Suppose I have a divisor $D$ of degree $2g-2$ on a hyperelliptic curve of genus $g$. Then I can prove that either

a) $K_C\otimes\mathcal{O}(-D)=\mathcal{O}_C$, that is $K_C\cong \mathcal{O}(D)$, or

b) $K_C\otimes\mathcal{O}(-D)\neq \mathcal{O}_C$, but is a non-trivial degree 0 line bundle. Hence $h^0(C, K_C\otimes\mathcal{O}(-D))=0$. Hence $h^0(\mathcal{O}(D))=2g-2+1-g=g-1$, by Riemann Roch formula.

I am looking for an example of the following type. Say genus $g=2$. Let $i$ be the hyperelliptic involution. Can we find a divisor $D$ of the form $D=P+i(P)$ which satisfies (b). That is I want $h^0(C, K_C\otimes\mathcal{O}(-D))=0$. Is this possible.

Or more generally if $C$ is a hyperelliptic curve of genus $g$, is it possible to find a divisor $D=\Sigma_{j=1}^{g-1} p_j+\Sigma_{j=1}^{g-1} i(p_j)$ of degree $2g-2$ which satisfies b) or are all such divisors linearly equivalent to $K_C$?

I will be thankful for an example of the above type. Thanks in advance!

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    $\begingroup$ They are all linearly equivalent to the canonical bundle. $\endgroup$ – Mohan Dec 29 '15 at 19:32

As Mohan suggests, (b) cannot occur. Indeed, if $D$ is an effective divisor of degree $g-1$ on a hyperelliptic curve $C$, then $D + i(D)$ is a canonical divisor. Recall that the hyperelliptic class $H$ of $C$ is the divisor class of a fibre of the canonical map $C \to \mathbb{P}^1$, so it is represented by a point $P \in C$ plus its hyperelliptic involution $i(P)$. By Riemann-Roch, the canonical class $K_C$ is $(g-1)H$. Therefore, if $D = P_1 + \ldots + P_{g-1}$, then $$ D + i(D) = (P_1 + i(P_1)) + \ldots + (P_n + i(P_n)) = (g-1)H = K_C. $$ $$ \textrm{} $$

Edit: I've added an explanation of why the canonical class of a hyperelliptic curve is $g-1$ times the hyperelliptic class (the argument can also be found on this wiki).

Let $P \in C$ be a ramification point of the canonical map $C \to \mathbb{P}^{g-1}$, then the divisor class of $2P$ is the hyperelliptic class. Take the short exact sequence for $2P$, twist by $\mathcal{O}_C(2kP)$ for any $k \geq 1$, and take global sections to get $h^0(C,2(k+1)P) \geq h^0(C,2kP) + 1$. In particular, $h^0(C,(2g-2)P) \geq g$. (We have not yet used anything about the point $P$.)

Now, Riemann-Roch says that $$ h^0(C, (2g-2)P) - h^0(C,K_C - (2g-2)P) = \deg((2g-2)P) -g + 1 = g-1, $$ so $h^0(K_C - (2g-2)P) \leq 1$, but it cannot be equal to zero, otherwise the Riemann-Roch equality would not hold. It follows that $h^0(K_C - (2g-2)P) = 1$, but any divisor of degree-$0$ and $h^0 = 1$ is principal (see this question), hence $K_C$ and $(2g-2)P$ are linearly equivalent. That is, $K_C$ is $g-1$ times the divisor class of $2P$.

  • $\begingroup$ Thanks @msteve! Can you tell me how is it that we get $K_C$ is $(g-1)H$, using Riemann Roch? $\endgroup$ – gradstudent Dec 30 '15 at 1:58
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    $\begingroup$ It's briefly explained on this wiki but I'll add an edit later! $\endgroup$ – msteve Dec 30 '15 at 2:25
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    $\begingroup$ @Quasicoherent thank you - fixed now! $\endgroup$ – msteve Apr 30 '17 at 18:24

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