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In a vague sense, what I want to ask is to what extent can the most general Riemann-Roch theorems be regarded as adelic duality statements?

For example, the Riemann-Roch theorem for a (proper, irreducible) curve follows from self duality of the adeles of its function field. This duality can either be proved directly for a completely adelic proof, or viewed as a simple consequence of Serre duality. Direct proofs (through residues) often require the assumption that the base field is perfect, but I'd like to neglect that here.

Self-duality of the additive group of the adeles of proper, irreducible surface $S$ over a field is sufficient to imply the following statement for a Cartier divisor $D$ on $S$:

$(D,C-D)=2(\chi(\mathcal{O}_S)-\chi(\mathcal{O}_S(D)))$,

where $C$ is the canonical divisor. This statement differs from the Hirzebruch-Riemann-Roch theorem by the following formula (Noether's formula):

$\chi(\mathcal{O}_S)=\frac{1}{12}(C^2+c_2)$,

where $c_2$ is the second Chern class. Is it reasonable to expect that this can be interpreted as a duality statement, perhaps of the K-groups of the adeles?

What then, of higher dimensions, and the Grothendieck-Riemann-Roch theorem?

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There is a link between duality for coherent sheaves and Grothendieck-Riemann-Roch. Consider the Lefschetz-Verdier formula for coherent sheaves (see SGA 5, Exp. III, Appendix) and (with the terminology of the appendix) let $X_1=X_2=X$, $Y_1=Y_2=Y$ and $C'=C''=$diagonal, $D'=D''$=diagonal. Then the formula gives (although this is difficult to show...) Grothendieck-Riemann-Roch with values in Hodge cohomology, ie $\oplus_{p}H^p(Y,\Omega^p)$. The Lefschetz-Verdier formula depends only on duality. –  Damian Rössler Jun 4 '13 at 19:06

1 Answer 1

The closest I have seen to what you are asking, is the paper

Hübl, Reinhold; Yekutieli, Amnon Adelic Chern forms and applications. Amer. J. Math. 121 (1999), no. 4, 797–839.

Not quite Riemann-Roch, but probably the right approach.

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