Please take a look at the following:
Let $(M,g)$ be a four-dimensional oriented Riemannian manifold. The Hodge star operator $*$ obeys $**=Id$ acting on 2-forms. This allow us to decompose the vector space of 2-forms into eigenspaces of $*$: $$\Lambda^2(M)=\Lambda^2_{+}(M)\oplus\Lambda^2_{-}(M)$$ where a 2-form $\omega\in\Lambda^2_{\pm}(M)$ iff $*\omega=\pm\omega$
My questions are:
- How is this decomposition possible?
- Why are the eigenvalues $1$ and $-1$?
Thanks
The eigenvalues of any $\Bbb R$-linear involution have to be $\pm 1$. If $Tx=\lambda x$ and $T^2=Id$, then $x=T^2(x)=\lambda^2 x$ implies $\lambda=\pm 1$. (Actually this should remain true in any field with characteristic not $2$.)
The decomposition in this case is easy to work out. You have two eigenspaces $V=\{x\mid Tx=x\}$ and $W=\{x\mid Tx=-x\}$. Note that $y-Ty\in W$ for all $y\in M$, and $y+Ty\in V$ for all $y$. Thus $y=\frac12((y-Ty) + (y+Ty))\in W+V$ for arbitrary $y$. Finally it's easy to check that $W\cap V=\{0\}$. Thus $W\oplus V$ is a direct sum.