MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 $3$) (?) – Damian Rössler Oct 1 '11 at 16:47
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. – DZN Oct 1 '11 at 17:16
Typo in title of question – 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).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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