Let $(N,g_N)$ be Riemannian manifold, $I \subset \mathbb{R}$ open with coordinate $r$ and $M:= I \times N$ with metric $g=dr^2+f^2 g_N$.
Let $c_p: I \rightarrow M, (t \mapsto (t,p))$ for a fixed $p$ be a curve in $M$.
I want to show that $c$ is a geodesic.
To do that, I first look at local coordinates for $M$, define $(y^1,….,y^{n+1})=((t,q) \mapsto t, x^1,…,x^n)$ where $x^i$ are local coordinates in $N$. Now:
$y^1 \circ c_p= (t \mapsto t)$ and $y^i \circ c_p=(t \mapsto x^i(p))$ for $i>1$. I can conclude that $$\dfrac{\partial (y^1 \circ c_p)}{\partial t} =1,\quad \dfrac{\partial^2 (y^i \circ c_p))}{\partial t^2}=0 \ \forall i,\quad \dfrac{\partial (y^i \circ c_p)}{\partial t} =0 \ \forall i>1.$$
Now the equation for $\nabla_{c^{'}} c^{'}$ reduces to:
$$\nabla_{c^{'}} c^{'}= \sum\limits_{i=1}^{n+1} \Gamma^i_{11} \dfrac{\partial}{\partial y_i} = \nabla_{\partial_1} \partial_1$$
Is it correct, that $ \nabla_{\partial_1} \partial_1=0$ (then my proof would be complete)? And if yes, why?
Best Answer
$F(t,x)=(t,\phi (x))$ is chart for $M$ where $\phi$ is a chart for $N$.
Hence $E_t=\partial_t,\ E_\alpha= d\phi\ e_\alpha$ are coordinate field. Note that $g(E_t,E_t)=1 ,\ g( E_t,E_\alpha)=0,\ g(E_\alpha,E_{ \alpha'})=f^2 g_N(E_\alpha,E_{\alpha'})$
So we must show that $\nabla_t E_t=0$ : $g(\nabla_tE_t,E_t)=0$ and $ g(\nabla_tE_t,E_\alpha)=E_tg(E_t,E_\alpha) -\frac{1}{2}E_\alpha g(E_t,E_t) =0$