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¸ is injective.
Orientability of M is built into the definition of admissible. So now assume M
is non-orientable with M its orientable double covering admissible. Let G0 be the
connected component of the identity of G. For case (i), the image of the evaluation
homomorphism is trivial. Therefore, G0 lifts to M and must be a torus acting
injectively on M . The action of G0 can be lifted to the universal covering M of
M, but it cannot be lifted to the universal covering of M , unless G0 is trivial.
Since the universal coverings of M and M are identical, G0 must be trivial. Now
as before, let K = ker(¨). Then, as above, × Zp, Zp ‚" K, acts effectively on M,
and Zp commutes with À1(M , x) ‚" À1(M, x). Therefore, Zp must be in the center
of À1(M , x ) since M is admissible. Consequently, Zp cannot act effectively on M
and we have a contradiction, implying that K is trivial.
For part (ii), once again G0 is trivial, since G0 lifts to M , with fixed points, and
M is admissible. As before, À1(M, x) × Zp acts effectively on M. Zp commutes
48 1. TRANSFORMATION GROUPS
with the action of À1(M , x ). Since M is admissible, Zp will have to be in the
center of À1(M , x ). This will contradict that Zp acts effectively on M.
Examples of non-orientable closed manifolds whose orientable double covers are
admissible, other than the obvious ones, are all the closed non-orientable aspherical
manifolds.
question If M is admissible and M ’! M" is a (finite) covering, is M" ad-
missible? If M is admissible and M ’! M is a finite (regular) covering, is M
admissible?
Second question: problematic If M ’! M" is a finite regular covering and M"
is admissible, is M admissible? Let us not assume M" is orientable, just assume
that every cyclic group action on the universal covering M" and commutes with
À1(M") must be in the center of the covering transformations. Then if C is some
cyclic group actiong on M, the universal covering of M, and commuting with the
covering transformations of À1(M) on M, then we must show it belongs to the center
of À1(M) (as covering transformations). So our problem is to somehow construct
or extend the induced action or part of it on M to one on M" at least. But it is
not likely that we are to find that 11(M) in À1(M") is invariant in general. So I
think this direction is diffeicult (or tricky at least). It is an interesting question by
it would seem to take some thought to settle it one way or another.
newly added exer
1.17.13 Exercise. Let M ’! M" be a finite regular covering with M admissible.
Then each effective action of a cyclic group on the universal covering of M", M",
that commutes with the covering transformations of M" is a covering action. (That
is, M" is also admissible if M" is orientable, or satisfies the crucial condition of
admissibility if M" is not orientable. Is the orientable assumption of M really
necessary to obtain the crucial condition?)
Proof. leave out? Let C be a cyclic group acting effectively on M". If C
commutes with the covering transformations of M", À1(M")m then C commutes
with the subgroup À1(M) of covering transformations. Since M is admissible, C
belongs to the covering transformations of M. Thus it also belongs tot he covering
transformations of M".
1.17.14 Exercise. Let X be a closed aspherical manifold and ¸ : N(À1(X, x)) ’!
Aut(À1(X, x)), where N(À1(X, x)) is the normalizer of À1(X, x) in TOP(X), be as
in section 1.8.1. Then ¸ is injective on any torsion subgroup. (The proof uses the
fact that À1(X, x) is torsion free).
1.17.15 Example. Let G be a compact Lie group acting effectively on the m-torus
m s
so that ¦ : G ’! Out(À1(T )) = GL(m, Z) is trivial. Then G = T × F , and it acts
m m-s
freely on T , where F is finite abelian and isomorphic to a subgroup of T , for
m
some 0 d" s d" m. Furthermore, F acts as covering transformations and F \T is
s
again a topological torus while the T action splits into a product action.
Proof. The connected component of the identity must be an s-torus for some
m s
0 d" s d" m, and acts injectively. Suppose there is a y " T so that Ty contains
s
an element h of order p. Let g(t) be a path in T = G0 from 1 to h. Then
1.17. MANIFOLDS ON WHICH ONLY TORI CAN ACT 49
h(t)y is a loop at y. For any loop (t) based at y, the loop (t) is homotopic to
m
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