Having calculated lower bounds for the eigenvalues of each irreducible representation, we have done the bulk of the work to make a statement about the second eigenspace, E2. We begin with the following
Lemma 3.21 The dimension of E2 is a multiple of 4.
Proof: Recall our upper bound on λ2 in Section 3.3. The only irreducible representations that can appear in this eigenspace are χ8 and the four dimensional ones. To rule out χ8, we will return to Proposition 3.14. The eigenfunction corresponding to the second positive eigenvalue of the domain in Figure 3.13 has 2 nodal domains. So we have two Dirichlet boundary problems each with the same eigenvalue. The entire region has area 2π, so unless the nodal line splits it equally, one of the nodal domains will have area less than π. Therefore, we can take π as an upper bound on the area of at least one of the nodal domains, and use the Faber-Krahn inequality so conclude that the eigenvalue of the nodal domain, and therefore the whole region, is greater than νπ and therefore greater than 6.
This means that only the four dimensional irreducible representations can appear in E2.
Theorem 3.22 The dimension of E2(B) is 4.
Proof: By Lemma 3.21, we already know that the multiplicity of λ2 is a multiple of 4.
To prove that it is exactly four, we will prove that the multiplicity is less than 8. We will use Selberg’s trace formula in the same way as for the multiplicity bound above, using our upper bounds on the first and second positive eigenvalues. Again, we use the test function
h(t) = sin(tL) tL
4
,
so that the Fourier transform of h(t) is supported on [−4L, 4L]. We choose 4L to be less than the length of the shortest closed geodesic of the Bolza surface to avoid contribution from the length spectrum. Recall that the shortest closed geodesic has length
2 arccosh 1 +√
2
≈ 3.057141839.
Choosing
L = 3/4, Selberg’s trace formula becomes
X
j
h(tj) = 2 Z ∞
0
h(t) tanh(πt)tdt.
Figure 3.20: h(t) on [0, 4]
1 2 3 4
0.2 0.4 0.6 0.8 1.0
Now we use the straightforward inequality tanh(x) < 1 to state
X
j
h(tj) < 2 Z ∞
0
h(t)tdt.
The integral on the right hand side can be explicitly calculated; for L = 3/4, we have 2
Z ∞ 0
h(t)tdt = 32 log(2)
9 .
On the other hand, by Theorem 3.9 we have
h(t0) + 3h(t1) + mh(t2) < X
j
h(tj),
where m is the dimension of E2. We therefore have
h(t0) + 3h(t1) + mh(t2) < 32 log(2)
9 ,
and a simple rearrangement gives m <
32 log(2)
9 − h(t0) − 3h(t1) h(t2).
Since h(t) is decreasing on the interval [0, 4L], we can apply our bounds on the first two eigenvalues to state
h(t1) > h
r116469 28089 + 1/4
!
and
h(t2) > h
r1408244 252552 + 1/4
! . We conclude that
m <
32 log(2)
9 − h(i/2) − 3hq
116469
28089 + 1/4 hq
1408244
252552 + 1/4
=
4070947840281 csc4
√2017659
10523
8
32 log(2)
9 −
39897665536 sin4 3
√48643 3121 8
!
63885819123 −409681 sinh4 38
453564534784
≈ 7.1091.
Chapter 4
The Klein Quartic
4.1 Systole of the Klein quartic
The systole of the Klein quartic is described in Klein’s original paper [48] on the prop-erties of the quartic curve, and is expounded upon in [44, 80]. We will use K to denote the surface. Consider a tessellation of K by heptagons (these can be barycentrically sub-divided into 14 (2, 3, 7) triangles, see Figure 4.2). Starting with the central heptagon, join the midpoints of two adjoining edges with a geodesic. Extend this geodesic in both directions until it meets the boundary of the fundamental 14-gon. On its way, it passes through the midpoints of a further 3 and a half heptagon midpoints in either direction; a total of 8 midpoints. For this reason, Karcher and Weber coin it an “eight step geodesic”.
Note that the two boundary edges that it meets are actually the same edge once the side associations are made; that is, we have a simple closed geodesic. In fact, the length of this curve turns out to be the systole of the surface [44]. An order 7 rotation around the centre of the 14-gon maps this geodesic onto six other copies.
We obtain two more geodesics of length equal to the systole by creating curves that pass through adjoining edges of heptagons at varying distances from the centre; these are shown in Figure 4.2. There are 7 heptagons surrounding the central heptagon. The magenta curve passes through two edges of one of these. This time, when it reaches the boundary, it does not form a closed curve since the two sides are not associated with one another. We make the relevant side association and continue the geodesic from here, following the pattern of passing through midpoints of heptagon edges, until the geodesic is closed. Since we could have chosen any of the 7 heptagons, we get a further 7 closed geodesics. The orange (and green) curves pass through two adjoining edges of a heptagon in the second closest group from the centre. Again, there will be 7 of these. They follow the same pattern through 8 heptagon edge midpoints, thus having the same length. The multiplicity of the systole is 21.
Figure 4.1: Calculating 18 of the systole
c b a
We calculate the systole using basic hyperbolic trigonometry to give a closed formula for one of the “steps”, and then multiply this by 8. Similar to the formulae in Theorem 2.30 for right-angled hyperbolic triangles, we have
sinh(a) = sin(α) sinh(c) (4.1)
for a right-angled hyperbolic triangle where c is the length of the hypotenuse and α is the angle opposite the side of length a [13]. Also recall from Section 2.2.3 that the side of medium length in a (2, 3, 7) triangle has length
arccosh 1
Consider the blue geodesic in Figure 4.2. We can zoom in on its centre “step” to get the triangle shown in Figure 4.1. Since step is symmetric with respect to b, we can divide the larger triangle into two smaller right-angled triangles. Note that the medium side of a (2, 3, 7) triangle is the hypotenuse of the right-angled triangle in Figure 4.1. Let a be half the length of the blue line, that is, the shortest side of the triangle on the left. We calculate the length of a using Equation (4.1), where α is the angle opposite a and is equal to π7. We have
This is half of the length of one step of the geodesic, and there are eight steps altogether, therefore
Remark 4.1 Note that there is more than one way to calculate a closed formula for the systole; [72] came up with
8 arccosh 3
2− 2 sin2π 7
≈ 3.93594624883.
Figure 4.2: Geodesics of length equal to the systole
Remark 4.2 Numerical calculations of the other values in the geodesic length spectrum of the Klein quartic are given in the doctoral thesis of Vogeler [81], along with multiplicities.
Vogeler also gives similar length spectra calculations for many other Hurwitz surfaces, including the Macbeath surface mentioned in the introduction to this work.