Euler-Maclaurin's formula transforms the integral $I=\int_a^b f(x)dx$ into the finite sum $S=\sum_a^b f(x)$, for two integers $a,b$. As Dmitri pointed out, in 1993 Khovanskii and Pukhlikov gave a multi-dimensional generalization of Euler-Maclaurin which, in particular says the following:

Let $P$ be an $n$-dimensional polytope in $\mathbb R^n\supset\mathbb Z^n$ with integral vertices, and further assume that $P$ defines a nonsingular toric variety (i.e. $P$ is simplicial and at every vertex the integral generators of the edges give a basis in $\mathbb Z^n$). Let us say the facets of $P$ are defined by the inequalities $l_j(x)\le a_j$ for some primitive integral linear functions $l_j(x_1,\dots,x_n)$. Denote by $P(h_j)$ P(h)$ the polytope defined by the inequalities $l_j(x)\le a_j+h_j$. Finally, let $$ I(f,h)= \int_{P(h)} f(x)dx, \quad S(f)= \int_{P\cap \mathbb Z^n} f(x).$$ Then for any quasipolynomial $f(x)$ (a sum of products of polynomial and exponential functions) one has $$ S(f) = \prod_j Td(\partial / \partial h_j)\ I(f,h)\ |_{h_j=0}.$$

Here is how the Hirzebruch-Riemann-Roch for the sheaf $\mathcal F=\mathcal O(d)$ on $X=\mathbb P^n$ follows from the Khovanskii-Pukhlikov's version of Euler-Maclaurin's formula:

Taking $P$ to be a simplex of side $d$ and $f(x)=1$, the Khovanskii-Pukhlikov's formula gives $$ h^0(\mathbb P^n, \mathcal O(d)) = \prod_{j=0}^n Td(\partial/\partial h_j) \frac{(d+h_0+\dots+h_n)^n}{n!} \ |_{h_j=0}$$ which by making a substitution $y=d+h_0+\dots+h_n$ transforms into $Td(\partial/\partial y)^{n+1} (y^n/n!)\ |_{y=d}.$

The usual Hirzebruch-Riemann-Roch, on the other hand, gives says that $h^0(\mathbb P^n,\mathcal O(d))$ is the coefficient of $x^n$ in the expression $Td(x)^{n+1} e^{dx}$. So why is this the same? Because $$ Td(x)^{n+1} e^{dx} = Td(\partial/ \partial y)^{n+1} e^{yx}\ |_{y=d}$$ (here I we used the fact that $(\partial/ \partial y)^k e^{yx} = x^k e^{yx}$) and the coefficient of $x^n$ in $e^{yx}$, expanded as a power series in $x$, is $(y^n/n!)$. QED

Now that wasn't so hard, but why isn't this written somewhere? Or am I missing a reference?

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So what does this suggest conceptually about the meaning of Hirzebruch-Riemann-Roch? I think, clearly, it suggests that

1. The pushforward $$ f_!:K(X)\to K(pt)=\mathbb Z, \qquad \mathcal F\mapsto \chi(\mathcal F) = h^0(F)-h^1(F)+\dots$$ between the K-groups should be considered to be the "discrete summation" of a "function" $f=f(\mathcal F)$. Indeed, for say a toric variety $X$ and an ample line bundle $\mathcal F$ we are just counting integral points in a polytope $P$. So that fits.

2. The pushforward $$ f_*: A(X)_Q\to A(pt)_Q=\mathbb Q $$ between the Chow groups should be considered to be a "continuous" version, an integral. Indeed, for any homogeneous a cycle on $X$ this its pushforward can be interpreted as, and computed by, an integral of a corresponding differential form. So this makes perfect sense as well.

So now the Riemann-Roch, = the Euler-Maclaurin for this situation, transforms the integral into a the sum, by multiplying it by the differential operator given by the Todd class. This also explains why in HRR the Todd class of $T_X$ appears and not, say, of $\Omega^1_X$. The tangent bundle is the place where the derivations $\partial/\partial z$ live.

Euler-Maclaurin's formula transforms the integral $I=\int_a^b f(x)dx$ into the finite sum $S=\sum_a^b f(x)$, for two integers $a,b$. As Dmitri pointed out, in 1993 Khovanskii and Pukhlikov gave a multi-dimensional generalization of Euler-Maclaurin which, in particular says the following:

Let $P$ be an $n$-dimensional polytope in $\mathbb R^n\supset\mathbb Z^n$ with integral vertices, and further assume that $P$ defines a nonsingular toric variety (i.e. $P$ is simplicial and at every vertex the integral generators of the edges give a basis in $\mathbb Z^n$). Let us say the facets of $P$ are defined by the inequalities $l_j(x)\le a_j$ for some primitive integral linear functions $l_j(x_1,\dots,x_n)$. Denote by $P(h_j)$ the polytope defined by the inequalities $l_j(x)\le a_j+h_j$. Finally, let $$ I(f,h)= \int_{P(h)} f(x)dx, \quad S(f)= \int_{P\cap \mathbb Z^n} f(x).$$ Then for any quasipolynomial $f(x)$ (a sum of products of polynomial and exponential functions) one has $$ S(f) = \prod_j Td(\partial / \partial h_j)\ I(f,h)\ |_{h_j=0}.$$

Here is how the Hirzebruch-Riemann-Roch for the sheaf $\mathcal F=\mathcal O(d)$ on $X=\mathbb P^n$ follows from the Khovanskii-Pukhlikov's version of Euler-Maclaurin's formula:

Taking $P$ to be a simplex of side $d$ and $f(x)=1$, the Khovanskii-Pukhlikov's formula gives $$ \prod_{j=0}^n Td(\partial/\partial h_j) \frac{(d+h_0+\dots+h_n)^n}{n!} \ |_{h_j=0}$$ which by making a substitution $y=d+h_0+\dots+h_n$ transforms into $Td(\partial/\partial y)^{n+1} (y^n/n!)\ |_{y=d}.$

The usual Hirzebruch-Riemann-Roch, on the other hand, gives the coefficient of $x^n$ in the expression $Td(x)^{n+1} e^{dx}$. So why is this the same? Because $$ Td(x)^{n+1} e^{dx} = Td(\partial/ \partial y)^{n+1} e^{yx}\ |_{y=d}$$ (here I used the fact that $(\partial/ \partial y)^k e^{yx} = x^k e^{yx}$) and the coefficient of $x^n$ in $e^{yx}$, expanded as a power series in $x$, is $(y^n/n!)$. QED

Now that wasn't so hard, but why isn't this written somewhere? Or am I missing a reference?

---

So what does this suggest conceptually about the meaning of Hirzebruch-Riemann-Roch? I think, clearly, it suggests that

1. The pushforward $$ f_!:K(X)\to K(pt)=\mathbb Z, \qquad \mathcal F\mapsto \chi(\mathcal F) = h^0(F)-h^1(F)+\dots$$ between the K-groups should be considered to be the "discrete summation" of a "function" $f=f(\mathcal F)$. Indeed, for say a toric variety $X$ and an ample line bundle $\mathcal F$ we are just counting integral points in a polytope $P$. So that fits.

2. The pushforward $$ f_*: A(X)_Q\to A(pt)_Q=\mathbb Q $$ between the Chow groups should be considered to be a "continuous" version, an integral. Indeed, for any homogeneous cycle on $X$ this pushforward can be interpreted as, and computed by, an integral of a corresponding differential form. So this makes perfect sense as well.

So now the Riemann-Roch, = the Euler-Maclaurin for this situation, transforms the integral into a sum, by multiplying it by the differential operator given by the Todd class. This also explains why in HRR the Todd class of $T_X$ appears and not, say, of $\Omega^1_X$. The tangent bundle is the place where the derivations $\partial/\partial z$ live.

Euler-Maclaurin's formula transforms the integral $I=\int_a^b f(x)dx$ into the finite sum $S=\sum_a^b f(x)$, for two integers $a,b$. As Dmitri pointed out, in 1993 Khovanskii and Pukhlikov gave a multi-dimensional generalization of Euler-Maclaurin which, in particular says the following:

Let $P$ be an $n$-dimensional polytope in $\mathbb R^n\supset\mathbb Z^n$ with integral vertices, and further assume that $P$ defines a nonsingular toric variety (i.e. $P$ is simplicial and at every vertex the integral generators of the edges give a basis in $\mathbb Z^n$). Let us say the facets of $P$ are defined by the inequalities $l_j(x)\le a_j$ for some primitive integral linear functions $l_j(x_1,\dots,x_n)$. Denote by $P(h_j)$ the polytope defined by the inequalities $l_j(x)\le a_j+h_j$. Finally, let $$ I(f,h)= \int_{P(h)} f(x)dx, \quad S(f)= \int_{P\cap \mathbb Z^n} f(x).$$ Then for any quasipolynomial $f(x)$ (a sum of products of polynomial and exponential functions) one has $$ S(f) = \prod_j Td(\partial / \partial h_j)\ I(f,h)\ |_{h_j=0}.$$

Here is how the Hirzebruch-Riemann-Roch for the sheaf $\mathcal F=\mathcal O(d)$ on $X=\mathbb P^n$ follows from the Khovanskii-Pukhlikov's version of Euler-Maclaurin's formula:

Taking $P$ to be a simplex of side $d$ and $f(x)=1$, the Khovanskii-Pukhlikov's formula gives $$ \prod_{j=0}^n Td(\partial/\partial h_j) \frac{(d+h_0+\dots+h_n)^n}{n!} \ |_{h_j=0}$$ which by making a substitution $y=d+h_0+\dots+h_n$ transforms into $Td(\partial/\partial y)^{n+1} (y^n/n!)\ |_{y=d}.$

The usual Hirzebruch-Riemann-Roch, on the other hand, gives the coefficient of $x^n$ in the expression $Td(x)^{n+1} e^{dx}$. So why is this the same? Because $$ Td(x)^{n+1} e^{dx} = Td(\partial/ \partial y)^{n+1} e^{yx}\ |_{y=d}$$ and the coefficient of $x^n$ in $e^{yx}$, expanded as a power series in $x$, is $(y^n/n!)$. QED

Now that wasn't so hard, but why isn't this written somewhere? Or am I missing a reference?

---

So what does this suggest conceptually about the meaning of Hirzebruch-Riemann-Roch? I think, clearly, it suggests that

1. The pushforward $$ f_!:K(X)\to K(pt)=\mathbb Z, \qquad \mathcal F\mapsto \chi(\mathcal F) = h^0(F)-h^1(F)+\dots$$ between the K-groups should be considered to be the "discrete summation" of a "function" $f=f(\mathcal F)$. Indeed, for say a toric variety $X$ and an ample line bundle $\mathcal F$ we are just counting integral points in a polytope $P$. So that fits.

2. The pushforward $$ f_*: A(X)_Q\to A(pt)_Q=\mathbb Q $$ between the Chow groups should be considered to be a "continuous" version, an integral. Indeed, for any homogeneous cycle on $X$ this pushforward can be interpreted as, and computed by, an integral of a corresponding differential form. So this makes perfect sense as well.

So now the Riemann-Roch, = the Euler-Maclaurin for this situation, transforms the integral into a sum, by multiplying by the differential operator given by the Todd class.