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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.

Incidentally, $1/(1-e^{-D})$ also appears in the Euler-Maclarin formula which connects a discrete sum $\sum_{i=1}^n f(i)$ with the integral $\int_0^{n} f(x)dx$. (The Wikipedia page for this formula does not make this obvious, but this formula is explained very well e.g. in "Concrete Mathematics" by Graham-Knuth-Patashnik).

I always thought that there must be a connection with Riemann-Roch, and for example it should become clear when looking at the case $X=\mathbb P^n$, $Y=pt$ and $\mathcal F=\mathcal O(d)$. But somehow the proof in this case is different. It uses a residue computation and does not look to me in any sense similar to the Euler-Maclaurin formula. I certainly would like to know if someone here knows if there is a connection!

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.

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.

Incidentally, $1/(1-e^{-D})$ also appears in the Euler-Maclarin formula which connects a discrete sum $\sum_{i=1}^n f(i)$ with the integral $\int_0^{n} f(x)dx$. (The Wikipedia page for this formula does not make this obvious, but this formula is explained very well e.g. in "Concrete Mathematics" by Graham-Knuth-Patashnik).

I always thought that there must be a connection with Riemann-Roch, and for example it should become clear when looking at the case $X=\mathbb P^n$, $Y=pt$ and $\mathcal F=\mathcal O(d)$. But somehow the proof in this case is different. It uses a residue computation and does not look to me in any sense similar to the Euler-Maclaurin formula. I certainly would like to know if someone here knows if there is a connection!

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.

Incidentally, $1/(1-e^{-D})$ also appears in the Euler-Maclarin formula which connects a discrete sum $\sum_{i=1}^n f(i)$ with the integral $\int_0^{n} f(x)dx$. (The Wikipedia page for this formula does not make this obvious, but this formula is explained very well e.g. in "Concrete Mathematics" by Graham-Knuth-Patashnik).

I always thought that there must be a connection with Riemann-Roch, and for example it should become clear when looking at the case $X=\mathbb P^n$, $Y=pt$ and $\mathcal F=\mathcal O(d)$. But somehow the proof in this case is different. It uses a residue computation and does not look to me in any sense similar to the Euler-Maclaurin formula. I certainly would like to know if someone here knows if there is a connection!

You look at the case when $X=D$ is a Cartier divisor on $Y$ (so that the relative tangent bundle is the normal bundle $\mathcal O_Y(D)|_D$ (conveniently a line bundle, so 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 $$ \frac{D}{1-e^{-D}}$$ is what you are after. Now you have discovered the Todd class.

Incidentally, $1/(1-e^{-D})$ also appears in the Euler-Maclarin formula which connects a discrete sum $\sum_{i=1}^n f(i)$ with the integral $\int_0^{n} f(x)dx$. (The Wikipedia page for this formula does not make this obvious, but this formula is explained very well e.g. in "Concrete Mathematics" by Graham-Knuth-Patashnik).

I always thought that there must be a connection with Riemann-Roch, and for example it should become clear when looking at the case $X=\mathbb P^n$, $Y=pt$ and $\mathcal F=\mathcal O(d)$. But somehow the proof in this case is different. It uses a residue computation and does not look to me in any sense similar to the Euler-Maclaurin formula. I certainly would like to know if someone here knows if there is a connection!

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.

Incidentally, $1/(1-e^{-D})$ also appears in the Euler-Maclarin formula which connects a discrete sum $\sum_{i=1}^n f(i)$ with the integral $\int_0^{n} f(x)dx$. (The Wikipedia page for this formula does not make this obvious, but this formula is explained very well e.g. in "Concrete Mathematics" by Graham-Knuth-Patashnik).

I always thought that there must be a connection with Riemann-Roch, and for example it should become clear when looking at the case $X=\mathbb P^n$, $Y=pt$ and $\mathcal F=\mathcal O(d)$. But somehow the proof in this case is different. It uses a residue computation and does not look to me in any sense similar to the Euler-Maclaurin formula. I certainly would like to know if someone here knows if there is a connection!