Foundations of Hyperbolic Manifolds

Foundations of Hyperbolic Manifolds Third Edition Solution Manual John G. Ratclie 1 / 4

Chapter 1 Euclidean Geometry 1.1 Euclid's Parallel Postulate 1.2 Independence of the Parallel Postulate Exercise 1.2.1LetPbe a point outside a lineLin the projective disk model.Show that there exists two linesL1andL2passing throughPparallel toLsuch that every line passing throughPparallel toLlies betweenL1andL2. The two linesL1andL2are called theparallelstoLatP. All the other lines passing throughPparallel toLare calledultraparallelstoLatP. Conclude that there are innitely many ultraparallels toLatP.

Solution:The linesL1andL2are the two lines passing throughPthat end

at the two ideal endpoints ofL. See Figure 1.2.1. There are obviously innitely many lines in the projective disk model passing throughPlying betweenL1 andL2, and so there are innitely many ultraparallels toLatP.Exercise 1.2.2Prove that any triangle in the conformal disk model, with a vertex at the center of the model, has angle sum less than 180

.

Solution:Let be a triangle with verticesA; B; Cin the conformal disk model

withCthe center of the model. The sidesACandBCof are Euclidean line segments. LetLbe the hyperbolic line passing through the pointsAandB.ThenLis a circular arc that is orthogonal to the circle at innity. The hyperbolic half-plane bounded byLthat does not contain is Euclidean convex, since it is the intersection of two disks. Therefore, sideABof is a circular arc that is contained in the corresponding Euclidean triangle

with verticesA; B; C.The angle of atAis the angle between sideACand the Euclidean tangent lineTto the circular arcABatA. The angle between lineTand the Euclidean line segmentABatAis positive. Hence, the angle of atAis less than the angle of

atA. Likewise, the angle of atBis less than the angle of

at

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• Therefore, the sum of the angles of is less than the sum of the angles of

, and so the sum of the angles of is less than 180

.Exercise 1.2.3Letu; vbe distinct points of the upper half-plane model. Show how to construct the hyperbolic line joininguandvwith a Euclidean ruler and compass.

Solution:If Re(u) = Re(v), draw the vertical rayLstarting at Re(u) and

passing throughuandv. Otherwise, draw the Euclidean line segmentSjoining the pointsuandv. Construct the Euclidean perpendicular bisectorMof the line segmentS. Letwbe the point of intersection ofMand the real axis. Then juwj=jvwj. Draw the semi-circleLcentered atwpassing through the pointsuandvwith endpoints on the real axis. Then in either caseLis the hyperbolic line joiningutov.Exercise 1.2.4Let(z) = az+b cz+d witha; b; c; dinRandadbc >0. Prove that maps the complex upper half-plane bijectively onto itself.

Solution:Observe that

az+b cz+d

cz+d cz+d = acjzj 2 +adz+bcz+bd jcz+dj 2 = acjzj 2

• (ad+bc)Re(z) + (adbc)Im(z)i+bd

jcz+dj 2 Hence, we have Im

az+b cz+d

= (adbc)Im(z) jcz+dj 2

>0:

Thereforemaps the complex upper half-plane into itself.Dene real numbersa

; b

; c

; d

by the matrix equation

a

b

c

d

=

a b c d

1

:

Observe that a

a

z+b

c

z+d

+b c

a

z+b

c

z+d

+d =

aa

z+ab

c

z+d

+

bc

z+bd

c

z+d

ca

z+cb

c

z+d

+

dc

z+dd

c

z+d

= (aa

+bc

)z+ (ab

+bd

) (ca

+dc

)z+ (cb

+dd

) =z; since a b c d

a

b

c

d

=

• 0
• 1

:

Let (z) = a

z+b

c

z+d

• Then( (z)) =z, and by reversing the roles ofand , we

have ((z)) =z. Thereforemaps the complex upper half-plane bijectively onto itself with inverse .

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Exercise 1.2.5Show that the intersection of the hyperboloidx 2 y 2 z 2 = 1 with a Euclidean plane passing through the origin is either empty or a hyperbola.

Solution:The equation of a plane passing through the origin is

ax+by+cz= 0;with (a; b; c)6= (0;0;0):

Assume rst thata= 0. Thenby+cz= 0. Now, the hyperboloidx 2 y 2 z 2 = 1 is symmetric with respect to thex-axis. Hence, we can rotate the normal vector (b; c) in theyz-plane so thatb= 0. Thenz= 0 is the equation of the plane and the intersection with the hyperboloid is the hyperbolax 2 y 2 = 1 in the xy-plane.Now assumea6= 0. Then we can normalize the normal vector (a; b; c) so thata= 1. Then the plane has the equationx+by+cz= 0. Next, we rotate the normal vector (1; b; c) about thex-axis so thatc= 0. Then the plane has the equationx+by= 0. Hence, the intersection of the plane with the hyperboloid satises the equationb 2 y 2 y 2 z 2 = 1. Now, the equation (b 2 1)y 2 z 2 = 1 has a real solution if and only ifb 2 >1. The intersection of the plane with the hyperboloid is the set

f(by; y; z) : (b

2 1)y 2 z 2

= 1g:

Letu= ( b p b 2 +1 ; 1 p b 2 +1 ;0) andv= (0;0;1). Thenfu; vgis an orthonormal basis for the planex+by= 0. Letw= p b 2

• 1y. Then the intersection is the set

fwu+zv:

b 2 1 b 2 +1

w 2 z 2

= 1g:

Hence, the intersection is either empty or a hyperbola.

1.3 Euclideann-Space Exercise 1.3.1Letv0; : : : ; vmbe vectors inR n such thatv1v0; : : : ; vmv0 are linearly independent. Show that there is a uniquem-plane ofE n containing v0; : : : ; vm. Conclude that there is a unique 1-plane ofE n containing any two distinct points ofE n .Solution:The vectorsv0; : : : ; vmare contained in them-plane P=v0+ Spanfv1v0; : : : ; vmv0g: Supposev0; : : : ; vmare contained in them-planea+VwithVanm-dimensional vector subspace ofR n . Thenviv0is inVfor eachi= 1; : : : ; m. Hence V= Spanfv1v0; : : : ; vmv0g; sincev1v0; : : : ; vmv0are linearly independent. Asv0is ina+V, we have thatv0=a+vfor somevinV. Hencev0ais inV. Thereforea+V=v0+V.Thus, them-planeP=v0+Vis unique.

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