Let μ be a Haar measure on the additive group g.
a) There exists a unique Haar measure μG on G with the following property: if ωG and ωg are invariant differential forms of degree n on G and g, respectively, such that ωG(e)=ωg(0) and ∣ωg∣=μ, then ∣ωG∣=μG. The map μ→μG is a bijection from the set of Haar measures on g to the analogous set for G. b) Choose a scalar product (∣) invariant under g, and a Haar measure τ on t such that μ (resp. τ ) corresponds to the Lebesgue measure when g is identified with Rn (resp. t with Rr) by means of an orthonormal basis. Prove the formula
∫tπg(x)e−(x∣x)/2dτ(x)=(2π)n/2μG(G)τT(T)w(G)
c) Identify X(T) with a subset of the Hilbert space t∗ by means of the map (2πi)−1δ(§4, no. 2), and put P(x)=∏α∈R+⟨α,x⟩ for x∈t. With the notations
of Exerc. 1 and 2, show that
l
μG(G)τT(T)=(2π)Nw(G)−1(P∣P)=πN∏(α∣α)∏mi!
α∈R+i=1 d) With the notations of Exerc. 9 of §3, let gZ be the Z-Lie subalgebra of g generated by (2π)−1Γ(T) and the elements uα,vα for α∈R. Prove the equality
μG(G)=μ(g/gZ)2∏irπmiN+!r.
e) Assume that G is simply-connected. Show that l=r and that
μG(G)=2l/2π−Nf1/2∏α∈R+(α∣α)−1∏i=1l(αi∣αi)−1/2∏i(mi!)−1,
where {α1,...,αl} is a basis of R and f is the connection index of R.
f) Assume further that R is irreducible and that all its roots are of the same length; take the scalar product on g to be the opposite of the Killing form. Prove that μG(G)=(2π)N+r(2h)n/2f1/2∏i(mi!)−1.
Nothing.