# Equivalent definitions of the Hasse invariant

As probably many others before me, I got stuck in verifying all the nice properties of the Hasse invariant. Let me start by recalling one definition:

Let $E\to S$ be an elliptic curve in characteristic $p$ and $V:E^{(p)}\to E$ the Verschiebung isogeny. We get an induced map of sheaves $\text{tg}(V): \underline{\omega}_{E/S}\to\underline{\omega}_{E^{(p)}/S}\cong\underline{\omega}_{E/S}^{\otimes p}$ which is the Hasse invariant $A=\text{tg}(V)\in \text{Hom}_{O_S}(\underline{\omega}_{E/S},\underline{\omega}_{E/S}^{\otimes p})=\text{H}^0(S,\underline{\omega}_{E/S}^{\otimes p})$.

I like this definition because of its direct link to the behaviour on ordinary curves, but I don't know how to compute the q-expansion directly with it. The way Katz-Mazur compute it is by claiming that if the elliptic curve E is equipped with a basis $\omega$ of $\underline{\omega}_{E/S}$ and $D$ is the corresponding invariant derivation, then $D^p=A(E,\omega)D$ where $D^p$ is the $p$th iterate of $D$.

How do we prove that this two definitions are equivalent? Thank you in advance for your help.

• The formal completion of the Tate curve along 0 is (via Raynaud's construction, upon which the proofs of the black-boxed properties in K-M all depend) is the formal completion of $\mathbf{G}_m$ along its identity section in such a way that (by comparison on cotangent spaces along identity section) the 1-form "$dq/q$" on the Tate curve goes over to $dt/t$ on $\mu_p$, whose tangent space over an $\mathbf{F}_p$-algebra agrees with that of $\mathbf{G}_m$. To deduce $A$ has $q$-expansion 1, by functoriality and base change compatibility of $V_{G/S}$ we use that $V_{\mu_p/\mathbf{F}_p}={\rm{id}}$. – nfdc23 Apr 15 '16 at 1:23