Grothendieck-Riemann-Roch interpretation of a calculation Let $X$ be some smooth projective variety over $\mathbb{C}$ and let 
$$\mathscr{H}_{q}(X):=ch(\Omega_{X}^{q})Td(X).$$
For $Y$ a certain elliptic fibration 
$$\varphi:Y\to B$$ 
where $B$ is of arbitrary dimension, I have been computing 
$$\varphi_{*}\mathscr{H}_{q}(Y)$$ 
where $\varphi_{*}$ is the proper pushforward. The total space $Y$ is a subvariety of a projective bundle 
$$\mathbb{P}(\mathscr{O}\oplus \mathscr{L}\oplus \mathscr{L}\oplus \mathscr{L})$$ 
where $\mathscr{L}$ is a line bundle on $B$. I have computed that 
$\varphi_{*}Td(Y)=(1-e^{-L})Td(B)$
$$\varphi_{*}\mathscr{H}_{1}(Y)=(1-e^{-L})\mathscr{H}_{1}(B)+(-4-e^{-L}+3e^{-2L}+2e^{-3L})Td(B),$$
where $L=c_1(\mathscr{L})$. What I would like is an explanation of the result of these calculations in terms of Grothendieck-Riemann-Roch as I have only recently acquainted myself with GRR. Thanks everyone. 
 A: Let $T\phi$ be the relative tangent bundle. So 
we have an exact sequence $0\to T\phi\to TY\to \phi^*TB\to 0$. Now GRR and the projection formula gives 
$$
\phi_*{\rm Td}(Y)=\phi_*({\rm Td}(T\phi){\rm Td}(\phi^*TB))=
{\rm Td}(TB)\phi_*({\rm Td}(T\phi))={\rm ch}(1-R^1\pi_*({\cal O}_Y)){\rm Td}(TB)
$$
which suggests that ${\cal L}=R^1\pi_*({\cal O}_Y)^\vee=\pi_*(\Omega_\phi):=\pi_*(T\phi^\vee)$ 
(by Grothendieck duality).  I will take this for granted; up to $\otimes$ by a torsion bundle, it is forced upon you by the equation; if $\cal L$ and $\pi_*(\Omega_\phi)$ differ by a torsion line bundle, the calculations below still work. 
Furthermore, applying GRR and the projection formula again, we may compute
$$
\phi_*{\cal H}_1(Y)=\phi_*({\rm ch}(\Omega_Y){\rm Td}(TY))=
\phi_*(\ [\phi^*{\rm ch}(\Omega_B)+{\rm ch}(\Omega_\phi)]{\rm Td}(T\phi)\phi^*{\rm Td}(TB)\ )=
$$
$$
{\rm Td}(TB){\rm ch}(\Omega_B)\phi_*({\rm Td}(T\phi))+{\rm Td}(TB)\phi_*({\rm Td}(T\phi)
{\rm ch}(\Omega_\phi))=
$$
$$
{\rm Td}(TB){\rm ch}(\Omega_B)\phi_*({\rm Td}(T\phi))+{\rm Td}(TB){\rm ch}({\cal L}-
R^1\pi_*(\Omega_\phi))=
$$
$$
{\rm Td}(TB){\rm ch}(\Omega_B)\phi_*({\rm Td}(T\phi))+{\rm Td}(TB){\rm ch}({\cal L}-1)=
$$
$$
{\rm Td}(TB){\rm ch}(\Omega_B)(1-{\rm ch}({\cal L}^\vee))+{\rm Td}(TB){\rm ch}({\cal L}-1)=
(1-e^{-L}){\cal H}_1(B)+(e^{L}-1){\rm Td}(TB)\,\,\, (*)
$$
Now use the fact that $\cal L$ is actually a torsion bundle, because the discriminant modular 
form will trivialise ${\cal L}^{\otimes 12}$ (or possibly a higher power, if one needs to introduce 
level structures). This last fact is also a consequence of GRR, since 
$$
\phi_*({\rm Td}(T\phi))=\pi_*({\rm ch}(1))\phi^*\phi_*({\rm Td}(T\phi))=0={\rm ch}(1-R^1\pi_*({\cal O}_Y))
$$
(because $T\phi=\pi^*\pi_* T\phi)$. 
Hence, one gets, all in all, that 
$$
\phi_*{\rm Td}(TY)=0
$$
and in view of (*), that 
$$
\phi_*{\cal H}_1(Y)=0
$$
which is equivalent to the two equations you are considering, since ${\cal L}$ is a torsion line bundle (observe that the degree $0$ part of $−4−e^{-L}+3e^{−2L}+2e^{−3L}$ vanishes).
