# Block Decomposition of a linear map on $\Lambda^2TM$

I'm trying a exercise from Peter Petersen's book, and I did the following:

Let $*$ be the Hodge star operator, I know that $\Lambda^2TM$ decompose into $+1$ and $-1$ eigenspaces $\Lambda^+TM$ and $\Lambda^-TM$ for $*$.

I know that, if $e_1,e_2,e_3,e_4$ is an oriented orthonormal basis, then $$e_1\wedge e_2\pm e_3\wedge e_4\in\Lambda^{\pm}TM$$ $$e_1\wedge e_3\pm e_4\wedge e_2\in\Lambda^{\pm}TM$$ $$e_1\wedge e_4\pm e_2\wedge e_3\in\Lambda^{\pm}TM$$

What i can't prove is:

Thus, any linear map $L:\Lambda^2TM\to\Lambda^2TM$ has a block decomposition $$\begin{bmatrix} A&B \\ C&D \end{bmatrix}$$

$$A:\Lambda^+TM\to\Lambda^+TM$$ $$B:\Lambda^-TM\to\Lambda^+TM$$ $$C:\Lambda^+TM\to\Lambda^-TM$$ $$D:\Lambda^-TM\to\Lambda^-TM$$

## 1 Answer

Presumably $M$ is a $4$-manifold here, since you are working with $\Lambda^2 TM$. If $M$ is $2n$-dimensional, the Hodge star gives rise to a decomposition $\Lambda^n TM = \Lambda^+ TM \oplus \Lambda^- TM$.

First, note that any $\omega \in \Lambda^2 TM$ can be written as $$\omega = \tfrac{1}{2}(\omega + \ast \omega) + \tfrac{1}{2}(\omega - \ast \omega) = \omega^+ + \omega^- \in \Lambda^+ TM \oplus \Lambda^- TM.$$ If $L: \Lambda^2 TM \longrightarrow \Lambda^2 TM$ is linear, we can once again write $$L(\omega) = L(\omega^+ + \omega^-) = L(\omega^+) + L(\omega^-).$$ But now both $L(\omega^+)$ and $L(\omega^-)$ can be decomposed with respect to the direct sum: $$L(\omega^+) = \tfrac{1}{2}(L(\omega^+) + \ast L(\omega^+)) + \tfrac{1}{2}(L(\omega^+) - \ast L(\omega^+)) = L(\omega^+)^+ + L(\omega^+)^-$$ and similarly $$L(\omega^-) = L(\omega^-)^+ + L(\omega^-)^-.$$ Then the block decomposition of $L$ is simply $$L = \begin{pmatrix} A & B \\ C & D \end{pmatrix},$$ $$A(\omega^+) = L(\omega^+)^+, \quad B(\omega^-) = L(\omega^-)^+,$$ $$C(\omega^+) = L(\omega^+)^-, \quad D(\omega^-) = L(\omega^-)^-.$$