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Hello,

Let $H \subset G$ be reductive groups defined over $\mathbb{Q}$. I consider the spaces of automorphic forms of $G$ and $H$. One has a restriction map from the space of automorphic forms of $G$ to the space of automorphic forms of $H$. Let us denote by $\rho$ this map.

I want to know in which directions $\rho$ has been studied. I am especially interested in the question of surjectivity. More specifically, I consider the following situation:

Let $k_0$ be a totally real number field and $k$ a quadratic imaginary extension of $k_0$. Let $V$ be a $k$-vectorial space endowed with an anisotropic hermitian product (for the non-trivial Galois automorphism). Denote by $G$ the corresponding unitary group. It is a reductive group, defined over $k_0$ and compact at the archimedean places. The group $G$ contains a subgroup $H$ defined as the stabilizer of a decomposition $V = V_1 \oplus V_2$, where $V_i$ is one-dimensional. Thus, at an archimedean place, the inclusion $H \subset G$ is an inclusion $U(1) \times U(1) \subset U(2)$.

One uses Weil restriction to define these groups over the rational numbers and I want to know if the restriction map defined above is surjective.

Do you have any references for this kind of problem ?

Thanks.

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why is that obvious that $H(\mathbb{Q}) \backslash H(A)$ is a subset of $G(\mathbb{Q}) \backslash G(A)$ as adelic quotient spaces? Why is this map welldefined? –  Marc Palm Jun 3 '13 at 13:19
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I am not sure to understand your comment. There is an inclusion $H(A) \rightarrow G(A)$, sending $H(\mathbb{Q})$ to $G(\mathbb{Q})$, where I see the rational points of a group in the adelic points, by the diagonal embedding. Hence we get the indicated map. Dually, a smooth complex-valued function on $G(A)$ which is $G(\mathbb{Q})$-invariant, restricts to a $H(\mathbb{Q})$-invariant map on $H(A)$. Am I missing something ? –  Jeremy Daniel Jun 3 '13 at 15:53
    
That is of course correct. I am not sure what problem I saw:/ –  Marc Palm Jun 10 '13 at 9:38
    
I see also slight issues with $GL(1) \times GL(1) \subset GL(2)$, because the later cuspidal automorphic reps are infinite-dimensional and the former are one-dimensional. –  Marc Palm Jun 10 '13 at 9:49

2 Answers 2

An automorphic form on $G$ apart from being $G({\mathbb Q})$ invariant, is annihilated by an ideal of finite codimension in the enveloping algebra of the lie algebra of $G({\mathbb R})$. So, it is not true that restriction of an automorphic form on $G$ to $H$ gives an automorphic form on $H$. Can you clarify what you had in mind?

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I turned my comment to an answer, because the OP also asked for references. –  Marc Palm Jun 25 '13 at 15:58
    
I want to find a cuspidal automorphic form on $H$, which is orthogonal to the image of the restriction map. This should have a meaning even if the restricted maps are not automorphic forms, am I right ? –  Jeremy Daniel Jul 6 '13 at 9:28
    
In the co-compact case, the restriction of $G({\mathbb Q})$ invariant smooth functions to $H$ is onto, so that means that the restriction of the images of auto functions on $G$ to $H$ is dense in $H({\mathbb Q})$ invariant functions on $H$; so the orthogonal complemet is zero. –  Venkataramana Jul 12 '13 at 16:27

For semisimple algebraic groups, look at Proposition 1 in Akshay Venkatesh's article here:

http://math.stanford.edu/~akshay/research/bs.pdf

The results seems due to Clozel and Ullmo. The references point to a preprint of them which I couldn't track down.

For the special consideration indicated at the end, I have no clue whether things are surjective:( Certainly induction isn't, since cuspidal components for GL(n) at p-adic places can't be produced that way.

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