Circular inversions = hyperbolic isometries

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1 Circular inversions = hyperbolic isometries Recall how all isometries of the Euclidean plane are products of at most three line reflections. In the (Beltrami-Poincaré model of the) hyperbolic plane, we can easily describe the corresponding h-reflection across an h-line, provided such a line is one of the vertical open rays with endpoint X on the x-axis. If the Euclidean equations of such a ray are x = a, y > 0, then it maps any point P (x, y) to the point P (a x, y) (so that the midpoint of P P is (a, y), on the axis of the h-reflection). But what if the axis of the h-reflection is a Euclidean semicircle X ) Y with center O on the x- axis? Such an h-relection should fix all points on the semicircle (the h-line which is the axis of this h- reflection) and carry points interior to the semicircle to points exterior to it, and vice versa. This requires us to consider the concept of (Euclidean) circular inversion. Let O be center of a circle C in the Euclidean plane of radius r. Then we define inversion in C to take the arbitrary point P on a ray with endpoint O to that point P on the same ray so that O P OP = r. Theorem Inversion in a circle C with center O fixes all rays with endpoint O. //

2 Corollary Inversion in a circle C with center O fixes all lines passing through O. // Theorem Let H be a half-plane determined by a line l. Then inversion in any circle C with center O on l fixes all points on the semicircle that lies within H and carries points inside this semircle to points outside, and vice versa. Proof If P is any point in H, then its image P under the inversion lies on the ray O P r, hence is in H as well, and O P = r OP, so OP = r O P = r; OP < r O P > r; and OP > r O P < r. // This is enough to suggest that inversion in a circle centered at a point O on the x-axis is the right way to define h-reflection in the h-line which corresponds to that portion of the circle that lies in the upper half plane. The remainder of our discussion here will be to show that the other isometric properties we expect of h-reflections hold in the model under this definition. To accomplish this, we will use analytic geometry (the algebra of equations for curves). We coordinatize the plane so that our x-axis corresponds to the x-axis of the Beltrami-Poincaré model and that the origin is the center O of our circle. Then the equation of the circle is

3 x + y = r and inversion r in the circle carries the point P (x, y) on a ray O P to a point P (, y ) on the same ray, whence (, y ) = (kx,ky ) for some positive constant k. Since O P OP = r, we have O P OP = r 4 ( )( x + y ) = r 4 ((kx) +(ky) )( x + y ) = r 4 k (x + y ) = r 4 so k = r x + y, whence x = kx = r x x + y, y = ky = r y x + y. These last two equations define inversion in the circle in terms of coordinates. Note also that the image of P under the inversion is P, so we have r x =, y = r y. We know that lines through O are fixed by inversion in C. Any other line l has an equation of the form ax +by +c = 0 with c 0. Substituting the last set of equations into that for the line gives

4 ar + br y +c = 0 c( ) +ar +br y = 0 + ar c + br c y = 0 The last of these is the equation of a circle (since we can complete the square in and y ) that passes through O. This proves the first part of the Theorem Inversion in C carries any line l that does not pass through O to a circle that does pass through O with center on the perpendicular from O to l. Moreover, if l is exterior to C, then its image circle is interior to C. If l is tangent to C, then so is its image circle, and at the same point; and if l is secant to C, cutting it in points A and B, then its image circle is the circle determined by the noncollinear points A, B and O. // Corollary Inversion in C carries any circle L that passes through O to a line that does not pass through O perpendicular to the line through O and the center of L. If L is interior to C, then its image line is exterior to C. If L is tangent to C, then so is its image line, and at the same point; and if L is secant to C, cutting it in points A and B, then its image line is the one determined by A and B. //

5 Now suppose that a circle L does not pass through O. Then it has an equation of the form x + y +ax +by = c with c 0. To find its image we substitute the equations x = r + r y r 4 ( ) ( ) + ar r, y = + ar + +br y r y : br y = c = c r 4 +ar +br y = c( ) ar c br c y = r 4 and the last of these is the equation of another circle! Theorem Inversion in C carries any circle L that does not pass through O to another circle L that does not pass through O. The centers of L and L are collinear with O. //

6 Recall that the angle measure between h-lines at a vertex P is defined to be the standard curvilinear angle measure at P, that is, the measure between the tangent lines to the h-lines at P. We now show that Theorem An h-reflection (circular inversion) preserves angle measure. Proof Suppose h-lines L 1 and L meet at P and that t 1 and t are the respective tangent lines there, meeting at an angle θ. Then the images of L 1 and L under inversion in C carries L 1 and L into h-lines L 1 and L meeting at P and carries t 1 and t into circles t 1 and t meeting at P in the same curvilinear angle θ that L 1 and L meet in at P. But t 1 and t also meet at O and in the same angle θ, and the lines s 1 and s tangent to t 1 and t at O are carried by the inversion to circles s 1 and s that meet at P and have t 1 and t as tangent lines. Now the radial lines from O through the centers of the circles t 1 and t are perpendicular to the lines t 1 and t, so t 1 and t are parallel to the lines s 1 and s, whence the angle θ between t 1 and t equals the angle θ between s 1 and s. That is, the angle between L 1 and L is the same as that between L 1 and L. //

7 Now to show that h-reflections preserve h-distance. Lemma Let A, B be the respective images of A, B under inversion in circle C centered at O. Then ΔO A B ~ ΔOBA. Proof O A OA = O B OB, since both equal the square of the radius of C. Thus O A / O B = OA / OB. Since the angles at O in both triangles ΔO A B and ΔOBA are congruent, we have ΔO A B ~ ΔOBA by SAS similarity. // Theorem Circular inversions are conformal transformations, i.e., they preserve cross ratios. Proof Let A, B, C, D be the images under inversion of A, B, C, D. Then ΔO A C ~ ΔOCA by the Lemma, so A C = AC( A O / AO); by the same logic, B D = BD( B O / BO), B C = BC( B O / BO), and A D = AD( A O / AO). Therefore, ( A B, C D ) = A C B D B C A D = AC( A O / AO) BD( B O / BO) BC( B O / BO) AD( A O / AO) AC BD = BC AD = (AB,CD). //

8 Theorem An h-reflection (circular inversion) preserves hyperbolic distance. That is, if A, B are the respective images of points A, B under an h-reflection, then A B * = AB *. Consequently, h-reflections are hyperbolic isometries. // Corollary [ABCD Property] H-reflections preserve hyperbolic Angle measure, Betweenness, Collinearity, and hyperbolic Distances. //

9 Hyperbolic trigonometry In Euclidean geometry, the ratios of the lengths of the sides of a right triangle ΔABC with right angle at C define the trigonometric quantities sin A = a c, cos A = b c, tan A = a b. Bolyai and Lobachevsky discovered that these relations do not hold in hyperbolic geometry; similar identities do hold, however, but they take the form sin A = sinha sinhc, where tanhb cos A = tanhc, tanha tan A = sinhb, sinha = 1 (e a e a ), cosh a = 1 (e a +e a ), tanha = sinha cosha define the hyperbolic trigonometric functions. (We similarly define cscha = 1 sinha, secha = 1 cosha, cotha = 1 tanha. )

10 There are many striking similarities between the hyperbolic trigonometric functions and the trigonometric functions, among which are sinh0 = 0, cosh0 = 1, tanh0 = 0, cosh x sinh x = 1, sinh(x + y) = sinh x cosh y + cosh x sinh y, but there are some surprises as well: the hyperbolic trigonometric functions are not periodic, nor are sinhx and coshx bounded by 1 and 1 (in fact, coshx 1 for all x). The Pythagorean Theorem, which in Euclidean geometry has the familiar form a +b = c, takes the form cosha coshb = coshc in hyperbolic geometry. For the hyperbolic law of cosines, see Exercise

1 Solution of Homework

Math 3181 Dr. Franz Rothe February 2, 2012 All3181\4080_spr12h1.tex Name: Use the back pages for extra space due date: February 10 1 Solution of Homework Figure 1: A neutral construction of the tangents