algebraic geometry - Help to translate this theorem to a more accessible language

I'm trying to understand the chapter 2 of this article.

I'm stuck in this part:

The theorem he mentioned is from this book and it is the following:

I need help to translate this theorem to a weaker one so that I could understand these comments in the article. My only background in Algebraic Geometry is Fulton's algebraic curves book.

Thanks.

Answer

The map

$$

\varphi: V \otimes H^0(C,\mathscr{F}) \to H^0(C, \mathscr{F} \otimes L)
$$

is just the map that sends $s \otimes t$ to the corresponding section of $\mathscr{F} \otimes L$ where $s$ is a section of $L$ contained in $V$ and $t$ is a section of $\mathscr{F}$. This is usually just written as a product $st$.

Now we get an exact sequence

$$
0 \to \ker \varphi \to V \otimes H^0(C, \mathscr{F}) \to \operatorname{im}\varphi \to H^0(C, \mathscr{F} \otimes L) \to H^0(C,\mathscr{F} \otimes L)/\operatorname{im}\varphi \to 0
$$

By the definition of the map the image is just

$$
H^0(C,\mathscr{F})s_1 + H^0(C,\mathscr{F})s_2
$$

and by the base-point-free pencil trick,

$$

\ker \varphi = H^0(C, \mathscr{F} \otimes L^{-1}(B)).
$$

So taking dimensions of the terms in the exact sequence, we have

$$
\operatorname{codim} (H^0(C,\mathscr{F})s_1 + H^0(C, \mathscr{F})s_2) =
$$
$$
\dim H^0(C,\mathscr{F} \otimes L) - \dim (V \otimes H^0(C, \mathscr{F})) + \dim H^0(C, \mathscr{F} \otimes L^{-1}(B)).

$$

since $V$ is two dimensional, $\dim (V \otimes H^0(C, \mathscr{F})) = 2 \dim H^0(C, \mathscr{F})$ so we get

$$
\operatorname{codim} (H^0(C, \mathscr{F})s_1 + H^0(C, \mathscr{F})s_2) =
$$
$$
h^0(C, \mathscr{F} \otimes L) - 2 h^0(C, \mathscr{F}) + h^0(C, \mathscr{F} \otimes L^{-1}(B))
$$

where $h^0(\ldots) = \dim H^0(\ldots)$.

Now this is applied to the special case $\mathscr{F} = \Omega^{n-1}(F - D)$ and $L = \Omega^1(D)$ where $D := \inf\{\operatorname{div}(\omega_g),\operatorname{div}(\omega_{g-1})\}$. By the definition of $D$, we have that $\omega_g, \omega_{g-1} \in H^0(C,L)$ and so we take $V$ to be the span of these two.

Then we have that $\mathscr{F} \otimes L = \Omega^{n}(F - D + D) = \Omega^n(F)$ since the product of a one-form and an $n-1$ form is an $n$-form, and if $\operatorname{div}(s_1) \geq F - D$ and $\operatorname{div}(s_2) \geq D$, then $\operatorname{div}(s_1 s_2) = \operatorname{div}(s_1) + \operatorname{div}(s_2) \geq F - D + D = F$.

And so the codimension of $\Omega^{n-1}(F - D)\omega_{g-1} + \Omega^{n-1}(F - D)\omega_g$ in $\Omega^{n}(F)$ is equal to

$$

\dim \Omega^n(F) - 2 \dim \Omega^{n-1}(F - D) + \dim \Omega^{n-1}(F - D) \otimes L^{-1}(B)
$$

by the statement on dimensions above (and some abuse of notation). Now $B = D$ by the definition of $D$ as the infimum of the divisors of $w_g, w_{g-1}$ and

$$
L^{-1}(D) = (\Omega^1(D) \otimes \mathcal{O}(D))^{-1} = (\Omega^1 \otimes \mathcal{O}(2D))^{-1} = (\Omega^1)^{-1} \otimes \mathcal{O}(-2D)
$$

and so

$$
\Omega^{n-1}(F - D) \otimes L^{-1}(B) = \Omega^{n-1}(F - D) \otimes (\Omega^1)^{-1} \otimes \mathcal{O}(-2D) = \Omega^{n-2}(F - 2D)
$$

as giving the required formula.

Now all of this stuff about differentials $\Omega^n(F)$ etc seem to be in Fulton so I hope that makes sense. The only thing I don't see there is the base locus, so the base locus $B$ of sections $s_1$ and $s_2$ is exactly the divisors where $s_1$ and $s_2$ both vanish.

In the case of $\omega_g, \omega_{g-1}$ above, this is the locus where both differentials have order of vanishing greater than $0$ and this is exactly $D$ as defined.