Consider the following observation:

"The importance of tensors is due to the fact that they offer the opportunity of the coordinate-independent description of geometrical and physical laws. Any tensor can be viewed as a geometric object, independent on coordinates. Let us consider, as an example, the mixed $(1, 1)$-type tensor with components $A^j_{i}$. The tensor (as a geometric object) can be presented as a contraction of $A^j_{i}$ with the basis vectors \begin{equation} A = A^j_{i} e_{i}\otimes e_{j} . (1.18) \end{equation} The operation ⊗ is called “direct product”, it indicates that the basis for the tensor A is composed by the products of the type $e_{i}\otimes e_{j}$ . In other words, the tensor A is a linear combination of such “direct products”. The most important observation is that the Eq. (1.18) transforms as a scalar. Hence, despite the components of a tensor are dependent on the choice of the basis, the tensor itself is coordinate-independent."

There are two points that I didn't understand well:

1) when the autor says "can be presented as a contraction of $A^j_{i}$".

Where are the contraction operation?

2) And then: "eq. (1.18) transforms as a scalar. Hence, despite the components of a tensor are dependent on the choice of the basis, the tensor itself is coordinate-independent."

I mean, the object is a Tensor, not a scalar right? I don't know what the author wants to show saying "transforms as a scalar (...) the tensor itself is coordinate-independent."

Reference: Lecture notes on vector and tensor algebra and analysis by Shapiro I.L.

  • $\begingroup$ $\otimes$ is not the direct product, but rather the tensor product. There's a big difference (and many physicists seem unable to get it... Probably because they were never properly taught about these things :( ) $\endgroup$ – Danu Sep 18 '17 at 13:22
  • $\begingroup$ I know that. But as the text isn't mine I maintained direct product. $\endgroup$ – M.N.Raia Sep 18 '17 at 16:57

1) This is poor use of standard terminology. Usually a contraction means that you take the components of the object and perform a sum over them to reduce the rank of the tensor. So, for the tensor above it would mean taking $A_i^j$ and summing over the indices to create $A_i^i$ which is a scalar. Here the author seems to refer to the sum over the basis vectors as a contraction. This is non-standard terminology.

2) The author is using non-standard language here. When you say that something transforms as a scalar, vector, or type (1,1) tensor, what you really mean is that the components of that object mix among themselves in a certain way when you change the basis in which you express it. It doesn't make sense to talk about the transformation properties of the tensorial object itself. In fact, this is what the text is trying to emphasize: tensors are coordinate independent objects, only the components depend on the basis in which you express them. I think this is what the author is trying to say when they say that it transforms like a scalar (i.e. doesn't change), but this is confusing and non-standard language to use.

Finally, another thing to point out is that the author seems to be inconsistent in how they refer to basis vectors and dual vectors. The indices on $A$ are properly raised/lowered as to refer to a rank (1,1) tensor. However, the basis vectors both have their indices lowered. Normally this would have been written as $A = A_i^j e_i\otimes \omega^{\,j}$. In other words, the $e_i$ and $\omega^j$ belong to different spaces, one is a vector and one is a dual vector.

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For a simpler but analogous case, look at a vector, expanding into it basis components:

${\bf \vec{a}} = a_x{\bf\hat{e}}_x + a_y{\bf\hat{e}}_y + a_z{\bf\hat{e}}_z = a_i{\bf\hat{e}}_i$.

The final representation $a_i{\bf\hat{e}}_i$ is manifestly covariant, and since all indices are summed over, it is a frame independent object, which transforms like a scalar. Of course, it is a vector, and not a scalar-so calling it a scalar does cause confusion. The problem is, it is not a sum over numbers, it's a sum over numbers times basis vectors.

The important take away here, and it is a point of confusion in many explanations of scalar, vectors, and tensors, is that a scalar is NOT A NUMBER. It is a geometric object that is the same in all coordinates. Most of the time, it looks like a number and acts like a number, but it is not a number.

A real example would be the square of the electron spin--it's the square of a vector, so it's a scalar--and that's a number, right? Not so fast:

$ {\bf\vec{S}} = \frac{\hbar}{2}{\bf\vec{\sigma}}$


$ {\bf \vec{\sigma}} = \sigma_x {\bf\hat{e}}_x + \sigma_y {\bf\hat{e}}_y + \sigma_z {\bf\hat{e}}_z $,

and the $\sigma_i$ are 2 x 2 complex Pauli matrices with $\sigma_i^2 = \delta_{ij}$, so that:

${\bf\vec{S}}^2 = \frac{3}{4} \hbar^2 \delta_{ij}$

which is a 2 x 2 matrix, and definitely not a number, but it is still a scalar.

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