# Splitting of the 2-forms on a 4-dimensional vector space specifies unique conformal class

The problem I'm facing is the following.

Let $V$ be a 4-dimensional real vector space with orientation $\omega\in\Lambda^4V\setminus\{0\}$. Suppose that $U_-,U_+\subset \Lambda^2V$ are linear subspaces with the following properties:

1. $\Lambda^2V=U_-\oplus U_+$
2. $U_-$ and $U_+$ are wedge-orthogonal, i.e. $\alpha\wedge\beta=0$ for all $\alpha\in U_-$ and $\beta\in U_+$
3. $\alpha\wedge\alpha$ is a positive multiple of $\omega$ for all $\alpha\in U_+\setminus\{0\}$
4. $\alpha\wedge\alpha$ is a negative multiple of $\omega$ for all $\alpha\in U_-\setminus\{0\}$

In 'The Wild World of 4-manifolds' by Alexandru Scorpan (p. 364) it is now claimed that there exists a unique conformal class of Riemannian metrics on $V$ such that $U_-=\Lambda_-^2V$ and $U_+=\Lambda^2_+V$, where $\Lambda^2_{\pm}V$ denotes the self-dual/anti-self-dual 2-forms on $V$.

I haven't been able to prove or find a proof of this. I also wonder whether it is possible to explicitly construct a metric in the mentioned conformal class. Any help with this would be much appreciated!