# Comparing the norm of a trace of a curvature tensor with the full norm

Let $V$ and $E$ be complex vector spaces of dimensions $n$ and $r$, equipped with hermitian inner products $\omega$ and $h$ respectively. Let $R$ be a curvature-type tensor, that is an element of $\bigwedge^{1,1}V^* \otimes \operatorname{End} E$ that is hermitian. (This is just a local model of a curvature tensor of a rank $r$ holomorphic vector bundle $E \to X$, where $X$ is a complex manifold of dimension $n$.) I want to know if we have a Cauchy-Schwarz-type inequality $$\lvert \operatorname{tr}_\omega R \rvert_h^2 \leq \kappa \lvert R \rvert^2,$$ where $\kappa$ is some constant that depends on $n$ and $r$?

The reason I care is because I'm trying to write a coordinate-free proof of the Kobayashi-Lubke inequality for Hermite-Einstein vector bundles. I've got it down to writing the square of the norm of $R$ as a linear combination of the Chern classes $c_2(E)$, $c_1(E)^2$, $(\operatorname{tr}_\omega c_1(E))^2$ (all times the right power of $\omega$) plus that thing above. I now hope that all that's left is to estimate the norm of that trace and out will pop the Kobayashi-Lubke inequality.

If we pick orthonormal bases $(z_j)$ and $(e_\alpha)$ of $V$ and $E$ and write $$R = \sum_{j,k,\alpha,\beta} c_{j,k,\alpha,\beta} \; dz_j \wedge d\bar z_k \otimes e^*_\alpha \otimes \bar e^*_\beta$$ (so the hermitian condition on $R$ means that $\overline{c_{j,k,\alpha,\beta}} = c_{k,j,\beta,\alpha}$) then we get $$\lvert \operatorname{tr}_\omega R \rvert_h^2 = \sum_{j,k,\alpha,\beta} c_{j,j,\alpha,\beta}\, \overline{c_{k,k,\alpha,\beta}} \quad\hbox{and}\quad \lvert R\rvert^2 = \sum_{j,k,\alpha,\beta} c_{j,k,\alpha,\beta} \overline{c_{j,k,\alpha,\beta}},$$ but I'm not exactly sure how best to proceed.

I also don't like this coordinate-based approach and would much prefer to get my inequality by an abstract application of Cauchy-Schwarz, maybe to some weird "inner product" with values in a vector space. Any ideas here?

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