Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.


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 ?


share|improve this question
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? –  plusepsilon.de Jun 3 '13 at 13:19
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:/ –  plusepsilon.de 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. –  plusepsilon.de Jun 10 '13 at 9:49
add comment

2 Answers

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?

share|improve this answer
I turned my comment to an answer, because the OP also asked for references. –  plusepsilon.de 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. –  Aakumadula Jul 12 '13 at 16:27
add comment

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


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.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.