As an exercise for myself I wanted to check GRR in the following situation. Consider $P:X \rightarrow B$ to be an Weierstrass elliptic fibration with a section, and $X\times\_B X$ be the fiber product with $\pi\_{1,2}$ be the corresponding projections to the first and second factor. Then from GRR we have, 


$ch(R\pi\_{2\*} \mathcal{O}\_{X\times\_B X})=\pi\_{2\*} \big(ch(\mathcal{O}\_{X\times\_B X})\cdot \pi\_1^\* Td (X/B)\big) = \\ \pi\_{2\*} \big[\pi\_1^\*\big(1-\frac{1}{2}P^\* c\_1(B)+\frac{13}{12} P^\* c\_1^2(B)+\sigma P^\*c\_1(B)-\frac{1}{2}\sigma P^\* c\_1^2(B)\big)\big]$.


where $\sigma$ is the section. So obviously the left-hand side is 1, but I cannot show why the right-hand side should be 1.


**As mentioned in the comments, LHS is not 1 necessarily...**


Base $B$ is a surface (can be projective space or Hirtzebruch), and maybe I should mention that $\sigma^2=-c\_1(B)\sigma$


Let's be more specific, if I'm right we can use the following commutative diagram,


$\require{AMScd}
\begin{CD}
X\times\_B X @>{\pi\_1}>> X\\ @V{\pi\_2}VV @VV{P\_1}V\\
X @>>{P\_2}> B
\end{CD}$


Then $\pi\_{2\*}o \pi\_1^\* \sim P\_{2}^\* o P\_{1\*} $. So we can use the projection formula. Now the specific questions are, 


$P\_{\*}(\sigma)=?$


$P\_{\*}(1)=?$.


Naively both should be 1, but it cannot be true...



A question on Grothendieck Riemann Roch
Thanks to Jim Bryan for mentioning my error in writing GRR, and also his hints about the pushforward of cycles, the problem is solved.


$ch(R\pi\_{2\*}\mathcal{O}\_{X\times\_BX})=ch(\pi\_{2\*}\mathcal{O}\_{X\times\_BX})-ch(R^1\pi\_{2\*}\mathcal{O}\_{X\times\_BX})=1-ch(R^1\pi\_{2\*}\mathcal{O}\_{X\times\_BX})$.


Now note that 


$R\mathcal{Hom}(R\pi\_{2\*}\mathcal{O}\_{X\times\_BX},\mathcal{O}\_X)=R\pi\_{2\*}R\mathcal{Hom}(\mathcal{O}\_{X\times\_BX},\pi\_2^!\mathcal{O}\_X)=R\pi\_{2\*}R\mathcal{Hom}(\mathcal{O}\_{X\times\_BX},\omega\_{X/B}[1])$.


I forgot to mention that X is Calabi-Yau, so $\omega\_{X/B}=P^\*K\_B^{-1}$. Therefore since $\mathcal{Ext}^1(\mathcal{O},\mathcal{O})=0$ we have,


$(R^1\pi\_{2\*}\mathcal{O}\_{X\times\_BX})^\*=P^\*K\_B^{-1}$


So


$ch(R^1\pi\_{2\*}\mathcal{O}\_{X\times\_BX})=1-P^\*c\_1(B)+\frac{1}{2} P^\* c\_1^2(B)$.


This is consistent with the RHS of the above question if $P\_\*\sigma=B$, and $P\_\* 1=0$.


