Properties of Aff(n)

1. F preserves triangles: For if ABC is a triangle in Eⁿ then {AF, BF, CF} is a set of three distinct points in Eⁿ. Suppose BF is in the line L(AF, CF). Since L(A, C)F is the unique line containing AF and CF, there exists D in L(A,C) such that DF = BF. But since F is 1-1, D = B, a contradiction. Hence AF BF CF is a triangle in Eⁿ.

 

 

A configuration in Eⁿ is an arrangement of lines and their intersections. Clearly F preserves configurations. So far we have avoided specific axioms for Eⁿ, relying instead on intuition. But since we need it in the next proof we state explicitly Pasch's Axiom: Let P be a plane in Eⁿ and let ABC be a triangle in P. Then 

P={D in Eⁿ: there exists a line through D meeting at least two sides of ABC}
 
 

2. F preserves planes: Let P be a plane and ABC a triangle in P. Let D be any point in P. The configuration consisting of the lines L(A,B), L(A,C), L(B,C) and a line through D meeting two sides of triangle ABC exists by Pasch's Axiom and is preserved by F. Hence DF is in the plane of triangle AF BF CF, so PF is a plane.

 

 

(In fact F preserves k-dimensional subspaces and their cosets but we don't need to prove that in this course.)
 
 

3. F preserves parallels: If lines L and M are parallel, then L 0 M and hence LF 0 MF are empty. Furthermore, the plane containing L and M is mapped by F into the plane containing LF and MF by 3., so LF is parallel to MF.
5. Aff(n) is a group: clearly the composite of line preserving bijections is a line preserving bijection. Suppose F is in Aff(n) but F^-1 is not. Then F^-1 does not preserve all lines, so there is a line L in Eⁿ and points A, B and C on L such that AF^-1, BF^-1, CF^-1 is a triangle. But F preserves triangles, so ABC is a triangle, a contradiction. Hence Aff(n) is closed under products and inverses, so is a group.
6. If F fixes 0, then F is a linear transformation: Let u and v be non-zero vectors in Rⁿ and let r be a scalar. The proof that (u + v)F = uF + vF and r(uF) = (ru)F is exactly the same as the proof of Theorem 17.1, replacing E² by the plane containing u and v.
7. F is a unique product R.S where R is linear and S is a translation.
8. If F = R.S_A and G = R'.S_B are in Aff(n), then F.G = R.R'.S_C where C = AR' + B: For any u in Eⁿ, uFG = (uR + A)G = (uR + A)R' + B = uRR' + (AR'+B).
9. A function F : Eⁿ -> Eⁿis said to preserve ratios if for all r in R and for all u and v in Eⁿ , (ru + (1-r)v)F = r(uF + (1-r)vF. F preserves ratios since linear transformations and translations do.

1.  F 1  Aff(n).
2. F preserves triangles.
3. F preserves ratios.

 2. -> 1.: Suppose F preserves triangles. We show that F^-1 is affine. If not, then there is a line L containing points A, B and C such that AF^-1, BF^-1, CF^-1 is a triangle. Hence ABC is a triangle, a contradiction. Hence F^-1 is affine, so by 5. above, F is affine.

 3. -> 1.: It suffices to show that if some point w is on a line L, then wF is on the line LF. So let w be on L(u,v). Then w = ru + (1-r)v for some scalar r. Since F preserves ratios, wF is on L(uF,vF). 

For many applications it is useful to show that a group is a subgroup of a linear group. This is true for Aff(n).

Theorem 18.2 There is a monomorphism of Aff(n) into GL(n+1,R).

Proof. Let B_n and B_n+1 be the standard bases for Rⁿ and Rⁿ⁺¹. Let F = R.S 1 Aff(n) and let r = (r_ij) be the standard matrix of R and a = (a_i) the component vector of A. Map F into the n+1 x n+1 matrix whose first n rows are the rows of r with 0 appended and whose n+1 row is a with 1 appended.

 It is routine to check that this matrix is the standard matrix of an element of GL(n+1,R) and that the map is a homomorphism with kernel I.

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 Last update Jan 11, 2000

Author: Phill Schultz, schultz@maths.uwa.edu.au