By Penrose R.

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**Extra info for A lecture on 5-fold symmetry and tilings of the plane**

**Example text**

Consequently, G ×H X is a free G-space if and only if X is a free H-space. Proof. We leave the proof as an easy exercise for the reader. 1 Let X = [−1, +1] and Z2 act on X as multiplication by ±1. We identify Z2 with the subgroup {0, π} of SO(2) = R/2πZ. The twisted product SO(2) ×Z2 X is homeomorphic to the Mobius band. The induced SO(2)-action on SO(2) ×Z2 X has one singular orbit — the centre circle of the Mobius band. 1 Construct an SO(2)-action on the Klein bottle which has precisely two singular orbits, both consisting of points with isotropy Z2 .

3 Orbit stratiﬁcation for standard D4 action embeds as a subgroup of the symmetric group Sn (elements of Dn permute the vertices of the regular n-gon). In case n = 3, we have D3 = S3 . For n > 3, Dn = Sn . (The equilateral triangle is the only regular n-gon, n > 2, such that the distance between all distinct vertices is equal. 6 Representations We continue to assume that G is a compact topological group. Let F denote the ﬁeld of real or complex numbers and V be a ﬁnite dimensional vector space over F.

2) di | |G|, i ∈ k. 24–27]. 2 (1) For n ≥ 1, we have previously shown that there are n inequivalent C-representations of Zn , all 1-dimensional. 2) implies there are no other C-representations of Zn . (2) There are ﬁve distinct irreducible C-representations of D4 . 1(3), we gave a 2-dimensional irreducible C-representation of D4 . 2) implies there are exactly ﬁve irreducible Crepresentations of D4 . 1 applies to show that if X is an H-representation and Z a G-representation then G LG (Z, iG H X) ≈ LH (resH Z, X) (Frobenius reciprocity).