I'm having the same problem as it was questioned here.

I can't get throught the step where I need to show that $\nabla_{E_i}E_j (p)=0$. It only leads to $$ \nabla_{E_i}E_j(p)=\sum_{lk}^n a_{il}(0)\dfrac{\partial b_jk}{\partial x_l}(0)\partial x_k(p) $$ which I can't show that equals zero.

On the linked question, there is also the indication that $\nabla_{E_i}E_j (E_k)=\Gamma^k_{ij}$. This could lead that $\nabla_{E_i}E_j (E_k)(p)=\Gamma^k_{ij}(p)=0$, but $\nabla_{E_i}E_j (E_k)(p)$ is not equal to $\nabla_{E_i}E_j (p)$ which is the one I need.

Can someone help me?


2 Answers 2


That more or less follows from definition. Note that $E_i(q)$ is given by parallel transport along the geodesic $\gamma$ with $\gamma(0) = p$ and $\gamma(1)=q$. Let $\gamma$ be a geodesic such that $\gamma(0) = p$ and $\gamma'(0) = E_i(p)$. Then we have $\gamma'(t) = E_i(\gamma(t))$ and so

$$\nabla_{E_i} E_j (\gamma(t)) = \nabla_{\gamma'} E_j (\gamma(t)) = 0$$

for all $t$, since $E_j$ is by definition the parallel transport along $\gamma$. In particular if you set $t=0$, we have $\nabla_{E_i} E_j (p) = 0$.

  • $\begingroup$ I'll check this later, I'm really sleepy right now. Thanks! :) $\endgroup$
    – Marra
    May 16, 2014 at 3:01

Let $v=\sum_iv_iE_i(p)\in T_pM$. Then, by the linearity of $g$, $$ g(\nabla_{E_i}E_j(p),v)=\sum_kv_kg(\nabla_{E_i}E_j(p),E_k(p))=\sum_kv_k\Gamma_{ij}^k(p)=0,$$ which implies that $\nabla_{E_i}E_j(p)=0$ by the arbitrary of $v$.


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