# Different ways to construct maps and the tensor products of line bundles

Let $C$ be a curve. Then I know of two ways to create morphisms. To get morphisms from $C$, take a line bundle of any degree $L$ and use the linear system it determines to get a map into projective space, which may or may not be injective, so you get a map to another curve. To get a morphism to $C$, a particular case is to take a line bundle $\mu$ of order 2, which gives us a curve $\tilde{C}$ as a double cover of $C$.

How do these two methods interact? For instance, given a curve $C$, and the tower $\tilde{C}\to C\to \bar{C}$ with the first map given as a double cover by $\mu$, and the second map given by a line bundle $L$, presumably the geometry of this tower is connected to the line bundle $L\otimes\mu$, but it's not obvious to me how to connect the two notions to get any actual information.

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The genus of $\tilde{C}$ is 2(g(C)-1)+1. Unless $C$ has genus $1$, it's unlikely that you have a tower of the form you describe. Is the second $\tilde{C}$ different from the first? – Keerthi Madapusi Pera Feb 4 '11 at 15:42
The second is a $\bar{C}$ (C-bar, not C-tilde), and is the image of the map given by a line bundle. For instance, it may be $\mathbb{P}^1$ if the line bundle is a $g^1_d$. – Charles Siegel Feb 4 '11 at 16:07
I don't really see what you mean by "the geometry of this tower", could you give an example of the kind of thing you are looking for? You seem to be asking "What is the interaction between maps from a curve to a projective space and étale double covers of the same curve?". It doesn't seem to me that there should be any interaction at all between these two things. – Dan Petersen Feb 4 '11 at 17:48
Well, I'm asking about, say, what does knowing the geometry here (the double cover and the map to projective space) tell us about the tensor product of the two line bundles. As for what the geometry of such a tower might look like, there's work from the 80s by Recillas, Donagi, Debarre and others that study it when $\bar{C}=\mathbb{P}^1$ and the map is a $g^1_3$ or $g^1_4$, and the set up (called the n-gonal construction) is quite useful for a lot of things. Anyway, ideally, I'd like to be able to use an understanding of the two maps separately to compute the cohomology of $L\times\mu$. – Charles Siegel Feb 4 '11 at 18:49
That should be $L\otimes \mu$ – Charles Siegel Feb 4 '11 at 18:49

## 1 Answer

A possible answer is the following. Assume

$h^0(L)=n, \quad h^0(L \otimes \mu)=m$,

set

$\phi \colon \tilde{C} \to C, \quad \psi \colon C \to \bar{C} \subset \mathbb{P}^{n-1}$

and let $f=\psi \circ \phi \colon \tilde{C} \to \mathbb{P}^{n-1}$ be the composition.

Then $f^*\mathcal{O}_{P^n}(1)=\phi^*L$. On the other hand, by projection formula we have

$h^0(\phi^*L)=h^0(L) + h^0(L \otimes \mu)=n+m$,

and this shows that the complete linear system $|\phi^*L|$ induces a map $g \colon \tilde{C} \to \mathbb{P}^{n+m-1}$. Moreover, the map $f$ is obtained by composing $g$ with the projection from the linear subspace $\mathbb{P}^{m-1} \subset \mathbb{P}^{n+m-1}$ corresponding to the natural inclusion $\phi^* H^0(L \otimes \mu) \subset H^0(\phi^*L)$.

The easiest example is perhaps the following. Assume that $C$ is a genus $2$ curve and take $L=K_C$. Then $\bar{C}=\mathbb{P}^1$, $\tilde{C}$ is a hyperelliptic (this can be proven) genus $3$ curve and $\phi^*L=K_{\tilde{C}}$. We have

$h^0(L)=2, \quad h^0(L \otimes \mu)=1$

and the map $g \colon \tilde{C} \to \mathbb{P}^2$ is precisely the canonical map of $\tilde{C}$, which is a double cover of a conic $D \subset \mathbb{P}^2$. The map $f \colon \tilde{C} \to \mathbb{P}^1$ is a quadruple cover, obtained by composing $g$ with the projection of $D$ from the point in $\mathbb{P}^2$ corresponding to the non-zero section of $L \otimes \mu$.

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Thank you! This wasn't exactly what I was looking for, but in going through and confirming what you posted, I realized how what I am specifically looking at works. – Charles Siegel Feb 5 '11 at 3:53