# kahler differential on hyperelliptic curves

Suppose $X$ is a projective smooth, geometrically connected, hyperelliptic curves over a field k, we may ask $X(k)\neq\varnothing$. I want to know how to compute the $H^{0}(X,\Omega_{X}^{1})$.

And there is theorem of Noether, which claims that: Let $X$ be a non-hyperelliptic curve, then the map: $H^{0}(X,K_{X})^{\otimes2}\rightarrow H^{0}(X,K_{X}^{\otimes 2})$ is surjective.

Can someone explain intuitively why the theorem is true and why hyperelliptic curves are excluded.

• The failure of surjectivity is related to the failure of the canonical map $X \to \mathbb{P}(H^0(X, K_X))$ to be an embedding. If you start with a curve in the form $y^2 = f(x)$, the easiest way to see that this map is not an embedding is to show directly that the differential forms $\frac{dx}{y}, \ldots, \frac{x^{g-1} dx}{y}$ form a basis of $H^0(X,K_X)$ and then compute the canonical map directly. On the other hand, for non-hyperelliptic curves, some basepoint computations show that the canonical map is an embedding. – Jesse Silliman Sep 4 '16 at 6:23
• Thanks a lot! I am quite confused by the differential forms at the point over infinity, how to see it is ramified or not ramfied by the equation? – Bin Wang Sep 5 '16 at 6:33
• If you pull back those differential forms from the singular model $y^2 = f(x)$ to the smooth curve $X$, they are meromorphic sections $\omega$ of $K_X$, and we need to determine their behavior at the one (degree of $f$ is odd) or two (degree of $f$ is even) points at infinity. If there is one point, compute the degree of $\omega$. If there are two, also note that the differential forms are fixed by the hyperelliptic involution, so have the same order of vanishing at the two points at infinity. – Jesse Silliman Sep 5 '16 at 6:49

I just wanted to mention another way to recover the basis of holomorphic differentials and the failure of surjectivity given in the comments : an hyperelliptic curve of genus $g\geq 2$ can be expressed as a 2-to-1 cover of $\mathbb{P}^1$ ramified along $2g+2$ points, $f\colon X\to \mathbb{P}^1$. In particular, using double covers formulas we see that $$f_*\mathcal{O}_X = \mathcal{O}_{\mathbb{P}^1}\oplus \mathcal{O}_{\mathbb{P}^1}(-g-1) \qquad K_X \cong f^*K_{\mathbb{P}^1}\otimes f^*\mathcal{O}_{\mathbb{P}^1}(g+1) \cong f^*\mathcal{O}_{\mathbb{P}^1}(g-1)$$ so that $$H^0(X,K_X) = H^0(\mathbb{P}^1,f_*K_X) = H^0(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^1}(g-1)\otimes f_*\mathcal{O}_X) = H^0(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^1}(g-1))$$ and by the same reasoning $$H^0(X,K^2_X) = H^0(\mathbb{P}^1,f_*K^2_X) = H^0(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^1}(2g-2)\otimes f_*\mathcal{O}_X) = H^0(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^1}(2g-2)) \oplus H^0(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^1}(g-3))$$ Hence, the multiplication map $H^0(X,K_X)^{\otimes 2}\to H^0(X,K^2_X)$ can be interpreted as $$H^0(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^1}(g-1))^{\otimes 2} \to H^0(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^1}(2g-2)) \oplus H^0(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^1}(g-3))$$ which is clearly not surjective as soon as $g\geq 3$, whereas it is surjective for $g=2$. However, one can prove in the same way that the map $H^0(X,K_X)^{\otimes 3} \to H^0(X,K_X^{\otimes 3})$ is not surjective when $g=2$.