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Jason Starr
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Edit. The odd Bernoulli numbers are zero, of course! So the first attempt below is wrong. The revised examples use complete subvarietes of moduli spaces of curves.

Examples from complete subvarieties of moduli spaces of curves. Kodaira constructed complete curves in the (uncompactified) moduli spaces $M_g$ of smooth, projective curves of genus $g.$ This was generalized to higher-dimensional complete subvarieties by E. Y. Miller in the 1980s. I found a very readable reference for higher-dimensional complete subvarieties written by Zaal.

Christiaan Zaal
Complete subvarieties of moduli spaces of algebraic curves.
Thesis, Universiteit von Amsterdam
https://pure.uva.nl/ws/files/3915897/36235_Thesis.pdf

For every integer $n\geq 0$, there exists an integer $g(n)\geq 0$ and a projective, $n$-dimensional, closed subvariety $B$ of the moduli space $M_g$ of smooth, projective, genus $g$ curves (over $\mathbb{C}$, for simplicity). In his thesis, Zaal gives a very explicit construction for such subvarieties. (In the opposite direction, I believe that the theorem of Steven Diaz remains the best upper bound on the maximal dimension $n(g)$ of a complete subvariety of $M_g$.)

Up to base change of $B$ by a finite cover, the subvariety of $M_g$ arises from a family of curves, $$\rho:C\to B.$$ The class in K-theory of $R\rho_*\mathcal{O}_C$ equals $1-[(\rho_*\omega_\rho)^\vee]$. Thus, it suffices to show that the Chern character of the Hodge bundle, $\mathbf{E}_\rho=\rho_*\omega_\rho$, has nonzero term in degree $n$.

Mumford computed the Chern character of the Hodge bundle on the compactified moduli space $\overline{M}_g$. When restricted to the interior, Mumford's formula says that the $r^{\text{th}}$ graded piece of the Chern character is zero if $r$ is even, and for $r=2s-1$ odd, it equals $$\text{ch}_{2s-1}(\mathbf{E}_\rho)= (-1)^{s+1}\frac{|B_{2s}|}{(2s)!}\cdot \kappa_{2s-1}.$$ By Theorem 6.33 of Harris-Morrison, Moduli of Curves, the divisor class $c_1(\omega_\rho)$ is ample on $C$ (I will try to find the original source of this theorem later). Thus, for every irreducible subvariety $Z\subset B$ of odd dimension $r=2s-1$ no greater than $n$, the intersection number $\langle c_1(\omega_\rho)^{2s},\rho^{-1}Z \rangle$ is positive. Thus, the intersection number $\langle \kappa_{2s-1},Z \rangle$ is positive. So $\text{ch}_{2s-1}(\mathbf{E}_\rho)$ has positive intersection pairing against $Z$.

These are examples of families of curve, not of surfaces, and they only account for odd degree pieces of the Chern character. However, it is straightforward to get a similar result for a family of surfaces $\pi:S\to B$. Let $B'$ be a smooth, projective curve, and let $\rho':C'\to B'$ be a smooth, projective morphism of relative dimension $1$ whose fibers are curves of genus $g\geq 2$ such that $\kappa_1(\rho')$ is a divisor class on $B'$ of nonzero degree. Now, with respect to the definitions above, form the product target scheme $B\times B'$, for the product domain scheme $S=C\times C'$, and define $\pi$ to be $\rho\times \rho'$. Then by Künneth's formula, $$\pi_*\mathcal{O}_S = \text{pr}_1^*(\rho_*\mathcal{O}_C)\otimes \text{pr}_2^*(\rho'_*\mathcal{O}_{C'}).$$ By the multiplicative property of the Chern character, $$\text{ch}_{2s-1}(\pi_*\mathcal{O}_S) = (1-g)\text{pr}_1^*\text{ch}_{2s-1}(\rho_*\mathcal{O}_C),$$ and this is nonzero by the previous paragraph. Also, $$ \text{ch}_{2s}(\pi_*\mathcal{O}_S) = \text{pr}_1^*\text{ch}_{2s-1}(\rho_*\mathcal{O}_C)\cup \text{pr}_2^*\kappa_1(\rho').$$ Since $\kappa_1(\rho')$ has nonzero degree, this class is also nonzero.

First attempt with ridiculous error because the odd Bernoulli numbers equal zero. You can create many such examples with isotrivial families. The simplest isotrivial morphism is a smooth, projective morphism of relative dimension $1$ whose geometric fibers are isomorphic to the projective line, $$\rho:C\to X.$$ The pushforward of $T_\rho =\textit{Hom}_{\mathcal{O}_C}(\Omega_\rho,\mathcal{O}_C)$ is a locally free $\mathcal{O}_X$-module of rank $3$ whose total Chern class equals $1-a$ for a cycle class $a$ of degree $2$. For a locally free $\mathcal{O}_X$-module $E$ of rank $2$, if $C/X$ represents the functor on $X$-schemes of invertible quotients of the pullback of $E$, then the class $a$ equals $$a = c_1(E)^2 - 4c_2(E).$$ Up to scaling and torsion, this is the unique polynomial expression in Chern classes of $E$ of cohomological degree $2$ that is compatible with arbitrary pullback and that is preserved by the operation of tensoring $E$ with invertible sheaves.

The first Chern class $c = c_1(T_\rho)$ has square equal to a pullback, $$c^2 = \rho^* a.$$ Thus, the relative Todd class of $\rho$ equals $$\text{Td}(\rho) = [C]\cap\left(\sum_{m\geq 0} \frac{B_{2m}}{(2m)!}\rho^*(a^m) - \sum_{m\geq 0}\frac{B_{2m+1}}{(2m+1)!}c\cup \rho^*(a^m) \right). $$ Thus, the pushforward of the Todd class equals, $$\rho_*\text{Td}(\rho) = -2[X]\cap \sum_{m\geq 0} \frac{B_{2m+1}}{(2m+1)!} a^m.$$

In particular, for a second smooth, projective morphism of relative dimension $1$, $$\sigma:X\to B,$$ for $S$ defined to be $C$, and for the composition $\pi=\sigma\circ \rho$, the pushforward of the relative Todd class equals $$\pi_*\text{Td}(\pi) = \sigma_*\left(-2\text{Td}(\sigma)\cap \sum_{m\geq 0}\frac{B_{2m+1}}{(2m+1)!} a^m \right).$$ In the very simplest case that $X$ equals $\mathbb{P}^1_B$, $\sigma$ is the natural projection, and for any choice of constant cross-section $s:B\to \mathbb{P}^1_B$ (the $\infty$-section, for instance), this gives $$\pi_*\text{Td}(\pi) = -[B]\cap \left(\sum_{m\geq 0} \frac{B_{2m+1}}{(2m+1)!} (s^*a)^m +2\sum_{m\geq 0} \frac{B_{2m+1}}{(2m+1)!}\sigma_*(a^m) \right).$$ To be very explict, if $E$ is the locally free sheaf $\sigma^*\mathcal{L}\oplus \mathcal{O}_{\mathbb{P}^1}(1)$ for some choice of invertible $\mathcal{O}_B$-module $\mathcal{L}$ with first Chern class $c_1(\mathcal{L})=\lambda$, this evaluates to $$\pi_*\text{Td}(\pi) = -[B]\cap \left( \sum_{m\geq 0} \frac{B_{2m+1}}{(2m+1)!}\lambda^{2m} + 4\sum_{m\geq 1} \frac{B_{2m+1}}{(2m+1)!} m\lambda^{2m-1}\right).$$ If $\mathcal{L}$ is ample, then $\lambda^r$ is nonzero for every integer $r=0,\dots,\text{dim}(B).$

Jason Starr
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