Question

Suppose for some reason one would be expecting a formula of the kind

$$\mathop{\text{ch}}(f_!\mathcal F)\ =\ f_*(\mathop{\text{ch}}(\mathcal F)\cdot t_f)$$

valid in $H^*(Y)$ where

• $f:X\to Y$ is a proper morphism with $X$ and $Y$ smooth and quasiprojective,
• $\mathcal F\in D^b(X)$ is a bounded complex of coherent sheaves on $X$,
• $f_!: D^b(X)\to D^b(Y)$ is the derived pushforward,
• $\text{ch}:D^b(-)\to H^*(-)$ denotes the Chern character,
• and $t_f$ is some cohomology class that depends only on $f$ but not $\mathcal F$.

According to the Grothendieck–Hirzebruch–Riemann–Roch theorem (did I get it right?) this formula is true with $t_f$ being the relative Todd class of $f$, defined as the Todd class of relative tangent bundle $T_f$.

So, let's play at "guessing" the $t_f$ pretending we didn't know GHRR ($t_f$ are not uniquely defined, so add conditions on $t_f$ in necessary).

Question. Expecting the formula of the above kind, how to find out that $t_f = \text{td}\, T_f$?

You don't have to show this choice works (that is, prove GHRR), but you have to show no other choice works. Also, let's not use Hirzebruch–Riemann–Roch: I'm curious exactly how and where Todd classes will appear.

Answer 1

You look at the case when $X=D$ is a Cartier divisor on $Y$ (so that the relative tangent bundle -- as an element of the K-group -- is the normal bundle $\mathcal N_{D/X}=\mathcal O_D(D)$ (conveniently a line bundle, so is its own Chern root), and $\mathcal F=\mathcal O_D$. And the Todd class pops out right away.

Indeed from the exact sequence $0\to \mathcal O_Y(-D) \to \mathcal O_Y\to \mathcal O_D\to 0$, you get that $$ch(f_! \mathcal O_D)=ch(O_Y(D))= ch(\mathcal O_Y) - ch(\mathcal O_Y(-D)) = 1- e^{-D}.$$ And you need to compare this with the pushforward of $[D]$ in the Chow group, which is $D$. The ratio $$ \frac{D}{1-e^{-D}} = Td( \mathcal O(D) )$$ is what you are after. Now you have just discovered the Todd class. I suspect that this is how Grothendieck discovered his formula, too -- after seeing that this case fits with Hirzebruch's formula, that the same Todd class appears in both cases.

Answer 2

Here is another sanity check. Consider the map $f$ from $X$ to a point. The (topological) Euler characteristic of $X$ is $$\sum (-1)^k h^k(X) = \sum (-1)^{p+q} h^{pq}(X) = \sum (-1)^{p+q} h^q(X, \Omega^p) = \sum (-1)^p f_{!} \Omega^p.$$

If the Chern roots of the tangent bundle are $r_1$, $r_2$, ... $r_n$, then the Chern character of $\Omega^p$ is $e_p(e^{-r_1}, e^{-r_2}, \ldots, e^{-r_n})$, where $e_p$ is the $p$-th elementary symmetric function.

So the Euler characteristic of $X$ is $$\sum (-1)^p f_* \left( e_p(e^{-r_1}, e^{-r_2}, \ldots, e^{-r_n}) t_f \right) = f_* \left( \prod(1-e^{-r_i}) t_f \right).$$

Now, what has class $u$ has the property that $f_*(u)$ is the Euler characteristic of $X$? The top chern class of $T_X$, in other words, $\prod r_i$. So it seems very likely that we should have $$\prod(1-e^{-r_i}) t_f= \prod r_i$$ and $$t_f = \prod \frac{r_i}{1-e^{-r_i}}.$$

That's not a proof, because there are other classes with the same pushforward and because $\prod(1-e^{-r_i})$ is a zero-divisor, but I find it persuasive.