Bourbaki

Exercise 28, § 4

Integration, VIII, p. 72

Bourbaki marks this one as harder than the rest.

Let G1,G2G_1, G_2 be two locally compact groups, β1,β2\beta_1, \beta_2 left Haar measures on G1,G2G_1, G_2, and A1A_1 (resp. A2A_2) the topological algebra (over R\mathbf{R}) L1(G1,β1)L^1(G_1, \beta_1) (resp. L1(G2,β2)L^1(G_2, \beta_2)). Let TT be an algebra isomorphism of A1A_1 onto A2A_2, such that the relation f≥0f \geq 0 almost everywhere is equivalent to T(f)≥0T(f) \geq 0 almost everywhere. a) Show that TT is an isomorphism of topological algebras. (First note that if a decreasing sequence (fn)(f_n) of elements of A1A_1 tends to 0 almost everywhere, then the same is true of the sequence of the T(fn)T(f_n); deduce from this that for every sequence (fn)(f_n) that is bounded above in A1A_1, T(sup⁡(fn))=sup⁡(T(fn))T(\sup(f_n)) = \sup(T(f_n)), and conclude with the help of Fatou’s lemma.) b) Show that TT may be extended, in only one way, to an isomorphism of the topological algebra M1(G1)\mathscr{M}^1(G_1) onto the topological algebra M1(G2)\mathscr{M}^1(G_2), and that there exist a topological isomorphism uu of G1G_1 onto G2G_2 and a continuous homomorphism χ\chi of G1G_1 into R+∗\mathbf{R}_+^* such that T(εs)=χ(s)εu(s)T(\varepsilon_s) = \chi(s)\varepsilon_{u(s)} (make use of Exer. 15 a) and Exer. 19 b); to prove that uu and χ\chi are continuous, observe that TT defines an isomorphism of the algebra L(A1)\mathscr{L}(A_1) of continuous endomorphisms of the topological vector space A1A_1 onto the analogous algebra L(A2)\mathscr{L}(A_2), and that this isomorphism is bicontinuous for the topology of pointwise convergence; on the other hand, observe that s↦δ(s−1)s \mapsto \delta(s^{-1}) is an isomorphism of G1G_1 onto a multiplicative subgroup of L(A1)\mathscr{L}(A_1).

c) Deduce from b) that, for every f∈A1f \in A_1,

(T(f))(t)=χ(u−1(t))f(u−1(t)) (T(f))(t) = \chi(u^{-1}(t))f(u^{-1}(t))

for all t∈G2t \in G_2.

Recall (A, III, §2, Exer. 9) that there exist finite groups G1,G2G_1, G_2 such that the algebras L1(G1,β1)L^1(G_1, \beta_1) and L1(G2,β2)L^1(G_2, \beta_2) are isomorphic without G1G_1 and G2G_2 being so.

Solution

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Transcribed from the printed volume. Not translated.