Orthonormal basis expression for ordinary contraction of a tensor

[Shirish]

I'm reading Semi-Riemannian Geometry by Stephen Newman and came across this theorem:

Orthonormal Basis Expression for Ordinary Contraction: Let $(V,g)$ be a scalar product space with signature $(\epsilon_1, \ldots, \epsilon_n)$, and let $(e_1,\ldots,e_n)$ be an orthonormal basis for $V$. If $A\in\mathcal{T}^1_s(V)$ (where $s\geq 2)$ and $l<s$, then the tensor $C^1_l(A)\in\mathcal{T}^0_{s-1}(V)$ is given by: $$C^1_l(A)(v_1,\ldots,v_{s-1})=\sum_i\epsilon_i\langle\mathcal{R}^{-1}_s(A)(v_1,\ldots,v_{l-1},e_i,v_{l+1},\ldots,v_{s-1}),e_i\rangle$$

For context, $\mathcal{R}_s:Mult(V^s,V)\to\mathcal{T}^1_s$ is the representation map, which acts like this:
$$\mathcal{R}_s(\Psi)(\eta,v_1,\ldots,v_s)=\eta(\Psi(v_1,\ldots,v_s))$$
I don't understand the importance of the relation that the author is trying to prove here - it seems kind of irrelevant. Judging from the name of the theorem, shouldn't it be sufficient to figure out $C^1_l(A)(e_{j_1},\ldots,e_{j_{s-1}})$, i.e. the components of the contracted tensor w.r.t. different combinations of orthonormal basis vectors? And indeed this is what the author does in the first 2 lines of the proof.

But why are we going overboard to get the result in the theorem statement? Does it mean something significant?

 

[martinbn]

But why are we going overboard to get the result in the theorem statement? Does it mean something significant?

It gives you the contraction evaluated at any vectors. Of course it is enough to know it for basis vectors, but this is more general and equally easy.

 

[Shirish]

It gives you the contraction evaluated at any vectors. Of course it is enough to know it for basis vectors, but this is more general and equally easy.

Sorry if I sound ignorant since I'm completely new to these topics. To me the most intuitive form of the general case is:
$$C^1_l(A)(v_1,\ldots,v_{s-1})=A(\theta^i,v_1,\ldots,v_{l-1},e_i,v_l,\ldots,v_{s-1})$$
where $\theta^i$ is the dual basis vector corresponding to $e_i$.

I guess the RHS in the equation in the theorem is because we don't even want to see $\theta^i$, so we use the elaborate construction to get rid of it and just have an RHS in terms of $e_i$'s?