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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.

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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
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1 Answer

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).

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