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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