In mathematicsHaboush's theorem, often still referred to as the Mumford conjecture, states that for any semisimple algebraic groupG over a fieldK, and for any linear representation ρ of G on a K-vector space V, given v ≠ 0 in V that is fixed by the action of G, there is a G-invariant polynomialF on V, without constant term, such that The polynomial can be taken to be homogeneous, in other words an element of a symmetric power of the dual of V, and if the characteristic is p>0 the degree of the polynomial can be taken to be a power of p. When K has characteristic 0 this was well known; in fact Weyl's theorem on the complete reducibility of the representations of G implies that F can even be taken to be linear. Mumford's conjecture about the extension to prime characteristic p was proved by W. J., about a decade after the problem had been posed by David Mumford, in the introduction to the first edition of his book Geometric Invariant Theory.
Applications
Haboush's theorem can be used to generalize results of geometric invariant theory from characteristic 0, where they were already known, to characteristic p>0. In particular Nagata's earlier results together with Haboush's theorem show that if a reductive group acts on a finitely generated algebra then the fixed subalgebra is also finitely generated. Haboush's theorem implies that if G is a reductive algebraic group acting regularly on an affine algebraic variety, then disjoint closed invariant sets X and Y can be separated by an invariant function f. C.S. Seshadri extended Haboush's theorem to reductive groups over schemes. It follows from the work of, Haboush, and Popov that the following conditions are equivalent for an affine algebraic groupG over a field K:
G is reductive.
For any non-zero invariant vector in a rational representation of G, there is an invariant homogeneous polynomial that does not vanish on it.
For any finitely generated K algebra on which G act rationally, the algebra of fixed elements is finitely generated.
Proof
The theorem is proved in several steps as follows:
Finite groups are easy to deal with as one can just take a product over all elements, so one can reduce to the case of connected reductive groups. By taking a central extension which is harmless one can also assume the group G is simply connected.
Let A be the coordinate ring of G. This is a representation of G with G acting by left translations. Pick an element v′ of the dual of V that has value 1 on the invariant vector v. The map V to A by sending w∈V to the element a∈A with a = v′. This sends v to 1∈A, so we can assume that V⊂A and v=1.
The structure of the representation A is given as follows. Pick a maximal torus T of G, and let it act on A by right translations. Then A splits as a sum over characters λ of T of the subrepresentations Aλ of elements transforming according to λ. So we can assume that V is contained in the T-invariant subspace Aλ of A.
The representation Aλ is an increasing union of subrepresentations of the form Eλ+nρ⊗Enρ, where ρ is the Weyl vector for a choice of simple roots of T, n is a positive integer, and Eμ is the space of sections of the line bundle over G/B corresponding to a character μ of T, where B is a Borel subgroup containing T.
If n is sufficiently large then Enρ has dimension N where N is the number of positive roots. This is because in characteristic 0 the corresponding module has this dimension by the Weyl character formula, and for nlarge enough that the line bundle over G/B is very ample, Enρ has the same dimension as in characteristic 0.
If q=pr for a positive integer r, and n=q−1, then Enρ contains the Steinberg representation of G of dimension qN. The Steinberg representation is an irreducible representation of G and therefore of G, and for r large enough it has the same dimension as Enρ, so there are infinitely many values of n such that Enρ is irreducible.
If Enρ is irreducible it is isomorphic to its dual, so Enρ⊗Enρ is isomorphic to End. Therefore, the T-invariant subspace Aλ of A is an increasing union of subrepresentations of the form End. However, for representations of the form End an invariant polynomial that separates 0 and 1 is given by the determinant. This completes the sketch of the proof of Haboush's theorem.
