Index attached to derivative operator

Question

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.

Answer 1

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: $t^a \nabla_a f$ is the natural pairing of a tangent vector $t^a$ with a cotangent vector $\nabla_a f = df$. You can view this as the contraction of the $(1,1)$-tensor $t^a \nabla_b f = t \otimes df$.

Answer 2

I'm dealing with Penrose's abstract index notation (from his and Rindler's book Spinors and Spacetime), and what I have gotten from it is indeed contraction (i.e. a contracted product), but no summation whatsoever (since the $a$ has no numerical value. With the abstract index notation, you take whatever set with structure of interest (set of derivations, world-vector, (p,q)-tensors) and you make copies of this set, using an (infinite) abstract indexing set $\mathcal{L}=\{a,b,\ldots,a_0,b_0,\ldots\}$. For example if $\mathcal{I}^\cdot$ is the $C^\infty(M)$-module consisting of world-vector fields, then we get copies $\mathcal{I}^a,\mathcal{I}^b$, etc., and every $V\in\mathcal{I}^\cdot$ has corresponding elements $V^a\in\mathcal{I}^a$ etc.. A lot of usual "summation convention" type expressions still hold, but there are no summations since these elements are abstract vectors/tensors, and not merely the components. I hope Wald mentioned that $V^aV_a(\text{abstractly})=V^{\mathbf{a}}V_{\mathbf{a}}(\text{components})$. Now, as you also mention, $df$ is an element of $(\mathcal{I}^\cdot)^*$, and so has a canonical image in $\mathcal{I}_a$, which is then defined to be denoted by $\nabla_af$. Now the canonical image of $t$ in $\mathcal{I}^a$ is $t^a$, and so we choose to express $df(V)=V(f)$ as $V^a\nabla_af$, thereby stating that $V^a\nabla_a$ is the canonical image of $V$ in $\mathcal{I}^a$.

I am myself struggling with all this notation and just gave you my understanding of it, I hope it helps somewhat, maybe we can shed more light on this.