# Exercise 5, chapter 4 in Do Carmo's “Riemannian Geometry”

This is probably a silly question and maybe in the end the answer is trivial but I can't see it.

The problem is the following.

Let $M$ be a Riemannian manifold, $\gamma \colon[0,l]\to M$ be a geodesic and let $X \in \mathcal{X}(M)$, where $\mathcal{X}(M)$ is the space of all the smooth vector fields on $M$. Assume that $X(\gamma(0)) = 0$. I want to show $$\nabla_{\gamma'}(R(\gamma',X)\gamma')(0) = (R(\gamma', X')\gamma')(0),$$ where $X' = \frac{DX}{dt}$ and $R$ is the Riemann curvature.

In the book there is the following hint.

Hint: Let $R$ be the curvature tensor. Observe that, for all $Z \in \mathcal{X}(M)$, and $t=0$,

\begin{align} 0 &= (\nabla_{\gamma'}R)(\gamma', X, \gamma',Z) \\ &=\frac{d}{dt}\langle R(\gamma',X)\gamma', Z \rangle - \langle R(\gamma', X')\gamma',Z\rangle - \langle R(\gamma',X)\gamma', Z'\rangle \\ &=\langle\nabla_{\gamma'}(R(\gamma',X)\gamma'),Z\rangle - \langle R(\gamma',X') \gamma', Z \rangle. \end{align}

The only thing I can't understand is the first equality. How do we know that $(\nabla_{\gamma'}R)(\gamma', X, \gamma',Z) = 0$ ?

## 1 Answer

I asked this question also in the forum MathOverflow even though this question is so basic. Indeed somone solved my problem almost immediately in the comments. I write the answer here because maybe can be useful to someone in a future.

The answer (as I espected) was obvious and I had this problem just because tensors are a new thing for me.

The object $\nabla_{\gamma'}R$ is a $4$-tensor and since $X(\gamma(0))=0$, then it's clear that $(\nabla_{\gamma'}R)(\gamma', X, \gamma', Z)=0$ in $t=0$, because $$(\nabla_{\gamma'}R)(\gamma', X, \gamma', Z)|_{t=0}= (\nabla_{\gamma'}R)(\gamma'(0), X(\gamma(0)), \gamma'(0), Z(\gamma(0))) = (\nabla_{\gamma'}R)(\gamma'(0), 0, \gamma'(0), Z(\gamma(0))) = 0.$$