Haboush's Theorem - Proof

Proof

The theorem is proved in several steps as follows:

• We can assume that the group is defined over an algebraically closed field K of characteristic p>0.
• 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 (as the connected component has finite index). By taking a central extension which is harmless one can also assume the group G is simply connected.
• Let A(G) 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(G) by sending w∈V to the element a∈A(G) with a(g) = v′(g(w)). This sends v to 1∈A(G), so we can assume that V⊂A(G) and v=1.
• The structure of the representation A(G) is given as follows. Pick a maximal torus T of G, and let it act on A(G) by right translations (so that it commutes with the action of G). Then A(G) splits as a sum over characters λ of T of the subrepresentations A(G)λ of elements transforming according to λ. So we can assume that V is contained in the T-invariant subspace A(G)λ of A(G).
• The representation A(G)λ is an increasing union of subrepresentations of the form E_λ+nρ⊗E_nρ, where ρ is the Weyl vector for a choice of simple roots of T, n is a positive integer, and E_μ 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 E_nρ has dimension (n+1)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 n large enough that the line bundle over G/B is very ample, E_nρ has the same dimension as in characteristic 0.
• If q=pr for a positive integer r, and n=q−1, then E_nρ contains the Steinberg representation of G(F_q) of dimension qN. (Here F_q ⊂ K is the finite field of order q.) The Steinberg representation is an irreducible representation of G(F_q) and therefore of G(K), and for r large enough it has the same dimension as E_nρ, so there are infinitely many values of n such that E_nρ is irreducible.
• If E_nρ is irreducible it is isomorphic to its dual, so E_nρ⊗E_nρ is isomorphic to End(E_nρ). Therefore the T-invariant subspace A(G)λ of A(G) is an increasing union of subrepresentations of the form End(E) for representations E (of the form E_(q−1)ρ)). However for representations of the form End(E) an invariant polynomial that separates 0 and 1 is given by the determinant. This completes the sketch of the proof of Haboush's theorem.