Riemann-Roch sur la courbe

Let $C/\mathbf{Q}_p$ be a complete algebraically closed extension, and let $X = X_{C^\flat}$ be the Fargues-Fontaine curve associated with $C^\flat$ . If $\mathcal{E}$ is any vector bundle on $X$ , the cohomology groups $H^i(X,\mathcal{E})$ vanish for all $i>1$ and are naturally Banach-Colmez Spaces for $i=0,1$ . Recall that the latter things are roughly “finite-dimensional $C$ -vector spaces up to finite-dimensional $\mathbf{Q}_p$ -vector spaces”. By a hard and wonderful theorem of Colmez, these Spaces form an abelian category, and they have a well-defined Dimension valued in $\mathbf{N} \times \mathbf{Z}$ which is (componentwise-) additive in short exact sequences. The Dimension roughly records the $C$ -dimension and the $\mathbf{Q}_p$ -dimension, respectively. Typical examples are $H^0(X, \mathcal{O}(1)) = B_{\mathrm{crys}}^{+,\varphi = p}$ , which has Dimension $(1,1)$ , and $H^1(X,\mathcal{O}(-1)) = C/\mathbf{Q}_p$ , which has Dimension $(1,-1)$ .

Here I want to record the following beautiful Riemann-Roch formula.

Theorem. If $\mathcal{E}$ is any vector bundle on $X$ , then $\mathrm{Dim}\,H^0(X,\mathcal{E}) - \mathrm{Dim}\,H^1(X,\mathcal{E}) = (\mathrm{deg}(\mathcal{E}), \mathrm{rk}(\mathcal{E}))$ .

One can prove this by induction on the rank of $\mathcal{E}$ , reducing to line bundles; the latter were classified by Fargues-Fontaine, and one concludes by an explicit calculation in that case. In particular, the proof doesn’t require the full classification of bundles.

So cool!

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