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In what follows, $C$ will be a projective, geometrically regular, geometrically integral curve over a field $k$, and $\mathcal{L}$ is an invertible sheaf on $C$.

$19.2.A.$ EXERCISE. Suppose $\mathcal{L}$ is a degree $2g - 2$ invertible sheaf. Show that it has $g - 1$ or $g$ sections, and it has $g$ sections if and only if $\mathcal{L}\cong \omega_C$.

What does it mean $\mathcal{L}$ to have $g-1$ or $g$ sections?

Is it $h^{0}(C, \mathcal{L})= g-1$ or $g$? Would it be this?

Because by applying Riemann-Roch directly, we get

$h^0(C, \mathcal{L}) - h^1(C, \mathcal{L})= \text{deg} \mathcal{L} -g+1= g-1$.

But what about $h^1(C, \mathcal{L})$?

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    $\begingroup$ You are correct that you are trying to show $h^0(C,\mathcal L) = g-1$ or $g$. Riemann-Roch reduces you to showing $h^1(C,\mathcal L) = 0$ or $1$. This is what Serre Duality is for: it tells us that $h^1(C,\mathcal L) = h^0(C,\omega_C\otimes\mathcal L^\vee)$, where $\vee$ denotes the dual line bundle. Notice that $\deg (\omega_C\otimes\mathcal L^\vee) = 0$. Now, you should have proved a lemma at some point by now that a degree $0$ line bundle has a unique section if and only if it is isomorphic to $\mathcal O_C$, which in this case is equivalent to $\mathcal L \cong \omega_C$. $\endgroup$ – Tabes Bridges 2 days ago

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