a monotonicity theorem
Suppose that
(1) K is the intersection of an i-line with
the disk,
(2) l(1),..,l(m) are fixed positive numbers, and
(3) A(0),..A(m)
lie in order on the minor (or finite) arc
A(0)A(m)
of K such that l(i) = d(A(i-1),A(i)), i = 1,..,m.
Then, as the size of K increases,
L = d(A(0),A(m)) increases
continuously, and tends to l(1)+..+l(m).
proof
We shall assume that A(0) = O and that the euclidean circle determined by
K
has its centre on the positive real axis, and euclidean radius R. This can
be
achieved by mapping A(0) to O and rotating about O. By reflecting in the
real axis if neccesary, we can assume that A(1),..,A(m) lie above the real
axis since they lie in order on the minor or finite arc A(0)A(m).
For 2R < 1, K is a hyperbolic circle of radius r = arctanh(2R).
This increases with R, and tends to infinity as 2R tends to 1.
Let P be its hyperbolic centre (then P is also on the real axis).
Observe that, if A,B on K with x = d(A,B) and the hyperbolic angle APB is θ,
then sinh(½x) = sinh(½r)sin(½θ). Let θ(i) be the hyperbolic angle A(i-1)PA(i).
Then sinh(½L) = sinh(½r)sin(θ(1)+..+θ(m)).
Using the trigonometric identity,
sinh(½L) = ΣiP(m,i)S(i)sinh(½l(i)),
From our assumption about the minor arc, ½θ(1)+..+½θ(i) < ½π
so the cosines in P and S are all positive. As R increases, so does r, so the
θ(i) decrease to zero. Hence the P and S decrease to 1.
Thus, as R tends to ∞ sinh(½L) decreases with limit sinh(½l(1))+..+sinh(½l(m)).
For 2R = 1, we have a horocycle at 1.
Then, by the horocycle theorem
sinh(½L) = sinh(½l(1))+sinh(½l(m))
For 2R > 1, we get a hypercircle, meeting the boundary at z and z*. Either
by calculation, or from a picture, the width d of the hypercircle decreases
as 2R increases. It tends to infinity as 2R tends to 1 (as z,z* tend to 1), and
tends to 0 as 2R tends to infinity, as K tends to the imaginary axis.
Let H be the hyperbolic line joining z and z*.
Let a(i) denote the hyperbolic length of projection of A(i-1)A(i) on H,
and a = a(1)+..+a(m). This is the length of the projection of A(0)A(m) on H.
From the saccheri theorem, sinh(½L) = cosh(d)sinh(½a) and, for each i,
sinh(½l(i)) = cosh(d)sinh(½a(i)). By the hyperbolic trigonometric identity,
sinh(½a) = sinh(½a(1)+..+½a(m)) = ΣiPh(m,i)Sh(i)sinh(½a(i)), where the
Ph and Sh depend on values of cosh(½a(i)+..½a(i)). Multiplying by cosh(d),
sinh(½L) = ΣiPh(m,i)Sh(i)sinh(½l(i))
As R increases towards ∞, d decreases towards 0, so that cosh(d) decreases
towards 1. It follows that each a(i) increases towards l(i) and the Ph and Sh
increase towards their values when each a(i) is replaced by l(i). Thus the limit
for sinh(½L) is sinh(½l(1)+..+½l(m)) by the identity again. Thus, L tends to
l(1)+..+l(m).
Finally, suppose that 2R decreases towards 1. Then d increases towards ∞.
For each i, sinh(½a(i) = sinh(½l(i))/cosh(d), so each a(i) decreases towards 0.
It follows that the Ph and Sh decrease toward 1, so sinh(½L) decreases towards
sinh(½l(1))+sinh(½l(m)) - the value for 2R = 1.
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