tangent space of the derivative: if $v$ is a vector field and $\gamma(t)$ is a curve, the covariant derivative of $v$ along $\gamma$ is $$ \nabla_{\gamma'(t)}v(t) = {v(\gamma(t))'}^{\perp} $$ In the general setting, one define the acceleration of a curve $\gamma$ to be the vector field along $\gamma$ defined by $\nabla_{\gamma'}\gamma'$. One can show a geodesic is a curve that is solution to the equation of geodesics $$\nabla_{\gamma'}\gamma'=0$$ Once again, if $M\subset \mathbb{R}$, a curve $\gamma : I \to M$ is a geodesic if and only if its acceleration is orthogonal to $M$, if and only if $M$ is a solution of the equation of geodesics. Thus, saying $\nabla_{\gamma'}\gamma'=0$ is a generalisation of saying the acceleration of $\gamma$ is orthogonal to $M$: if there was an ambiant space, it would be.
If for example $M = \mathbb{S}^n \subset \mathbb{R}^{n+1}$, a geodesic is a great circle of the sphere. It can be parametrized by $$\gamma(t) = \cos(t\|v\|)p + \sin(t\|v\|)\frac{v}{\|v\|}$$ where $p\in \mathbb{S}^n$ and $v \in T_p\mathbb{S}^n = p^{\perp}$. The usual differentiation of $\gamma$ in the ambiant space shows that $$ \gamma''(t) = -\|v\|^2\cos(t\|v\|)p -\|v\|^2\sin(t\|v\|)\frac{v}{\|v\|}=-\|v\|^2\gamma(t) $$ which is colinear to $\gamma(t)$ and thus, orthogonal to $T_{\gamma(t)}\mathbb{S}^n$. Its orthogonal projection onto $T_{\gamma(t)}\mathbb{S}^n$ is then $0$.