The group Aff(2)

 Let Aff(2) be the set of collineations, i.e. the set of all bijections of E² which map lines to lines. Aff(2) has the following properties. Let F in Aff(2): 

1. F preserves parallel lines: for if l is parallel to m then l 0 m is empty. Since F is one to one, lF 0 mF is empty.
2. F preserves triangles: for if ABC is a triangle, AFBF, BFCF and CFAF are distinct lines which intersect in three distinct points.
3. Aff(2) is a group: it is clear that the composite of two collineations is a collineation. Suppose F is a collineation and l is a line. If lF^-1 is not a line, it contains a triangle AF^-1BF^-1CF^-1, where A, B and C are on l, whose image under F is a triangle ABC, a contradiction. Thus F^-1 is a collineation, so Aff(2) is closed under composites and inverses.
4. Aff(2) is the set of all bijections which preserve triangles: Let F in Sym(2) preserve triangles. To show that F in Aff(2), let l be a line. If lF^-1 is not a line, it contains a triangle AF^-1BF^-1CF^-1 whose image under F is a triangle ABC contained in l, a contradiction. Hence F^-1 is a collineation, so F is too.
5. Consider E² with a coordinate system. F is uniquely a product of a collineation fixing 0 and a translation: for F is a unique product R₀.S of a symmetry R₀ fixing zero and a translation S, by Theorem 16.2. But every translation is a collineation, so S and S^-1 are in Aff(2) and hence R₎ is also in Aff(2).
6. T is a normal subgroup of Aff(2) and hence F = S'.R₀ for some translation S'.

In the following theorem, L(u,v) means the line through u and v where u and v are point vectors.

Theorem 17.1 If F in Aff(2) fixes 0, the F is a non-singular linear transformation.

Proof. Let u and v be point vectors. The lines L(0,v) and L(u,u+v) are parallel, so the lines L(0,vF) and L(uF, u+vF) are parallel. Similarly, the lines L(0,uF) and L(vF, u+vF) are parallel.

 Hence if {u, v} is linearly independent, then {0, uF, (u+v)F, vF} is a parallelogram. But {0, uF, (uF+vF), vF} is also a parallelogram, so uF+vF = (u+v)F.

 On the other hand, if either u or v = 0 then clearly uF+vF = (u+v)F, so assume v = ru for some non-zero scalar r and let w be independent of u. By the previous parafraph, we have (u+v)F+wF = (u+v+w)F = uF+(v+w)F = uF+vF+wF, so uF+vF = u+v.

 Now to show that scaling is preserved by F, let u and v be linearly independent and let r be a scalar. Since ru is in L(0,u), (ru)F is in L(0,uF), say (ru)F=r'(uF) for some scalar r' with r' = 0 if and only if r = 0. Similarly (rv)F=r''(vF) and we need to show r = r' = r'' when all are non-zero.

 Since L(ru,rv) is parallel to L(u,v), L((ru)F,(rv)F) is parallel to L(uF,vF). Hence L( r'(u)F, (r''(v)F) = L(uF,vF), so L(0, r'(u)F - r''(v)F) = L(0, uF-vF). Hence r'(u)F - r''(v = s(uF-vF) for some scalar s and hence r' = r''.

 Now let w be in R². Either {u, w} or {v, w} is linearly independent, so for all r in R, (rw)F = r'(wF). Let us examine properties of the function r -> r' of R into R.

 
1. (r+s)' = r'+s': for all u in R², (r+s)'uF = r'uF + s'uF = (ru)F + (su)F = (ru + su)F = (r+s)uF = (r+s)'(uF).
2. (rs)' = r's' : for all u in R², ((r's')u)F = r'(su)F = (rsu)F = (rs)'uF
3. r -> r' is 1-1 and onto: let r be in R and u in R². Since (L(0,u))F = L(0, uF), (ru)F = s(uF) for some s in R, so s = r'. If r -> r' and s -> r', then (ru)F = (su)F, so r = s.
4. We have shown r -> r' is an automorphism of the field R. All the proof so far works for any field. The rest is specific to R-- r = r'. Let 
5. H: R -> R be an automorphism. Since H: 1 -> 1, H: n -> n for all integers n, so H: r -> r for all rationals r. Let s be in R. For all rationals r < s, s - r > 0, so s - r = t² for some t in R. Hence sH - rH = sH - r = (tH)² so sH > r. Similarly, r > s, then r > sH. So if {r_i} is a sequence of rationals converging to s, then {r_i} is a sequence of rationals converging to sH, so s = sH.

Corollary 17.2 Every affine transformation is uniquely a product of a linear transformation and a translation. Thus Aff(2) is the split product of GL(2,R) by T.

 We say a function F: R² -> R² preserves ratios if for all u and v in R², and for all r in R, (ru + (1-r)v)F = ruF + (1-r)vF. Since linear transformations and translations preserve ratios, the point of the next Theorem is the converse:

Theorem 17.3 Let F be in Sym(2). F is in Aff(2) if and only if F preserves ratios.

Proof. Let l be any line and u and v two points on l. Then l = L(u,v) and every point w on l can be uniquely represented as w = ru + (1-r)v for some r in R. Hence wF = ruF + (1-r)vF so w is on L(uF,vF). Thus F is a collineation.

 The product of affine maps: we know Aff(2) is a group, but we need an explicit description of FG for F and G in Aff(2).

Theorem 17.4 Let F = RS and G = UT, where R and U are linear transformations, S is translation by s and T is translaion by t. Then FG = RUV, where V is translation by sU + t.

 Proof. For all u in R², (uF)G = (uR + s)UT = (uRU + sU) + t = uRU + (sU + t).
 

Higher dimensional affine groups

1. Aff(n) is the group of line preserving symmetries of Eⁿ.
2. Aff(n) is the group of triangle preserving symmetries of Eⁿ.

Aff(n) is the group of ratio preserving symmetries of Eⁿ.
3. Each F in Aff(n) can be uniquely represented as F = R.S where R is a linear transformation and S a translation.

Theorem 17.5 There is a monomorphism 
     X: Aff(n) -> GL(n+1,R).

Proof. Let B_n = {e₁,e₂,..,e_n} be the standard basis of Rⁿ and B_n+1 = {e₁,e₂,..,e_n+1} the standard basis of Rⁿ⁺¹.

 Let F = RS be in Aff(n), where R is linear and S is a translation by u. Let r = (r_ij) be the matrix of R and let u = (u_k) be the component vector of u.

 Define FX to be the element of GL(n+1,R) whose partitioned matrix M has top n rows (r,0) and bottom row (u,1). Then det M = det r so M is invertible and FX is the identity matrix if and only if F = I. It is routine to check that X preserves multiplication, so X is a monomorphism 

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 Last update Dec 10, 1999

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