I've been reading about the Riemann tensor and I've realised that I've convinced myself of two mutually exclusive facts. Suppose we have a manifold with the standard Levi-Civita connection $\nabla$, and let $U$, $V$ and $W$ be vectors at some point.
First, $f(U,V,W) = \nabla_U \nabla_V W$ is a tensor, because the covariant derivative of a tensor is a tensor. In particular $f$ is a vector when all of its arguments are given.
However, $f$ does not appear to be linear in its arguments. If $U = U^ie_i$ and $V = V^je_j$, then
\begin{align} f(U, V, W) &= \nabla_{U^i e_i} \nabla_{V^j e_j} W \\ &= U^i \nabla_{e_i} \left(V^j \nabla_{e_j} \right) W \\ &= U^iV^j \nabla_{e_i} \nabla_{e_j} W + U^i V^j_{ \ , i}\nabla_{e_j} W. \end{align} But this is not the same as $U^i V^j f(e_i, e_j, W)$, which is what is demanded by linearity.
Could anyone point out what it is that I'm missing?
Thanks
The Problem is that $\nabla(\nabla W)$ is indeed a tensor field, but it is not given by $(U,V)\mapsto \nabla_U\nabla_V W$. It is true that $\nabla W$ maps $V$ to $\nabla_VW$, but $\nabla W$ is a $\binom11$-tensor field. Hence $(\nabla\nabla W)(U,V)=(\nabla_U(\nabla W))(V)=\nabla_U(\nabla W(V))-(\nabla W)(\nabla_UV)=\nabla_U\nabla_V W-\nabla_{\nabla_UV}W$. For the latter expression it is easy to verify directly that is is tensorial.