Index attached to derivative operator So in "General Relativity", Wald introduces a derivative operator $\nabla$ on a smooth manifold $M$ that sends $(k, l)$ tensors to $(k, l + 1)$ tensors. One of their properties he says is that if $f \in C^{\infty}(M)$ and $t^a \in T_pM$ ($t^a$ is written with a superscript because Wald uses abstract index notation), then $t(f) = t^a \nabla_a f$. What does the notation $t^a \nabla_a f$ mean? It looks like a contraction to me, but Wald says that it is just for notational convenience. EDIT: It would make sense to me if $t$ were a vector field on $M$, for then $tf \in C^{\infty}M$ and then $t^a \nabla_a f$ could be looked at as $\nabla_a(tf)$. But if $t$ is just a single tangent vector, then I am still confused.

One important property of the action of vector fields on functions is that it is _local_ : the value of $X(f)$ at $p$ depends only on the value of $X$ at $p$. (This is a consequence of the $C^\infty(M)$-linearity of vector fields.) Thus it makes sense to talk about $t(f)$ when $t \in T_pM$ is just a single tangent vector - simply extend $t$ to a vector field $X$ such that $X(p) = t$ and define $t(f) = (Xf)(p)$, and the locality tells you that this is in fact independent of the extension you choose.

Thus tangent vectors are operators $C^\infty (M) \to \mathbb R$, so $t(f)$ should just be a real number. This agrees with the other side of the equation: $t^a \
abla_a f$ is the natural pairing of a tangent vector $t^a$ with a cotangent vector $\
abla_a f = df$. You can view this as the contraction of the $(1,1)$-tensor $t^a \
abla_b f = t \otimes df$.