Answer by Julien Grivaux for Why do Todd classes appear in Grothendieck-Riemann-Roch formula?

2011-03-08T12:20:44Z

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* $.

User julien grivaux - MathOverflow

Answer by Julien Grivaux for Why do Todd classes appear in Grothendieck-Riemann-Roch formula?