I am currently reading about the non-Abelian Hodge correspondence. Let $(X,\omega)$ be a compact Kahler manifold. Given a Higgs bundle $(E, D\_0)$ on $X$, we want to construct the corresponding flat bundle, which after fixing a Hermitian metric $K$ on $E$ requires looking at when the curvature $\mathbf{F}\_K$ of the associated connection vanishes.  
  

I have seen multiple claims (c.f. [Simpson](http://www.numdam.org/item/PMIHES_1992__75__5_0.pdf) p16, [Garcia-Raboso--Rayan](https://arxiv.org/pdf/1406.1693.pdf) p21) that the Hodge-Riemann bilinear relations imply that there exist non-zero constants $C\_1,C\_2$ such that
$$\int\_X\mathrm{Tr}(\mathbf{F}\_K\wedge\mathbf{F}\_K)\wedge\omega^{n - 2} = C\_1\|\mathbf{F}\_K\|\_{L^2}^2 + C\_2\|\Lambda\mathbf{F}\_K\|\_{L^2}^2.$$
I am struggling to show this.  
  

The closest I have gotten is that if one defines
$$Q(\alpha,\beta) = \int\_X \mathrm{tr}(\alpha\wedge\beta^\dagger)\wedge\omega^{n-k}$$
on $A^k(\mathrm{End}E)$ (where $(-)^\dagger$ is defined on $A^0(\mathrm{End}E)$ through the metric $K$ on $E$, then extended to $A^k(\mathrm{End}E)$ by acting on the $\mathrm{End}E$ component) then the Riemann bilinear relations ought to apply as usual. Since $\mathrm{F}\_K^\dagger = -\mathrm{F}\_K$, it suffices to show that the above holds for $Q(\mathrm{F}\_K,\mathrm{F}\_K)$. One may decompose $\mathbf{F}\_K$ into its type composition, and then its Lefschetz decomposition. $\mathbf{F}\_K^{2,0}$ and $\mathbf{F}\_K^{0,2}$ are primitive by degree reasons, and $\mathbf{F}\_K^{1,1}$ has Lefschetz decomposition
$$\mathbf{F}\_K^{1,1} = (\mathbf{F}\_K^{1,1} - \frac{1}{n}L\Lambda \mathbf{F}\_K^{1,1}) + \frac{1}{n}L\Lambda\mathbf{F}\_K^{1,1}.$$
I then tried to use the orthogonality of $Q$ wrt the type and Lefschetz decompositions, and then use its agreement up to scalar with the $L^2$-norm on primitive pure type forms to get the result, but could not successfully deduce the claim. Any help would be appreciated; apologies if this is obvious.



Formula in non-Abelian Hodge theory - Hodge-Riemann bilinear relations
I have figured out the issue. I mistakenly claimed that $F\_h^\dagger = - F\_h$. In fact, what we really have is
$$(F\_h^{1,1})^\dagger = -F\_h^{1,1},\quad (F\_h^{2,0})^{\dagger} = F\_h^{0,2}.$$
Thus defining the Hermitian inner product
$$H(\alpha,\beta) = i^k\int\_X\mathrm{tr}(\alpha\wedge\beta^\dagger)\wedge\omega^{n - k}$$
on $A^k(\mathrm{End}E)$, we have that the Lefschetz and type decompositions are orthogonal for $H$, and that when $\beta$ is a primitive pure type form of type $(p,q)$, $k = p + q$, then
\begin{equation}\tag{1}\label{eq:eq1}
H(\alpha,\beta) = (-1)^{\frac{k(k+1)}{2}}i^{k + q - p}(n - k)!\langle \alpha,\beta\rangle\_{L^2}.
\end{equation}
Thus
\begin{equation\*}
\begin{aligned}
\int\_X\mathrm{tr}(F\_h\wedge F\_h)\wedge \omega^{n - 2} &= H(F, F^{2,0} + F^{0,2} - F^{1,1}) \\
&= H(F^{2,0}) + H(F^{0,2}) - H(F^{1,1} - \frac{1}{n}L\Lambda F^{1,1}) - \frac{1}{n^2}H(L\Lambda F^{1,1}) \\
&= (n - 2)!\left\{\|F - \frac{1}{n}L\Lambda F\|\_{L^2} - \frac{n - 1}{n}\|\Lambda F\|\_{L^2}\right\}
\end{aligned}
\end{equation\*}
where $H(\alpha)$ denotes $H(\alpha,\alpha)$ and throughout we have used the type and Lefschetz orthogonality relations, as well as \eqref{eq:eq1} and the fact that $\Lambda F = \Lambda F^{1,1}$ for degree reasons.  
  

Finally, using that $\Lambda$ is the adjoint of $L$ with respect to the $L^2$-norm, one can go from the last line of the above to the desired form.  
  

Moreover, for anyone interested in the final computation, this gives
$$\mathrm{tr}(F\_h\wedge F\_h).[\omega]^{n - 2} = (n-2)!\left\{\|F\_h\|\_{L^2} - \|\Lambda F\_h\|\_{L^2}\right\}.$$


