Toy Universes

The transpose of a tensor

In our previous post, you may have wondered why we have chosen to compute

instead of using a different order in the slots of T, say:

.

The answer comes clearly if you try to see what happens if you do so. Write:

.

Then

.

It means that thecomponent of the vector-valued functionis. If you use the matrix representation of tensors by writing the expression for thatcomponent in terms of a square matrix multiplied by a column matrix, the result, as compared to the previous way of expressing the components, is that the computed matrices are transposed in relation to one another in terms of columns and rows.

So you see, the vector valued function is the transpose of . This can be generalized to the bilinear form, e.g.:

.

So the final answer to the above question is simply: there was no reason for chosing one specific order in the slots. But once you have chosen one, you must be consistent throughout on what is the transpose of what.

Symmetries of tensors

Matrices can be classified as symmetric and antisymmetric, and since tensors can be represented by matrices, what can we say about symmetric and antisymmetric properties of tensors?

First, recall that a symmetric matrix is a square matrix such that it is equal to its transpose. And an antysymmetric one is equal to the negative of its transpose.

Now, since:

,

we see that the above set of components, which form a matrix, will be equal to its transposed set if and only if:

, for all and.

Similarly,

, for antisymmetric tensors.

The above equations mean, respectively:

, and .

You may want to extend the above concepts to tensors of higher rank. For example, consider a third rank tensor, viewed as a 3-linear functional. If you want to make that tensor a completely symmetric one, you will have to impose the following conditions:

, etc,

that is, to all permutation of the input vectors. For a completely antisymmetric one:

, etc,

and I leave as an exercise to you to derive when you use the minus or plus signs above.

As a final note, you may wonder whether it its possible to consider symmetry properties in mixed tensors. Yes, it is. You may just consider symmetry on interchange of the arguments in a given pair of input slots (say, the 1st. and the 2nd.), and antisymmetry on another (say, the 2nd. and the 3rd.), and so on.

Ref.: [Nea10].

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].