Unlike the ping pong lemma, finding valid intervals for generalized ping pong leaves us with a bit more work to do. As covered in the Generalized Ping Pong section, we are left with a bound on the size of the kernel instead of knowledge that that kernel is just 0. After finding valid intervals, we can theoretically compute what this upper bound is and then simply check all words of length up to a number related to that bound in order to officially verify the faithfulness of the representation.
The paper 'Uniform Hyperbolic Finite-Valued SL(2,R)-Cocycles' outlines this bound in Chapter 2 but does not give a means of explictly computing it. In order to get a numerical bound as a function of our valid intervals, we need to solve for two separate values: \(\lambda\) and \(C\). We've found this to be quite a lot of work to do, so we've finished some of the calculation for \(\lambda\) and left finding \(C\) for future work.
Calculating an Explicit Value for \(\lambda\)
The theorem in the paper states that \(\lambda = \min_{\alpha, i} ( \lambda(M_{\alpha, i}, L_{\alpha, i} ))\) where \(M_{\alpha, i}\) is the i-th component of the interval associated to the generator \(\alpha\) and \(L_{\alpha, i}\) are the images of components which need to be contained in \(M_{\alpha, i}\). The function \(\lambda(M_{\alpha, i}, L_{\alpha, i})\) is a ratio of the Hilbert metrics for these two nested intervals which we know to be greater than 1. If we can find an explicit value for this function of two nested intervals, we can simply loop across all pairs of components and images within them to find the minimum ratio across all of them.
To find this value computationally is a bit of a challenge. The idea is to find a matrix \(A \in SL(2,\mathbb{R})\) which centers \(M_{\alpha, i}\) about the origin. From there we want to find a \(B \in SL(2,\mathbb{R})\) which fixes the endpoints of \(AM_{\alpha, i}\) and slides \(AL_{\alpha, i}\) to also be centered about the origin. Then we can make a symmetry argument that \(x=y=0\) must be where the minimum occurs for \(\frac{d_{L_{\alpha, i}}(x, y)}{d_{M_{\alpha, i}}(x, y)}\). We save interval endpoints as angles \(a,b \in [0, \pi]\) which we first need to take to vectors \(u,v\) in \(\mathbb{RP}^1\) using the function rp1_interval. From here, we want to map our interval endpoints to be 'centered around 0' by taking \(u\) to \(P^{-1}(-1)\) and \(v\) to \(P^{-1}(1)\) where \(P: \mathbb{R}P^1 \rightarrow \mathbb{R} \cup \{\infty\}\). We have that:
\( P^{-1}(-1) = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ -1 \end{bmatrix} \quad P^{-1}(1) = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix}\)
So then we should have an \(A \in SL(2, \mathbb{R})\) that applies the desired transformation and is given by:
\( A = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} u & v \end{bmatrix}^{-1}\)
Then, we want to find a \(B \in SL(2, \mathbb{R})\) which fixes these points, and slides our sub-interval to also be centered about 0. Since \(B\) must fix both of the new interval endpoints, it can be written in the form:
\( B = \begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix}^{-1}\)
where \(\lambda_1, \lambda_2 \in \mathbb{R}\) are the eigenvalues of \(B\). Since we want \(B \in SL(2, \mathbb{R})\), it follows that we must have \(\lambda_1 \lambda_2 = 1\), and further, \(\lambda_{2} = \frac{1}{\lambda_{1}}\), yielding the one parameter family of matrices \(B_{\lambda_1}\) which fix the endpoints \(Au, Av\) of our interval:
\( B_{\lambda_{1}} = \dfrac{1}{2 \lambda_{1}} \begin{bmatrix} \lambda_{1}^{2} + 1 & 1 - \lambda_{1}^{2} \\ 1 - \lambda_{1}^{2} & \lambda_{1}^{2} + 1 \end{bmatrix}\)
We now want to find some \(\lambda_1 \in \mathbb{R}\) which will center the vectors corresponding to the internal interval \((c, d) \Subset (a, b)\). That is, if \(w = P^{-1}(c)\) and \(z = P^{-1}(d)\), we want to find \(\lambda_1\) so that \(P(B_{\lambda_1}Aw) + P(B_{\lambda_1}Az) = 0\). Avoiding showing all the details of the calculation, we will simply give that:
\(\lambda_1 = \pm \sqrt{\frac{-\alpha \pm \sqrt{\alpha^2-8\beta cd}}{2\beta}}\)
where \(\alpha = 4cd-2c-2d\), \(\beta = (c-1)(d-1)(\sqrt{1+c^2} + \sqrt{1+d^2})\), and \(a,b,c,d \in \mathbb{R} \cup \{\infty\}\) are the interval enpoints of \(M\) and \(L\) respectively.
This is where we stopped our calculations to continue work on the interval search algorithm. From here we would need to find the limit as x tends to 0 of the ratio mentioned earlier. Work also needs to be done to find \(C\).