3
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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 $T'(A, B) = T(A, g'(B))$.

That is what we're doing when we raise or lower indices. We might have some multilinear function that takes particular arguments (a specific combination of vectors and forms), and we're converting it to take different arguments using the metric.

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.