Let $X$ be some smooth projective variety over $\mathbb{C}$ and let


For $Y$ a certain elliptic fibration

$$\varphi:Y\to B$$

where $B$ is of arbitrary dimension, I have been computing


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



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.

  • $\begingroup$ You need to provide more information about the embedding of $Y$ into the projective bundle. It doesn't seem to be an embedding locally given by Weierstrass equations, for otherwise the projective bundle would be of relative dimension $2$ (and not $3$) (?) $\endgroup$ – Damian Rössler Oct 1 '11 at 16:47
  • $\begingroup$ The generic fiber is a complete intersection of two quadrics in $\mathbb{P}^3$ and $[Y]=(2H+2L)^2\in A^{*}\mathbb{P}(\mathscr{E})$, where $H$ is the hyperplane class in $A^{*}\mathbb{P}(\mathscr{E})$. But I don't see how this will give insight to how the results of the calculation are reflecting properties of GRR. $\endgroup$ – DZN Oct 1 '11 at 17:16
  • $\begingroup$ Typo in title of question $\endgroup$ – Yemon Choi Oct 1 '11 at 23:49

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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.