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I am working on a particular case of the following problem.

Let $X$ be a projective algebraic surface, $L$ a base point free invertible sheaf on $X$ and $\varphi:X\rightarrow \mathbb{P}^n$ the morphism induced by the complete linear system $|L|$. We denote by $W$ the image of $\varphi$. I was wondering if there is any way to compute de dimension of $W$.

This is what I have thought: If we take $m>>0$ then the dimension of $W$ coincides with the degree on $m$ of the Hilbert polynomial $h^0(\mathcal{O}_W(m))$ of $W$. Imagine that we are able to compute $h^0(L^{\otimes m})$. Since $\varphi^*\mathcal{O}_W(1)=L$, we obtain $h^0(\varphi^*\mathcal{O}_W(m))$.

So, my question is: Is there any relation between $h^0(\varphi^*\mathcal{O}_W(m))$ and $h^0(\mathcal{O}_W(m))$ (or at least if we impose certain conditions on our surface or our sheaf)? In case there is not, how could we approach this problem?

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    $\begingroup$ The way you say it, I think you do not want to assume that $X$ is projective. But if $X$ is, say affine, then $h^0(\phi^*\mathcal{O}_W(m))$ is not even a finite dimensional vector space, so I do not understand your question. $\endgroup$ – Mohan Jan 10 '18 at 19:32
  • $\begingroup$ Dear @Mohan, I want to assume that $X$ is projective as it is stated in the question now. Thank you. $\endgroup$ – Zé Pequeno Jan 11 '18 at 9:15
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    $\begingroup$ Once $X$ is projective, the image $\phi(X)$ is already closed, so not necessary to take the closure. The dimension can be easily determined by the self intersection. If $L=\mathcal{O}_X$, $\dim W=0$. If not, then $\dim W=1$ if $L^2=0$ and $\dim W=2$ if $L^2>0$. $L^2$ can not be negative. $\endgroup$ – Mohan Jan 11 '18 at 15:08
  • $\begingroup$ @Mohan I see that if $\dim W=2$, then we must have $L^2>0$. I also understand that $\dim W=0$ iff $L$ is trivial. Nevertheless I do not see why $L^2=0$ is enough for us to say that $\dim W=1$. Thank you. $\endgroup$ – Zé Pequeno Jan 12 '18 at 11:59
  • $\begingroup$ So, the only case to be avoided is when $\dim W=2$ and $L^2=0$. This says that $\phi$ is generically finite and so let $Z\subset W$ be the closed set where fiber dimension is positive. Let $H_1,H_2$ be general sections intersecting outside $Z$. Then easy to check that $0<\phi^*(H_1\cdot H_2)=(\phi^*H_1\cdot\phi^* H_2)=L^2$. $\endgroup$ – Mohan Jan 12 '18 at 14:45

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