## Homework 15

For this homework, you can submit up to 3 times.

#### Q15.1 \(\quad\) \(C_3\) factorization

If a system has a discrete symmetry, then the contribution of a periodic orbit to the dynamical zeta function and spectral determinant can be factorized into several terms. Each of them corresponds to an irreducible representation. Let us work out the factorization of the dynamical zeta function for a system whose dynamics is invariant under the cyclic group \(C_3\).The following is the character table of \(C_3\): \begin{bmatrix} & e & C^{1/3} & C^{2/3} \\ A & 1 & 1 & 1 \\ \Gamma_1 & 1 & \omega & \omega^2 \\ \Gamma_2 & 1 & \omega^2 & \omega \\ \end{bmatrix} Here, \(\omega = e^{i2\pi/3}\). It has 3 irreducible representations, and all of them are 1-dimensional. As stated in ChaosBook, for 1-dimensional irreducible representation, \[ \det(1 - D^{(\mu)}(g)t_{\hat{p}}) = 1 - \chi^{(\mu)}(g)t_{\hat{p}} \] So for a periodic orbit \(p\) with symmetry \(C^{1/3}\) (i.e., built from three repeats of the relative periodic orbit \(\hat{p}\)), its contribution to dynamical zeta function is factorized as \(1-t_{\hat{p}}\), \(1-\omega t_{\hat{p}}\), and \(1-\omega^2 t_{\hat{p}}\) corresponding to irreducible representation \(A\), \(\Gamma_1\) and \(\Gamma_2\) respectively. The product of these 3 terms gives the total contribution, which is \(1-t_{\hat{p}}^3\). This makes sense because if we ignore symmetry, the contribution should be \(1-t_p\), and \(t_p = t_{\hat{p}}^3\). Here \(\hat{p}\) is the relative periodic orbit in the full state space, periodic in the fundamental domain. This result corresponds to the second row of the following table. Each row represents the contribution from periodic orbits with a given symmetry, and each column represents the contribution to a given irreducible representation. The second column is the total contribution. \begin{bmatrix} & & A & \Gamma_1 & \Gamma_2 \\ e : & ? & & & \\ C^{1/3} : & 1-t_{\hat{p}^3} = & 1-t_{\hat{p}} & 1-\omega t_{\hat{p}} & 1-\omega^2 t_{\hat{p}} \\ C^{2/3} : & & & & * \end{bmatrix} So you can fill out the remaining entries in the above table. What are the terms at position (?) and (*)

\( 1 - t_{\hat{p}}^3, 1 - \omega t_{\hat{p}} \) \( 1 - t_{\hat{p}}^3, 1 - \omega^2 t_{\hat{p}} \) \( 1 - t_{\hat{p}}^3, 1 - t_{\hat{p}} \) \( (1 - t_{\hat{p}})^3, 1 - \omega^2 t_{\hat{p}} \) \( (1 - t_{\hat{p}})^3, 1 - \omega t_{\hat{p}} \) \( 1 - \omega t_{\hat{p}}^3, 1 - t_{\hat{p}} \) \( 1 - \omega^2 t_{\hat{p}}^3, 1 - t_{\hat{p}} \) \( (1 - t_{\hat{p}})^3, 1 - t_{\hat{p}} \)

#### Q15.2 \(\quad\) \(C_{4v}\) factorization

Chaosbook shows how dynamics equivariant under symmetries of a triangle, group \(C_{3v}\) is factorized. Now let us turn to dynamics invariant under symmetries of a square: \(C_{4v} = \{ e, C^{1/4}, C^{2/4}, C^{3/4}, \sigma, \sigma C^{1/4}, \sigma C^{2/4}, \sigma C^{3/4} \} \). \(C_{nv}\) turns out to have different irreducible representation structures for \(n=odd\) and \(n=even\). This case is even, and \(C_{4v}\) has 5 different irreducible representations: \(\{A_1, A_2, B_1, B_2, E\}\). The first 4 are 1-dimensional, and \(E\) is 2-dimensional (verify for yourself that this satisfies the dimensions rule for the decomposition of the regular representation ).\(A_1\) and \(A_2\) are symmetric under quarter-circle rotations \(C^{1/4}, C^{3/4}\), while \(B_1\) and \(B_2\) are antisymmetric.

First, figure out the character table of \(C_{4v}\). Then factorize the dynamical zeta function into different irreducible representations for periodic orbits with different symmetries. What is the contribution of a periodic orbit with symmetry \(\sigma C^{2/4}\) to dynamical zeta function corresponding to irreducible representation \(E\) ?

\( 1 - t_{\hat{p}}^2 \) \( ( 1 - t_{\hat{p}}^2 )^2 \) \( ( 1 - t_{\hat{p}} )^2 \) \( ( 1 - t_{\hat{p}} )^4 \) \( ( 1 + t_{\hat{p}} )^4 \) \( ( 1 + t_{\hat{p}}^2 )^2 \) \( ( 1 + t_{\hat{p}} )^2 \)

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