Symmetric sums and Representations of SO(3) I had tried to help someone on math.StackExchange to prove the identity:
$$ (1-Tr(A))^2+\sum_{1\le i\le j\le 3}(a_{ij}-a_{ji})^2=4$$
I guess you could argue the left hand side is independent of basis.  Then we can diagonalize.  But I couldn't come up with an invariant way of expressing the 2nd term.  
Someone came up with $(1-\mathrm{Tr}\\,A)^2- \frac{1}{2}\mathrm{Tr}\left[(A-A^T)^2\right]=4$ and prove it.
This look a bit Pythagoras' theorem since $1 - Tr A = 2 \cos \varphi$ so the other term must be $\sum_{1\le i\le j\le 3}(a_{ij}-a_{ji})^2 = 2 \sin \varphi$ which I also couldn't prove.  
Is this sum related to a direct sum of representations of SO(3)?
 A: First, I think there is a typo in your question.  Maybe you mean $\sum_{1\le i\le j\le 3}(a_{ij}-a_{ji})^2 =4\sin^2\phi$?
Anyway, I think it is easier to work out these identities in $\mathsf{SU}(2)$, the double cover of $\mathsf{SO}(3)$.  Up to conjugation, $A=\mathrm{diag}(e^{i\phi},e^{-i\phi})$ in $\mathsf{SU}(2)$. When projecting to $\mathsf{SO}(3)$, one has to add 1 to the trace since the covering map sends $A$ into $\mathsf{SO}(2)\times 1$.  So $(1-\mathrm{tr}A)^2=(1-(1+e^{-i\phi}+e^{i\phi}))^2=4\cos^2\phi.$
Since the transpose of $A$ corresponds in $\mathsf{SU}(2)$ to the inverse $A^{-1}=\mathrm{diag}(e^{-i\phi},e^{i\phi})$ which simultaneously diagonalizes with $A$, the other part is $-\frac{1}{2}\mathrm{tr}(A-A^{-1})^2 =-\frac{1}{2}\mathrm{tr}(\mathrm{diag}(e^{i\phi},e^{-i\phi})-\mathrm{diag}(e^{-i\phi},e^{i\phi}))^2$ $=-\frac{1}{2}\mathrm{tr}\mathrm{diag}(2i\sin\phi,-2i\sin\phi)^2=-\frac{1}{2}\mathrm{tr}\mathrm{diag}(-4\sin^2\phi,-4\sin^2\phi)=4\sin^2\phi.$  Note that we did not have to add 1 to the trace  as in the first paragraph since we introduced a 1 in each of $A$ and $A^{-1}$ and subtracted them; resulting in the 1's canceling.
Thus, the Pythagorean theorem gives the result $(1-\mathrm{tr}A)^2-\frac{1}{2}\mathrm{tr}(A-A^{T})^2=4\cos^2\phi+4\sin^2\phi=4.$
Lastly, $\sum_{1\le i\le j\le 3}(a_{ij}-a_{ji})^2=-\frac{1}{2}\mathrm{tr}(A-A^{T})^2$ for any 3x3 matrix.  This follows from an easy computation observing that $A-A^T$ is anti-symmetric.
EDIT:  Just in case it was not clear.  The direct answer to your question is "no".  The sum is artificial in the problem since it holds for any 3x3 matrix; it has nothing to do with orthogonal matrices.
A: Using the parametrization
$$
  A=  \begin{pmatrix} a^2+b^2-c^2-d^2&2bc-2ad &2bd+2ac \cr
 2bc+2ad &a^2-b^2+c^2-d^2&2cd-2ab \cr
 2bd-2ac &2cd+2ab &a^2-b^2-c^2+d^2\\ \end{pmatrix}, 
$$
which comes from quaternions of unit length we have
$$
(1-tr (A))^2-\frac{1}{2} tr ((A-A^t)^2)-4=
$$
$$
(9a^2 + b^2 + c^2 + d^2 + 3)(a^2 + b^2 + c^2 + d^2 - 1)=0,
$$
since $a^2+b^2+c^2+d^2=1$.
