Nakano positivity

Remarks

Taking $\tau_{j\lambda} = \xi_j v_\lambda$ recovers the condition of Griffiths positivity, so Nakano positive bundles are Griffiths positive. For $r = 1$ or $n = 1$ every tensor $\tau$ is of this form, and the two notions agree.

Example: $E = L^{\oplus r}$, where $L$ is a positive line bundle with curvature $i \sum a_{jk} dz_j \wedge d\bar z_k$, with the direct sum metric. Then $c_{jk\lambda\mu} = a_{jk} \delta_{\lambda\mu}$, and the sum is $\sum_\lambda \sum_{j,k} a_{jk} \tau_{j\lambda} \bar\tau_{k\lambda}$. Each inner sum is $\ge 0$ because $(a_{jk})$ is positive definite, and some inner sum is $> 0$ because $\tau \ne 0$.

Non-example: a trivial bundle $X \times \C^r$ with the standard metric has $\Theta = 0$, so the sum is $0$ for every $\tau$.

Nakano's vanishing theorem: if $X$ is compact Kähler and $E$ is Nakano positive, then $H^q(X, K_X \otimes E) = 0$ for $q \ge 1$, where $K_X$ is the canonical bundle.