Independent components of trace?

Question

A $3 \times 3$ symmetric matrix has $6$ independent components: \begin{equation} \{ S_{ij} \} = \begin{pmatrix} S_{11} & S_{12} & S_{13} \\ S_{12} & S_{22} & S_{23} \\ S_{13} & S_{23} & S_{33} \end{pmatrix} \end{equation} If we decompose this matrix in a symmetric and traceless part: \begin{equation} \{ T_{ij} \} \equiv \begin{pmatrix} S_{11} & S_{12} & S_{13} \\ S_{12} & S_{22} & S_{23} \\ S_{13} & S_{23} & -(S_{11} + S_{22}) \end{pmatrix} \end{equation} then it has $5$ independent components. I believe this means that the trace: \begin{equation} S_{ii} = S_{11}+S_{22}+S_{33} \end{equation} must only have $1$ independent component? However, I do not see why this would be true, because if I would look (naively) at the above equation, then I would assume the trace has $3$ independent components. Can anybody explain me this apparent contradiction?

Thanks!

P.S. The motivation for this question is based on the decomposition of $\mathrm{SO(3)}$ representation in the antisymmetric part, the traceless symmetric part and the trace: \begin{equation} 3 \otimes 3 = 3 \oplus 5 \oplus 1 \end{equation}

Answer 1

To determine the matrix $S$, we need 6 independent components (degrees of freedom), where 5 out of 6 comes from $S_{ij}$ and the other is the trace. What we concern is how many degrees of freedom of the trace provides. And obviously it's 1 because $\mathrm{tr}S\in\mathbb F$, no matter how many independent components are required to determine the trace. Yes, the trace itself has 3 independent componets. The point is, those three components contributes to only one out of 6 numbers for $S$.

Answer 2

Your initial analysis about a symmetric matrix being decomposable into the trace and a trace-less symmetric matrix is correct. But, your "naive" description of the trace is not correct. First of all, the trace is independent of the basis, and is a sum of the eigenvalues. This can be seen by considering a diagonal matrix, $D$, and its similar matrix, $A$, by the transform, $O$, then

$$D = S^{-1}AS$$

and

$$\begin{align}\operatorname{tr}{D} &= \operatorname{tr}{(O^{-1}AO)} \\ &= \operatorname{tr}{AO^{-1}O} \\ &= \operatorname{tr}{A}. \end{align} $$

So, while your matrix $S$ has six independent components, they are not the matrix components directly. The trace constrains the diagonal elements, as you show by eliminating $S_{33}$ from the traceless component of $S$. $S$ could just as easily be written

$$\begin{align} S &= \frac{1}{3} I \operatorname{tr}{S} + \begin{pmatrix} S_{11} - \frac{1}{3}\operatorname{tr}{S} & S_{12} & S_{13} \\ S_{12} & S_{22} - \frac{1}{3}\operatorname{tr}{S} & S_{23} \\ S_{13} & S_{23} & S_{33} - \frac{1}{3}\operatorname{tr}{S} \end{pmatrix} \\ &= \frac{1}{3} I \operatorname{tr}{S} + \frac{1}{3} \begin{pmatrix} 2S_{11} - S_{22}-S_{33} & 3S_{12} & 3S_{13} \\ 3S_{12} & 2S_{22} - S_{11}-S_{33} & 3S_{23} \\ 3S_{13} & 3S_{23} & 2S_{33} - S_{11}-S_{22} \end{pmatrix} \end{align} $$

where the second form is more familiar in physics.

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To clarify this a bit. The trace is just a number, defined as the sum of the diagonal elements. In effect, you have one equation with four unknowns, implying that one of them is superfluous. Without further information, you can choose to eliminate any of the four.

The trace, however, has a nice property: it is basis independent, as mentioned above. Geometrically, this means the trace has spherical symmetry, corresponding to the s-shell in chemical orbitals, or mathematically, it is a one dimensional representation of $SO(3)$. None of the other diagonal components have this property; they only have it in aggregate through the trace.