Tensors are geometric objects

Posted on 2025/12/10 by arkajad

Originally, while writing the previous post, I was convinced that this one should be devoted to the Grassmann (or exterior) algebra.  I even began drafting it and managed about half a page before realizing that, if the promise of a truly “gentle” approach is to be taken seriously, there is an earlier stop on this journey. Before exterior algebra can comfortably appear on stage, one really ought to say a few clear words about tensors in general.​

Moreover, since the plan is to reach Maxwell’s equations at some point, we will also need pseudo-tensors and tensor densities—those slightly more exotic geometric objects that insist on transforming with a twist. So, in this post, tensors and their “relatives” will make their entrance; Grassmann will simply have to wait his turn in the queue of structures. After all, even in mathematics, good manners suggest introducing the family before discussing the exterior.

When considering transformations of bases and components of vectors, tensors, etc., it is convenient to use a notation that avoids writing A^{-1} in transformation laws. Namely, if e_i is a basis in V, a new basis, previously denoted e'_i will be written as e_{i'}. Then the transformation to the new basis will be written as

(1)   \[ e_{i'}=A_{i'}^i e_i, \]

and the inverse transformation will be written as

(2)   \[ e_{i}=A_{i}^{i'}e_{i'}.\]

The lower index is the column number.

Thus

    \[A^i_{i'}A^{i'}_j=\delta^i_j}\]

and

    \[A^i_{i'}A_i^{j'}=\delta^{j'}_{i'}.\]

Then, for vectors x\in V, if

(3)   \[ x=x^ie_i=x^{i'}e_{i'},$\]

then

(4)   \[ x^{i'}=A^{i'}_ix^i,\]

and

    \[x^{i'}=A^{i'}_ix^i.\]

Similarly for covectors, elements of V^*: if u=u_ie^i=u_{i'}e^{i'}, then

    \[u_{i'}=A^i_{i'}u_i,\quad u_i=A^{i'}_iu_{i'}.\]

This notation, simple and useful, is, however, somewhat old-fashioned. Since we will need tensors and tensor densities, it is good to get acquainted with the modern formulation via the principal bundle of frames and associated vector bundles. This approach is usually discussed within the framework of differential manifolds. Here we have just one vector space V. It plays the role of the tangent space to a manifold at a single point. With this in mind, let me explain this more modern point of view.

Let \mathcal{B} denote the set of all linear bases of V.  The group \GL (n) acts on \mathcal{B} from the right. If A\in \GL (n) and e\in \mathcal{B}, then e'=eA stands for the primed basis as discussed above. The action of \GL (n) on \mathcal{B} is transitive, effective, and free, so that \mathcal{B} is almost identical to \GL (n), except that \GL (n) has a distinguished “origin” – the identity matrix, while \mathcal{B} is a homogeneous space, with no distinguished point. With that in mind we can now proceed to define  simple “geometric objects” (tensors, tensor densities, etc.).

Let X be any set on which \GL (n) acts from the left. Denote by \rho this action. Thus, for \xi\in X,\, A\in \GL (n), we will  have \xi'=\rho(A)\xi.

Consider the Cartesian product \mathcal{B}\times X. This is the set of ordered pairs (e,\xi). Define the equivalence relation in \mathcal{B}\times X as follows:

(e,\xi)\sim (e',\xi') if and only if there exist  A\in\GL (n) such that e'=eA  and \xi'=A^{-1}\xi.

In other words

    \[ (e,\xi)\sim (eA,A^{-1}\xi).\]

Exercise 1. Verify that \sim defined above is indeed an equivalence relation.

The set of equivalence classes is denoted \mathcal{B}\times_\rho X. The elements of \mathcal{B}\times_\rho X.  are called “geometric objects of type \rho”. So, a geometric object is, loosely speaking, something that is represented, in every basis,  by an element of X, usually  a set of numbers, with a given consistent transformation rule which tells us how these numbers change when we change the basis.

Examples

Example 1. This is the basic example. Take X=\RR^n. Here \xi is a column of n real numbers. If A is in \GL (n), let \text{def} be the natural action of n\times n matrices on column vectors (the defining representation). What is \mathcal{B}\times_\rho \RR^n for \rho=\text{def}?

Exercise 2. Use Eqs. (1) and (4) to define a natural isomorphism between V and— \mathcal{B}\times_\rho \RR^n as in Example 1.

The next example is even simpler.

Example 2. Consider the trivial representation of \GL (n) on \RR. For \xi\in\RR, let \rho(A)\xi=\xi for all A\in \GL (n).

Another important example is by taking the contragradient representation.  Thus, the contragradient representation is given by composing with the inverse transpose.  Let X=\RR^n, but this time think of elements of \RR^n as rows of numbers, rather than columns, as it was in Example 1. We denote it \RR^{n*}. Let \text{def}^* be defined by:

    \[ \text{def}^*(A)\lambda = \lambda A^{-1},\quad \lambda\in\RR^{n^\star}.\]

Exercise 3. Verify that indeed we have a representation of \GL(n).

Exercise 4. With \rho=\text{def}&*, show that \mathcal{B}\times_\rho \RR^{n^\star} can be naturally identified with V^*.

To be continued…

 

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16 Responses to Tensors are geometric objects

  1. Bjab says:
    2025/12/11 at 13:58

    The upper index is always the column number. ->
    The lower index is always the column number.

  2. Bjab says:
    2025/12/11 at 14:01

    we will \xi'=\rho(A)\xi\, in X ->
    ?

  3. Bjab says:
    2025/12/11 at 14:03

    Define the equivalence relation in \mathcal{E}\times X as follows ->
    B

  5. Bjab says:
    2025/12/11 at 14:54

    have \xi'=\rho(A)\xi\, in X ->
    What is n X?

  6. Bjab says:
    2025/12/11 at 14:56

    In other words

        <span class="ql-right-eqno"> </span><span class="ql-left-eqno"> </span><img src="https://arkadiusz-jadczyk.eu/blog/wp-content/ql-cache/quicklatex.com-98ca5fe8f5f72cfe3577c476f902ef24_l3.png" height="26" width="175" class="ql-img-displayed-equation " alt="\[ (e,\xi)\sim (Ae,A^{-1}\xi).\]" title="Rendered by QuickLaTeX.com"/>

    ->
    Ae or eA?

  7. Bjab says:
    2025/12/11 at 14:58

    Strange things happen.

  8. Anna says:
    2025/12/14 at 11:34

    Ark, exercises are actually very useful but… a bit tedious 🙁
    By now I could do only the Exercise #3, the easiest one:
    Show that def*(A) is a representation of GL(n).

    We need to shaw that def* of composition A1A2 is the composition of def*(A1) and def*(A2). So,
    def*(A1A2)λ = λ(A1 A2)^-1 = λ(A2)^-1 (A1)^-1 = def*(A2) λ(A1)^-1 = def*(A2)def*(A1)

    Now thinking how to show rigorously that mathcal{B}times_rho RR^n for rho=text{def} is actually V and how to define a natural isomorphism between V and— mathcal{B}times_rho RR^n. Loosely speaking, it is just an isomorphism of vector spaces V–>V and the usual matrix action on a column gives one-to-one correspondence between a vector and its image.

  9. bayak says:
    2025/12/15 at 13:50

    Просто в тот момент мне показалось, что комментарий будет вам интересен. Но я ещё думаю надэтой темой.

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