Decomposition of a Cartesian tensor

Question

In "Modern Quantum Mechanics" by Sakurai J.J. he gives an example of a Cartesian tensor of rank $2$ which is a dyadic formed out of two vectors $\mathbf{U}$ and $\mathbf{V}$, i.e. $T_{ij} \equiv U_i V_j$. It obviously has $3^2=9$ components because each index can run from $1$ to $3$. However then they show that this dyadic can be decomposed into the following sum:

$$U_i V_j = \frac{\mathbf{U}\cdot\mathbf{V}}{3} \delta_{ij} + \frac{U_iV_j - U_j V_i }{2} + \left( \frac{U_i V_j + U_j V_i}{2} - \frac{\mathbf{U}\cdot\mathbf{V}}{3} \delta_{ij}\right)$$

It is then claimed that the first term is a 0-rank tensor (one component, scalar, invariant under rotation), the second term - tensor of rank 1 (three components) and the third term - traceless tensor of rank 2 (five components).

A couple of questions (I'm new to tensors, so pardon me if the questions are trivial):

1. Why the second term is a tensor of rank 1? Sure, since the only difference between $U_i V_j - U_j V_i$ and $U_j V_i - U_i V_j$ is the sign, one can claim that it has only three independent components (the binomial coefficient $\binom{3}{2} = 3$ doesn't care about the order). However that's not how the rank of tensor is usually defined. Why is it wrong to say that $(U_i V_j - U_j V_i)$ has 9 components? Three of them are zero, and the rest six are pairs of numbers with opposite sign. Or maybe 6 components, if we neglect the zeros?

2. Similarly, why the last term is a rank 2 tensor? First of all, from what I learned we can only add/subtract tensors of the same rank. But from what I see $(U_i V_j + U_j V_i)$ has at least six components ($(1,1), (1,2), (1,3), (2, 3), (2, 2), (3, 3)$) whereas $\frac{\mathbf{U}\cdot\mathbf{V}}{3} \delta_{ij}$ is the same scalar (0-rank tensor) which appears in the beginning of the decomposition.

3. The same question applies to the general composition: if these are tensors of different rank isn't it meaningless to add them? Maybe in reality each term should be multiplied (as in outer product) by some sort of a unit tensor (which increases the rank/order of the tensors) to make this decomposition mathematically precise?

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The same terminology is used in the book by A. Messiah. He writes:

$\mathbf{U} \otimes \mathbf{V}$ is reducible. The nine-dimensional space in which it is defined is the direct sum of three irreducible invariant subspaces (with respect to rotations), having respectively 1, 3 and 5 dimensions. The projections of $\mathbf{U} \otimes \mathbf{V}$ onto each of these subspaces are therefore irreducible tensors; they are, to within a constant, the scalar product $\mathbf{U} \cdot \mathbf{V}$, the vector $\mathbf{U} \times \mathbf{V}$ and the irreducible 5-component tensor which transforms under rotation like the harmonic polynomials of second degree.

Notice that he says "5-component" tensor. I assume that in reality the tensor has 9 components (i.e. it's a rank-2 tensor) but only 5 of them are independent (which actually has nothing to do with the rank of the tensor).

Answer 1

This is just different terminology to refer to transformation properties under rotation, rather than the total degree.

Basically $$ x^2+y^2+z^2 $$ is called a scalar, or rank-0 spherical tensor, even if it is quadratic in the components. In that way the nomenclature for spherical tensors is slightly different from that of general Cartesian tensors.

Thus a vector is a collection of 3 (non-zero) objects that transform under conjugation as $L=1$ states do under rotations.

Quoting Wigner in his Group Theory book:

.. an irreducible tensor operator of the $\omega$th degree... is defined by the condition that its $2\omega+1$ component $\textbf{T}^{(\rho)}$ transform under rotation of the axes as follows: $$ \textbf{O}_R^{-1} \textbf{T}^{(\rho)} \textbf{O}_R =\sum_{\sigma=-\omega}^{\omega} \boldsymbol{{\cal D}}^{\omega}(R)_{\rho\sigma} \textbf{T}^{(\sigma)}$$

where clearly Wigner uses $\rho$ and $\sigma$ to index components.