Let $F$ be a number field and $G$ (resp. $H$) an odd orthogonal (resp. metaplectic group) over $F$.

Let $v$ be a finite place of $F$ and $\sigma_v$ a supercuspidal representation of $G_v(F_v)$. Let $\pi_v=\theta_v(\sigma_v)$ be the local theta lift of $\sigma_v$ to $H_v(F_v)$ and $\pi$ be a irreducible cuspidal representation of $H(\mathbb{A})$ such that the localization of $\pi$ at $v$ is $\pi_v$.

Consider the global theta lift $\Theta(\pi)$ of $\pi$ to $G(\mathbb{A})$ and assume that it is nonzero.

In some literature, it is written that $\Theta(\pi)$ is cuspidal because $\sigma_v$ is supercuspidal.

I can't understand this because we don't know that $\Theta(\pi)$ is irreducible.

I am wondering whether the statement in the literature is really true.

(ps: My guess is that suppose that $\rho$ is an irreducible constituent of $\Theta(\pi)$. Then there is a nonzero linear map $l:\omega_v \to \pi_v \otimes \rho_v$, where $\omega_v$ is the local Weil representation. By the maximal semi-simplicity of the local theta lift $\theta(\pi_v)$, $\theta(\pi_v) \simeq \rho_v$. Therefore $\rho_v=\sigma_v$ and so $\rho$ is cuspidal.

But I am not sure $\Theta(\pi)$ is cuspidal if all of its irreducible constituents are cuspidal.


1 Answer 1


If all the irreducible components of $\Theta(\pi)$ are components of the space of cusp forms, then $\Theta(\pi)$ is a component of the space of cusp forms and so is cuspidal. This should also follow more directly from writing the theta lift and its local component and looking at what being a subquotient induced from a parabolic would mean.


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