Lemma 20.2.56 (Reverse Bertini theorem)

Lemma 20.2.56 (Reverse Bertini theorem).A7BAA368EEF9 Let $X$ be a complex manifold, $f\colon X\rightarrow \Delta ^*$ be a projective surjective morphism to the punctured unit disk $\Delta ^*$. Let $(L,h)$, $(L,h')$ be Hermitian pseudo-effective line bundles on $X$ with the same underlying line bundle. Assume that there is a biholomorphic $S^1$-action on $(X,L)$ making $f$ equivariant and such that $h$ and $h'$ are invariant under this action. Assume that for quasi-every $z\in \Delta ^*$, $X_ z$ is smooth and $h|_{X_ z}\sim _{\mathcal{I}}h'|_{X_ z}$. Then $h\sim _{\mathcal{I}}h'$.

Proof. We need to show that $\mathcal{I}(kh)=\mathcal{I}(kh')$ for every positive integer $k$. It suffices to prove the case $k=1$, since our assumptions are invariant after replacing $h$ and $h'$ by $h^ k$ and $h'^ k$. We will therefore prove $\mathcal{I}(h)=\mathcal{I}(h')$. First observe that it suffices to prove that

133BE510F62F 20.2.57

\begin{equation} f_*\Bigl (K_{X}\otimes L\otimes \mathcal{I}(h)\Bigr )=f_*\Bigl (K_{X}\otimes L\otimes \mathcal{I}(h')\Bigr ) \end{equation}

as subsheaves of $f_*(K_{X}\otimes L)$. In fact, suppose that (20.2.57) holds. Let $H$ be a $f$-ample invertible sheaf on $X$. Then (20.2.57) also holds with $L\otimes H^ m$ in place of $L$. It follows from Grauert–Remmert's version of Serre vanishing theorem [Theorem 2.1(A), BS76] that $\mathcal{I}(h)=\mathcal{I}(h')$.

It remains to prove (20.2.57). Observe that both sides of (20.2.57) are locally free by [Corollary 1.5, Mat16]. We claim that it suffices to show that

C42AE95D23AF 20.2.58

\begin{equation} f_*\Bigl (K_{X}\otimes L\otimes \mathcal{I}(h)\Bigr ) \otimes \mathbb {C}(z)=f_*\Bigl (K_{X}\otimes L\otimes \mathcal{I}(h')\Bigr ) \otimes \mathbb {C}(z) \end{equation}

for quasi-every $z\in \Delta ^*$. In fact, this implies that the same holds outside a countable subset of $\Delta ^*$. But the set where (20.2.58) fails has to be $S^1$-invariant, it has to be empty.

By cohomology and base change together with the analytic Bertini theorem 20.2.48, for quasi-every $z\in \Delta ^*$, we have

\[ \begin{split} f_*\Bigl (K_{X}\otimes L\otimes \mathcal{I}(h)\Bigr ) \otimes \mathbb {C}(z)=\mathrm{H}^0\Bigl (X_ z,K_{X}|_{X_ z}\otimes L|_{X_ z}\otimes \mathcal{I}(h|_{X_ z})\Bigr ),\\ f_*\Bigl (K_{X}\otimes L\otimes \mathcal{I}(h')\Bigr ) \otimes \mathbb {C}(z)=\mathrm{H}^0\Bigl (X_ z,K_{X}|_{X_ z}\otimes L|_{X_ z}\otimes \mathcal{I}(h'|_{X_ z})\Bigr ). \end{split} \]

But we assumed that for quasi-every $z$, $h|_{X_ z}\sim _{\mathcal{I}}h'|_{X_ z}$, it follows that for quasi-every $z\in \Delta ^*$, (20.2.58) holds. The proof is complete. $\square$

Lemma 20.2.56 (Reverse Bertini theorem)`A7BAA368EEF9`

Comments

Lemma 20.2.56 (Reverse Bertini theorem)A7BAA368EEF9

Comments

Lemma 20.2.56 (Reverse Bertini theorem)`A7BAA368EEF9`