A question on self-dual differential 2-forms This question is from Lemma 2 in Derdzinski's paper.
Let $$\omega=e_1\wedge e_2+e_3\wedge e_4, \eta=e_1\wedge e_3+e_4\wedge e_2, \theta=e_1\wedge e_4+e_2\wedge e_3$$
be a basis for self-dual two-forms on a four-manifold, where $\{e_1,e_2,e_3,e_4\}$ is an orthonormal basis of the tangent space. Derdzinski said that
$$\omega^2=\eta^2=\theta^2=-id,\ \omega\eta=\theta=-\eta\omega.$$ 
I was wondering what $\omega^2=-id$ and $\omega\eta=\theta$ mean? Is this wedge product or other operation? Thank you very much.
 A: The key to interpreting these formulas is the isomorphism Derdzinski mentions in part (b) of Lemma 2: 
$$
\Lambda^2 T_xM \cong \mathfrak s \mathfrak o(4).
$$
Note that $\Lambda^2 T_xM$ refers to the space of $2$-vectors (alternating contravariant $2$-tensors), not $2$-forms (which would be denoted $\Lambda^2 T_x^*M$ in his notation). 
What this isomorphism means is that a $2$-vector can be canonically interpreted as a skew-symmetric endomorphism of the tangent space. The endomorphism corresponding to a $2$-vector $\omega$ is the $(1,1)$-tensor $\omega^\sharp$ obtained by raising one index of $\omega$. That $\omega^\sharp$ is skew-symmetric is easiest to see in an orthonormal basis, where the matrix of $\omega^\sharp$ is either same as that of $\omega$ or its negative (depending on which index is raised). 
Using this canonical isomorphism, the products that Derdzinski writes are compositions of endomorphisms.  You can check this by writing the matrices of $\omega,\eta,\theta$ in terms of the basis $\{e_1,\dots,e_4\}$ and computing their matrix products. (You'll have to figure out which is the correct index to raise in order to get the signs right in these products.)
