# raising/ lowering indices

Here is my understanding of tensors:

There is more than one way to think about tensors.

One way is be thinking about tensors as objects with components which obey some transformation laws. For instance ${T^{abc}}_{def}$ is one component of the type $(3,3)$ tensor $T$. To raise or lower indices you multiply by the metric tensor, like $g_{ah}g^{ei}{T^{abc}}_{def}={{{{{T_h}^{bc}}_d}^i}_f}$.

Another way is to think of a type $(p,q)$ tensor as a multilinear function from the Cartesian product of $p$ copies of the dual space $V^*$ and $q$ copies of the vector space $V$ to the reals: $T: \underbrace{V^* \times \cdots \times V^*}_{\text{p times}} \times \underbrace{V \times \cdots \times V}_{\text{q times}} \to \Bbb R$.

The way to recover the components of the tensor from the multilinear function is just by evaluating the tensor at the basis one-forms and vectors. For instance, if $\{\omega^a\}$ is the standard orthonormal basis of one-forms and $\{v_a\}$ is the standard orthonormal basis of vectors, then $T(\omega^a, \omega^b, v_c, v_d) = {T^{ab}}_{cd}$.

My question is:

What corresponds to the idea of raising and lowering indices for the multilinear function form of a tensor?

The metric specifies a canonical isomorphism between $V$ and $V^*$--and therefore an invertible map $g' : V \to V^*$.
Consider some $T: V \times V^* \to \mathbb R$.
Now consider $T': V \times V \to \mathbb R$ such that $T'(A, B) = T(A, g'(B))$.
In this picture the map $T'$ is technically distinct from $T$, but people don't observe this distinction when talking about the components (perhaps because there's no possibility of confusion there).