may be modified by spectral calculus to obtain an element z A
with rf(z) a rank-one projection for K near K . As in [Di,4.4.4],
z defines a non-trivial continuous-trace ideal of A, and since z
is G-invariant, so is this ideal. This proves the theorem in the
case G = T.
Next we consider the case G a compact Lie group. Let G
denote the connected component of the identity, and consider its
action on A. We may assume that A is liminary, as before. By the
previous case, and reasoning as in [Di, 4.4.5], we see that for
every closed subgroup H of G with HST, A contains a dense open
H-invariant Hausdorff subset. But G contains a finite number of
one-parameter subgroups H1,...,H , each isomorphic to T, which
generate G algebraically (no closures needed). So choosing such
a dense open H.-invariant Hausdorff subset U. of A for each i and
letting U = U„A.. . fl U , we obtain a dense G -invariant open
1 x o
Hausdorff subset of A. (Density of U follows from the fact that
A is a Baire space.)
Choose representatives g. for the (finitely many) cosets of
G in G, and let Y = Ag.U. Again by the Baire property, Y is a
dense open G-invariant Hausdorff subset of A and corresponds to
some non-trivial G-invariant ideal J. Let C be some
continuous-trace ideal in J and let D be its G-saturation. Then D
is a non-trivial G-invariant ideal with Hausdorff spectrum and D
has local rank-one projections, thus D is continuous-trace. This
completes the proof of the theorem when G is a compact Lie group.

THEOREM 2.8. Let G be a compact Lie group, not necessarily
connected. If F is a collection of G-algebras and if each
continuous-trace G-algebra A with A = G/H (H running over closed
subgroups of G, G acting by translation) may be constructed from
F as described above, then F is a CQ-fundamental family.
Similarly, if F is the collection of commutative G-algebras of
the form C(G/H), then F is an AG-fundamental family.
PROOF: We shall prove only the statement about C -fundamental
families. The corresponding statement about Ap-fundamental
families is much easier, and uses (a proper subset of) exactly
the same arguments.
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