Why do Todd classes appear in Grothendieck-Riemann-Roch formula?

A conceptual answer to "exactly how and where Todd classes will appear:" You don't have to use Todd classes if you stick to Hochschild homology. This might not be the best reference, but I learned this from 1.2.1 in https://arxiv.org/pdf/1804.00879.pdf You can stick to just the Chern character and factor through HH(Perf(X)); the Todd class corrects the fact that the HKR isomorphisms $HH(Perf(X)) simeq H^*(X, bigoplus Omega_X^p)$ are NOT natural in $X$ until correction by Todd.

My apologies -- I'm sure there's a more down-to-earth reference of which I'm not aware.

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* $.

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.

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 $0to mathcal O_Y(-D) to mathcal O_Yto mathcal O_Dto 0$, 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.