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Consider algebraic group G over some field "k". Consider the SET of all its complex irreducible representations (I think I need unitary also).

Question Is there some way to put algebraic structure (manifold, stack, algebraic space, whatever) on this set ?

Motivations: 1) In handful examples you always see that irreps parametrized by some parameters (k1, k2, ...) something like highest weight for semisimples.

2) Let us think of orbit method - the set of irreps is the same as set of orbits $g^*/G$ which seems to carry algebraic structure, although I am not very clear how to describe it in a good way ( see MO 110584 ).

Moreover there is natural way to put topology on the set of irreps and there is Brown's theorem for nilpotent groups that orbit method bijection is continuous with respect to this topology and factor topology on the set of orbits. See e.g.

Examples 1) Abelian groups - vector spaces over k, products of k^* - unitary duals coincide with themselves so we see algebraic structure. Actually what happens with elliptic curves over F_q ?

2) Consider unitriangular groups of 3x3 matrices (Heisenberg group). xy=cyx. Irreps can be described like this:

$V_a$, $c-> a \ne 0$ we have standard unitary irrep x-> exp(q) y->exp(ad/dq).

1-dimensional $V_{st}: c->0, x->s , y->t$.

So the set of irreps is $a \in k \backslash 0$ glued with $(s,t) \in k \oplus k$. The plane $(s,t) \in k \oplus k$ is glued to line $a \in k \backslash 0$ instead of zero which we drop out from $k$. This is typical "ugly" topology on the set of irreps considered in the orbit method theory.

I am not sure I expose this example clearly, if someone is interested I can try to give more details.

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"Manifold" is an algebraic structure? – Qiaochu Yuan Oct 28 '12 at 17:40
In my meaning yes. Is something wrong ? But it will not be manifold, as clear from examples and since it is something like M/G as a SET – Alexander Chervov Oct 28 '12 at 17:58
Isn't it, as a stack, a disjoint union of points mod a trivial action of $\mathbb G_m$? (Because the automorphism group of an irrep is $\mathbb G_m$) – Will Sawin Oct 28 '12 at 18:12
@Will "it" is what ? In general it is NOT disjoint union - in Heisenberg example we GLUE the whole 2d-plane into 1d-line instead of point zero in line... – Alexander Chervov Oct 28 '12 at 18:20
If it would be stcasethen some irreps should have some internal structure like autmorphidms it seems it is not the case. – Alexander Chervov Oct 28 '12 at 20:09

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