ON NON-PARALLEL s-STRUCTURES

Using algebraic topology, we find out the number of all non-parallel s-structures which an n-dimensional Euclidean space En admits. The obtaining results are generalized on a manifold M which is CW-complex.


INTRODUCTION
Let E be an n-dimensional Euclidean space and f E E an automorphism Ifzo E E then the expresmon f(:co) + Df_ (:czo) is the linear approximation off at zo We assume that z0 s a fixed point of f and the Jacobian matrix Df s an orthogonal matrix Then, if in a closed neighborhood of zo (under the usual topology) there is no other fixed point, f is called s- symmetry on :co and it is written where the Jacobian A: belongs to 0(r) {I}, (if A I, then every point of E will be a fixed point) A family { f0 zo E E } of s-symmetries is called an s-structure on E An s-structure is called regular if f0"f,0 f,0"f0, where u0 fo (#0) An s-structure is called parallel if A0 is constant e does not depend on :co It is clear that a parallel s-structure is also a regular one If f0 is an orthogonal transformation at the origin without fixed vectors and to is a translation on E such that t0 (0) z0 then Lo to fo o t;o are the only parallel s-structures on E Therefore the following question arises Do there exist non-parallel regular s-structures on E 9 O Kowalski in 1] proved that the Euclidean spaces E , E and E admit only parallel regular s- structures and found out a non-parallel regular s-structare on E S Wegrzynowski in [2] obtained the same results using analytical calculations on Lie algebras In the present paper, we will give a complete classification of the Euclidean spaces of arbitrary dimension admitting non-parallel s-structures and we will give the number of these ones as well Finally we will generalize the meaning of an s-structure on every mamfold wt.ch is a CW-complex and we will solve the analogous problem on these manifolds We will prove that the aumber of the non-parallel s- structures is a dimensional-invariant

I. Euclidean Spaces
In the present paragraph we will prove the following THEOREM I.In an n-dimensional Euchdean space E" the number N of the non-parallel s- structures s gven by 0 for n=2,3,4 N-2" 1(2"-1) for n_>5 PROOF.Let f,, be an s-symmetry on E", e an isometry with an isolated fixed point x0 If 3;o and x are two fixed points, then we can find positive numbers and e' such that (under the usual topology in E') xo N(x'o, e') and Xo N(xo, e), where X(x'o, (N(x'o, ')) is the closure of the open neighborhood of xo(x'o) with radius e(e') So, we can substitute the fixed point x0 with the neighborhood N(xo, e) preserving the geometrical properties off, Then, f-0 becomes f'o E' N(xo, e) E N(xo, e) where N(xo, e) is lnvariant under the action of Denoting ,, () E" N(3;o, e), fxo takes the form fxo. o() o().
A A:o We observe that this action can be decomposed to a sum of mutually independent parts Choosing the j-th part where 1 _< j _< n, we shall prove that if z0, z are two different fixed points in ,n-1 we can pass from/)-(e, j) to (e' j) for every 3, where, :L'J .:l(e,,j)Azo(Ez,o (e,,j)).
Taking/'-(e* 3) with e* min{e, e'} we have the following commutative diagram sn-l(* where, aj x0 :g (*, 3), and h, are homeomorphisms, ql I is the natural map, q2 is the quotient map, q (id.qs,) is the quotient map and Sn-(e*) is the (n 1)-sphere with center xo and radius e* under quotient topology (V C S n-1 is open if and only if q-(V) is open) Repeating the above diagram for every j, it turns out that the existence of C depends on the existence of % which are classified by definition from the n-1 homotopy group of E,-l(e, j) But :gO /,-1 (e, 3) is of the same homotopy type with S:go -(e), hence z0 71"n_l (.':gno-1 (,.)) "Kn_ (Sn-2) Finally, we obtain that the mapping.
Hence, it is clear that the spaces E2, E 3 and E admit only parallel z-structures because a's exist and belong to the same homotopy equivalence class.
(S,, For n > 5 we have 7r,_ Z, so the spaces E for n > 5 admit non-parallel s-structures. To find out the number of the non-parallel z-structures of a Euclidean space E (n > 5) xe consider the case n 5 Composing again the 4-dimensional spaces we have o(e): Eo(e, 1) o If 0 and are the classes of Z2 then every/4 (e, 3) and E, (e, 3) corresponds to 0 or Hence :gO and