Recall that $$ \otimes^2 E = \wedge^2 E \oplus S^2_0E\oplus {\bf R}$$
Clearly this is an $O(n)$-decomposition. Irreducibility can be checked from the following property:
Let $Ae_1=e_k,\ Ae_2=e_l,\ Ae_k=e_1,\ Ae_l=e_1,\ Ae_i=e_i\ (i\neq 1,\ 2,\ k,\ l)$ Then $$ A(e_1\otimes e_2 + e_2\otimes e_1)= e_k\otimes e_l+ e_l\otimes e_k$$
But some book [Gallot-Hulin-LaFontaine, Riemannian Geometry (1987), p.150] introduces classical invariant theory so that there exists following notations $$ \sum t_{ij}t^{ij},\ \sum t_{ij}t^{ji},\ ({\rm trace}\ (t))^2 $$
So they concluded that there exist at most three irreducible components. And note that this argument is used to decompose Riemannian curvature tensor.
How can we prove that?