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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 $2$.

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  • $\begingroup$ The answer depends on the embedding. This will not work if it is given by the global sections of $L$, since then $S_M$ is cut down by the hyperplane corresponding to $s$. Even worse if it is given by the global sections of $L^k$ for $k>1$, since then all sections which are multiple of $s$ give hyperplanes containing $S_M$. So if $\operatorname{Pic}(M)=\mathbb{Z}\cdot [L] $, the answer to your question is negative. $\endgroup$ – abx Jun 15 '18 at 7:03
  • $\begingroup$ Crossposted on MSE: math.stackexchange.com/q/2820345. $\endgroup$ – Alex M. Jun 22 '18 at 12:03
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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 $d$.

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  • $\begingroup$ You are perfectly right about my comment, I considered only complete linear systems. However I have some doubts about your last paragraph: having a lot of sections is not always enough to get an embedding. $\endgroup$ – abx Jun 16 '18 at 4:34
  • $\begingroup$ Is it obvious that $D$ induces an embedding and not only a birational map? $\endgroup$ – Francesco Polizzi Jun 16 '18 at 7:21
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    $\begingroup$ In the last paragraph, $M$ is already embedded by $O(1)$, so it is certainly embedded by $O(d)$, $d >> 1$ - this is just composing $M$ with the Veronese map on $P^n$. And once embedded, a generic projection to $P^N$, $N=2dim(M)+1$, is also embedded. So yes, it is obvious that $D$ (in the case of $P^2$, where $N=5$) and a generic $D$ in $|O(d)|$ of dimension $\ge N$ (in the general case) gives an embedding of $M$. $\endgroup$ – Curtis McMullen Jun 17 '18 at 17:19
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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.

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  • $\begingroup$ I'm confused. What if $\mathcal{L} = \mathcal{L'}^{\otimes 2} $ $\endgroup$ – meh Jun 15 '18 at 16:50
  • $\begingroup$ Well, I think it is still ok. For instance, suppose that $M$ is a curve of genus at least $3$, $\mathcal{L}=2K_M$ and $\mathcal{L}'=K_M$. Then no canonical divisor can entirely contain a bicanonical one, for instance by degree reasons. Am I missing something? $\endgroup$ – Francesco Polizzi Jun 15 '18 at 17:22
  • $\begingroup$ I'm sure I'm missing something obvious about grammar in your answer. As I'm reading your first statement, my comment is that the condition is true for the right choice of $\mathcal{L}$ on any surface. That seems to contradict what you and abx are saying. I'm sure at this point that I'm mis-reading your answer. $\endgroup$ – meh Jun 15 '18 at 17:59
  • $\begingroup$ I do not completely understand your last comment. We start from a fixed line bundle $\mathscr{L}$ (so there is no choice for this, the choice is only on $\mathscr{L}'$ ). Now, regarding your comment, we can find a $\mathscr{L}'$ such that $\mathscr{L}=\mathscr{L}' ^{\otimes 2}$ if and only if $\mathscr{L}$ is $2$-divisible in the Picard group, that is not always the case. For instance, this is surely impossible when $\mathscr{L}$ generates the Picard group, as in abx's comment. $\endgroup$ – Francesco Polizzi Jun 15 '18 at 18:51

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