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.
Best Answer
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.