# moving a basis along a curve (parallel transport)

I'm considering a Riemannian Manifold $M^m$ and a Basis $\{X_1 ,...,X_n\}$ of the tangent space $T_pM$.

When I consider now the parallel transport $E_i$ of the vectors $X_i$ along a curve c, then the Vector fields $E_i$ form a basis in every point. (i.e. $\{(E_1)_{|p} ,...,(E_n)_{|p}\}$ is a Basis)

Why is this true? I couldn't prove it. Maybe someone can help me!

Regards

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Are you aware of the fact that parallel transport is a isometry between tangent spaces? –  matgaio May 14 '12 at 20:29
Thank you very much! Now I know the answer! –  Braten May 14 '12 at 20:41
I've written an answer, so I will put here, just to complete: Consider $f(t)=\|P_\gamma^t(v)\|^2$. Then by compatibility, \begin{eqnarray} \frac{d}{dt}f(t)&=&\frac{d}{dt}\langle P_\gamma^t(v),P_\gamma^t(v)\rangle\\ &=&2\langle \nabla_{\gamma'}P_\gamma^t(v),P_\gamma^t(v)\rangle\\ &=&0 \end{eqnarray} because $\nabla_{\gamma'}P^t(v)=0$. So $f$ is constant and once $P_\gamma^0=\textrm{id}$, we have $\|f(v)\|=\|v\|$. You can check that the parallel transport is linear. Then, it is linear and preserves norm, wich implies that it is a isometry between tangent spaces. Your result follows from this. –  matgaio May 14 '12 at 20:43