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Let $k$ be a number field, and $G$ a connected, reductive group over $k$. Let $\omega$ be a unitary character of $Z_G(\mathbb A_k)/Z_G(k)$. An irreducible subspace $(\pi, V)$ of $L^2(G(k) \backslash G(\mathbb A), \omega)$ is said to be cuspidal if

$$\int\limits_{N(k) \backslash N(\mathbb A_k)} \pi(n)w \space dn = 0 \tag{1}$$ for all $w \in V$ and all unipotent radicals $N$ of all proper parabolic subgroups of $G$.

Let $v$ be a finite place of $k$, and let $(\pi_v, V_v)$ be a smooth irreducible representation of $G(k_v)$. We say that $\pi_v$ is cuspidal, or supercuspidal, if the Jacquet module of $\pi_v$ is trivial, i.e.

$$\int\limits_{N(k_v)} \pi(n_v)w \space dn_v = 0 \tag{2}$$ for every unipotent radical $N$ of every proper parabolic subgroup of $G \times_k k_v$. Formally, (2) is not really a vector valued integral over all $N(k_v)$; the cuspidality condition just states that the integral is zero when integration over $N(k_v)$ is replaced by integration over a suitaly large open compact subgroup.

Is there a connection between local and global notions of cuspidal, other than the vanishing of those integrals along unipotent radicals?

(1) is an integral over $N(k) \backslash N(\mathbb A_k)$, not over $N(\mathbb A_k)$. If it was over $N(\mathbb A_k)$, and we decomposed $\pi$ as a tensor product of representations $\pi = \otimes_v \pi_v$, then I would like to do something like

$$\int\limits_{N(\mathbb A_k)} \pi(n)w dn = \prod\limits_v \int\limits_{N(k_v)} \pi_v(n_v) w_v dn_v$$ and the vanishing of one of these integrals in the product would imply the cuspidality of $\pi$. On the other hand, I know there are cuspidal automorphic representations of $GL_2(\mathbb A_k)$ which are not supercuspidal at any finite place.

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    $\begingroup$ Supercuspidal at a finite place implies globally cuspidal, though of course the converse is not true. Other than that, I don't think much else can be said. $\endgroup$ Commented Sep 10, 2019 at 22:04
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    $\begingroup$ As Peter points out, locally supercuspidal implies globally cuspidal, but the converse is not true. The argument is exactly what you give, except we have to observe that $\int_{N(k) \backslash N(\mathbb A_k) } \int_{U} \pi(n) \pi(n_v) w dn_v dn= \int_{N(k) \backslash N(\mathbb A_k) } \pi(n) w dn$ for $U$ a compact open subgroup of $N(k_v)$, which is easy. $\endgroup$
    – Will Sawin
    Commented Sep 11, 2019 at 1:58

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