# Random fun fact with two totally different proofs

Here’s a fun fact: let $X$ be a smooth proper surface over a perfect field $k$ of positive characteristic $p > 0$, if there is no global $2$-form on $X$, then the Hodge-to-de-Rham spectral sequence (for $X$) degenerates at $E_1$ page. Below we will give two proofs that are totally different which builds on many (very) nontrivial results.

Here’s one proof. Look at the conjugate spectral sequence. It starts at $E_2$ page and is contained in the $[0,2] \times [0,2]$ box. By Serre duality,  both $E_2^{0,2}$ and $E_2^{2,0}$ are zero. Therefore by plain formalism of spectral sequence, it degenerates at $E_2$ page. Recall that the Hodge-to-de-Rham spectral sequence (for $X$) degenerates at $E_1$ page is equivalent to the conjugae spectral sequence degenerates at $E_2$ page, therefore we are done.

Here’s another proof. Again by Serre duality and the configuration of the spectral sequence, it suffices to show that the differential map $H^1(\mathcal{O}_X) \to H^1(\Omega^1_X)$ is zero. By Serre duality, we see that $H^2(\mathcal{O}_X) = 0$ which implies that the Picard variety of $X$ is reduced. Therefore the pullback map $H^1(\mathrm{Alb}(X), \mathcal{O}) \to H^1(X,\mathcal{O}_X)$ is an isomorphism, where $\mathrm{Alb}(X)$ is the Albanese variety of $X$. Since the differentials on any abelian variety is zero, we get what we want.

Maybe I should propose a vote with the question “which proof is better” and let people choose between “the first one”/”the second one”/”neither because you use a steam-hammer to crack nuts, fool!”.