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The Euclidean Group
1 fR,u(fR′,u′^ (x)) = fR,u(R′x + u′) = R(R′x + u′) + u = (RR′)x + (Ru′^ + u) = fRR′,Ru′+u(x), so (R′′, u′′) = (RR′, Ru′^ + u) will work and is uniquely determined by the composed function.
2 If x′^ = fR,u(x) = Rx + u, then x = R−^1 (x′^ − u) = R−^1 x′^ + (−R−^1 u) = fR− (^1) ,−R− (^1) u(x), so (R′, u′) = (R−^1 , −R−^1 u) will work and is uniquely determined by the composed function.
3 There is of course a group of all invertible functions on Rn, the permutation group Rn!. The previous problems prove that E(n) is a subset of Rn! that is closed under the operations of composition and in- version. Furthermore, f 1 , 0 (x) = 1x + 0 = x, so E(n) also owns the identity function. Therefore, E(n) is a subgroup of Rn!, in particular a group.
4 G is constructed as the range of a function from Rn^ to E(n), which is invertible by the fundamental prop- erty of ordered pairs. First, (1, u)(1, u′) = (11, 1 u′^ + u) = (1, u + u′), so G is closed under multiplication and the correspondence between G and Rn^ preserves that operation. Next, (1, u)−^1 = (1−^1 , − 1 −^1 u) = (1, −u), so G is closed under inverses and the correspondence between G and Rn^ preserves that opera- tion. Finally, the identity element of E(n) is (1, 0), so G owns the identity element and it corresponds to the identity element 0 of Rn. Therefore, G is a subgroup of E(n) that is isomorphic to Rn. Also, (R, u′) × (1, u)(R, u′)−^1 = (R 1 , Ru + u′)(R−^1 , −R−^1 u′) = (RR−^1 , −RR−^1 u′^ + Ru + u′) = (1, Ru), which belongs to G, so the subgroup G is normal. H is constructed as the range of a function from O(n) to E(n), which is invertible by the fundamental property of ordered pairs. First, (R, 0)(R′, 0) = (RR′, R0 + 0) = (RR′, 0), so H is closed under multiplica- tion and the correspondence between H and O(n) preserves that operation. Next, (R, 0)−^1 = (R−^1 , −R−^1 0) = (R−^1 , 0), so H is closed under inverses and the correspondence between H and O(n) preserves that oper- ation. Finally, the identity element of E(n) is (1, 0), so H owns the identity element and it corresponds to the identity element 1 of O(n). Therefore, H is a subgroup of E(n) that is isomorphic to O(n). (1, u)(R, 0) = (1R, 10 + u) = (R, u), so by the fundamental property of ordered pairs, every element of E(n) is a unique product of an element of G and an element of H.
5 Let u be f (0), and let Rx be f (x) − u for x ∈ Rn. Note that R0 = u − u = 0 and
|Rx − Ry| = |(f (x) − u) − (f (y) − u)| = |f (x) − f (y)| = |x − y|,
so R preserves the origin and lengths. Therefore, R must be an element of O(n). Since f (x) = Rx + u, the desired result follows.
The Galilei Group
6 Set f = fR,u and f ′^ = fR′,u′^. Then
Ff,v,s(Ff ′,v′,s′ (x, t)) = Ff,v,s(R′x + u′^ + v′t, t + s′) = (R(R′x + u′^ + v′t) + u + v(t + s′), t + s′^ + s) = ((RR′)x + (Ru′^ + u + vs′) + (Rv′^ + v)t, t + (s′^ + s)) = FfRR′,Ru′+u+vs′ ,Rv′+v,s′+s,
so (f ′′, v′′, s′′) = (fRR′,Ru′+u+vs′ , Rv′^ + v, s′^ + s) will work and is uniquely determined by the composed function.
7 Set f = fR,u. Then if (x′, t′) = Ff,v,s(x, t) = (f (x) + vt, t + s) = (Rx + u + vt, t + s), then t = t′^ − s = t′^ + (−s) and x = R−^1 (x′^ − u − vt) = R−^1 (x′^ − u − v(t′^ − s)) = R−^1 x′^ + (sR−^1 v − R−^1 u) + (−R−^1 v)t′, so (x, t) = FfR− (^1) ,sR− (^1) v−R− (^1) u,−R−^1 v,−s, so (f ′, v′, s′) = (fR−^1 ,sR−^1 v−R−^1 u, −R−^1 v, −s) will work and is uniquely determined by the composed function.
8 There is of course a group of all invertible functions on Rn+1, the permutation group Rn+1!. The previous problems prove that G(n + 1) is a subset of Rn+1! that is closed under the operations of composition and inversion. Furthermore, F 1 , 0 , 0 (x, t) = (x + 0t, t + 0) = (x, t), so G(n + 1) also owns the identity function. Therefore, G(n + 1) is a subgroup of Rn+1!, in particular a group.
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9 Given points p, q, r ∈ Rn+1, de ne vectors v = q − p and w = r − p tangent to Rn+1^ at p. Write v = (vx, vt) and w = (wx, wt). I claim that Galilean transformations preserve the time dierence wt − vt and the rela- tive speed |wx/wt − vx/vt|. Now, the latter is not de ned if vt or wt is 0 , but since these denominators are also time dierences, |vtwx − wtvx| should be preserved as well. Furthermore, I claim that any transforma- tion that preserves these is Galilean. Thus in terms of the original points p = (px, pt), q = (qx, qt), and r = (rx, rt), the claimed complete invariant is the ordered pair (|qtrx − ptrx − rtqx + ptqx + rtpx − qtpx|, rt − qt). First, let me verify that this is an invariant. Applying Ff,v,s, where f = fR,u, the rst component of the claimed invariant becomes |(qt + s)(Rrx + u + vrt) − (pt + s)(Rrx + u + vrt) − (rt + s)(Rqx + u + vqt)
- (pt + s)(Rqx + u + vqt) + (rt + s)(Rpx + u + vpt) − (qt + s)(Rpx + u + vpt)| = |qtRrx − ptRrx − rtRqx + ptRqx + rtRpx − qtRpx| = |R(qtrx − ptrx − rtqx + ptqx + rtpx − qtpx)| = |qtrx − ptrx − rtqx + ptqx + rtpx − qtpx|,
since R ∈ O(n) is linear and preserves lengths. Meanwhile, the second component becomes (rt + s) − (qt + s) = rt − qt. Thus, this is indeed an invariant. Now suppose that F : Rn+1^ → Rn+1^ preserves this invariant. Let s be the second component of F (0, 0). Setting q = (0, 0) and r = (x, t), I see from the second component of the invariant that t + s is the second component of F (x, t). Thus let Fx(x, t) be the rst component of F (x, t), so that F (x, t) = (Fx(x, t), t + s). Now let u be Fx(0, 0), let v be Fx(0, 1) − u, and let Rx be Fx(x, 1) − u − v for x ∈ Rn. Note that R0 = (v + u) − u − v = 0. Setting p = (0, 0), q = (y, 1), and r = (x, 1),
|Rx − Ry| = |(Fx(x, 1) − u − v) − (Fx(y, 1) − u − v)| = |Fx(x, 1) − Fx(y, 1)| = |(1 + s)Fx(x, 1) − sFx(x, 1) − (1 + s)Fx(y, 1) + sFx(y, 1) + (1 + s)u − (1 + s)u|. Applying the rst component of the invariant, this equals | 1 x − 0 x − 1 y + 0y + 1 · 0 − 1 · 0 | = |x − y|. Thus, R preserves the origin and lengths, so it must be an element of O(n). Now setting p = (0, 0), q = (x/t, 1), and r = (x, t), the rst component of the invariant says that | 1 x − 0 x − t(x/t) + 0(x/t) + t 0 − 1 · 0 | = |(1 + s)Fx(x, t) − sFx(x, t) − (t + s)(R(x/t) + v + u) + s(R(x/t) + v + u) + (t + s)u − (1 + s)u|, or 0 = |Fx(x, t) − tR(x/t) − tv − u|. Since every element of O(n) is linear, it follows that Fx(x, t) must be (Rx + u + vt, t + s). Thus F = FfR,u,v,s ∈ G(n + 1).
The Free Particle Since our Galilean transformations have been passive, not active, time translation acts as s(x, p) = (q − sp/m, p), not (q + sp/m, p).
10 In G(n + 1), (f, v, s) = (1, v, s)(f, 0 , 0) = (1, 0 , s)(1, v, 0)(f, 0 , 0), while I already know that (R, u) = (1, u) × (0, R) in E(n). That is, (R, u, v, s) = (1, 0 , 0 , s)(1, 0 , v, 0)(1, u, 0 , 0)(R, 0 , 0 , 0). Then (R, u, v, s)(q, p) = (1, 0 , 0 , s)(1, 0 , v, 0)(1, u, 0 , 0)(R, 0 , 0 , 0)(q, p) = (1, 0 , 0 , s)(1, 0 , v, 0)(1, u, 0 , 0)(Rq, Rp) = (1, 0 , 0 , s)(1, 0 , v, 0)(Rq + u, Rp) = (1, 0 , 0 , s)(Rq + u, Rp + mv) = (Rq + u − s(Rp + mv)/m, Rp + mv) = (Rq + u − sRp/m − sv, Rp + mv).
11
(R, u, v, s)
(R′, u′, v′, s′)(q, p)
= (R, u, v, s)(R′q + u′^ − s′R′p/m − s′v′, R′p + mv′) = (R(R′q + u′^ − s′R′p/m − s′v′) + u − sR(R′p + mv′)/m − sv, R(R′p + mv′) + mv) = (RR′q + Ru′^ − s′RR′p/m − s′Rv′^ + u − sRR′p/m − sRv′^ − sv, RR′p + mRv′^ + mv) = (RR′q + Ru′^ + u + s′v − s′RR′p/m − sRR′p/m − s′Rv′^ − s′v − sRv′^ − sv, RR′p + mRv′^ + mv) = ((RR′)q + (Ru′^ + u + s′v) − (s′^ + s)(RR′)p/m − (s′^ + s)(Rv′^ + v), (RR′)p + m(Rv′^ + v)) = (RR′, Ru′^ + u + s′v, Rv′^ + v, s′^ + s)(q, p) =
(R, u, v, s)(R′, u′, v′, s′)
(q, p).
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