# A question on Grothendieck Riemann Roch

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...

• The lefthand side should have a $R^1\pi_{2*}\cal{O}$ term as well. Also, you have to be careful computing the relative Todd class because your total space isn't smooth (you will get singularities over the discriminant locus in $B$. – Jim Bryan Mar 8 '18 at 6:27
• $P_*(\sigma) = B=1$ and $P_*(1)=0$. Note that $P_*$ reduces the cohomological degree by 2. You are using interchangeably a cohomology class of degree $d$ and the corresponding cycle of co-dimension $d$. The image of the co-dimension cycle $\sigma$ under $P$ is the co-dimension 0 cycle $B$ which corresponds to the cohomology class 1. – Jim Bryan Mar 8 '18 at 6:37
• Thanks, Jim. About the singularity of $X\times_B X$, I agree with you. – Mohsen Karkheiran Mar 8 '18 at 6:45
• Also thanks for mentioning the higher direct image mistake. – Mohsen Karkheiran Mar 8 '18 at 6:51

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 $P_* 1=0$.

• Glad to see you solved it, and thank you for taking the time to write the solution for everybody! Please mark your post as an answer, so it appears as a "solved" question to the community. – Tintin Mar 10 '18 at 14:06