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L(0) = L(−x) + (1/2)(L(x) − L(−x)),
or more succinctly (since L(0) = 0):
L(−x) = −L(x). (2)
Further, if r ∈ (0, 1) and x is again some nonzero vector, then 0, rx and x are distinct colinear points with
L(rx) = L(0) + r(L(x) − L(0)) = rL(x).
If r > 1, then using the above we have
L(x) = L(r−^1 (rx)) = r−^1 L(rx).
The above in conjunction with (2) show that for any r ∈ R and x ∈ Rn^ we get
L(rx) = rL(x). (3)
Finally, take x and y distinct. Then (1) gives
L((x + y)/2) = L(x) + (1/2)(L(y) − L(x)),
which, when combined with (3) yields
L(x + y) = L(x) + L(y). (4)
Equations (4) and (3) show that L is linear. The function L is linear, so there is a matrix R ∈ Mn(R) such that L(x) = Rx. Recall the polarization identity for the inner product on Rn:
(u, v) = (1/4)(|u + v|^2 − |u − v|^2 ).
The fact that L is an isometry gives:
(Rx, Ry) = (x, y)
for all x, y ∈ Rn. But then:
(x, (RT^ R − I)y) = 0
for all x and y, and so RT^ R = I and we see that R ∈ O(n). We wrap all this up by noting that
f (x) = L(x) + f (0) = Rx + f (0)
and we wipe our hands and call it a day.
The Galilei Group
We define a map
i : G(n + 1) → M
where M denotes the collection of functions from Rn+1^ to itself, by
X = (R, u, v, s) 7 → FX with FX (x, t) = (Rx + u + tv, s + t).
Let’s show that this map is an injection. Given X = (R, u, v, s) and Y = (R′, u′, v′, s′) with i(X) = i(Y ) we see that
FX (0, 0) = (u, s) = (u′, s′) = FY (0, 0)
so u = u′^ and s = s′. Computing
FX (x, 0) = (Rx + u, s) = (R′x + u′, s′) = FY (x, 0)
we see that Rx = R′x for all x ∈ Rn^ and we have R = R′. Finally, noting the previously obtained equalities together with the fact that FX (0, 1) = FY (0, 1) yields v = v′. Thus X = Y and our map is an injection. We will henceforth identify G(n + 1) with its image in M.
- Given two elements X = (R, u, v, s), Y = (R′, u′, v′, s′) ∈ G(n + 1), compute
FX ◦ FY
and show that the composition is again in G(n + 1).
Back to computation:
FX ◦ FY (x, t) = FX (R′x + u′^ + tv′, s′^ + t) = (R(R′x + u′^ + tv′) + u + (s′^ + t)v, s + s′^ + t) = (RR′x + (Ru′^ + u + s′v) + t(Rv′^ + v), (s + s′) + t) = FZ (x, t)
where Z = (RR′, Ru′^ + u + s′v, Rv′^ + v, s + s′).
Thus FX ◦ FY is in the image of G(n + 1) in M. Because of remarks made above, Z must be unique.
- Compute Y −^1 for Y ∈ G(n + 1).
We are looking for a X ∈ G(n + 1) such that FX satisfies FY ◦ FX = (I, 0 , 0 , 0) = FX ◦ FY. Using the notation above for X and Y this amounts to solving the following system of equations for R, u, v and s.
RR′^ = I Ru′^ + u + s′v = 0 Rv′^ + v = 0 s + s′^ = 0
The solutions are readily obtained to be R = R′T^ , s = −s′, v = −R′T^ v′, and u = −R′T^ u′^ + s′R′T^ v′. So we claim that
To see this, note that π 1 gives maps that shift the first coordinate of (q, p) by a fixed vector, π 3 gives maps that shift the second by a fixed vector and π 4 yields maps which mix the two coordinates so that both q and p appear in the first slot in the output; π 2 gives maps which rotate both coordinates. It is simply impossible for any three types of these of maps to accomplish the operation of the remaining type of map unless all maps being considered are the identity map! Think this through a bit... Define the set G = Rn^ × O(n) × Rn^ × R. Let σ be any permutation on the set of four elements and define Fσ : G → Aut(X) by
Fσ (x 1 ,... , xn) = πσ(1) (xσ(1)) ◦ · · · ◦ πσ(4)(xσ(4)).
Suppose
Fσ (x 1 ,... , xn) = Fσ (y 1 ,... , yn).
Then it would follow that
πσ(1)
xσ(1)y− σ(1)^1
◦ · · · ◦ πσ(4)
xσ(4)y− σ(4)^1
= I.
Now, the above requires that the inverse of any given factor lie in the product of the images of the other three. By the argument about the images of the πi, this is impossible unless all of the factors are the identity map. But then xi = yi since each πi was an imbedding. Whence Fσ is an injection from G to Aut(X). Finally, give G group structure by pullback along Fσ.^1 2 With this structure we see that Fσ : G → Aut(X) is now an action of G on X and further that G can be identified with its image in Aut(X) which, by the opening comments of this diatribe, comes with a de facto action on X. Oh, based on the work I have done, here is the formula:
(R, u, v, s)(q, p) = Fσ (R, u, v, s)(q, p)
which we admit, is not very helpful. So it goes...
(^1) What’s really going on here is that we’re saying it doesn’t matter which particular way we choose to combine the actions—there will be a group structure on G which will be compatible with this action, and not only that, but will reduce to the original four actions when we consider, say O(n) as a subgroup of G. (^2) What’s really going on here (I think) is that we are implicitly defining the Galilei group to be any one of the isomorphic copies of the semi direct product of Rn, Rn, R and O(n) found in AutX.
