Exact holomorphic hyperplane sequences for negative twists

On a chart $Z_j\mathrm{\ne }0$ with $j\mathrm{\ne }0$, put $w=Z_0\mathrm{/}{Z}_j$. In local frames of $\mathcal{O}(m-1)$ and $\mathcal{O}(m)$, multiplication by $Z_0$ is multiplication by $w$ up to the invertible frame factor. The holomorphic local ring has no zero divisors, so this map is injective on stalks. The quotient map is restriction to $w=0$. If a holomorphic germ $f(u,w)$ restricts to zero there, then $f(u,0)=0$ and $f(u,w)=wg(u,w),g(u,w)={\int }_0^1\frac{\mathrm{\partial }f}{\mathrm{\partial }w}(u,tw)dt\mathrm{.}$ On a sufficiently small polydisc the integrand is holomorphic in $(u,w)$; its locally uniformly convergent power series may be integrated termwise, so $g$ is holomorphic and the germ lies in $(w)$. Thus the kernel of restriction is precisely the image of multiplication. On a chart not meeting $H$, the section $Z_0$ is nowhere zero, so multiplication is an isomorphism and the quotient sheaf is zero. These stalk calculations prove exactness of the first sequence for every $m$.

Tensor the exact $m=0$ sequence by the locally free line bundle $\mathcal{O}(-1)$. Tensoring is exact because on each stalk this bundle is a free rank-one module. By the transition rule in [F2], its restriction to $H$ is ${\mathcal{O}}_H(-1)$; hence $i_{\ast }{\mathcal{O}}_H\otimes \mathcal{O}(-1)\cong i_{\ast }{\mathcal{O}}_H(-1)$ on each stalk. The resulting exact sequence is $0\to {\mathcal{O}}_{{\mathbb{P}}^r}(-2)\to {\mathcal{O}}_{{\mathbb{P}}^r}(-1)\to i_{\ast }{\mathcal{O}}_H(-1)\to 0,$ as claimed.