Friday, 23 August 2013

ag.algebraic geometry - Why do Todd classes appear in Grothendieck-Riemann-Roch formula?

Yes you are right! You can in fact prove that the Todd class is the only cohomology class satisfying a GRR-type formula.



Indeed, assume that for any smmoth quasiprojective variety $X$, you have an invertible cohomology class $alpha(X)$ satisfying that:



(i) for any proper morphism $f colon X rightarrow Y$ between smooth quasi-projective morphism and for any bounded complex $mathcal{F}$ of coherent sheaves on $X$, $f_{*}(ch(mathcal{F})alpha(X))=ch(f_{!}mathcal{F})alpha(Y)$.



(ii) for any $X$ and $Y$, $alpha(X times Y)=pr_1^*alpha(X) otimes pr_2^*alpha(Y) $
(this is a kind of base change compatibility condition).



Then for any $X$, $alpha(X)$ is the Todd class of $X$.
In fact, it is sufficient to know (i) for closed immersions and (ii) for $X = Y$.



Here is a quick proof:



1-First you prove GRR for arbitrary immersions. This is done in two steps:



(a) $Y$ is a vector bundle over $X$ and f is the immersion of $X$ in $Y$, where $X$ is identified with the zero section of $Y$. Then $mathcal{O}_X$ admits a natural locally free resolution on $Y$ which is the Koszul resolution. Then a direct computation gives you that $ch(mathcal{O}_X)$ is the Todd class of $E^* $, which is therefore the Todd class of the conormal bundle $N^*_{X/Y}$. Thus the Todd class pops out this computation just like in the divisor case.



(b) For an arbitrary closed immersion $f colon X rightarrow Y$, a standard deformation technique (which is called deformation to the normal cone) allows to deform $f$ to the immersion of $X$ in its normal bundle in $Y$, and then to use part (a)



2-Then you compare the two GRR formulas you have for the diagonal injection $delta$ of $X$ in $X times X$: the one with Todd classes and the ones with alpha classes. It gives you the identity $delta_* (td(X) delta^* td(X times X)^{-1}) = delta_* (alpha(X) delta^* alpha(X times X)^{-1})$, so that $delta_* td(X)^{-1} = delta_* alpha(X)^{-1}$.
Then you get $alpha(X)=td(X)$ by applying $pr_1* $.

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