Consider bianchi map $$ b(T)(x,y,z,t) = \frac{1}{3}(T(x,y,z,t)+T(y,z,x,t) + T(z,x,y,t))$$ where $T\in S^2(\wedge^2 E)$
I already checked that $b(b(T))=b(T)\in S^2(\wedge^2 E)$
But how can we derive the following ?
$$ S^2(\wedge^2 E) = {\rm Ker}\ b \oplus {\rm Im}\ b$$ and
$${\rm Im}\ b = \wedge^4 E$$
The splitting $V = \mathrm{ker\ } T \oplus \mathrm{im\ } T$ occurs for any projection operator (i.e. idempotent) $b$ on a finite-dimensional vector space $V$. To see this note that any $v\in V$ can be written $v = (v-b v) + b v$ and apply the rank-nullity theorem.
To show that $\mathrm{im\ }b = \bigwedge^4 E$, you just need to show that $3b(T)_{ijkl} = T_{ijkl} + T_{jkil} + T_{kijl}$ is completely antisymmetric when $T \in S^2 \bigwedge^2 E$. Since you already know $b(T)$ has some antisymmetries (it is still in $S^2 \bigwedge^2 E$), you really just need to check that you pick up a sign change when you swap one of the first two indices with one of the last two. (This shows $\mathrm{im\ }b \subset \bigwedge^4 E$; for the other direction just show that $b$ acts as the identity on $\bigwedge^4 E$.)