are unions of some of the blocks in £.
Because c p is a PL action, T can always be chosen so that c p is a
simplicial action, which shall be assumed henceforth. The reader can
easily verify the following:
Lemma 1.2. The action cp: TL xN - N leaves invariant each block b.(£)of £.
Lemma 1.3. The semi-free action cp: TL xN - * N can be reconstructed from
the free action tp: TL - £.
Proof of 1.3: Let
Xj ^ (TN^RT) U (Ue),
where the union Ue is taken over all cells e of C which are of dimension
_ j. The proof proceeds by induction, over the nested sequence
N^R = X c Xm - c -c N, m-,
1 *
where k = dimension (K). Let e be any cell of C dual to a k-dimensional
simplex of K. To extend cp: TL x (eflR) - » eflR to cp: TL xe - * e cone the first
t- ^ n y n
action, noting that cone (eflR) = e, and that c p leaves eflR invariant
because eflR = b.(.£) for some 3jx. Doing this for all e with dimension
fe) = m-k,togives , the extension of cp: TL -+ r to cp: TL xX , - + X , . To
v n ^ ^ n m- k m- k
extend this last action to cp: Z xX , . -* X , _, , note that for any cell
n m-k+1 m-k+1 J
e in C dual to a simplex in K of dimension k - 1 , e DX , = de. So cone
r m-k
(enxm~,) = e, allowing cp:Znx (eflXm ,) - » eC\X , to be extended to
K m- K m- K
cp: TL xe - e by coning. All these extensions give cp: TL xX , ,
- X , ,-.
n ' ° ° n m-k+1 m-k+1
In the same way this last action extends to cp: ZxX , ^~ - » X , ^~ by
' m-k+z m-k+z J
coning, and then to cp: TL*X * X , ,_-, etc.,and finally to cp: TL xN-»N.
' n
k 3 m-k+3 ' J n
Now let us return to the problem stated in the introduction. Let
K c N be a compact, PL, subset of N as in 0.7, T, R, R, £, b.(£)is
described above. The same argument used to prove 1.3 above yields the
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