Given F = \bigl(\begin{smallmatrix} \alpha & \beta \\ \gamma & \delta \end{smallmatrix}\bigr), let X' := D_{\alpha,\gamma} and Z' := D_{\beta,\delta} where D_\p is the displacement operator from Definition 5.46. Observe from Lemma 5.48 that \begin{aligned} Z' X' &= \tau^{\langle(\beta,\delta),(\alpha,\gamma)\rangle} D_{\alpha+\beta,\,\gamma+\delta}, \\ X' Z' &= \tau^{\langle(\alpha,\gamma),(\beta,\delta)\rangle} D_{\alpha+\beta,\,\gamma+\delta}. \end{aligned} Note from Definition 3.2 that \langle(\beta,\delta),(\alpha,\gamma)\rangle = \alpha\delta - \beta\gamma = \det F = 1. Using Lemma 3.3, \langle(\alpha,\gamma),(\beta,\delta)\rangle = -1. Putting these two observations together: Z' X' = \tau D_{\alpha+\beta,\,\gamma+\delta} = \tau^2 X' Z'. Recall from Lemma 2.11 that \omega = \tau^2 so Z' X' = \omega X' Z'.

Define \ket{f_0} := \frac{1}{\sqrt{d}} \sum_{r \in \mathbb{Z}_d} (Z')^r \ket{0}. By Lemma 5.52, (Z')^d = I, so the sum is d-periodic and Z' \ket{f_0} = \ket{f_0}. Next, define \ket{f_r} := (X')^r \ket{f_0} for all r \in \mathbb{Z}_d and note from (X')^d = I (Lemma 5.52) that X' \ket{f_r} = \ket{f_{r+1}} where r+1 is computed mod d. Applying the above commutation relation inductively gives Z' \ket{f_r} = \omega^r \ket{f_r} for all r.

We next find an explicit formula for \ket{f_0}. By Lemma 5.49 and Definition 5.46, (Z')^r = D_{r\beta,r\delta} = τ^{\beta \delta r^2} X^{r\beta} Z^{r\delta}. Note from Definition 4.32 that Z^{r\delta} \ket{0} = \ket{0} and from Definition 4.29 that X^{r\beta} \ket{0} = \ket{r\beta}, so (Z')^r \ket{0} = τ^{\beta \delta r^2} X^{r\beta} Z^{r\delta} \ket{0} = \tau^{\beta\delta r^2} X^{r\beta} \ket{0} = \tau^{\beta\delta r^2} \ket{r\beta}. Writing \tau^{\beta\delta r^2} = \tau^{\beta^{-1}\delta(\beta r)^2} (which holds since \tau^d = 1 from Lemma 2.12) and substituting into the defining sum: \ket{f_0} = \frac{1}{\sqrt{d}} \sum_{r \in \mathbb{Z}_d} \tau^{\beta^{-1}\delta(\beta r)^2} \ket{r\beta}. Since d is prime and \beta \neq 0, as r ranges over \mathbb{Z}_d so does \beta r, giving \ket{f_0} = \frac{1}{\sqrt{d}} \sum_{t \in \mathbb{Z}_d} \tau^{\beta^{-1}\delta t^2} \ket{t}.

Let us now find a formula for \ket{f_r}, for general r. Applying Lemma 5.49 to X' = D_{\alpha,\gamma} gives (X')^r = D_{r\alpha,r\gamma} (the same calculation as above for Z'), so \ket{f_r} = (X')^r\ket{f_0} = D_{r\alpha,r\gamma}\ket{f_0}. By Definition 5.46, Definition 4.29, and Definition 4.32, the action on each basis state is D_{r\alpha,r\gamma}\ket{s} = \tau^{r^2\alpha\gamma + 2r\gamma s}\ket{s + r\alpha}. Substituting into the sum for \ket{f_0}, re-indexing with t = s + r\alpha, and expanding the exponent using \alpha\delta - \beta\gamma = 1: \ket{f_r} = \frac{1}{\sqrt{d}} \sum_{t \in \mathbb{Z}_d} \tau^{\beta^{-1}(\delta t^2 - 2rt + \alpha r^2)} \ket{t}. Comparing with the definition of V_F gives V_F = \sum_{r \in \mathbb{Z}_d} \ket{f_r}\bra{r}, hence V_F \ket{r} = \ket{f_r}.

We now verify that \{\ket{f_r}\}_{r \in \mathbb{Z}_d} is orthonormal. Since \ket{f_r} and \ket{f_s} are eigenvectors of Z' with distinct eigenvalues \omega^r and \omega^s (which are distinct because \omega has order exactly d as d is prime), we have \langle f_r | f_s \rangle = 0 for r \neq s. Unit norms follow from the explicit formula: \braket{f_0 | f_0} = \frac{1}{d}\sum_t |\tau^{\beta^{-1}\delta t^2}|^2 = 1 since |\tau| = 1, and \braket{f_r | f_r} = \braket{f_0|(X')^{-r}(X')^r|f_0} = \braket{f_0|f_0} = 1 since X' is unitary. Hence V_F = \sum_{r}\ket{f_r}\bra{r} is unitary.

Continuing with the same notation as before, note that V_F \ket{r} = \ket{f_r} and hence V_F Z V_F^\dagger \ket{f_r} = V_F Z \ket{r} = \omega^r V_F \ket{r} = \omega^r \ket{f_r} = Z' \ket{f_r}, so V_F Z V_F^\dagger = Z', and similarly V_F X V_F^\dagger = X'. Using Definition 5.46, Lemma 5.49, and Lemma 5.48: \begin{aligned} V_F D_\p V_F^\dagger &= \tau^{p_1 p_2}(V_F X V_F^\dagger)^{p_1}(V_F Z V_F^\dagger)^{p_2} \\ &= \tau^{p_1 p_2}(X')^{p_1}(Z')^{p_2} \\ &= \tau^{p_1 p_2} D_{p_1\alpha,p_1\gamma} D_{p_2\beta,p_2\delta}. \end{aligned} Note that (p_1\alpha,\,p_1\gamma) + (p_2\beta,\,p_2\delta) = \begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix} \begin{pmatrix} p_1 \\ p_2 \end{pmatrix} = F \p and \langle(p_1\alpha,\,p_1\gamma),(p_2\beta,\,p_2\delta)\rangle = p_1 p_2(\gamma\beta - \alpha\delta) = -p_1 p_2 \det F = -p_1 p_2. By Lemma 5.48 and Definition 3.2 we get V_F D_\p V_F^\dagger = \tau^{p_1 p_2 - p_1 p_2} D_{F\p} = D_{F\p}. In particular, V_F D_\p V_F^\dagger = D_{F\p} is a displacement operator for all \p \in \mathbb{Z}_d^2, so V_F \in \Cliff(d) by Definition 6.55.

By Theorem 8.3, V_F D_{\p} V_F^{\dagger} = D_{F\p}, so U D_{\p} U^{\dagger} = D_{\bchi} D_{F\p} D_{\bchi}^{\dagger}. Using D_{\bchi}^{\dagger} = D_{-\bchi} from Lemma 5.47 and two applications of Lemma 5.48, D_{\bchi} D_{F\p} D_{-\bchi} = τ^{\langle \bchi,\, F\p\rangle} D_{\bchi+F\p} D_{-\bchi} = \tau^{\langle \bchi,\, F\p\rangle + \langle \bchi+F\p,\, -\bchi\rangle} D_{F\p}. By Lemma 3.5, Lemma 3.8, Lemma 3.4, and Lemma 3.3, the exponent equals \begin{aligned} \langle \bchi,\, F\p\rangle + \langle \bchi + F\p,\, -\bchi\rangle &= \langle \bchi,\, F\p\rangle + \langle \bchi,\, -\bchi\rangle + \langle F\p,\, -\bchi\rangle \\ &= \langle \bchi,\, F\p\rangle + 0 + \langle \bchi,\, F\p\rangle \\ &= 2\langle \bchi,\, F\p\rangle. \end{aligned} Since \omega = \tau^2 by Lemma 2.11, U D_{\p} U^{\dagger} = \omega^{\langle \bchi,\, F\p\rangle} D_{F\p}. As D_{F\p} is a displacement operator (up to a scalar phase), U \in \Cliff(d) according to Definition 6.55.