RÜPPERT

{ geG|f^ ( g , 1 ) = f} of f. Then N is a closed normal subgroup of G and G/N satisfies conditions (i)-{vii). Moreover, ф(NX{1})z = z^(NX{1}) = {z} (note that f is central m S). From the universality of the weak almost periodic compacti- fication we deduce that the semigroup fS is topologically isomorphic with the weak almost periodic compactification of (G/N)XH. It IS obvious that z = fz = lim fф (g ,h ) , so assertion (viii) holds if and only if the corresponding sertion holds for G/N. The minimal ideal fф (Fix (z) X{ 1 } ) of Ф(Fix(z) X {1 } ) IS a compact subgroup of H(f), so condition (vii) implies that fip (Fix(z) X { I }) с f ip(GX{1}K Thus we are allowed to replace G by G/N and this will enforce that the assumption of Reduction 2 is satisfied.

Before entering the actual proof of our claim ez = z we introduce a Lemma on ad-compact elements of a Lie algebra. Here we call an element x m a Lie algebra L ad-compact if the closure T of {e |te Ж } in AutL (endowed with the usual operator norm) is compact. The Killing form is denoted by Kill( , ).

Lemma . Let x be an ad-compact element in a Lie algebra L and suppose that Kill(x,x)> 0. Then x i^ central in L.

Proof . Since T is a torus group, its action on L is semi-

simple , so adx is a semisimple endomorphism of L and

dun L 0 < Kill(x,x) = Trace (ad x)^ = "I o-u '

k=l ^

where <ia, 11 <к<dimL> is an enumeration of the eigen- JC — —

values of adx. It follows that a, = 0 for all k, whence ad X = 0.

We now prove z = ez. Under the assumption of Reduction 2 this IS equivalent to e = 1, z€ф(GX{1}); that is, it is equivalent to the existence of lim g m G. Write G for the Lie algebra associated with the Lie group G and let a: G—»"AutG, g-» (x-»g.x), be the adjoint representation of G. Then a is infective, since the center of G is trivial. Then by (vii), a(G) is closed m Aut G, so a is a topological embedding.

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