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I know that the automorphic representation can be defined only for reductive algebraic group.

What property of algebraic group makes it hinder to define for all algebraic group and what nice property of reductive enable us define automorphic representation to it?

And I also wondering whether unitary group is reductive group. I think it does so.

But for me, it is not clear why they are reductive. It's hard to guess it just from the definition of reductiveness.

Would you shed me light on this?

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  • $\begingroup$ @PraphullaKoushik, bumping a 5.5-year-old question for a typo? $\endgroup$
    – LSpice
    Commented Apr 18, 2020 at 3:24
  • $\begingroup$ @LSpice meta.mathoverflow.net/questions/4515/… :) $\endgroup$
    – Praphulla Koushik
    Commented Apr 18, 2020 at 3:25
  • $\begingroup$ @LSpice I have edited 15 posts with that typo. I agree that it is too much. It was necessary, though not an emergency.. :) $\endgroup$
    – Praphulla Koushik
    Commented Apr 18, 2020 at 3:32

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It is not at all that automorphic representations can be defined only for reductive algebraic groups, but, rather, that the essential difficulties arise in that case... as opposed to (for example) abelian unipotent (e.g., additive groups). Indeed, "Jacobi modular forms" are modular forms (in effect) for parabolic subgroups of reductive groups.

There is no obstacle to ask about irreducible subrepresentations (e.g.) of $L^2(G_k\backslash G_\mathbb A)$ or related things for general $G$, rather than merely reductive: the question makes sense.

Unitary groups are indeed reductive, as can be verified by testing their (algebraic) Lie algebras. No, the definition of reductiveness itself is not directly helpful. The fact that the complexification of $U(p,q)$ is $GL(p+q,\mathbb C)$ is more-or-less a proof that $U(p,q)$ is reductive, since "we know that" $GL(p+q)$ is reductive.

Generally, the "definition"s of "reductive", "parabolic subgroup", and so on, are not so easy to check for the classical groups and the corresponding subgroups... Indeed. So if/when one finds this difficulty, it should not be surprising, but it is not evidence for any genuine issue, since the notions of "reductive", etc., were formed many decades after these classical (reductive or semi-simple) groups and their "parabolic" subgroups were very well understood.

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  • $\begingroup$ @garret, Thanks for you kind comment. Would you please let some heuristic example that is not reductive and what property hinders it from being reductive? Since the definition of the reductive group touch not my heart and so if you suggest one one couterexample, that would help me. $\endgroup$
    – Monty
    Commented Oct 2, 2014 at 0:56
  • $\begingroup$ As Paul Garrett knows, unitary groups can be defined rather more generally for certain hermitian forms relative to a separable quadratic extension of any field at all (even in char > 0, where Lie algebras are insufficient). Reductivity is a property of an algebraic group, rather than a Lie-group concept (though it is not so distant from Lie theory), and as such it is a geometric property, testable over an algebraic closure of the ground field (unlike pseudo-reductivity, for example). Since a unitary group becomes a general linear group after scalar extension to an algebraic closure... $\endgroup$
    – user27920
    Commented Oct 2, 2014 at 1:27
  • $\begingroup$ @user52824, Thanks for your illumination. The geometric property of unitary group is the key to be reductive. $\endgroup$
    – Monty
    Commented Oct 2, 2014 at 5:59
  • $\begingroup$ $SU(n)\times G_a$ as the same Lie algebra as the unitary group $U(n)$ but is not reductive. $\endgroup$
    – YCor
    Commented Oct 2, 2014 at 9:35
  • $\begingroup$ @YCor, Thank you for your penetrating example. $\endgroup$
    – Monty
    Commented Oct 3, 2014 at 12:27

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