is symmetric in A, B and C. The expression for sin2 y shows that D 0.
Now observe that if we multiply both numerator and denominator of cos a cos fi + cos -y
sin a sin fi by the positive value of
(A2 — 1)"2(B2 — 1)"2(C2 — 1), we obtain
cos a cos J3 -4-- cos y [(BC — A)(CA — B) + (AB — C)(C2 — 1)]
sinasinf3
D= C.
EXERCISE 7.12
1. For a general triangle, prove that a b c if and only if a /3 y. [Use the Sine Rule and the Corollary of Theorem 7.13.1.]
2. Showthat a triangle is an equilateral triangle if and only if a = /3 =yand that in this case,
2 = 1.
3. Showthatfor a generaltriangle, the angle bisector at contains the mid-point of
[vb, if and only if b c (Isosceles triangles).
4. Prove that there exists an isometry mapping a triangle T1 onto a triangle T2 if and only if T1 and T2 have the same angles (or sides of the same lengths).
§7.13. The Area of a Triangle
Theorem 7.13.1. For any triangle Twith angles cz, f3 andy, h-area(T) = it — (a + /3 + y).
Corollary. The angle sum of a hyperbolic triangle is less than it.
§7.14. The inscribed Circle 151
H
Figure 7.13.1PROOF. Assume first that y = 0. We may assume that = and that Va andvblieon Izi = 1. ReferringtoFigure7.13.1 wefindthat
(.COSP
d 1
h-area(T) f dx
LJu_x2)h/2 J
= iv — + /3),
which is the desired result when = 0. In general, any triangle is the difference of two such triangles (continue the ray from Va through to w on the circle at infinity and consider T(Va, Vb, w)) and the general case follows easily. E
§7.14. The Inscribed Circle
This is the last section on hyperbolic trigonometry and we leave the reader to provide most of the details.
Theorem 7.14.1. Thethree angle bisectors of a triangle Tmeet at a point C in T.
PROOF. We may assume that y is the smallest angle so y < iv/2. Now construct angle bisectors at Va and Vb: these must meet at a point C in T(see Section 7.7).
Next, define and Y2 asin Figure 7.14.1. As /3/2, and Y2 areeach less than it/2, we can construct points Wa,Wband as in Figure 7.14.1 (and these points must lie on the open sides of T).
The Sine Rule applied to the two triangles with side Vb] gives
= p(C,we).
The same result holds with Wb instead of wa so the points Wa, and lie
on a circle with centre C. Moreover, elementary trigonometry now shows that
=
152 7. Hyperbolic Geometry
Va
The circle centre passing through wa, Wb and is called the inscribed circle of T.
Theorem 7.14.2. The radius R of the inscribed circle of T is given by
2 cos2cc+cos2fl+cos2y+2coscccosficosy-- 1
tanh R=
2(1 + cos cc)(1 + cos fl)(1 + cos y)
PROOF. Letx = p(Va, and y = Vb). Then
coscccosfl+cosy
= coshx cosh y + sinh x sinh y
sin cc sin /3 so
[(cos cc cos $ + cos y) — (sincc sinh x)(Sin /3 sinhy)]2
= [(1 — cos2 cc) + sin2 cc sinh2 x] [(1 — cos2 /3) + sin2 /3sinh2 y].
The identity
together with the relation
sin 0 = (1 + cos 8) tan(0/2)
yields
tanh R = sinhx tan(cc/2)
sin a sinh x = (1 + cos cc) tanh R.
A similar expression holds for /3,y and R and substitution yields (after some
simplification) the desired result. El
The next example is of interest.
Vb
Wg
Vc
Figure 7.14.1
§7.15. The Area of a Polygon 153
Example 7.14.3. For each in (0, it) we can construct a triangle T with
angles 0, 0. Then
4 R = + cos
=
=
In Euclidean geometry, a triangle may have a large area but a small inscribed circle. The next result shows that the situation in hyperbolic geometry is quite different: for a proof of this, see [10].
Theorem 7.14.4. The radius R of the inscribed circle of T satisfies
tanh R
and this lower bound is best possible for each value of h-area(T).
Example 7.14.3 shows that this lower bound is best possible.
§7.15. The Area of a Polygon
A polygon P is the interior of a closed Jordan curve
[z1, z2] u [z2, z3] u u [zn, z1].
The interior angle of the polygon at is the angle determined by D n P for all sufficiently small discs D centered at Note that this is not neces- sarily the interior of the angle determined by the two sides of P leaving it is this interior angle if and only if 0 < < it. We allow the vertices to lie on the circle at infinity: if is such an infinite vertex, then = 0.
Theorem 7.15.1. If P is any polygon with interior angles ... , then
h-area(P) = (n — 2)rr — -I- + On).
PROOF. This has been proved for the case n = 3(Section 7.13) and from this it follows for convex polygons by subdivision of P into n — 2 triangles (the details are omitted). It is worth noting explicitly that Theorem 15.1 applies to all polygons whether convex or not.
The proof for non-convex polygons is also by subdivision into triangles:
the subdivision is less tractable but we can compensate for this by using Euler's formula. We begin by extending each side of P to a complete geodesic.
This provides a subdivision of the entire hyperbolic plane into a finite number of non-overlapping convex polygons (convex as each is the inter- section of half-planes).
154 7. Hyperbolic Geometry
We now consider only those polygons P1 of the subdivision which lie in the original polygon P. By convexity, each P1 can be subdivided into triangles.
We have now subdivided P into non-overlapping triangles such that each vertex of P is a vertex of some i and each side of a is either a side of some other or is part of a side of P (and not of any other
Let this triangulation of P have N triangles, E edges, V vertices and let there be E0 edges which lie in the sides of P. Euler's formula for the sphere yields
(N + 1) — E
+ V =
2.As each of the N triangles has three sides we count sides in different ways and obtain
3N = E0 + 2(E — E0).
Elimination of E now gives
N — 2V
+ E0 =
—2. (7.15.1)We can now compute areas. Of the V vertices in the subdivision, n occur as vertices of P, E0 — noccur at points lying interior to a side ofF and V — E0
occur inside P. Thus area(P)
=
=
(n — —+ ...
+ by virtue of (7.15.1).Remark. For a Euclidean polygon, of course, we have
§7.16. Convex Polygons
We establish two results concerning convex polygons. The first is a necessary and sufficient condition for a polygon to be convex: the second establishes the existence of convex polygons with prescribed angles.
Theorem 7.16.1. Let P be a polygon with interior angles 01,
...,
Then P is convex andonly eachsatisfies 0 it.
§7.16. Convex Polygons 155
This is an immediate consequence of Theorem 7.5.1 Observe that Theorem 7.15.1 shows that a necessary condition for the existence of a polygon with interior angles .. ., 8,,is
01+..+B,,<(n—2)ir.
In fact, for convex polygons (and possibly for all polygons) this is also
sufficient.
Theorem 7.16.2. Let 0,,
be any ordered n-tu pie with 0
< it, j= 1, ..., ii. Then there exists a polygon P with interior angles 0k,. 0,,,occurring in this order around ÔP, andonly if
(7.16.1)
In fact, we shall construct a polygon P with these angles and with an inscribed disc touching all sides of P.
PROOF. Given . . .,0,,satisfying (7.16.1) and each lying in [0, it), construct
quadrilaterals Q1,..., Q,, each
with one vertex at the origin in
as in Figure 7.16.1. The length d can take any positive value and is to be determined later: note that Q3 isdetermined (to within a rotation about the origin) by d and It is clear that we can construct the desired polygon P as the union of non-overlapping Q1provided that0 Figure 7.16.1
(7.16.2)
= it.
d
156 7. Hyperbolic Geometry
Now (Theorem 7.11.3)
cos(U./2)
coshd (7.16.3)
and so it is appropriate to examine the function g(t) = sin-1(cos(&i12))
where t 0 and where sin 1 takes values in [0, iz/2].
Clearly, g is continuous and decreasing and g(t) -+ 0as t —+ + Also, g(0) =
= — (01
+ ...
+ Or)]>Tt
because of (7.16.1). The Intermediate Value Theorem guarantees the existence of a positive d with g(d) = it andwith defined by (7.16.3), we see that (7.16.2) holds.
As an application of Theorem 7.16.2, observe that there exists a polygon with n sides and all interior angles equal to ir/2 and only
5.
§7.17. Quadrilaterals
It is a direct consequence of Theorem 7.16.2 that there exist quadrilaterals with angles it/2, it/2, ir/2, 4) if and only if 0 4> < ,t/2: such a quadrilateral
is illustrated in Figure 7.17.1. This quadrilateral is known as a Lainbert quadrilateral (after J. H. Lambert, 1728-1777). If we reflect across one side we obtain a quadrilateral with angles 11/2, it/2, 4>, 4>: this quadrilateral (illustrated in Figure 7.17.2) was used by G. Saccheri (1667—1733) in his study of the parallel postulate and is known as the Saccheri quadrilateral.
The next theorem refers to Figure 7.17.1.
a1 Figure 7.17.1 a2
§7.17. Quadrilaterals 157
Theorem 7.17.1. (i) sinh a1 sinh a2 = cos
(ii) cosh a1 = cosh b1 sin
The proof depends on two useful preliminary results.
Lenuna 7.17.2. Let L be a hyperbolic geodesic in A with Euclidean centre and radius r and let w be the point on L which is nearest to the origin. Then
sinh p(O,w) 1/r, coshp(O,w) =
PROOF. Clearly, I
= w
+ r and orthogonality gives 1 + r2. Using (7.2.4) we obtain sinhp(O, w)
1 — 1w12 r
The value for cosh follows immediately.
Lemma 7.17.3. Let L and L' be geodesics in the hyperbolic plane. Then the inversive product (L, L') is
cosh p(L, L'), 1, cos
according as L and are disjoint, parallel or intersecting at an angle where o q5 ir/2.
PROOF. It is not difficult to see that disjoint geodesics have a common orthog- onal geodesic (see Section 7.22) and (for the moment) p(L, L') is defined to be the length of this orthogonal segment between L and L'. By the usual invariance arguments we need only consider the cases
(i) L, L' are in H2 and are given by Izi = r, Izl = R;
(ii) L, L' are in H2 and are given by x = 0,
x =
(iii) L, L' are Euclidean diameters of A.b1
Figure 7.17.2
158 7. Hyperbolic Geometry
In all these cases, the formula for (L, L') given in Section 3.2 yields the desired result.
PROOF OF THEOREM 7.17.1. We may suppose that the quadrilateral in Figure 7.17.1 has the sides a1 and a2 lying on the positive real and imaginary axes.
Suppose that the sides labelled b1 and b2 lie on the circles
z—ivl=R, z—uI=r,
respectively, where u, v, r and R are positive. Then by Lemma 7.17.2, sinh a1 sinh a2 = l/rR.
Lemma 7.17.3 implies that
(L, L') =cos and from Section 3.2 we have
(L,L') =
+ R2—ju —— r2 + R2 — u2 —
2rR
= 1/rR because, for example, u2 = 1 + r2.
To prove (ii) we relocate the polygon so that the vertex with angle is at the origin and the side labelled b2 is on the positive real axis: see Figure 7.17.3.
Now reflect the quadrilateral in the real axis: let L be the geodesic con- taining the side labelled a2 and let L' be its reflection in the real axis. By Lemma 7.17.3 we have
(L, L') = cosh(2a1). (7.17.1)
Figure 7.17.3
§7.18. Pentagons 159
If L (viewed) as a Euclidean circle) has centre det" and radius r, then L' has centre de and radius r and clearly,
Thus
— > 2r.
— — 2r2I
(L,L')
= 2r2
— 2d2 sin2 — r2
r2
= 2 cosh2 b1 sin2 c/ — 1 by virtue of Lemma 7.17.2. This with (7.17.1) yields (ii).
EXERaSE 7.17
I. Derive Lemma 7.17.2 directly from Theorem 7.9.1 (Lemma 7.17.2 is simply a re- statement of the Angle of Parallelism formula).
§7.18. Pentagons
We shall examine the metric relationships which exist for the pentagon illustrated in Figure 7.18.1 where 0 < it.
Theorem 7.18.1. (i) cosh a cosh c + cos 4, =sinh a cosh b sinh c.
(ii) If 4, = n/2 then
tanh a cosh b tanh c = 1, sinh a sinh b = coshd.
(7.18.1) (7.18.2) PROOF. It is easy to see that there is a geodesic through the vertex with angle 4, which meets and is orthogonal to the side of length b. Let b1 and b2 be the
Figure 7.18.1
b
1. Hyperbolic Geometry
lengths as illustrated and let 4)2be the subdivision of 4); being on the same side of this geodesic as the side of length b1. By Theorem 7.17.1, we have
cosh a =coshh sin cosh c = cosh h sin 4)2, sinh asinh b1 = cos
sinh c sinh b2 = cos4)2.
It follows that
(cosh a cosh c — sin sin 4)2)2
= (cosh a cosh c — sin sin 4)2)2 — (cosh a sin 4)2 —cosh c sin 4)i)2
= (cosh2 a —sin2 4)1)(cosh2 c — sin2 4)2)
(sinh2 a + cos2 4)1)(sinh2 c + cos2 4)2)
= (sinh2 a cosh2 b1)(sinh2 c cosh2 b2) and so, taking positive square roots,
cosh a cosh c — sin sin 4)2 =sinh a sinh c cosh b1 cosh b2.
This leads directly to (i) as
cosh a cosh c + cos 4) = cosh a cosh C — sin + COS
= sinha sinh c (cosh b1 cosh b2 + sinh b1 sinh b2) sinh a sinh c cosh b.
Putting 4) = it/2 in (i), we obtain (7.18.1). To prove (7.18.2), we apply (7.18.1) to the triple b, c, d and eliminate c from the resulting expression and (7.18.1).
§7.19. Hexagons
We shall only consider the right-angled hexagon illustrated in Figure 7.19.1.
If we join the end-points of the sides labelled a1 and b1 to form a quadrilateral Q,wefind that each interior angle of Q is less than ir/2. This implies that the sides labelled a1 and b1 have a common orthogonal of length, say t, as illustrated.
Theorem 7.19.1.
sinh a1 — sinha2 —sinh a3 sinh b1 — sinhb2 — sinh b3
PROOF. From Theorem 7.18.1 we obtain
sinh b2 sinh a3 =cosh
t =
sinha2 sinh b3and the result follows by symmetry considerations.
§7.19. Hexagons 161
Theorem 7.19.2.
cosh b1 sinh a2 sinh a3 = cosha1 + cosh a2 cosh a3 PROOF. From Theorem 7.18.1 we obtain the identities
sinh x sinh a2 = cosh u;
sinh y sinh a3 = coshv;
sinh u sinh t = cosh a2;
sinh v sinh t = cosh a3.
Next, we obtain the identity
(cosh2 a2 + sinh2 u)(cosh2 a3 + sinh2 v)
= (cosha2 cosh a3 + sinh u sinh v)2 by expressing both sides as functions of u, a and t. Thus
cosh b1 sinh a2 sinh a3
= (coshx cosh y + sinh x sinh y)sinh a2 sinh a3
= coshx cosh y sinh a2 sinh a3 + cosh u cosh v
= (coshx sinh a2)(cosh y sinh a3) + cosh u cosh v
= (sinh2 a2 + cosh2 u)112(sinh2 a3 + cosh2 v
= (cosh2 a2 + sinh2 u)112(cosh2 a3 + sinh2 v)'12 + cosh u cosh v
= cosh a2 cosh a3 + sinh u sinh v + cosh u cosh v. EJ
Remark. Theorem 7.19.2 shows that the lengths of all sides of the hexagon are determined by the lengths a,, a2 and a3.
Figure 7.19.1
162 7. Hyperbolic Geometry
§7.20. The Distance of a Point from a Line
For each point z and each geodesic L, define p(z, L) = inf{p(z, w): we L}.
There is a unique geodesic L1 through z and orthogonal to L and p(z, L) is the distance from z to L measured along L1.
We work in H2 and we may assume that L is the positive imaginary axis.
Then
L1 = = zI}
and we are asserting that
p(z, L) = p(z,iIzI). (7.20.1)
Each point on L is of the form it (t > 0) and from Theorem 7.2.1, x2 + y2 + t2
cosh p(z, it)
= 2 (z = x + iy)
=
—I— + —
t2y\t
Izi(7.20.2) y
As equality holds here if and only if t = IzL thisverifies (7.20.1).
With 8 as in Figure 7,20.1, we can use (7.20.2) and cosh p(z, L) = 1/cos 0;
sinh p(z, L) = tan 0; (7.20.3)
tanh p(z, L) = sin 8.
L
iIzI
0 x
Figure 7.20.1
i neDistance of a Point from a Line 163
As an application of these formulae, note that the regions {zEH2: p(z,L) <k} (k > 0) are precisely the hypercyclic regions described in Section 7.5.
We can also obtain a formula for p(z, L) when L is the Euclidean semi- circle {w: wi = r} in H2: see Figure 7.20.2.
Suppose first that I z < r. With 0 as in Figure 7.20.2, the Euclidean circle through z, rand —rhas centre —ir(tan and radius r/cos 0. Thus
and so
Jz + ir(tan8)12 = r2/cos2 U
r2 — 1z12
tan0 =
2yr (z = x + iy).
1z12 —
sinh p(z, L) = 2yr Figure 7.20.2
A similar formula holds for z1 when Izi I > r with izi 2 — r2 replacing
r2 — Thusif L is given by wi = rwe obtain from (7.20.3),
(7.20.4) We shall also need a formula for the model when L is the real diameter (—1, 1) of In this case we show that for all w in
sinhp(w, L) = (7.20.5)
First, there is a unique geodesic L' through w and orthogonal to L. Let L and L' meet at then there is an isometry g of A which fixes —1and 1, which maps to 0 and which leaves L invariant. Now g maps V to the segment (— i, i) and so g(w) = it for some real t. The relationship between w and t is
164 7. Hyperbolic Geometry
best found by noting that as g preserves both differentials IdzI/y
and— 1z12)we have
— Im(w)
1 — I —
On the other hand,
p(w, L) = p(w,
= p(it,0)
=
/1 +
—
andthis gives (7,20.5).
EXERCISE 7.20
1 Let L be the geodesic (— e°, in Findan explicit formula for sinh p(z,L),zE ii
§7.21.
The Perpendicular Bisector of a Segment
Let z1 and z2 be distinct points and let w be the mid-point of Izj, z2]. We
shall prove that
{z: p(z, z1) = p(z,z2)}
is the unique geodesic through w and orthogonal to [z1, z2]: this is the perpen- dicular bisector of [z1, z2].
We work in H2 and assume that z1 = I and z2 = r2i where r> 1: thus w = ri.From Theorem 7.2.1,
cosh p(z, z1) = coshp(z, z2)
ifand only if
Iz — z112 — — z2j2
y — r2y
and this simplifies to z = r.
In the model thedirect isometries are of the form
az+ë
2 2g(z) = —,
aj — cl = 1.
cz + a
§7.22. The Common Orthogonal of Disjoint Geodesics 165
Using (7.2.4) we find that z is on the perpendicular bisector of [0, gO] if and only if
1z12 — gOi2
(1 — z12)(1 — 012) (1— —
or, equivalently,
— =
As
Iãz — — Icz = — 1,
wesee that the perpendicular bisector of [0, go] is the isometric circle of g
EXERCISE 7.21
1. Showthattheperpendicularbisectorofthetwopointsz,= xj + (j = 1,2)in H2is
L = — = — z1j2}.
Deduce that for any z1 andany compact subset K of R2, L n K =
0
when 221issufficiently large.
§7.22. The Common Orthogonal of Disjoint Geodesics
If L1 and L2 are disjoint geodesics then there exists a unique geodesic which is orthogonal to both L1 and L2.
The assertion remains invariant under isometries so we may assume that L1 and L2 are in H2 with equations
x=0, (x—a)2+y2r=r2,
respectively, where a > r > 0. The only geodesics orthogonal to L1 are those with equations Iz = t and such a geodesic is orthogonal to L2 if and only if a2 = r2 + t2. As a > r there is a unique positive t satisfying this equation.
EXERCISE 7.22
1. Prove that if two distinct geodesics have a common orthogonal then they are disjoint.
7. Hyperbolic Geometry
§7.23. The Distance Between Disjoint Geodesics
For disjoint geodesics L1 and L2 define
p(L1, L2) = inf{p(z,w): z eL1, weL2}.
The distance p(L1, L2) between L1 and L2 is the distance measured along their common orthogonal.
We work in H2 and assume that the common orthogonal is the positive imaginary axis. Then L1 and L2 are given by zi = r, z = R, say and the result follows from (7.20.4) (see also Section 5.4).
There are other convenient expressions for p(L1, L2), for example Lemma 7.17.2. Also, p(L1, L2) can be expressed as a cross-ratio: if L1 has end-points
z1 and22 and if L2 has end-points w1 and w2, these occurring in the order Zi, w1, z2 aroundthe circle at infinity, then
[z1, W1, 22] . L2)] = 1. (7.23.1)
EXERCISE 7.23
1. Verify (7.23.1) by working in H2 and taking = 0, = I and :2 =
§7.24.
The Angle Between Intersecting Geodesics
The angle 0, 0 < 6 < it, between intersecting geodesics can be expressed both in terms of the inversive product (Lemma 7.17.2) and the cross-ratio.
If L1 (Zi, z2)
and L2 =
(w1, w2) with the end-points occurring in the order z1, w1, z2, w2 around the circle at infinity, then[z1, w1, z2,w2] sin2(0/2) = 1.
For the proof, use and L1 =(—1, 1), L2 = (e10,
§7.25. The Bisector of Two Geodesics
Let L1 and L2 be distinct geodesics: the bisector of L1 and L2 is
L =
{z: p(z, L1) = p(z,L2)}.We show that L is one or two geodesics.
Case 1: L1 and L2 are parallel.
In this case, take L1 and L2 to be x = a
and x =
—a in H2. From (7.20.3) we see thatzis onLifand only if — = x+ a,equivalently,x =0.
§7.26. Trarisversals 167
Case2:L1 and L2 are disjoint.
Take L1 and L2 to be = 1 and z is on L if
and only if
(1z12 r4)2 = — 1)2
andthis reduces to z = r.
Case 3:L1 and L2 are intersecting.
Take
L1 = (e—j0, — e_iO),
L2 = — e10)
in A where 0 < U < it/2.Let L' = (—1,1): then z is on L if and only if L') = p(z, L1)
= p(z, L2)
= p(e °z, L').
Using (7.20.5) with z = re" this becomes [sin(U + t)]2 [sin(U —
whichgives L as the union of the two geodesics (—1, 1) and (—i, i).
§7.26. Transversals
Let L1 and L2 be disjoint geodesics. A geodesic L is
a O-transrersal(0<0< it/2)ofL1 andL2 ifandonlyifLmeetsbothL1 andL2atanangle
0. As an example of how O-transversals arise naturally consider the isometry g(z) =kz(k > 0) of H2 and the geodesic L given by x =0.If L1 is any geodesic meeting L, the L is a 6-transversal of L1 and g(L1). We need to investigate the metric relations which exist for O-transversals.The common orthogonal of L1 and L2 is h the unique ir/2-transversal of L1 and L2. We shall see that for all other values of 6 there are exactly four 6-transversals. Let L0 be the common orthogonal of L1 and L2 and let L*
be the bisector of L1 and L2. We shall work in A and we assume that L0 = (— 1, 1), L* = (— i,i).
The situation is then as illustrated in Figure 7.26.1 where the four transversals are shown, two in each case. We omit the proofs (which are not difficult) that any 0 in (0, can be attained in this way and that there are no other O-transversals.
With an obvious reference to Euclidean geometry, we call the 6-transversals in Case (i) the alternate transversals: those in Case (ii) are the complementary transversals. Let t0 denote the length of the segment of a 0-transversal which
168 7. Hyperbolic Geometry
7.11.2,
sinh L2) = sin 0:
for complementary transversals, Theorem 7.17.1 yields
EXERCISE 7.26
cosh L2) = 0.
1. For a given 0, are the alternate 0-transversals longer or shorter than the comple- mentary O-transversals?
2. Let the two alternate £l-transversals meet L1 at z1 and z2: let the complementary O-transversals meet L1 at w1 and w2. Which of p(z1, z2) and p(w1, w2) is the greater?
§7.27. The General Theory of Pencils
Much of the hyperbolic geometry required for a detailed discussion of Fuchsian groups is best described in terms of pencils of geodesics. For example, we see that circles, horocycles and hypercycles are simply varia- tions of the same idea and this brings a greater unity into the subject. We shall also see that the classification of pencils leads naturally to the classification of isometries which is more illuminating than that given in Section 4.3. In this section we merely describe the notion of a pencil and list its main prop- erties: the details occur in the next three sections.
Each pair of geodesics, say L and L', lie in a geometrically defined one- parameter family of geodesics called the pencil determined by L and L'.
Case (i) Figure 7.26.1 Case (ii)
lies between L1 and L2. For alternate transversals we have from Theorem