# What is the Grossencharacter of this CM curve?

Let $E$ be an elliptic curve with complex multiplication by $\mathcal{O}$, the ring of integers of some imaginary quadratic field $K$. Then, the CM theory says that $E$ is related to a Grossencharacter over $K$.

My question is, if I take $E$ to be the curve given by $\mathbb{C}/\mathcal{O}$, what is the corresponding Grossencharacter? Is there an easy way to write down its values?

First of all, the curve $E = \mathbf{C}/\mathcal{O}$ is only defined over $K$ if $K$ has class number one. So, unless you make this assumption, the Grossencharacter will only be defined over $H$, the Hilbert class field of $K$.

Also, there are various ways to think about what a Grossencharacter is, I assume you are interested in a map

$$\chi: K^{\times} \backslash \mathbb{A}^{\times}_K \rightarrow \mathbf{C}^{\times}.$$

Suppose that $K$ is an imaginary quadratic field of class number one, and suppose for convenience that the number of roots of unity $w_K = 2$, so $\Delta_K \ne -3, -4$. This actually implies that $\Delta_K = - p$ for some prime $p \in \{7,11,19,43,67,163\}$ which are all $3 \pmod 4$. In particular, $-1$ is not a square modulo $p$.

$\chi$ is certainly determined by its restriction to $K^{\times}_v$ for all $v$. We will have:

$$\chi_{\infty}: \mathbf{C}^{\times} \rightarrow \mathbf{C}^{\times}$$

is given by $z \mapsto z$. On the other hand, suppose that $v$ is a finite prime different from $p$. Then

$$\chi_{v}: K^{\times}_v \rightarrow \mathbf{C}^{\times}$$

is trivial on $\mathcal{O}^{\times}_v$, and thus is determined by the image of $\pi_v$. Since $K$ has class number one, there is a global $\pi \in \mathcal{O}_K$ such that $(\pi)$ is the prime above $v$. This is unique up to units, so unique up to $\pm 1$. Let $\chi_v(\pi_v)$ denote the choice of $\pm \pi^{-1}$ which is a quadratic residue modulo $\sqrt{p}$.

Finally, for $v = p$, we have $K_v = \mathbf{Q}_p(\sqrt{-p})$. Now let $\chi_p(\sqrt{-p}) = 1/\sqrt{-p}$. To finish the description of $\chi_p$, we need to define its values on $\mathcal{O}^{\times}_v$. Unlike for other primes, it is no longer trivial. Instead, let $\chi_p$ on this space be the quadratic character map

$$\chi_p: a \in \mathcal{O}^{\times}_v \mapsto \left(\frac{a}{p}\right) \in \pm 1 \subset \mathbf{C}^{\times}.$$

This completes the description of $\chi$. However, we may also use thi description to define $\chi$. We certainly get a character of $\mathbb{A}^{\times}_K \rightarrow \mathbf{C}^{\times}$, with the property that on the finite ideles it has conductor $\sqrt{p}$. To show that this is a map on $K^{\times}\backslash \mathbb{A}^{\times}_K$, it suffices to show that $\chi$ vanishes on $K^{\times}$. It certainly vanishes on $\sqrt{-p}$, since, if $\alpha = (\sqrt{-p},\sqrt{-p},\ldots )$, then

$$\chi_v(\sqrt{-p}) = 1, v \ne p,\infty, \ \chi_p(\sqrt{-p}) = 1/\sqrt{-p}, \ \chi_{\infty}(\sqrt{-p}) = \sqrt{-p},$$

and the product of these numbers is trivial. Anything in $K^{\times}$ can be written as a power of $\sqrt{-p}$ times some $x \in K^{\times}$ which has trivial valuation at $p$. By unique factorization, write

$$x = \epsilon \prod \pi_i,$$

where $\pi_i$ are chosen (unqiuely) to be squares modulo $p$, and $\epsilon = \pm 1$. Note that

$$\left(\frac{x}{p}\right) = \epsilon.$$

If we evaluate $\chi$ on $x$, we find that:

1. The contribution from $\chi_{\infty}$ is $x$,
2. The contribution from all primes away from $p$ is $\prod (\pi_i)^{-1} = \epsilon/x$,
3. The contribution from $p$ is $\chi_p(x) = \displaystyle{\left(\frac{x}{p}\right) = \epsilon}$.

The product of all these terms is $1$, so we do get a Hecke character. The other Hecke character $\overline{\chi}$ is the conjugate of this one.

As an application, we compute $a_{\ell}$ for the corresponding elliptic curve for primes $\ell$ which split in $K$. We need to compute $\chi(\mathrm{Frob}_{\pi}) + \overline{\chi}(\mathrm{Frob}_{\pi})$. Using the required choice of Frobenius (geometric versus arithmetic) and the appropriate normalization of local class field theory, one sees that the answer is $\pi + \overline{\pi}$, where $(\pi/p) = + 1$. In otherwords, write $\ell = a^2 + p b^2$ with $(a/p) = + 1$, then $a_p = 2a$.

• Thank you for the detailed response. That's exactly what I was looking for. – user119481 Nov 26 '17 at 14:50