By Hiram Paley

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**Example text**

The reflection 4J in P(a, t) (or, as we sometimes say, in (x . a) = t) is defined in the usual way; that is 4J(x) = x + Aa, where the real parameter A is chosen so that t(x gives the explicit formula + 4J(x» is on P(a, t). This 4J(x) = x - 2[(x . Rnand, of course, 4J( (0) = 00. Again, 4J acts on iRn, 4J2(X) = x for all x in iRn and so 4J is a 1-1 map of iRn onto itself. Also, 4J(x) = x if and only if x E P(a, t). It is clear that any reflection 4J (in a sphere or a plane) is continuous in iRn except at the points 00 and 4J -1( (0) where continuity is not yet defined.

First, we consider the case when L is the plane Xn = 0 in ~n. Let L' = Sea, r) where a ELand r > O. As 00 E L, ¢ fixes 00: thus ¢ maps L' to a Euclidean sphere, say Lit = S(b, t). As a E L we have (L, L') = O. 3 yields (L, Lit) = 0 and so bEL: thus an = bn = O. Each point of L n L' is fixed by ¢, thus (Xl - a l )2 + ... + (x n - b l )2 + ... + (x n - l - a n _I)2 = r2, if and only if (Xl - l - bn _ I )2 = t 2 • We conclude that a = band t = r: hence ¢ maps L' onto itself. Next, we select any X not in L and let y = ¢(x).

Now let Y = (Yo, ... , Yn) be any smooth curve on Q. Thus for all t, YO(t)2 = Y1(t)2 + ... + Yn(t)2 + 1, so differentiating, Yo(t)Yo(t) = Y1(t)Y1(t) + ... + Yn(t)Yn(t), (more briefly, q(y, y) = 1 so q(y, y) = 0). We dedllce that . Y) -_ (Y1Y1 + ... 2) q( y, Y1 + . . + Yn Yo S (I yJ)(I yJ)jy~ - (L: yJ) = -(L: YJ)/Y~ sO, the summations being over j = 1, ... , n. Observe also that a strict inequality holds unless Y1 = ... = Yn = 0 in which case, Yo = 0 also. It follows that we can construct a metric on Q in the usual way by the line element + ...