Fix an imaginary quadratic field $K$, and let $\mathcal{O}\_K$ be its ring of integers. A Hermitian modular form of genus 1 (i.e., an automorphic form on $GU(1,1)$) of weight $(k\_1,k\_2)$ on a congruence subgroup $\Gamma \subset U(1,1)(\mathbb{Z})$ is a function $f \colon \mathfrak{h} \to \mathbb{C}$ satisfying an analytic condition and a transformation law:


$$ f\left(\frac{az+b}{cz+d}\right) = (cz+d)^{k\_1}(\overline{c}z+\overline{d})^{k\_2} f(z), \qquad \forall \begin{pmatrix} a & b \\ c & d\end{pmatrix} \in \Gamma. $$


Here $\Gamma \subset GU(1,1)(\mathbb{Z})$ is a group of matrices with entries in $\mathcal{O}\_K$, so conjugating the coefficients makes sense. At this stage these are very similar to classical modular forms, and in fact they have the same shape $q$-expansions and simply restricting the matrices that are allowed to act on them allows them to be viewed as classical modular forms of weight $k\_1+k\_2$ on the congruence subgroup $\Gamma \cap SL\_2(\mathbb{Z})$.


I have not found anywhere what the (complex) $L$-functions for these things are. For a classical modular cusp form, we have
$$ f(z) = \sum\_{n=1}^\infty a\_n q^n, \qquad \Rightarrow \qquad L(f,s) = \sum\_{n=1}^\infty \frac{a\_n}{n^s}. $$


Is there a similar formula for what the $L$-function of a hermitian cusp form of genus 1 should be? Is it the same, maybe modulo the Euler factors at infinity? How are the $L$-functions of the hermitian modular form of weight $(k\_1,k\_2)$ and the classical modular form of weight $k\_1+k\_2$ related?


Is there a similar relationship between the hermitian modular forms over a CM field $F$ and the Hilbert modular forms over its totally real subfield $F^+$?




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Follow up: how should I view the fact that every Hermitian modular form yields a classical modular form, while an elliptic curve over $\mathbb{Q}$ yields an elliptic curve over $K$? If I equate "elliptic curve over $\mathbb{Q}$" to "classical modular form on $\Gamma\_0(N)$," and "elliptic curve over $K$" to "hermitian modular form on a similar $\Gamma$," these feel like they go in opposite directions from each other.



Complex L-functions for Hermitian modular forms?
To define an "L-function" for an automorphic representation $\pi$ on a group $G$ you usually have to make a choice of a representation $\rho$ of the dual group. Only then can you define $L(\pi,\rho,s)$. If $G = \mathrm{GL}(n)$ then there is an obvious choice of representation so it is often ommitted in that case. In the cases you are looking at, the most natural $L$-function will coincide on the nose with the $L$-function of the modular form.


Yes, the situation over totally real fields is very similar.


Followup: First, you need to be more precise. An elliptic curve over $\mathbf{Q}$ should correspond to a modular form on $\Gamma\_0(N)$ of *weight two* and with *rational coefficients*.


Second, (and this is the real problem), an elliptic curve over $K$ defintely *won't* in general give rise to an Hermetian modular form, it should give rise to a form on the (quite different) group $\mathrm{GL}(2)/K$. The motives you *do* get from Hermetian modular forms have the property that they are *conjugate self dual* (up to a specific range of twists). Since $2$-dimensional representations are always self dual up to twist anyway (since $\mathrm{GL}(2) = \mathrm{GSp}(2)$), this means the corresponding motives are both conjugate self-dual and self-dual up to twist, which means that (if they are not induced) they are isomorphic to their conjugates up to twist. As it happens, this basically implies that the representations themselves up to twist descend to $\mathbf{Q}$, and so it turns out they have to come from $\mathrm{GL}(2)$ anyway. In particular, you get *both* fewer motives and fewer autormophic forms for $U(1,1)$.


One final wrinkle: you might say: given an elliptic curve $E/\mathbf{Q}$, the restriction $E/K$ certainly has an isomorphic Tate module to its conjugate. But why is that the only thing that can happen? After all, we are only requiring that the Tate module of $E/K$ is isomorphic to a *twist* of the Tate module of $E^c/K$. This could happen, for example, if $E$ is isogenous to $E^c$ over some finite extension $L/K$. (The Tate conjecture, proved in this case by Faltings, means that this is the only way this can happen.) Well, you would be correct, and indeed these elliptic curves do exist! In this case, one calls $E$ a $\mathbf{Q}$-curve. The $\mathbf{Q}$-curves *do* give rise to Hermetian modular forms even though they need not come from $\mathbf{Q}$. But this is not inconsistent with what I said previously, it's just that the modular forms corresponding to these $\mathbf{Q}$-curves - while still of weight $2$ - don't have rational coefficients (and they might have level $\Gamma\_1(N)$ as well rather than $\Gamma\_0(N)$).


