samples/texts/1693838/page_5.md

Note also that $h^0(E(1)) = \text{syz}_1 B^t - k$ and $h^0(E^*(1)) = \text{syz}_1 A - k$.

Remark 1.11. In the same way we obtain

$$h^1(E\otimes E^{\ast }(-1))={\text{syz}}_{0}Ch^1(E\; \backslash otimes\; E^*(-1))\; =\; \backslash text\{syz\}\_0\; C$$

$$h^2(E\otimes E^{\ast }(-1))=2k(nk-2n-k)+{\text{syz}}_{0}C\mathrm{.}h^2(E\; \backslash otimes\; E^*(-1))\; =\; 2k(nk\; -\; 2n\; -\; k)\; +\; \backslash text\{syz\}\_0\; C.$$

2. Example on $\mathbb{P}^5$.

Let $(a, b, c, d, e, f)$ be homogeneous coordinates in $\mathbb{P}^5$.

Example 2.1. ($k=3$) Let

The corresponding monad gives a special symplectic instanton bundle on $\mathbb{P}^5$ with $k=3$. With the notation of remark 1.10, using [BaS] we can compute $\text{syz}_0 C = 14$, $\text{syz}_1 C = 174$. Hence $h^2(E \otimes E^*) = 3$ from the formulas of Remark 1.10. Moreover $h^0(E(1)) = 4$.

Example 2.2. ($k=3$) Let $B^t$ as in the Example 2.1 and

We have $\text{syz}_0 C = 10$, $\text{syz}_1 C = 171$. Hence $h^2(E \otimes E^*) = 0$. We can compute also the syzygies of $B^t$ and $A$ and we get $h^0(E(1)) = 4$, $h^0(E^*(1)) = 3$, hence $E$ is not self-dual.

Example 2.3. ($k=4$) Let