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(partly motivated by this question, but different: Degree of a Variety)

For a hyperelliptic curve $C$ of genus $g$ (over an algebraically closed field of characteristic not two) what is the smallest $d$ for which $C$ can be embedded in some $\mathbb{P}^n$ (I guess $n=3$ wlog) as a smooth curve of degree $d$? Does it depend only on $g$? Can anything be said for an arbitrary curve?

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Felipe, sorry what do you mean by "(I guess $n=3$ wlog)"? – Dmitri May 2 '11 at 22:58
I guess he means that you can always embed it in $\mathbb P^3$ without loss of generality. – diverietti May 2 '11 at 23:07
..since generic projections only decrease the degree. – J.C. Ottem May 2 '11 at 23:09
up vote 11 down vote accepted

Felipe, I believe the answer here is d=g+3. To see that you can embed your curve in this degree is straightforward - just choose a generic line bundle of degree g+3 and it will work.

In the case of hyperelliptic curves, I don't think you can do better. The key point is that any special linear series on a hyperelliptic curve comes from taking a multiple of the pullback of O(1) on P^1 together with some base points. (You can find this fact in Arbarello-Cornalba-Griffiths-Harris.) Since these cannot give rise to embeddings (either have base points or the associated map factors through the hyperelliptic involution) we conclude the the embedding line bundle has no H^1. The Riemann-Roch gives g+3 as the lower bound for having 4 sections.

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For $g=1$, one should get $d=3$. So I guess you are assuming $g>1$ ? – François Brunault May 2 '11 at 23:51
yes, that's hiding in the assumption that the embedding goes to $P^3$. A hyperelliptic curve of genus larger than 1 cannot be embedded in P^2, since a smooth plane curve of degree larger than 3 has very ample canonical bundle. – Tom Graber May 3 '11 at 6:09
@Tom G : thanks for the explanation. Even for $g=1$, it seems that $d=4$ is indeed the minimal degree among projective embeddings whose images are not planar. – François Brunault May 3 '11 at 10:57

I agree with Tom G in the case of hyperelliptic curves. Interestingly, I think the bound for a general curve of degree $g$ should be $d = (3/4)g+3$. More specifically, the Brill-Noether theorem tells us that, for this $d$, there is a line bundle with $\dim H^0(X, L) =4$. I would guess (but don't know a reference) that, for generic $X$, this line bundle gives an embedding $X \to \mathbb{P}^3$.

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According to Castelnuovo's bound (e.g. Theorem 6.4 in Chapter IV of Hartshorne), the degree of a smooth projective curve of genus $g$ in $\mathbb{P}^{3}$ (hyperelliptic or otherwise) is of degree at least $2\sqrt{g}+2$ (if the degree is even) or $\sqrt{4g+1}+2$ (if the degree is odd).

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I guess you mean non-plane curve here? – J.C. Ottem May 2 '11 at 23:24
Thanks for catching that! The curve is indeed assumed to be nonplanar. Also, the bound is attained for all degrees 3 or greater. – Yusuf Mustopa May 2 '11 at 23:27
Regarding David's answer: The relevant reference is Theorem 1.8 on p.216 of Arbarello-Cornalba-Griffiths-Harris, which implies that if $C$ is a general curve of genus $g$ and $r \geq 3,$ then a general point of $G^{r}_{d}(C)$ corresponds to an embedding of $C.$ Since all the line bundles in $W^{3}_{\frac{3}{4}g+3}(C)$ give complete linear series--this is because $W^{4}_{\frac{3}{4}g+3}(C)$ is the empty set--we have that $W^{3}_{\frac{3}{4}g+3}(C) \cong G^{3}_{\frac{3}{4}g+3}(C)$ for a general curve $C,$ so the aforementioned theorem applies. – Yusuf Mustopa May 3 '11 at 0:11
I should have said all the line bundles in $W^{3}_{\frac{3}{4}g+3}(C)$ give complete linear series \textit{of dimension 3}... – Yusuf Mustopa May 3 '11 at 0:15

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