Toy Universes

When discussing functionals and their relation to tensors for the first time, in this post, we mentioned the representation theorem  for linear functionals (see also our Theorems page at the side bar).

Recall that it states that ifis a linear functional,  then there is a unique vector such thatfor all. The importance of this theorem is that all linear functionals can be precisely put into this form. We will explore this theorem more concretely as our exposition advances.

Now we just want to check how this theorem fits with what we have been doing in the last few posts, namely, working with tensors in component form, that is, through indexes, in order to track how tensors behave in terms of their components in some specified basis. Recall that we are working with Euclidian tensors, and already know how to deal with them in orthogonal and non-orthogonal bases. Also recall that we made a change in notation in order to make distinct sub- and superscripts. (Review our previous posts in the Courses page at the sidebar).

What we claim now and show immediately is that the unique vectorthat is mentioned in the representation theorem is exactly the one computed from:

Let us prove that quickly. Just take; then:

As an exercise, prove the analogous result for

Let us expand that to bilinear functionals. Let be a bilinear functional. The representation theorem now thought in such terms will, instead of the unique vector, claim that there is a unique vector-valued function, such that:

.

Again, you can prove, in complete analogous way as in the linear functional case, that such unique vector-valued function is directly computed from:

.

You can extend that to multilinear functionals in a completely analogous way.

Such computations lead us to infer simply the following cases.

Ifand , then:

.

Ifand , then:

.

Ifand , then:

.

Etc. So you see the appearance of covariant (subscripts), contravariant (superscripts), and mixed (both sub- and superscripts) components, respectively.

As an example to put into context, consider a bilinear functional called the metric tensor:

.

In a manner consistent with what we have been working until now, we see that the linear vector function that is found by pulling off the from above is actually an identity function:

.

Indeed:

More on the metric tensor later.

Ref.: [Nea10].