algebraic proof of Atiyah-Bott fixed point formula? - MathOverflow most recent 30 from http://mathoverflow.net2013-05-23T13:51:44Zhttp://mathoverflow.net/feeds/question/80951http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/80951/algebraic-proof-of-atiyah-bott-fixed-point-formulaalgebraic proof of Atiyah-Bott fixed point formula?unknown2011-11-15T03:58:05Z2011-11-15T13:27:10Z
<p>Hi,</p>
<p>Atiyah and Bott apparently proved the following theorem:</p>
<ul>
<li>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 $x\in X$.</li>
</ul>
<p>Where can one find an algebraic proof of this result?</p>
<p>Thanks!</p>
http://mathoverflow.net/questions/80951/algebraic-proof-of-atiyah-bott-fixed-point-formula/80963#80963Answer by Niels for algebraic proof of Atiyah-Bott fixed point formula?Niels2011-11-15T08:18:45Z2011-11-15T08:18:45Z<p>I am not sure this is the best place to learn the subject, but at least this book is an algebraic reference:</p>
<p>Riemann-Roch algebra By William Fulton, Serge Lang</p>
<p>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.</p>
http://mathoverflow.net/questions/80951/algebraic-proof-of-atiyah-bott-fixed-point-formula/80964#80964Answer by Damian Rössler for algebraic proof of Atiyah-Bott fixed point formula?Damian Rössler2011-11-15T08:27:16Z2011-11-15T08:27:16Z<p>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.</p>