Serre's 1987 letter to Tate about mod p modular forms

Question

In what follows, we have a level $N \geq 3$, and the modular curve $X(N)$, and the invertible sheaf $\omega$ on $X(N)$ such that the global sections of $\omega^{\otimes k}$ correspond to modular forms of weight $k$ and level $N$.

In this letter, Serre looks at an exact sequence of sheaves:

$0 \rightarrow \omega^{k - p + 1} \rightarrow \omega^k \rightarrow \mathcal{S}_k \rightarrow 0$

where the second arrow is given by the multiplication with the Hasse invariant $A$, which is a modular form of weight $p-1$.

Question 1. Serre says that since $A$ vanishes with multiplicity 1 at the supersingular points, it follows that $\mathcal{S}_k$ is 0 outside of the supersingular points, and of dimension 1 at the supersingular points. What does he mean by that exactly? Does he means the stalks $\mathcal{S}_k, x$ at various points $x$ are either ${0}$ or $1$-dim $\mathbb{F}_p$-vector spaces?

Question 2. We let $S_k$ denote the global sections of $\mathcal{S}_k$. Serre shows that $S_k$ depends only on $k \pmod{p^2 - 1}$. He does it in the following way (or how i understand it): $S_k$ is supported at the supersingular locus, so he looks at the stalk at a point corresponding to a supersingular elliptic curve $E$. He says that $E$ has a canonical (and functorial) structure on $\mathbb{F}_{p^2}$ such that the Frobenius acts as $-p$. This means that the tangent space to $E$ also has a canonical $\mathbb{F}_{p^2}$ structure, and its $p^2-1$ tensor power has a canonical basis. This basis lets us identify the global differentials on $E$ of order $k$ with those of order $k + p^2 - 1$, and that this identification is respected by isogenies.

What I don't understand, is what he means here by "canonical basis", or how this basis lets us identify these two spaces of differentials $\omega^k(E)$ and $\omega^{k+p^2 -1}(E)$. Is it only $\omega^{p^2-1}(E)$ which has a "canonical" basis? Or why can't we simiarly identify $\omega^k$ and $\omega^{k'}$ for random values of $k$ and $k'$?

I would appreciate it if someone can explain this part of the proof to me.

Edit: This proof also appears in Edixhoven's paper on the weights in Serre's conjecture, and in Alex Ghitza's PhD thesis, but the explanations there do not give any further clarifications, they just repeat what Serre says. So maybe I am missing something very easy?

Answer 1

This kind of reasoning is now standard in the subject (that's a fact) but not easy (at least that's my opinion -- each time the word "canonical" is used in an essential way in an argument, things are not "easy").

For question 1, you understand well what Serre meant, and I have nothing to add. Concerning question 2, remember that all spaces considered have dimension 1, so a basis of them is just a non-zero element in them. Now while an elliptic curve $E$ has a tangent space $t_E$ of dimension 1 with a lot of non-zero elements (a.k.a bases), it has no canonical one. You need to make a choice to construct a non-tirvial element in it, for example, you can choose a Weierstrass equation for $E$ and consider the form $dx/y$ as an element of the dual $\omega_E$ of $t_E$ which gives you an element of $t_E$ (the dual basis). However, as Serre said, $\omega_E^{\otimes (p^2-1)}$ has a canonical basis, and so has its dual $t_E^{\otimes (p^2-1)}$, and thus also all their tensor powers (for if $x$ is a basis of vector space $V$, $x \otimes \dots \otimes x$ where $x$ appears $k$ times is a basis of $V^{\otimes k}$). The reason for this is that $Hom(\omega_E,\omega_E^{\otimes p^2})$ has a canonical non-zero element, the Frobenius. And this space is canonically equivalent to $\omega_E^{\otimes p^2} \otimes \omega_E^{\ast}=\omega_E^{\otimes (p^2-1)}$.