Harold T. Stokes and Branton J. Campbell, Brigham Young University,
Provo, Utah, USA
Sander van Smaalen, University of Bayreuth, Bayreuth, Germany
Express $(3+d)$D superspace symmetry operator $g$ as a $(3+d+1)$D affine matrix: $$g = \left( \begin{array}{cc|c} R & 0 & v \\ M & \epsilon & \delta \\ \hline 0 & 0 & 1 \end{array}\right),$$ where $R$ is a $3\times3$ matrix, $\epsilon$ is a $d\times d$ matrix, $M$ is a $d\times3$ matrix, $v$ is a 3-component vector, and $\delta$ is a $d$-component vector. Similarly, express $(3+d)$D superspace coordinate $x$ as an augmented $(3+d+1)$D vector: $$x = \begin{pmatrix} x_v \\ x_\delta \\ \hline 1 \end{pmatrix}, \text{ where } x_v = \begin{pmatrix} x_{1} \\ x_{2} \\ x_{3} \end{pmatrix} \text{ and } x_\delta = \begin{pmatrix} x_{3+1} \\ \vdots \\ x_{3+d} \end{pmatrix},$$ so that the operation of $g$ on $x$ can be expressed as $$y = g~x = \left( \begin{array}{cc|c} R & 0 & v \\ M & \epsilon & \delta \\ \hline 0 & 0 & 1 \end{array}\right) \begin{pmatrix} x_v \\ x_\delta \\ \hline 1 \end{pmatrix} = \begin{pmatrix} R ~ x_v + v \\ M ~ x_v + \epsilon ~ x_\delta + \delta \\ \hline 1 \end{pmatrix}.$$ Next, define a $(3+d+1)$D affine matrix $S$ that transforms the superspace coordinates from the current setting to a new setting: $$S = \left( \begin{array}{cc|c} S_R & 0 & S_v \\ S_M & S_\epsilon & S_\delta \\ \hline 0 & 0 & 1 \end{array}\right) ~~~ \text{and} ~~~ S^{-1} = \left( \begin{array}{cc|c} (S^{-1})_R & 0 & (S^{-1})_v \\ (S^{-1})_M & (S^{-1})_\epsilon & (S^{-1})_\delta \\ \hline 0 & 0 & 1 \end{array} \right),$$ where the component matrices and vectors within $S^{-1}$ are calculated to be $$(S^{-1})_R = S_R^{-1}$$ $$(S^{-1})_\epsilon = S_\epsilon^{-1}$$ $$(S^{-1})_M = -S_\epsilon^{-1} ~ S_M ~ S_R^{-1}$$ $$(S^{-1})_v = -S_R^{-1} ~ S_v$$ $$(S^{-1})_\delta = - S_\epsilon^{-1} ~ S_\delta \; + \; S_\epsilon^{-1} ~ S_M ~ S_R^{-1} ~ S_v$$ The transformation of superspace coordinate $x$ to the new coordinate system is computed as $$x' = S~x = \left( \begin{array}{cc|c} S_R & 0 & S_v \\ S_M & S_\epsilon & S_\delta \\ \hline 0 & 0 & 1 \end{array}\right) \begin{pmatrix} x_v \\ x_\delta \\ \hline 1 \end{pmatrix} = \begin{pmatrix} S_R ~ x_v + S_v \\ S_M ~ x_v + S_\epsilon ~ x_\delta + S_\delta \\ \hline 1 \end{pmatrix},$$ and the transformation of superspace operator $g$ to the new coordinate system is computed as $$g' = S ~ g ~ S^{-1} = \left( \begin{array}{cc|c} R' & 0 & v' \\ M' & \epsilon' & \delta' \\ \hline 0 & 0 & 1 \end{array} \right), ~ \text{where}$$ $$R' = S_R ~ R ~ S_R^{-1}$$ $$\epsilon' = S_\epsilon ~ \epsilon ~ S_\epsilon^{-1}$$ $$M' = S_M ~ R \; - \; S_\epsilon ~ \epsilon ~ S_\epsilon^{-1} ~ S_M ~ S_R^{-1} \; + \; S_\epsilon ~ M $$ $$v' = - \; S_R ~ R ~ S_R^{-1} ~ S_v \; + \; S_R ~ v \; + \; S_v$$ $$\delta' = - \; S_M ~ R ~ S_v \; + \; S_\epsilon ~ \epsilon ~ (S_\epsilon^{-1} ~ S_M ~ S_R^{-1} ~ S_v - S_\delta) \; - \; S_\epsilon ~ M ~ S_v \; + \; S_M ~ v \; + \; S_\epsilon ~ \delta \; + \; S_\delta$$ It is convenient to define column vector $\tilde{a}_s = \begin{pmatrix} {\bf a}_{s1} \\ {\bf a}_{s2} \\ {\bf a}_{s3} \\ \vdots \\ {\bf a}_{s(3+d)}\end{pmatrix}$, which contains each of the superspace basis vectors. The new superpace basis vectors are related to old according to $$\tilde{a}'_s = (S^{-1})^t \tilde{a}_s.$$ It is also convenient to define three column vectors: $\tilde{a} = \begin{pmatrix} {\bf a}_1 \\ {\bf a}_2 \\ {\bf a}_3 \end{pmatrix}$, $\tilde{a}^* = \begin{pmatrix} {\bf a}_1^* \\ {\bf a}_2^* \\ {\bf a}_3^* \end{pmatrix}$ and $\tilde{q} = \begin{pmatrix} {\bf q}_1 \\ \vdots \\ {\bf q}_d \end{pmatrix}$, where the ${\bf a}_i$ are the basis vectors of the 3-dimensional external space, the ${\bf a}_i^*$ are the corresponding reciprocal space vectors, and the ${\bf q}_i$ are the modulation vectors, so that the components of the modulation vectors can be expressed in terms of the 3D reciprocal-space basis vectors as $$\tilde{q} = \sigma ~ \tilde{a}^*.$$ We are then able to summarize the relationships between the new and old basis vectors and modulations vectors as follows: $$\tilde{a}' = (S_R^{-1})^t ~ \tilde{a}$$ $$\tilde{a}^{*}{'} = S_R ~ \tilde{a}^*$$ $$\tilde{q}' = S_M ~ \tilde{a}^* + S_\epsilon ~ \tilde{q}$$ $$\sigma' = (S_M + S_\epsilon ~ \sigma) ~ S_R^{-1}$$ The example transformation given was based on a non-standard setting of superspace-group 65.2.43.64 $Cmmm(0,b_1,1/2)000(1,0,g_2)0s0$ with the following generators: $(0,0,1/2,0,1/2); (1/2,1/2,0,1/2,0); (x,y,-z,-t,u); (1/2+x,-y,z,1/2+t,u); (-x,-y,-z,-t,-u)$. The transformation to the standard setting is $$S = \left( \begin{array}{cc|c} S_R & 0 & S_v \\ S_M & S_\epsilon & S_\delta \\ \hline 0 & 0 & 1 \end{array}\right) = \left( \begin{array}{cc|c} \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 2 \end{pmatrix} & \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} & \begin{pmatrix} 1/4 \\ 1/4 \\ 0 \end{pmatrix} \\ \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} & \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} & \begin{pmatrix} 0 \\ 0 \end{pmatrix} \\ \hline \begin{pmatrix} 0 & 0 & 0 \end{pmatrix} & \begin{pmatrix} 0 & 0 \end{pmatrix} & \begin{pmatrix} 1 \end{pmatrix} \end{array} \right)$$ The relationship between this matrix and the example input is evident. $$\tilde{a}^{*}{'} =\begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 2 \end{pmatrix} ~ \begin{pmatrix} {\bf a}_1^* \\ {\bf a}_2^* \\ {\bf a}_3^* \end{pmatrix} =\begin{pmatrix} {\bf a}_2^*\\ -{\bf a}_1^* \\ 2{\bf a}_3^* \end{pmatrix}$$ $$\tilde{q}' = \begin{pmatrix} {\bf q}_1' \\ {\bf q}_2' \end{pmatrix} = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} ~ \begin{pmatrix} {\bf a}_1^* \\ {\bf a}_2^* \\ {\bf a}_3^* \end{pmatrix} + \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} ~ \begin{pmatrix} {\bf q}_1 \\ {\bf q}_2 \end{pmatrix} =\begin{pmatrix} {\bf a}_3^* + {\bf q}_2 \\ {\bf a}_2^* + {\bf q}_1 \end{pmatrix}$$ $$\sigma' = \left[ \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} + \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} ~ \begin{pmatrix} 0 & 0 & 0.156 \\ 0.178 & 0 & 0 \end{pmatrix} \right] ~ \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 2 \end{pmatrix}^{-1} = \begin{pmatrix} 0 & -0.178 & 1/2 \\ 1 & 0 & 0.078 \end{pmatrix}$$ While we have the freedom to express the vector form of $\tilde{q}'$ in terms of the old reciprocal basis vectors and modulation vectors, $\sigma'$ is the component form of $\tilde{q}'$ in the new coordinate system. Thus, in the old coordinate system for 3D reciprocal space, the component form of $\tilde{q}' = \begin{pmatrix} {\bf a}_3^* + {\bf q}_2 \\ {\bf a}_2^* + {\bf q}_1 \end{pmatrix}$ is $\begin{pmatrix} 0.0178 & 0 & 1 \\ 0 & 1 & 0.156 \end{pmatrix}$, which must then be transformed to the new 3D reciprocal-space coordinate system by applying a factor of $S_R^{-1}$ on the right-hand side in order to obtain $\sigma'$. The component forms of the superspace origin of the old setting in the new coordinate system ($\tau_{old}'$) and the new origin in the old coordinate system ($\tau_{new}$) are $$\tau_{old}' = \left( \begin{array}{cc|c} S_R & 0 & S_v \\ S_M & S_\epsilon & S_\delta \\ \hline 0 & 0 & 1 \end{array}\right) ~ \begin{pmatrix} 0 \\ 0 \\ \hline 1 \end{pmatrix} = \begin{pmatrix} S_v \\ S_\delta \\ \hline 1 \end{pmatrix} = \begin{pmatrix} \begin{pmatrix} 1/4 \\ 1/4 \\ 0 \end{pmatrix} \\ \begin{pmatrix} 0 \\ 0 \end{pmatrix} \\ \hline 1 \end{pmatrix} ~~~ \text{and} ~~~ \tau_{new} = \begin{pmatrix} (S^{-1})_v \\ (S^{-1})_\delta \\ \hline 1 \end{pmatrix} = \begin{pmatrix} \begin{pmatrix} 1/4 \\ -1/4 \\ 0 \end{pmatrix} \\ \begin{pmatrix} 1/4 \\ 0 \end{pmatrix} \\ \hline 1 \end{pmatrix}$$ By expressing the second centering translation, $(1/2,1/2,0,1/2,0)$, in matrix form and transforming it to the new setting, we obtain $$\left( \begin{array}{cc|c} I_3 & 0 & v' \\ 0 & I_d & \delta' \\ \hline 0 & 0 & 1 \end{array}\right) = S \left( \begin{array}{cc|c} I_3 & 0 & v \\ 0 & I_d & \delta \\ \hline 0 & 0 & 1 \end{array}\right) S^{-1} = \left( \begin{array}{cc|c} I_3 & 0 & S_R ~ v \\ 0 & I_d & S_M ~ v + S_\epsilon \delta \\ \hline 0 & 0 & 1 \end{array}\right)$$ $$v' = \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 2 \end{pmatrix} ~ \begin{pmatrix} 1/2 \\ 1/2 \\ 0 \end{pmatrix} = \begin{pmatrix} 1/2 \\ -1/2 \\ 0 \end{pmatrix}$$ $$\delta' = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} ~ \begin{pmatrix} 1/2 \\ 1/2 \\ 0 \end{pmatrix} + \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} ~ \begin{pmatrix} 1/2 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}.$$ The new centering vector is then $(1/2,-1/2,0,0,1)$, which is mod(1) equivalent to $(1/2,1/2,0,0,0)$. We can also treat the second non-translational generator, $(1/2+x,-y,z,1/2+t,u)$, in a similar fashion to obtain $$g' = S ~ \left( \begin{array}{cc|c} \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{pmatrix} & \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} & \begin{pmatrix} 1/2 \\ 0 \\ 0 \end{pmatrix} \\ \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} & \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} & \begin{pmatrix} 1/2 \\ 0 \end{pmatrix} \\ \hline \begin{pmatrix} 0 & 0 & 0 \end{pmatrix} & \begin{pmatrix} 0 & 0 \end{pmatrix} & \begin{pmatrix} 1 \end{pmatrix} \end{array} \right) ~ S^{-1} = \left( \begin{array}{cc|c} \begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} & \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} & \begin{pmatrix} 1/2 \\ -1/2 \\ 0 \end{pmatrix} \\ \begin{pmatrix} 0 & 0 & 0 \\ -2 & 0 & 0 \end{pmatrix} & \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} & \begin{pmatrix} 0 \\ 1 \end{pmatrix} \\ \hline \begin{pmatrix} 0 & 0 & 0 \end{pmatrix} & \begin{pmatrix} 0 & 0 \end{pmatrix} & \begin{pmatrix} 1 \end{pmatrix} \end{array} \right)$$ The new operator is then $(-x+1/2,y-1/2,z,t,-2x + u + 1)$, which is mod(1) equivalent to $(-x+1/2,y+1/2,z,t,-2x + u)$.