# Explicit formula for the Poincare dual of a CM endomorphism of an elliptic curve

Let $E/\mathbf{C}$ be an elliptic curve with CM by the maximal order $\mathcal{O}_K$ of $K=\mathbf{Q}(\sqrt{-D})$ where $D$ is positive and square-free integer. To make it even more precise, let us assume that $E=\mathbf{C}/\Lambda_{\tau}$, where $\Lambda_{\tau}=\mathbf{Z}+\tau\mathbf{Z}$ and $\tau$ is in the Poincare upper half-plane. We endow $E$ with the usual flat metric coming from the complex plane, so $E$ comes equipped with a volume form. Consider the endomorphism $[\sqrt{-D}]:E\rightarrow E$. Let $\Gamma\subseteq E\times E$ be the graph of $[\sqrt{-D}]$. Then $\Gamma$ is isomorphic (as a complex manifold) to $E$, in particular it is a 2-cycle on $E\times E$. We endow $\Gamma$ with the induced metric coming from $E$ via the map $[\sqrt{-D}]$. We also endow $E\times E$ with the product metric coming from $E$. Now let $\eta$ be the Poincare dual of $\Gamma$: so if $\iota:\Gamma\hookrightarrow E\times E$ denotes the inclusion then for all $\omega\in A^2(E\times E)$, we have $\int_{\Gamma}\iota^*\eta=\int_{E\times E}\omega\wedge \eta$. Let $a,b\in H^1(E,\mathbf{Z})$ be the standard oriented basis of $E$: $a$ corresponds to cohomology class of the oriented vector joining $0$ to $1$, $b$ corresponds to the cohomology class of the oriented vector joining $0$ to $\tau$, so $a\cdot b=1$ (or $b\cdot a=-1$). From the Kuenneth formula we have a priviledged isomorphism $H^*(E\times E)\simeq\oplus_{i=1}^4 H^{i}(E)\otimes H^{4-i}(E)$. Via this isomorphism, there exists real numbers $c_i$ ($i=1\ldots 4$) such that $$\eta=c_1\cdot(a\otimes a)+c_2\cdot(a\otimes b)+c_3\cdot(b\otimes a)+c_4\cdot(b\otimes b).$$ I would expect in general the $c_i$'s to depend on $D$. On the other hand I don't know if the $c_i$'s will depend on the Picard class of $\Lambda_{\tau}$ in general.

Q: Is there a method to compute the numbers $c_i$'s in a systematic way, so that we can clearly see their dependence on $D$?

• I thought a bit a bout my problem and now I realized that the way I set it up is probably not ideal. It is probably better to work with homology since then one can make pictures. The group $H_2(E,\mathbf{Z})$ has $3$ natural $\mathbf{Z}$-linearly independant elements, namely $E_1=E\times\{0\}$, $E_2=\{0\}\times E$ and $\Delta$ (the diagonal). Intuitively we should have $E_2\cdot \Gamma=1$ and $E_1\cdot\Gamma=D$. Though the intersection $\Delta\cdot \Gamma$ seems to be more complicated to compute. – Hugo Chapdelaine Aug 6 '14 at 19:49

Among $E_1, E_2, \Gamma, \Delta$ we have the intersections:

$E_1 \cdot E_1 = E_2\cdot E_2 = \Delta \cdot \Delta =\Gamma \cdot \Gamma =0$

$E_1 \cdot E_2 = E_1 \cdot \Delta = E_2 \cdot \Delta =1$

$E_1 \cdot \Gamma = 1$

$E_2 \cdot \Gamma = D$

$\Delta \cdot \Gamma = 1+D$.

Presumably only the last computation will need explanation. The reason is just that you are coutining solutions to $(1-\sqrt{-D}) (x) \in \mathbb Z[\tau]$ modulo $\mathbb Z[\tau]$. The number of solutions is always the norm of $1-\sqrt{-D}$, or $1+D$.

Does this help?

• Hi @Will. I agree with everything you said except that I don't quite see why $\Delta\cdot\Gamma$ corresponds to $[(1-\sqrt{-D})^{-1}\mathcal{O}_K:\mathcal{O}_K]$ – Hugo Chapdelaine Aug 6 '14 at 23:14
• Because it counts solutions to the equation $x = E(x)$ where $E$ is multiplication by $\sqrt{-D}$. For each such $x$, $(x,x)$ is a transverse intersection point of the two curves. – Will Sawin Aug 6 '14 at 23:29
• Yes you are right! So this solves my question. Many thanks! – Hugo Chapdelaine Aug 7 '14 at 14:41