# 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. Commented Sep 4, 2016 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? Commented Sep 5, 2016 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. Commented Sep 5, 2016 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 $g=2$.