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Taking wedge products of (1.2) we have the exact sequence

0O(t+12n2k)α0O(tk1)α2n+k2O(tk)α2n+k12n+k1S(tk)0 \begin{align*} 0 \to \mathcal{O}(t+1-2n-2k)^{\alpha_0} \to \dots \to \mathcal{O}(t-k-1)^{\alpha_{2n+k-2}} \\ \qquad \to \mathcal{O}(t-k)^{\alpha_{2n+k-1}} \to \bigwedge^{2n+k-1} S(t-k) \to 0 \end{align*}

for suitable $\alpha_i \in \mathbb{N}$ and from this sequence we can conclude.

Ellia proves Theorem 3.5 in the case of $\mathbb{P}^3$ ([E], Prop. IV.1). He also remarks that the given bound is sharp. This holds on $\mathbb{P}^{2n+1}$ as it is shown by the following theorem, which points out that the special symplectic instanton bundles are the “furthest” from having natural cohomology. $\square$

Theorem 3.6. Let $E$ be a special symplectic instanton bundle on $\mathbb{P}^{2n+1}$ with $c_2 = k$. Then

h1(E(t))0 for 1tk2.h^1(E(t)) \neq 0 \text{ for } -1 \le t \le k-2.

Proof. For $n=1$ the thesis is immediate from the exact sequence

0O(t1)E(t)JC(t+1)00 \rightarrow \mathcal{O}(t-1) \rightarrow E(t) \rightarrow \mathcal{J}_C(t+1) \rightarrow 0

where $C$ is the union of $k+1$ disjoint lines in a smooth quadric surface. Then the result follows by induction on $n$ by considering the sequence

0E(t2)E(t1)2E(t)E(t)P2n100 \rightarrow E(t-2) \rightarrow E(t-1)^2 \rightarrow E(t) \rightarrow E(t)|_{\mathbb{P}^{2n-1}} \rightarrow 0

and the fact that, for a particular choice of the subspace $\mathbb{P}^{2n-1}$, the restriction $E|_{\mathbb{P}^{2n-1}}$ splits as the direct sum of a rank-2 trivial bundle and a special symplectic instanton bundle on $\mathbb{P}^{2n-1}$([ST] 5.9). $\square$

Remark 3.7. In [OT] it is proved that if $E_k$ is a special symplectic instanton bundle on $\mathbb{P}^5$ with $c_2 = k$ then $h^1(\mathrm{End},E_k) = 20k - 3$.

In the following table we summarize what we know about the component $M_0(k) \subset \mathrm{MI}_{\mathbb{P}^5}(k)$ containing $E_k$.

Table 3.10

h1(Ek ⊗ Ek*) h2(Ek ⊗ Ek*) dim M0(k) MIP5(k)
k = 1 14 0 14 open subset of P14
k = 2 37 0 37 smooth, irreduc., unirat.
k = 3 57 3 54 singular
k = 4 77 12 65 singular
k ≥ 2 20k - 3 3(k - 2)2 ? ?