MathOverflow is a question and answer site for professional mathematicians. It only takes a minute to sign up.
Sign up to join this community
Suppose you have a projective manifold $M$, a very ample bundle $\scr L$ and a transverse holomorphic section $s \in H^0(\scr L)$. Then the zero set of $s$ is a complex submanifold $S_M$.
Can we have a embedding of the the projective manifold $M$ in some projective space such that image of $S_M$ will not be contained in a hyperplane? You can assume $M$ has complex dimension $2$.
For $M=P^2$ the projective plane and $L=S_M$ a line in $P^2$, there are simple examples of projective embeddings of $P^2$ where $L$ does not lie in a hyperplane.
To see this consider e.g. a generic 5-dimensional linear system $D$ in the 20-dimensional space of quintic curves on P^2. Then $D$ provides an embedding of P^2 into $P^5$, and $D$ is disjoint from the 14-dimensional space of reducible quintics that contain $L$ (which is isomorphic to the space of quartic curves), because $5+14 < 20$. Therefore no hyperplane section (no divisor in $D$) contains $L$.
(Here $Pic(M)=Z$ so this contradicts the comment by abx; the point is the linear system embedding $M$ here is not be complete.)
The same reasoning applies to any divisor $S$ of any projective variety $M$ in $P^n$ with $dim(M)>1$. To see this pick a finite set $T$ in $S$ with $|T| > 2 dim(M)+1$. Then for $d$ sufficiently large, the space of divisors of degree $d$ on $M$ passing through $T$ has codimension $|T|$ in the space of all divisors of degree $d$. (This is elementary.) A generic linear system $D$ of degree $d$ and dimension $N= 2 dim(M)+1$ will then contain no such divisors, and hence no divisors containing $S$; and it will embed $M$ in $P^N$ since it gives a generic projection of the Veronese embedding $M$ of degree $d$.
What you want is true if and only if there exists another very ample line bundle $\scr L'$ on $M$ such that $H^0(M, \, \scr L' \otimes L^{-1})=\{0\}$.
In fact, this condition is equivalent to say that no linear section in the embedding provided by $\scr L'$ splits as $S_M$ plus an effective divisor, that is precisely your request.
In particular, the answer is negative when the ample cone of $M$ is generated by $\scr L$, as explained in abx's comment.
For an example in the positive direction, take as $M$ the Hirzebruch surface $\mathbb{F}_e$, denote by $C_0$ the unique section of negative self-intersection and by $\mathfrak f$ the fibre, and set $$\mathscr{L} = C_0 + n \mathfrak{f}, \quad \mathscr{L}' = C_0 + m \mathfrak{f},$$ with $n > m > e$ (as usual, I do not distinguish between divisors and line bundles).
Then both $\scr L$ and $\scr L'$ are very ample by [Hartshorne, Algebraic Geometry, Theorem 2.17 p. 379], and we have $$H^0(\mathbb{F}_e, \, \mathscr{L}' \otimes \mathscr{L}^{-1}) = H^0(\mathbb{F}_e, \, (m-n) \mathfrak{f})=\{0\},$$ because $m-n <0$.
Edit. The "only if" part actually holds if we consider embeddings induced by complete linear systems, see C. McMullen answer below.
