Tensors on a picnic

Posted on 2025/12/14 by arkajad

This series of posts is meant as a “gentle introduction to …”. It’s so gentle that some readers may even find it boring. They might say, “All this is well known—get to the point!” But I can’t think of another approach that would give me the same pleasure in writing for my small audience. That’s simply how I see it. And of course, there are always a few little mistakes here and there. I seem to suffer from a mild form of dyslexia: even when reading my own text, I often fail to notice my errors. I apologize in advance and thank my faithful readers for their understanding and help.

In the previous post, we introduced the simplest geometric entities: vectors and covectors. From these humble beginnings, we’ll start building more intricate structures—tensors—by taking tensor products and then pruning them through symmetry operations such as symmetrization and antisymmetrization. But before diving into that algebraic craftsmanship, let’s pause to make a few relevant remarks.

A note on higher-order objects.

Our main interest lies in conformal structures on manifolds. Roughly speaking, a conformal structure is an equivalence class of metrics, where each metric defines a scalar product on the tangent space at every point. Two metrics are considered equivalent if they differ only by a positive scalar factor—one that may vary smoothly from point to point. Since this idea repeats identically at each point, we can focus our attention on a single tangent space and treat the discussion as a matter of pure algebra. This is exactly what we’re doing here: studying “geometric objects” at a point.

However, the geometric landscape is richer than it first appears. Not all objects on a manifold fit neatly under our current umbrella.

Tensors

Vectors, covectors, tensors, and tensor densities all belong to the realm of first-order objects, whose transformation laws involve only the first derivatives of coordinate transformations. Yet there also exist higher-order objects, whose transformations depend on higher derivatives. The coefficients of an affine connection provide a classic example—they are not tensors, but rather second-order objects, transforming under a more intricate symmetry group than GL(n).

Consider for instance V\otimes V^\star\otimes V. In a basis e_i elements of this space are represented by their components  {{\xi^i}_j}^k, so that:

    \[ \xi ={{\xi^i}_j}^k\,e_i\otimes e^j\otimes e_k.\]

If we change the basis: e_{i'}=A^i_{i'}e_i, the components change according to the rule:

    \[ \xi^{i'\phantom{j'}k'}_{\phantom{i'}j'}=A^{i'}_iA^j_{j'}A^{k'}_k\,\xi^{i\phantom{j}k}_{\phantom{i}j}.\]

We recall here that the matrices A_i^{i'} and A_{i'}^i are inverse to each other. This transformation law corresponds to the representation \rho={\text{def}\otimes\text{def}^\star\otimes\text{def} of \GL(n) on V\otimes V^\star\otimes V.

We call the upper indices contravariant, the lower indices – covariant. The total number of indices is called the rank of a tensor. Tensors of zero rank are called scalars. We will be particularly interested in totally symmetric and totally antisymmetric,  tensors, usually covariant ones. On manifolds totally antisymmetric tensors represent differential (or exterior) forms.

Example 1.

Let consider a simple example of doubly covariant tensor. Suppose that V is equipped with a bilinear form B. For each basis e_i let b_{ij} be the matrix

    \[ b_{ij}= B(e_i,e_j).\]

If we change the basis, then b_{ij} transform as:

    \[b_{i'j'}=A_{i'}^iA_{j'}^j\,b_{ij}.\]

Therefore we have a symmetric tensor. We can think of it as an element b of V^\star\otimes V^\star, more precisely, of its symmetric part. The numbers b_{ij} are components of b in the basis (e_i). The bilinear form is symmetric if and only if b_{ij}=b_{ji} in one basis. And, if it is symmetric in one basis, it is symmetric in any other basis.

Exercise 1. Prove this last statement.

Example 2.

There is a very special tensor, we may call  Id, 1-covariant, and 1-contravariant. In every basis it has the same components \delta_i^j – the Kronecker delta.

Exercise 2. Show that if  Id has Kronecker delta as components in one basis, it has also Kronecker delta as components in any other basis.

If we have two tensors, S of rank s, and T of rank t, we can always multiply the components to produce a new tensor of rank s+t. For instance if S has components {S_i}^j, and T has components {T^i}_{jk}, we can form U with components

    \[ {{{U_i}^{jk}}_{lm}={S_i}^j {T^k}_{lm}\]

.
If we want, we can arrange the components of U differently and write {{{U_{ilm}^{jk}} on the left hand side. But the notation used should be consistent. This follows from the fact that if we have two vectors spaces V_1 and V_2, then, even though V_1\otimes V_2 and V_2\otimes V_1 are two different vector spaces, there is a natural isomorphism between them.-

Contraction

If we have a tensor, say U, of rank u,  with at least one covariant, and at least one contravariant index, we can take a sum over this index to get a new tensor of rank u-2. For example we can form

    \[ ${{{U_{lm}^{k}}$ ={{{U_{ilm}^{ik}}.\]

Exercise 3. Show that \delta^i_i=n, the dimension of V. Here n is understood as a tensor of rank zero, a scalar.

To be continued …

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21 Responses to Tensors on a picnic

  1. Bjab says:
    2025/12/15 at 09:41

    Is formula under:
    “components change according to the rule:”
    correct? (primes)

  2. Bjab says:
    2025/12/15 at 09:42

    The formula under:
    “then b_{ij} transform as:”
    has not been interpreted.

  3. Bjab says:
    2025/12/15 at 09:43

    we may call id Id, 1-covariant, ->
    id duplicated

  4. Bjab says:
    2025/12/15 at 09:44

    “Contraction” ->
    Title should be lower

  5. Bjab says:
    2025/12/15 at 14:07

    it has the same components \delta_j^j – the Kronecker delta. -?
    \delta_i^j

  6. John G says:
    2025/12/16 at 01:26

    Your gentle math is not something I can do but your gentle terminology review is helpful. I like to think of a Cl(4)xCl(4) tensor product for Hodge Star map purposes and at least for today before I forget again the second Cl(4) fits with covector, covariance, cotangent space/bundle and differential form for me.

    • John G says:
      2025/12/17 at 05:41

      Below from Tony Smith’s website has covector and commutation terms and you come across those other terms when looking into it; it kind of starts out like defining something with something else you need to define. As an electrical engineer, the bit-inversion part I could get into much more easily. Physics is kind of just bits to me. I don’t actually do algebra. Engineers do get basic matrix algebra but it’s kind of too basic.

      Position-Momentum Duality:
      Dennis W. Marks in his paper A Binary Index Notation for Clifford Algebras
      (revised 27 February 2003) said: “… Duality operations generate isomorphism
      between grades k and n-k. There are several different duals, including … the
      Clifford dual … and the Hodge dual … A dual distinct from the preceding duals is introduced by bit inversion that maps e_m -> e_mbar , where mbar is the bit inverse of m … In particular, bit inversion transforms vectors (grade 1) … into covectors (grade n-1) … The bit inverse of the bit inverse is the original element … Bit inversion does not …[ depend on the handedness of the base coordinate system ]… Complementarity between space-time and momentum-energy is achieved by bit inversion, which interconverts between position representation and momentum representation. Treating momentum as a Clifford covector has the virtue of automatically enforcing the Heisenberg commutation relation as a consequence of the commutation and anti-commutation properties of the Clifford elements. …”.

  7. Anna says:
    2025/12/21 at 10:39

    Ark, thank you for such a gentle dive into the depths of maths. This is exactly what is needed: every experienced diver moves smoothly, stopping to verify one’s instruments and sensations. So often we slide easily over the surface and don’t notice the most interesting and essencial underlying aspects of reality. The exercises you give in the post are very a good example: they are not standard at all and do highlight some pecularities i never noticed in my student years.

    As usual, I have started from the end, Ex. 3 is a piece of cake: \delta^i_i=n, t
    We just summarize n units and that is that.

    Handling Exs.2 and 1 now…

    • arkajad says:
      2025/12/21 at 10:51

      Anna,
      Thanks for your feedback. It is extremely useful for me. Working on a new post, it gives me some orientation concerning what to write and how to write. There is always this hesitation concerning what to write and what to skip. On one hand I would like to skip a lot, to get to the ,in point quickly. On the other hand I feel guilty for skipping what seems to be important, but it requires a detour from the main path. With your feedback I recall myself the saying “Happiness is a journey, not a destination.” Thanks.

  8. Anna says:
    2025/12/23 at 06:55

    One more test. I am going to use latex now

        \[x=y\]

        \[ y = z \]

    This proves that x=z

  9. Anna says:
    2025/12/23 at 12:02

    Exercise #1.
    I will use k for i’ and l for j’ in order to avoid primes.
    By definition:

        \[b_{kl} = A_{k}^i A_{l}^j b_{ij}\]

        \[b_{lk} = A_{l}^j A_{k}^i b_{ji}\]

    In order to prove that b_{kl} = b_{lk} when b_{ij} = b_{ji} we only need to show that

        \[A_{k}^i A_{l}^j = A_{l}^j A_{k}^i\]

    Let us insert A_{j}^k in between of the factors on both sides:

        \[A_{k}^i A_{j}^k A_{l}^j =? A_{l}^j A_{j}^k A_{k}^i (*)\]

    Since A_{i}^k A_{k}^j = A_{k}^i A_{j}^k = \delta_{i}^j,
    we can use it in (*) to obtain:

        \[\delta_{j}^i A_{l}^j =? \delta_{l}^k A_{k}^i\]

    Due to Kronecker symbol, on the left survives only A with j = i and on the right survives only A with k = l:
    So, we get

        \[A_{l}^i = A_{l}^i\]

    and that is that.

    • arkajad says:
      2025/12/23 at 12:41

      Really you can’ tuse primes? If you use {i’}, always in curly brackets, like that, it should work. Or is a prime on your keyboard different than my prime? Try it. I want to understand the problem,

  10. Anna says:
    2025/12/23 at 13:16

    Exercise #2. (Checking primes)

    By tranformation law:

        \[b_{i'}^j'= A_{i'}^iA_{j}^j'b_{i}^j\]

    If b_{i}^j = \delta_{i}^j then

        \[b_{i'}^j'= A_{i'}^iA_{j}^j'\delta_{i}^j=A_{i'}^iA_{i}^j'= \delta_{i'}^j'\]

  11. Anna says:
    2025/12/23 at 13:19

    Exercise #2. Checking primes – now with extra bracket around {j’}

    By tranformation law:

        \[b_{i'}^{j'}= A_{i'}^iA_{j}^{j'}b_{i}^j\]

    If b_{i}^j = \delta_{i}^j then

        \[b_{i'}^{j'}= A_{i'}^iA_{j}^{j'}\delta_{i}^j=A_{i'}^iA_{i}^{j'}= \delta_{i'}^{j'}\]

    • arkajad says:
      2025/12/23 at 13:23

      Thank you ! I feel somewhat better now.
      But I am feeling really depressed, because I am having a nasty contradiction with my calculations, and am fighting with it already third day in a row! And no light yet!

      • Anna says:
        2025/12/23 at 15:14

        You mean your calculations for the Blog?! If they are so intricate, skip them over, we cannot continue our picnic in a nasty mood. Expecially on the Christmas Eve! A few days off, and you’ll return to the contradictions and probably discover they have gone…

        • arkajad says:
          2025/12/23 at 15:39

          No, at the moment they are for a paper that I considered almost finished, just needed to remove what is unnecessary, and verify the spelling. A catastrophe.
          BTW , before sleep I am reading a book by some Andreev (Shevcov), Mir Tropy, that someone mentioned on the chat during the conference. Very interesting, about what it means being Russian, digging for the roots, talking about traditions. Is it a popular book among scientist?

          • Anna says:
            2025/12/23 at 16:14

            In my opinion, this folk magic exists in every nation, not just Russians. Something akin to Carlos Castaneda’s investigations. Rather interesting, indeed, thank you for showing it to me. As far as I can tell, it’s not very popular right now here, but who knows? Things are changing quickly.

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