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Hi,

Atiyah and Bott apparently proved the following theorem:

  • Let $X$ be a smooth projective complex variety and $L$ a line bundle on $X$. Let $f:X\to X$ be an automorphism of $(X,L)$ with finitely many fixed points $X^f$. Then $$ \sum_{i=0}^{\dim X}(-1)^itr(f, H^i(X,L)) = \sum_{x\in X^f}\frac{tr(f,L_x)}{\det(1-T_xf)} $$ where $T_xf : T_xX\to T_xX$ is the derivative of $f$ at $x\in X$.

Where can one find an algebraic proof of this result?

Thanks!

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2 Answers 2

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I am not sure this is the best place to learn the subject, but at least this book is an algebraic reference:

Riemann-Roch algebra By William Fulton, Serge Lang

more precisely VI \S 9 Lefschetz-Riemann-Roch . You can find your formula proven for an arbitrary vector bundle (not only a line bundle) under the name "fixed point formula". The machinery behind is quite heavy, tough, there is probably a more straightforward algebraic proof.

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    $\begingroup$ @Niels : in this reference, the formula is only proven for $f$ an automorphism of finite order. In that case, the formula can be proven without appealing to Grothendieck duality. $\endgroup$
    – Damian Rössler
    Commented Nov 15, 2011 at 8:29
  • $\begingroup$ @Damian : thanks for pointing this out. Do you have another reference for the case $f$ is of finite order besides Fulton-Lang and SGA 5 ? $\endgroup$
    – Niels
    Commented Nov 15, 2011 at 8:35
  • $\begingroup$ @Niels : there is Thomason, R.: "Une formule de Lefschetz en K-theorie equivariante algebrique". Duke Math. J. 68 and then the original article of Baum-Fulton-Quart, "Lefschetz-Riemann-Roch for singular varieties", Acta Math. 143. The first one uses a concentration theorem in equivariant K-theory, whereas the second one (like the book by Fulton-Lang) uses deformation to the normal cone. $\endgroup$
    – Damian Rössler
    Commented Nov 15, 2011 at 8:45
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Notice that you must assume that the graph of $f$ intersects the diagonal tranversally (otherwise some determinants in the formula might vanish). This transversality condition is automatic if $f$ has finite order. With that assumption, the above formula is a special case of the "Woods hole" formula, which is proven using Grothendieck duality in SGA 5 (Springer Lecture Notes in mathematics 589), Appendix to Exp. III, Cor. 6.12, p. 131.

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