Figure 4, as well as figure 5 and figures 6 to 8, also displays the KL divergence $D_{\text{KL}}(E_i \parallel E_{18})$ of the usual (diagonal) $E^3$ covering-space CMB temperature correlation matrix $C_{\ell m \ell' m'}^{E_{18};TT}$ given a true underlying correlation matrix $C_{\ell m \ell' m'}^{E_i;TT}$ of $E_i$ ($i \neq 18$) (assuming identical cosmological parameters). This represents the information potentially available from measured CMB temperature fluctuations to distinguish between the possibility that the Universe is an $E_i$ manifold and the possibility that it is the covering space. We have also plotted $D_{\text{KL}}(E_{18} \parallel E_i)$ in figures 4 and 5, and an alternative cosmic variance signal-to-noise statistic (5.16) which is more straightforward to calculate for large ranges of $\ell$, and which conveys the same message. We have not displayed these two statistics in figures 6 to 8 as they contain no clear new information.

These measures of the information content suggest that once the clone-to-last-scattering ratio — the ratio of the distance to an observer’s nearest topological clone to the diameter of the last scattering surface — is greater than 1, the information available to distinguish between an $E_i$ manifold and the covering space declines rapidly. In most cases, we find that the “standard” manifold parameter choices and observer locations minimize the available information and that other manifold parameters or observer positions are more favorable for detecting topology. Nevertheless, in all cases, by the time the clone-to-last-scattering ratio reaches somewhere between 1.1 and 1.3 the KL divergences of the CMB temperature fluctuations have fallen below 1 and there is no prospect of using those fluctuations to differentiate between topologies. On the one hand, this is disappointing given that we already know observationally (with 95% confidence) [26, 93] that this ratio is greater than 0.985 — we might have hoped to explore much larger values of the ratio. On the other hand, if a non-trivial cosmic topology is the heart of the explanation for large-angle anomalies in the CMB temperature fluctuations [13–18, 94–98], then there is significant information about the topology of the Universe in the CMB, teaching us that the clone-to-last-scattering ratio is likely $\lesssim 1.3$, assuming that the insights gained from $E^3$ apply more broadly to all manifolds.

The CMB is fundamentally poor at determining the amplitudes of individual Fourier modes because the CMB signal is dominated by the Sachs-Wolfe effect on the last scattering surface, and is thus a measure of the gravitational potential on that spherical shell centered on us. The projection from the statistically simple correlation between Fourier modes to the correlation between spherical harmonics is certainly not invertible and leads to a considerable loss of information. We are optimistic that three-dimensional observations (large-scale structure surveys, intensity mapping, etc.) will not experience the same precipitous drop in the KL divergence for clone-to-last-scattering ratios greater than 1. This will be explored in an upcoming paper [91].

## Acknowledgments

We thank Jeffrey Weeks and David Singer for valuable conversations, and Javier Carrón Duque for providing extensive constructive comments on a draft of this paper. J.R.E. acknowledges support from the European Research Council under the Horizon 2020 Research and Innovation Programme (Grant agreement No. 819478). Y.A. acknowledges support by the Spanish Research Agency (Agencia Estatal de Investigación)'s grant RYC2020-030193-I/AEI/10.13039/501100011033, by the European Social Fund (Fondo Social Europeo) through
the Ramón y Cajal program within the State Plan for Scientific and Technical Research and Innovation (Plan Estatal de Investigación Científica y Técnica y de Innovación) 2017-2020, by the Spanish Research Agency through the grant IFT Centro de Excelencia Severo Ochoa No CEX2020-001007-S funded by MCIN/AEI/10.13039/501100011033, and by the Richard S. Morrison Fellowship at CWRU. C.J.C., A.K., D.P.M., and G.D.S. acknowledge partial support from NASA ATP grant RES240737; G.D.S. from DOE grant DESC0009946; Y.A., G.D.S., P.P., S.S., Q.T., and O.G. from the Simons Foundation; Y.A., A.H.J., and G.D.S. from the Royal Society (U.K.); and A.H.J. from STFC in the U.K. A.T. is supported by the Richard S. Morrison Fellowship. Y.A. and G.D.S. thank the INFN (Sezione di Padova); J.R.E., S.A., D.P.M., G.D.S., and A.T. thank the IFT; and A.K. thanks CWRU and ICL for support and hospitality where part of this work was accomplished.

## A Construction of general generators

In this work, the most general allowed set of three generators $g_{aj}$ for each of the orientable $E_i$ has been provided. Here we describe a constructive, algebraic approach to determining the parametrization of these generators.

The topology of each orientable manifold can be set by the choice of elements of SO(3), i.e., by the matrices $M_a^{E_i}$. There is limited freedom to adjust the $M_a^{E_i}$: only the freedom to rotate the coordinate system to choose the orientations of the axes of rotation.<fnref n="21" /> The remainder and essence of our task will then be to determine the most general allowed vectors $T_{aj}^{E_i}$ associated with the $M_a^{E_i}$ of each $E_i$. Once the orientation of the coordinate system has been set, any remaining freedom will be used to simplify the three vectors $T_{aj}^{E_i}$. There are two principal tools available to constrain the components of $T_{aj}^{E_i}$:

1. Any finite sequence of generators and their inverses are group elements. Functionally, this is enforced by checking that any such sequence that is a pure translation is an integer linear combination of some “basis set” of three linearly independent translations, so that the group is a discrete group, $\Gamma^{E_i}$. In section 3 we call this basis set (the generators of) “the associated $E_1$”. Part of this process is the determination of the set of such pure translations that can be chosen as the basis set.

For example, consider two generators $g_{ai}^{E_n}$ and $g_{bj}^{E_n}$ of $\Gamma^{E_n}$ for one of the orientable Euclidean manifolds $E_n$, associated with the SO(3) elements $M_a^{E_n}$ and $M_b^{E_n}$, respectively, and let $T_k^{E_n}$ ($k = 1, 2, 3$) represent the pure translations of the associated $E_1$. Since $(M_a^{E_n})^{-1}(M_b^{E_n})^{-1}M_a^{E_n}M_b^{E_n} = 1$ for all $a$ and $b$ in all such $\Gamma^{E_n}$,<fnref n="22" /> we must insist that

$$ (g_{ai}^{E_n})^{-1} (g_{bj}^{E_n})^{-1} g_{ai}^{E_n} g_{bj}^{E_n} : x \to x + \sum_{k=1}^{3} m_k T_k^{E_n} \quad (A.1) $$

for some triplet of integers $m_k$.

<footnote n="21">²¹The set of matrices is not always unique even up to rotations of the coordinate system. The Hantzsche-Wendt and non-orientable manifolds can be described by multiple different sets of matrices. Further, for the non-orientable manifolds the same set of $O(3)$ elements $M_a^{E_i}$ can be associated with discretely different translation vectors $T_{aj}^{E_i}$ and thus generate topologically distinct manifolds.</footnote>

<footnote n="22">²²Note that this is not true for arbitrary elements of $O(3)$ since it is a non-abelian group. However, it is true for the particular elements of $O(3)$ used in the generators for each of the $E_n$.</footnote>
This may not be sufficient. Consider the case where $M_a^{E_n} \neq 1$ but $(M_a^{E_n})^2 = 1$ and where there are (at least) two generators associated with this matrix. Then the application of any two of these generators should result in a pure translation. In other words, all combinations of the form

$$g_{ai}^{E_n} g_{aj}^{E_n}, \quad g_{ai}^{E_n} (g_{aj}^{E_n})^{-1}, \quad (g_{ai}^{E_n})^{-1} g_{aj}^{E_n}, \quad (g_{ai}^{E_n})^{-1} (g_{aj}^{E_n})^{-1} \quad (A.2)$$

for each of $i, j \in \{1, 2\}$ must lead to pure translations that are integer linear combinations of the basis set. If this is true, then (A.1) is trivially satisfied. For some topologies two of these combinations will be used to define the translation vectors and the rest will lead to constraints that must be satisfied.

2. We must also ensure that the set of transformations (the group elements) consists only of freely acting transformations, i.e., no transformation (other than the identity transformation) has a fixed point.

Once these conditions have been enforced, we may find that there appear to be distinct sets of “solutions”, i.e., parametrizations of the $T_{aj}^{E_i}$ that cannot be transformed into one another by rotations, reorderings, or rescalings. We must still prove that two such sets do not generate the same lattice of clones for a given starting point. A simple way that this can happen is if the $T_{aj}^{E_i}$ of one set are just integer linear combinations of the vectors of the other set.

We will show individually that we are able to bring this program to a successful conclusion for each of the orientable manifolds.

## A.1 $E_1$: 3-torus

We have established in the main body of the paper that the three translations for $E_1$ can be parametrized most generally as

$$T_1^{E_1} = \begin{pmatrix} L_{1x} \\ 0 \\ 0 \end{pmatrix}, \quad T_2^{E_1} = \begin{pmatrix} L_{2x} \\ L_{2y} \\ 0 \end{pmatrix}, \quad T_3^{E_1} = \begin{pmatrix} L_{3x} \\ L_{3y} \\ L_{3z} \end{pmatrix}, \quad (A.3)$$

where the three 0 entries have been fixed through a rotation. Specifics on the ranges of the parameters are given in the main body.

## A.2 $E_2$: half-turn space

The half-turn space has one generator, $g_B$, that is a corkscrew motion, while the other two, $g_{A_j}$ ($j = 1, 2$), are pure translations. Here we develop the techniques employed in the rest of the orientable manifolds to algebraically determine an action for the most general generators.

To begin we first use two of the rotational degrees of freedom to align the corkscrew axis with the z-axis, i.e., $M_B = \text{diag}(-1, -1, 0)$, and use the remaining rotational degree of freedom to set $L_{1y} = 0$. Next, we can choose a special origin that allows us to set $L_{Bx} = L_{By} = 0$ (cf. eq. (2.8)). We will restore an arbitrary origin at the end of the calculation. With all
of this, we can choose our general starting case to be$^{23}$

$$ T_{A_1} = \begin{pmatrix} L_{1x} \\ 0 \\ L_{1z} \end{pmatrix}, \quad T_{A_2} = \begin{pmatrix} L_{2x} \\ L_{2y} \\ L_{2z} \end{pmatrix}, \quad T_B = \begin{pmatrix} 0 \\ 0 \\ L_{Bz} \end{pmatrix}. \qquad (A.4) $$

From these a set of pure translations, an associated $E_1$, can be constructed. The actions of both the $g_{A_i}$ are already pure translations, so two of the needed translation vectors can be chosen as $T_1 = T_{A_1}$ and $T_2 = T_{A_2}$. A third translation vector can be defined based on the fact that $(M_B^{E_2})^2 = 1$. Note that

$$ g_3^{E_2} = (g_B^{E_2})^2 : \boldsymbol{x} \to \boldsymbol{x} + (1 + \mathbf{M}_B^{E_2}) \boldsymbol{T}_B^{E_2}. \qquad (A.5) $$

From this the third translation vector is

$$ T_3^{E_2} = (1 + \mathbf{M}_B^{E_2}) T_B^{E_2} = \mathbf{M}_{02}^{E_2} T_B^{E_2}, \qquad (A.6) $$

for $M_{02}^{E_2} = \text{diag}(0, 0, 2)$ from (4.35). Thus the third translation vector of the associated $E_1$ for $E_2$ is as quoted in the main text (3.8),

$$ T_3^{E_2} = \begin{pmatrix} 0 \\ 0 \\ 2L_{Bz} \end{pmatrix}. \qquad (A.7) $$

For our subsequent development here we choose $T_3 = T_3^{E_2}$.

The remaining condition to impose (all other combinations of generators will either already be pure translations or reduce to pure translations of these) is that for $i$ and $j \in \{1, 2\}$

$$ g_B^{-1} g_{A_i}^{-1} g_B g_{A_j} : \boldsymbol{x} \to \boldsymbol{x} + T_{A_j} - \mathbf{M}_B T_{A_i} = \boldsymbol{x} + \sum_k m_k^{(i)} T_k, \qquad (A.8) $$

for some set of $m_k^{(i)} \in \mathbb{Z}$. Since $T_{A_j}$ is already one of the pure translations, $T_j$, this is equivalent to requiring that

$$ \mathbf{M}_B T_{A_i} = \sum_k m_k^{(i)} T_k, \quad \text{for some } m_k^{(i)} \in \mathbb{Z}, \qquad (A.9) $$

leading to the set of conditions

$$ \begin{pmatrix} -L_{1x} \\ 0 \\ L_{1z} \end{pmatrix} = \begin{pmatrix} m_1^{(1)} L_{1x} + m_2^{(1)} L_{2x} \\ m_2^{(1)} L_{2y} \\ m_1^{(1)} L_{1z} + m_2^{(1)} L_{2z} + m_3^{(1)} 2L_{Bz} \end{pmatrix}, \qquad (A.10) $$

$$ \begin{pmatrix} -L_{2x} \\ -L_{2y} \\ L_{2z} \end{pmatrix} = \begin{pmatrix} m_1^{(2)} L_{1x} + m_2^{(2)} L_{2x} \\ m_2^{(2)} L_{2y} \\ m_1^{(2)} L_{1z} + m_2^{(2)} L_{2z} + m_3^{(2)} 2L_{Bz} \end{pmatrix}. \qquad (A.11) $$

<footnote>$^{23}$Here and throughout this appendix we will refrain from adding topology-specific superscripts to quantities until the final results. All intermediate quantities should be recognized as being specific to the topology under consideration.</footnote>
From the x-components of these equations and the fact that $L_{2y} \neq 0$ (since at least one of the translation vectors needs to have a nonzero y-component) we immediately see that $m_2^{(1)} = 0$ and $m_2^{(2)} = -1$, which in turn means that $m_1^{(1)} = -1$ and $m_1^{(2)} = 0$, so that we are finally left with

$$L_{1z} = m_3^{(1)} L_{Bz} \quad \text{and} \quad L_{2z} = m_3^{(2)} L_{Bz}. \qquad (A.12)$$

Since $T_{3z} = 2L_{Bz}$, we can always subtract integer multiples of $T_3$, which is equivalent to subtracting $2nL_{Bz}$ for $n \in \mathbb{Z}$ from $L_{1z}$ and $L_{2z}$, meaning that we only need to consider $m_3^{(1)} \in \{0, 1\}$ and $m_3^{(2)} \in \{0, 1\}$. However, if either of $m_3^{(i)}$ are 1, then $g_{A_i}^{-1} g_B$ has a fixed point. Explicitly, consider the case $m_3^{(1)} = 1$ so that $L_{1z} = L_{Bz}$. Notice that

$$g_{A_1}^{-1} g_B : x \to \begin{pmatrix} -x - L_{1x} \\ -y \\ z \end{pmatrix}, \qquad (A.13)$$

and so $g_{A_1}^{-1} g_B : x \to x$, for the special location $x = (-L_{1x}/2, 0, z)^T$. Hence $g_{A_1}^{-1} g_B$ has a fixed point and this case is invalid. A similar argument for $m_3^{(2)} = 1$ leads to the requirement $L_{1z} = L_{2z} = 0$.

Restoring an arbitrary origin (which reintroduces $L_{Bx}$ and $L_{By}$), we are left with the generic result quoted in the main text (3.5),

$$T_{A_1}^{E_2} = \begin{pmatrix} L_{1x} \\ 0 \\ 0 \end{pmatrix}, \quad T_{A_2}^{E_2} = \begin{pmatrix} L_{2x} \\ L_{2y} \\ 0 \end{pmatrix}, \quad T_{B}^{E_2} = \begin{pmatrix} L_{Bx} \\ L_{By} \\ L_{Bz} \end{pmatrix}, \qquad (A.14)$$

or equivalently (3.6),

$$T_{A_1}^{E_2} = L_1 \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \quad T_{A_2}^{E_2} = L_2 \begin{pmatrix} \cos \alpha \\ \sin \alpha \\ 0 \end{pmatrix}, \quad T_{B}^{E_2} = L_B \begin{pmatrix} \cos \beta \cos \gamma \\ \cos \beta \sin \gamma \\ \sin \beta \end{pmatrix}. \qquad (A.15)$$

### A.3 $E_3$: quarter-turn space

The quarter-turn space is similar to $E_2$. Since there is a quarter turn, instead of a half turn, the computations are a bit more involved, but follow similar logic. Here we will explicitly impose the ordering conditions discussed in the main text.

To begin we first use the freedom to rotate the coordinate system such that the axis of rotation of $M_B$ is parallel to $\hat{e}_z$, and then to choose $T_{A_1}$ to be in the $xz$-plane. For $E_3$

$$M_B = R_{\hat{e}_z}(\pi/2) = \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \qquad (A.16)$$

which has the properties

$$(M_B)^2 = \mathrm{diag}(-1, -1, 1), \quad (M_B)^3 = M_B^T, \quad (M_B)^4 = 1. \qquad (A.17)$$
Once again $\mathbf{M}_A = 1$, so $g_{A_1}$ and $g_{A_2}$ are pure translations by $\mathbf{T}_{A_1}$ and $\mathbf{T}_{A_2}$, respectively. We therefore assign them the alternative labels $\mathbf{T}_1 \equiv \mathbf{T}_{A_1}$ and $\mathbf{T}_2 \equiv \mathbf{T}_{A_2}$. Finally, as in $E_2$ we can shift to the origin where $L_{Bx} = L_{By} = 0$ and again have as our starting case

$$ \mathbf{T}_{A_1} = \begin{pmatrix} L_{1x} \\ 0 \\ L_{1z} \end{pmatrix}, \quad \mathbf{T}_{A_2} = \begin{pmatrix} L_{2x} \\ L_{2y} \\ L_{2z} \end{pmatrix}, \quad \mathbf{T}_B = \begin{pmatrix} 0 \\ 0 \\ L_{Bz} \end{pmatrix}. \qquad (A.18) $$

Similar to $E_2$, an associated $E_1$ can be constructed from three pure linearly independent translations. We again have the obvious translation vectors $\mathbf{T}_1 \equiv \mathbf{T}_{A_1}$ and $\mathbf{T}_2 \equiv \mathbf{T}_{A_2}$. A third translation vector can be defined based on the fact that $(\mathbf{M}_B^{E_3})^4 = 1$. Noting that

$$ g_3^{E_3} \equiv (g_B^{E_3})^4 : \boldsymbol{x} \to \boldsymbol{x} + \mathbf{M}_{04}^{E_3} \mathbf{T}_B^{E_3}, \qquad (A.19) $$

where $\mathbf{M}_{04}^{E_3} = \mathrm{diag}(0, 0, 4)$ from (4.35), the third translation vector of the associated $E_1$ for $E_3$ is, as quoted in the main text (3.14),

$$ \mathbf{T}_3^{E_3} = \begin{pmatrix} 0 \\ 0 \\ 4L_{Bz} \end{pmatrix}. \qquad (A.20) $$

For our subsequent development here we choose $\mathbf{T}_3 \equiv \mathbf{T}_3^{E_3}$.

To proceed we can always choose $\mathbf{T}_{A_1}$ such that $|\mathbf{T}_{A_1 \perp}| \le |\mathbf{T}_{A_2 \perp}|$, where $\perp$ means the projection into the $xy$-plane. Also, by adding or subtracting integer multiples of $\mathbf{T}_1$ from $\mathbf{T}_{A_2}$, we can ensure that $|L_{2x}| < |L_{1x}|$; thus we generically have

$$ (L_{2x})^2 < (L_{1x})^2 \le (L_{2x})^2 + (L_{2y})^2. \qquad (A.21) $$

Similarly, shifting by integer multiples of $\mathbf{T}_3$ allows us to insist that $|L_{1z}| \le 2|L_{Bz}|$ and $|L_{2z}| \le 2|L_{Bz}|$.

Given the basis set for the associated $E_1$ of $\mathbf{T}_1$, $\mathbf{T}_2$, and $\mathbf{T}_3$ we can proceed to find the most general action of the generators. Whereas for $E_2$ we had one set of conditions (A.9) on the products of generators and their inverses being integer linear combination of $\mathbf{T}_i^{E_2}$, for $E_3$ we have two independent sets, which we can take for $i$ and $j \in \{1, 2\}$ to be

$$ g_B g_{A_i}^{-1} g_B^{-1} g_{A_j} : \boldsymbol{x} \to \boldsymbol{x} + \mathbf{T}_{A_j} - \mathbf{M}_B \mathbf{T}_{A_i} = \boldsymbol{x} + \mathbf{T}_{A_j} + \sum_k m_k^{(i)} \mathbf{T}_k, \qquad (A.22) $$

$$ (g_B^{-1})^2 g_{A_i}^{-1} (g_B)^2 g_{A_j} : \boldsymbol{x} \to \boldsymbol{x} + \mathbf{T}_{A_j} - (\mathbf{M}_B)^2 \mathbf{T}_{A_i} = \boldsymbol{x} + \mathbf{T}_{A_j} + \sum_k n_k^{(i)} \mathbf{T}_k. \qquad (A.23) $$

Just like in $E_2$, these reduce to

$$ \mathbf{M}_B \mathbf{T}_{A_i} = \sum_k m_k^{(i)} \mathbf{T}_k, \quad \text{for some } m_k^{(i)} \in \mathbb{Z}, \qquad (A.24) $$

$$ (\mathbf{M}_B)^2 \mathbf{T}_{A_i} = \sum_k n_k^{(i)} \mathbf{T}_k, \quad \text{for some } n_k^{(i)} \in \mathbb{Z}. \qquad (A.25) $$

Since $\mathbf{M}_B \mathbf{T}_{A_1} = \mathbf{T}_{A_2}$, and since we could have used $g_B^{-1} g_{A_i}^{-1} g_B g_{A_j}$ instead of $g_B g_{A_i}^{-1} g_B^{-1} g_{A_j}$ for the left-hand side of (A.22), which would have replaced $\mathbf{M}_B$ with $\mathbf{M}_B^T = \mathbf{M}_B^3$ on the
right-hand side of (A.22), we need to only consider (A.24). Thus, despite all the apparent freedom, this once again leads to only two sets of conditions:

$$ \begin{pmatrix} 0 \\ L_{1x} \\ L_{1z} \end{pmatrix} = \begin{pmatrix} m_1^{(1)} L_{1x} + m_2^{(1)} L_{2x} \\ m_2^{(1)} L_{2y} \\ m_1^{(1)} L_{1z} + m_2^{(1)} L_{2z} + m_3^{(1)} 4L_{Bz} \end{pmatrix}, \qquad (A.26) $$

$$ \begin{pmatrix} -L_{2y} \\ L_{2x} \\ L_{2z} \end{pmatrix} = \begin{pmatrix} m_1^{(2)} L_{1x} + m_2^{(2)} L_{2x} \\ m_2^{(2)} L_{2y} \\ m_1^{(2)} L_{1z} + m_2^{(2)} L_{2z} + m_3^{(2)} 4L_{Bz} \end{pmatrix}. \qquad (A.27) $$

As in $E_2$ we must have $L_{2y} \neq 0$. From the y-component of (A.26) we see immediately that

$$ L_{1x} = m_2^{(1)} L_{2y} \neq 0. \qquad (A.28) $$

To show this, suppose that $L_{1x} = 0$. From (A.28) this means that $m_2^{(1)} = 0$. Applying these relations to the x- and y-components of (A.27) leads to $L_{2y} = -(m_2^{(2)})^2 L_{2y}$. Since $L_{2y} \neq 0$ this would then require $m_2^{(2)} = 0$, but from the x-component of (A.27) this would force $L_{2y} = 0$, which is inconsistent. Thus, $L_{1x} \neq 0$ and we must have $m_2^{(1)} \neq 0$.

Proceeding, we note that $|T_{A_1\perp}| = |L_{1x}| = |m_2^{(1)}||L_{2y}|$ and $|T_{A_2\perp}| = \sqrt{L_{2x}^2 + L_{2y}^2} \le |L_{2y}|$. Thus $|T_{A_1\perp}| \le |T_{A_2\perp}|$ requires $|m_2^{(1)}| \le 1$, which means that $m_2^{(1)} = \pm 1$. Next, from the x-component of (A.26), we require $L_{2x} = -m_1^{(1)}L_{2y}$. However, we can always add or subtract integer multiples of $T_1$ from $T_{A_2}$, so $|L_{2x}|$ cannot be larger than $\frac{1}{2}L_{1x}$, but again, since $|T_{A_1\perp}| \le |T_{A_2\perp}|$, we must have

$$ 0 < L_{1x}^2 \le L_{2x}^2 + L_{2y}^2 = \frac{1 + (m_1^{(1)})^2}{(m_1^{(1)})^2} L_{2x}^2 \le \frac{1 + (m_1^{(1)})^2}{(m_1^{(1)})^2} \frac{L_{1x}^2}{4}. \qquad (A.29) $$

This is satisfied if and only if $m_1^{(1)} = 0$. Since $m_2^{(1)} \neq 0$, the x-component of (A.26) informs us that

$$ L_{2x} = 0. \qquad (A.30) $$

The y-component of (A.27) then requires that $m_2^{(2)} = 0$.

Next, the x-component of (A.27) together with (A.28) and $m_2^{(2)} = 0$ gives us that $m_1^{(2)}m_2^{(1)} = -1$, thus (since $m_2^{(1)} = \pm 1$) we have $m_1^{(2)} = -m_2^{(1)} = \mp 1$. We thus conclude that

$$ L_{2y} = \pm L_{1x}. \qquad (A.31) $$

The z-components of the two conditions now give us

$$ \begin{aligned} 0 &= -L_{1z} \pm L_{2z} + m_3^{(1)} 2L_{Bz}, \\ 0 &= \mp L_{1z} - L_{2z} + m_3^{(2)} 4L_{Bz}. \end{aligned} \qquad (A.32) $$
Thus $L_{1z}$ and $L_{2z}$ are both integer multiples of $L_{Bz}$. But, as in $E_2$, unless these integers are both 0, at least one of $g_{A_1}^{-1}(g_B)^n$ for $n \in \{1, 2, 3\}$ will have a fixed point. From this, we can show that we must have

$$L_{1z} = L_{2z} = 0. \qquad (A.33)$$

Putting all of this together, shifting to an arbitrary origin, and renaming $L_{1x}$ to $L_A$ we arrive at the general form from the text (3.11)

$$T_{A_1}^{E_3} = \begin{pmatrix} L_A \\ 0 \\ 0 \end{pmatrix}, \quad T_{A_2}^{E_3} = \begin{pmatrix} 0 \\ L_A \\ 0 \end{pmatrix}, \quad T_B^{E_3} = \begin{pmatrix} L_{Bx} \\ L_{By} \\ L_{Bz} \end{pmatrix}. \qquad (A.34)$$

Note that we might have thought we should put $\pm L_A$ in $T_{A_2}^{E_3}$, however this can be removed by rotating about the z-axis by $\pi/2$ or by replacing $T_{A_2}^{E_3}$ with $-T_{A_2}^{E_3}$.

### A.4 $E_4$: third-turn space

The third-turn space is similar to the previous two cases, particularly $E_3$, from which the development follows. We again begin by using the rotational freedom to set the axis of $M_B$ parallel to $\hat{e}_z$ and then to choose $T_{A_1}$ to be in the $xz$-plane. For $E_4$

$$M_B = R_{\hat{z}}(2\pi/3) = \begin{pmatrix} -1/2 & -\sqrt{3}/2 & 0 \\ \sqrt{3}/2 & -1/2 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \qquad (A.35)$$

which has the properties

$$(M_B)^2 = M_B^T \quad \text{and} \quad (M_B)^3 = \mathbf{1}. \qquad (A.36)$$

As in $E_2$ and $E_3$ and shifting the origin so that $L_{Bx} = L_{By} = 0$, we again start with

$$T_{A_1} = \begin{pmatrix} L_{1x} \\ 0 \\ L_{1z} \end{pmatrix}, \quad T_{A_2} = \begin{pmatrix} L_{2x} \\ L_{2y} \\ L_{2z} \end{pmatrix}, \quad T_B = \begin{pmatrix} 0 \\ 0 \\ L_{Bz} \end{pmatrix}. \qquad (A.37)$$

Similar to $E_2$, an associated $E_1$ can be constructed from three pure translations. We again have the obvious translation vectors $T_1 = T_{A_1}$ and $T_2 = T_{A_2}$. A third translation vector can be defined based on the fact that $(M_B^{E_4})^3 = \mathbf{1}$. Noting that

$$g_3^{E_4} = (g_B^{E_4})^3 : x \to x + M_{03}^{E_4} T_B^{E_4}, \qquad (A.38)$$

where $M_{03}^{E_4} = \text{diag}(0, 0, 3)$ from (4.35), the third translation vector of the associated $E_1$ for $E_4$ is as quoted in the main text (3.20),

$$T_3^{E_4} = \begin{pmatrix} 0 \\ 0 \\ 3L_{Bz} \end{pmatrix}. \qquad (A.39)$$

For our subsequent development here, we choose $T_3 = T_3^{E_4}$.
To proceed, we can always choose $T_{A_1}$ such that $|T_{A_1\perp}| \le |T_{A_2\perp}|$, where $\perp$ means the projection into the $xy$-plane. Also, by adding or subtracting integer multiples of $T_1$ from $T_{A_2}$, we can ensure that $|L_{2x}| < |L_{1x}|$; thus we generically have

$$ (L_{2x})^2 < (L_{1x})^2 \le (L_{2x})^2 + (L_{2y})^2. \qquad (A.40) $$

Since $M_B^2 = M_B^T$, we only need to consider the condition with one factor of $g_B$,

$$ g_B^{-1} g_{A_i}^{-1} g_B g_{A_j} : \boldsymbol{x} \to \boldsymbol{x} + \boldsymbol{T}_{A_j} - \boldsymbol{M}_B^T \boldsymbol{T}_{A_i} = \boldsymbol{x} + \sum_k m_k^{(i)} \boldsymbol{T}_k, \qquad (A.41) $$

and not the condition analogous to (A.23). Again this reduces to

$$ \mathbf{M}_B \mathbf{T}_{A_i} = \sum_k m_k^{(i)} \mathbf{T}_k, \quad \text{for some } m_k^{(i)} \in \mathbb{Z}, \qquad (A.42) $$

and again leads to the two sets of conditions

$$ \begin{pmatrix} -\frac{1}{2}L_{1x} \\ -\frac{\sqrt{3}}{2}L_{1x} \\ L_{1z} \end{pmatrix} = \begin{pmatrix} m_1^{(1)}L_{1x} + m_2^{(1)}L_{2x} \\ m_2^{(1)}L_{2y} \\ m_1^{(1)}L_{1z} + m_2^{(1)}L_{2z} + m_3^{(1)}3L_{Bz} \end{pmatrix}, \qquad (A.43) $$

$$ \begin{pmatrix} -\frac{1}{2}L_{2x} - \frac{\sqrt{3}}{2}L_{2y} \\ \frac{\sqrt{3}}{2}L_{2x} - \frac{1}{2}L_{2y} \\ L_{2z} \end{pmatrix} = \begin{pmatrix} m_1^{(2)}L_{1x} + m_2^{(2)}L_{2x} \\ m_2^{(2)}L_{2y} \\ m_1^{(2)}L_{1z} + m_2^{(2)}L_{2z} + m_3^{(2)}3L_{Bz} \end{pmatrix}. \qquad (A.44) $$

Similar to $E_3$, the $y$-component of (A.43) requires that $m_2^{(1)} \ne 0$. To see this, we note that $m_2^{(1)} = 0$ means that $L_{1x} = 0$. Applying these two relations to the $x$- and $y$-components of (A.44) leads to contradictions. Explicitly, if $m_2^{(2)} = 0$ then $L_{2y} = 0$, whereas if $m_2^{(2)} \ne 0$ then $(L_{2x}/L_{2y})^2 = -1$; neither of these are valid solutions. Thus we require $m_2^{(1)} \ne 0$.

Returning to the $x$- and $y$-components of (A.43) we have

$$ L_{2x} = -\frac{2m_1^{(1)} + 1}{2m_2^{(1)}} L_{1x} \quad \text{and} \quad L_{2y} = -\frac{\sqrt{3}}{2m_2^{(1)}} L_{1x}, \qquad (A.45) $$

for $m_1^{(1)} \in \mathbb{Z}$ and $m_2^{(1)} \in \mathbb{Z} \ne 0$. For (A.40) to be satisfied with these relations, we must have

$$ (m_1^{(1)})^2 + m_1^{(1)} + \frac{1}{4} \le (m_2^{(1)})^2 \le (m_1^{(1)})^2 + m_1^{(1)} + 1. \qquad (A.46) $$

The only (integer) solutions to these inequalities are

$$ m_1^{(1)} \in \{-1, 0\} \quad \text{and} \quad m_2^{(1)} = \pm 1. \qquad (A.47) $$

These translate to

$$ L_{2x} = \pm \frac{1}{2} L_{1x} \quad \text{and} \quad L_{2y} = \pm \frac{\sqrt{3}}{2} L_{1x}, \qquad (A.48) $$
with independent sign choices. Plugging the expressions for $L_{2x}$ and $L_{2y}$ into the x- and y-components of (A.44) leads to

$$m_1^{(2)} \in \{-1, 0\} \quad \text{and} \quad m_2^{(2)} = \pm 1. \qquad (A.49)$$

The many choices for the sets of integers $m_j^{(i)}$ with $i, j \in \{1, 2\}$, when plugged into the z-components of (A.43) and (A.44), lead to the requirements that $L_{1z}$ and $L_{2z}$ must be integer multiples of $L_{Bz}$. We can always make these integer multiples 0, 1, or 2 by adding or subtracting integer multiples of $T_3$. But, as in the previous cases, choosing either 1 or 2 leads to fix points. Thus $L_{1z} = L_{2z} = 0$.

The independent choices of signs for $L_{2x}$ and $L_{2y}$ are equivalent as it is immaterial whether we take $T_{A_2}$ or $-T_{A_2}$, since it is a pure translation. Using this, we can always choose $L_{2y} > 0$. Finally, we can always switch between $L_{2x} = -\frac{1}{2}L_{1x}$ and $L_{2x} = +\frac{1}{2}L_{1x}$ by adding (subtracting) $T_{A_1}$ to (from) $T_{A_2}$. We choose $L_{2x} = -\frac{1}{2}L_{1x}$.

Putting all of this together, shifting to an arbitrary origin, and renaming $L_{1x}$ to $L_A$ we arrive at the general form from the text (3.16),

$$T_{A_1}^{E_4} = \begin{pmatrix} L_A \\ 0 \\ 0 \end{pmatrix}, \quad T_{A_2}^{E_4} = \begin{pmatrix} -L_A/2 \\ \sqrt{3}L_A/2 \\ 0 \end{pmatrix}, \quad T_B^{E_4} = \begin{pmatrix} L_{Bx} \\ L_{By} \\ L_{Bz} \end{pmatrix}. \qquad (A.50)$$

### A.5 $E_5$: sixth-turn space

The sixth turn space is very much like $E_4$, though with a richer matrix structure. We begin by using the rotational freedom to set the axis of $M_B$ parallel to $\hat{e}_z$ and then to choose $T_{A_1}$ to be in the $xz$-plane. For $E_5$

$$M_B = R_z(\pi/3) = \begin{pmatrix} 1/2 & -\sqrt{3}/2 & 0 \\ \sqrt{3}/2 & 1/2 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \qquad (A.51)$$

which has the properties

$$(M_B)^2 = R_z(2\pi/3), \qquad (M_B)^3 = R_z(\pi) = \text{diag}(-1, -1, 1), \\ (M_B)^4 = R_z(4\pi/3) = (M_B^T)^2, \qquad (M_B)^5 = R_z(5\pi/3) = M_B^T, \qquad (M_B)^6 = 1. \qquad (A.52)$$

As in the previous cases, we shift the origin so that $L_{Bx} = L_{By} = 0$, and we again start with

$$T_{A_1} = \begin{pmatrix} L_{1x} \\ 0 \\ L_{1z} \end{pmatrix}, \quad T_{A_2} = \begin{pmatrix} L_{2x} \\ L_{2y} \\ L_{2z} \end{pmatrix}, \quad T_B = \begin{pmatrix} 0 \\ 0 \\ L_{Bz} \end{pmatrix}. \qquad (A.53)$$

An associated $E_1$ can again be constructed from three pure translations, starting with the obvious translation vectors $T_1 = T_{A_1}$ and $T_2 = T_{A_2}$. A third translation vector can be defined based on the fact that $(M_B^{E_5})^6 = 1$. Noting that

$$g_3^{E_5} \equiv (g_B^{E_5})^6 : x \to x + M_{06}^{E_5} T_B^{E_5}, \qquad (A.54)$$
where $\mathbf{M}_{06}^{E_5} = \text{diag}(0, 0, 6)$ from (4.35), the third translation vector of the associated $E_1$ for $E_5$ is as quoted in the main text (3.26),

$$ \mathbf{T}_3^{E_5} = \begin{pmatrix} 0 \\ 0 \\ 6L_{Bz} \end{pmatrix}. \qquad (A.55) $$

For our subsequent development here we choose $\mathbf{T}_3 \equiv \mathbf{T}_3^{E_5}$.

To proceed, we can again choose $\mathbf{T}_{A_1}$ such that $|\mathbf{T}_{A_1 \perp}| \le |\mathbf{T}_{A_2 \perp}|$, where $\perp$ means the projection into the $xy$-plane. Also, by adding or subtracting integer multiples of $\mathbf{T}_{A_1}$ from $\mathbf{T}_{A_2}$, we can ensure that $|L_{2x}| < |L_{1x}|$; thus we generically have

$$ (L_{2x})^2 < (L_{1x})^2 \le (L_{2x})^2 + (L_{2y})^2. \qquad (A.56) $$

Whereas for $E_4$ we had two sets of conditions, (A.41), on the products of generators and their inverses being integer linear combination of $\mathbf{T}_i^{E_4}$, which we were able to limit to one, for $E_5$ we have 5 sets, which we can limit to 3

$$ \mathbf{M}_B \mathbf{T}_{A_i} = \sum_k m_k^{(i)} \mathbf{T}_k, \quad \text{for some } m_k^{(i)} \in \mathbb{Z}, \qquad (A.57) $$

$$ (\mathbf{M}_B)^2 \mathbf{T}_{A_i} = \sum_k n_k^{(i)} \mathbf{T}_k, \quad \text{for some } n_k^{(i)} \in \mathbb{Z}, \qquad (A.58) $$

$$ (\mathbf{M}_B)^3 \mathbf{T}_{A_i} = \sum_k p_k^{(i)} \mathbf{T}_k, \quad \text{for some } p_k^{(i)} \in \mathbb{Z}. \qquad (A.59) $$

This follows since the condition for $(\mathbf{M}_B)^4$ is equivalent to that for $(\mathbf{M}_B)^2$ and the condition for $(\mathbf{M}_B)^5$ is equivalent to that for $\mathbf{M}_B$.

This leads to a daunting 6 sets of conditions. However, if we focus on the $(\mathbf{M}_B)^2$ equations (A.58) we see they are identical to the conditions from $E_4$ (A.41), thus the solution is the same and we have an immediate candidate for the general solution. By direct computation, we can verify that this solution also satisfies (A.57) and (A.59), so there are no additional constraints.

Finally, shifting to an arbitrary origin, as in $E_4$ we have the general form from the text (3.22),

$$ \mathbf{T}_{A_1}^{E_5} = \begin{pmatrix} L_A \\ 0 \\ 0 \end{pmatrix}, \quad \mathbf{T}_{A_2}^{E_5} = \begin{pmatrix} -L_A/2 \\ \sqrt{3}L_A/2 \\ 0 \end{pmatrix}, \quad \mathbf{T}_B^{E_5} = \begin{pmatrix} L_{Bx} \\ L_{By} \\ L_{Bz} \end{pmatrix}. \qquad (A.60) $$

## A.6 $E_6$: Hantzsche-Wendt space

In contrast to the other orientable, compact spaces, the Hantzsche-Wendt space has rotations around multiple axes. These are conventionally chosen as half turns about three axes which must be orthogonal (see, e.g., ref. [19]). Here, we use the freedom to rotate the coordinate system such that these three axes are aligned with the coordinate axes choosing $\mathbf{M}_A = \mathbf{R}_{\hat{\pi}}(\pi)$, $\mathbf{M}_B = \mathbf{R}_{\hat{\eta}}(\pi)$, and $\mathbf{M}_C = \mathbf{R}_{\hat{z}}(\pi)$. For $E_6$ we then have

$$ \mathbf{M}_A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{pmatrix}, \quad \mathbf{M}_B = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix}, \quad \mathbf{M}_C = \begin{pmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \qquad (A.61) $$
so that

$$ (\mathbf{M}_A)^2 = (\mathbf{M}_B)^2 = (\mathbf{M}_C)^2 = 1. \qquad (A.62) $$

This has used up all the rotational freedom. There is still the freedom to choose a special origin so as to simplify the translation vectors (as we have done in the previous cases), however, we will begin without using this freedom and thus start with the most general vectors

$$ \mathbf{T}_a = \begin{pmatrix} L_{ax} \\ L_{ay} \\ L_{az} \end{pmatrix} \quad \text{for } a \in \{A, B, C\}. \qquad (A.63) $$

An associated $E_1$ can again be constructed from three pure translations. In contrast to $E_2-E_5$, the action of each generator in $E_6$ includes a rotation. However, since the rotations are all half turns, we can choose the three translations as was done for $g_3^{E_2}$. From the properties of the matrices $\mathbf{M}_a$, we construct the pure translations from $(g_a)^2: \mathbf{x} \to \mathbf{x} + (1 + \mathbf{M}_a)\mathbf{T}_a$. This leads to the pure translation vectors quoted in the main text (3.31),

$$ \mathbf{T}_1 \equiv \mathbf{T}_1^{E_6} = \begin{pmatrix} 2L_{Ax} \\ 0 \\ 0 \end{pmatrix}, \quad \mathbf{T}_2 \equiv \mathbf{T}_2^{E_6} = \begin{pmatrix} 0 \\ 2L_{By} \\ 0 \end{pmatrix}, \quad \mathbf{T}_3 \equiv \mathbf{T}_3^{E_6} = \begin{pmatrix} 0 \\ 0 \\ 2L_{Cz} \end{pmatrix}. \qquad (A.64) $$

To proceed, we note that $\mathbf{M}_a\mathbf{M}_b = \mathbf{M}_c$ for any $\{a, b, c\}$ permutation of $\{A, B, C\}$, which means that we must have $g_a g_b = g_c + \mathbf{t}$ for $\mathbf{t}$ a pure translation constructed from integer linear combinations of the $\mathbf{T}_i$. This leads to the set of conditions

$$ \mathbf{M}_A \mathbf{T}_B + \mathbf{T}_A = \mathbf{T}_C + \sum_i m_i^C \mathbf{T}_i, \qquad (A.65) $$

$$ \mathbf{M}_B \mathbf{T}_C + \mathbf{T}_B = \mathbf{T}_A + \sum_i m_i^A \mathbf{T}_i, \qquad (A.66) $$

$$ \mathbf{M}_C \mathbf{T}_A + \mathbf{T}_C = \mathbf{T}_B + \sum_i m_i^B \mathbf{T}_i, \qquad (A.67) $$

for some $m_i^a \in \mathbb{Z}$. Written out explicitly, these conditions are

$$ \begin{pmatrix} L_{Bx} + L_{Ax} \\ -L_{By} + L_{Ay} \\ -L_{Bz} + L_{Az} \end{pmatrix} = \begin{pmatrix} 2m_1^C L_{Ax} + L_{Cx} \\ 2m_2^C L_{By} + L_{Cy} \\ 2m_3^C L_{Cz} + L_{Cz} \end{pmatrix}, \qquad (A.68) $$

$$ \begin{pmatrix} -L_{Cx} + L_{Bx} \\ L_{Cy} + L_{By} \\ -L_{Cz} + L_{Bz} \end{pmatrix} = \begin{pmatrix} 2m_1^A L_{Ax} + L_{Ax} \\ 2m_2^A L_{By} + L_{Ay} \\ 2m_3^A L_{Cz} + L_{Az} \end{pmatrix}, \qquad (A.69) $$

$$ \begin{pmatrix} -L_{Ax} + L_{Cx} \\ -L_{Ay} + L_{Cy} \\ L_{Az} + L_{Cz} \end{pmatrix} = \begin{pmatrix} 2m_1^B L_{Ax} + L_{Bx} \\ 2m_2^B L_{By} + L_{By} \\ 2m_3^B L_{Cz} + L_{Bz} \end{pmatrix}. \qquad (A.70) $$
From these, we first focus on the x-component of (A.69), the y-component of (A.70), and the z-component of (A.68). This gives

$$ 
\begin{aligned}
L_{Cx} - L_{Bx} &= -(2m_1^A + 1)L_{Ax}, \\
L_{Ay} - L_{Cy} &= -(2m_2^B + 1)L_{By}, \\
L_{Bz} - L_{Az} &= -(2m_3^C + 1)L_{Cz},
\end{aligned} 
\qquad (A.71) 
$$

reducing the original set to 6 parameters. Shifting by integer multiples of the translation vectors $T_i$ allows us to remove all the even multiples on the right-hand side of these expressions. In particular, the choice $m_1^A = m_2^B = m_3^C = -1$ reproduces the general form from the text (3.28),

$$ T_A^{E_6} = \begin{pmatrix} L_{Ax} \\ L_{By} + L_{Cy} \\ L_{Az} \end{pmatrix}, \quad T_B^{E_6} = \begin{pmatrix} L_{Bx} \\ L_{By} \\ L_{Cz} + L_{Az} \end{pmatrix}, \quad T_C^{E_6} = \begin{pmatrix} L_{Ax} + L_{Bx} \\ L_{Cy} \\ L_{Cz} \end{pmatrix}. \qquad (A.72) $$

The behavior of the generators under a shift of origin suggests an alternative form for the translation vectors. A shift in the origin preserves the difference between the lengths on the left-hand side of (A.71) but not their sums. (In fact, this is another way to understand why the three lengths on the right-hand sides appear in the translation vectors of the associated $E_1$ (3.31).) The sums of the lengths can be used to parametrize the generators in another form. Notice that the sum of the x-components of (A.68) and (A.70) gives

$$ L_{Bx} + L_{Cx} = 2(m_1^C + m_1^B)L_{Ax} + L_{Bx} + L_{Cx}, \qquad (A.73) $$

in other words that

$$ 2(m_1^C + m_1^B)L_{Ax} = 0. \qquad (A.74) $$

But this is precisely the freedom used above to shift by integer multiples of $T_1$. In other words, the sum of lengths $L_{Bx} + L_{Cx}$ is unconstrained by the set of conditions. Similar arguments hold for the sums $L_{Cy} + L_{Ay}$ and $L_{Az} + L_{Bz}$. This motivates the definition of new continuous, dimensionless parameters given by the average of these length pairs as

$$ 
\begin{aligned}
r_x &\equiv \frac{L_{Bx} + L_{Cx}}{2L_{Ax}} = \frac{2L_{Bx} + L_{Ax}}{2L_{Ax}}, \\
r_y &\equiv \frac{L_{Cy} + L_{Ay}}{2L_{By}} = \frac{2L_{Cy} + L_{By}}{2L_{By}}, \\
r_z &\equiv \frac{L_{Az} + L_{Bz}}{2L_{Cz}} = \frac{2L_{Az} + L_{Cz}}{2L_{Cz}},
\end{aligned} 
\qquad (A.75) 
$$

where the second equality in each line follows from the same choices that went into the general form (3.28). Solving for one of the lengths in each equation, we can write these as

$$ L_{Bx} = \left(r_x - \frac{1}{2}\right)L_{Ax}, \quad L_{Cy} = \left(r_y - \frac{1}{2}\right)L_{By}, \quad L_{Az} = \left(r_z - \frac{1}{2}\right)L_{Cz}. \qquad (A.76) $$
Finally, these give the alternative form quoted in the main text (3.30),

$$ T_A^{E_6} = \begin{pmatrix} L_{Ax} \\ (r_y + \frac{1}{2}) L_{By} \\ (r_z - \frac{1}{2}) L_{Cz} \end{pmatrix}, \quad T_B^{E_6} = \begin{pmatrix} (r_x - \frac{1}{2}) L_{Ax} \\ L_{By} \\ (r_z + \frac{1}{2}) L_{Cz} \end{pmatrix}, \quad T_C^{E_6} = \begin{pmatrix} (r_x + \frac{1}{2}) L_{Ax} \\ (r_y - \frac{1}{2}) L_{By} \\ L_{Cz} \end{pmatrix}. \quad (A.77) $$

### A.7 $E_{11}$: Chimney space

The general form of the generators for the chimney space can be determined in a manner similar to $E_1$, though it requires two generators, not three. Since $M^{E_{11}} = 1$, $E_{11}$ is homogeneous and we can use the rotational freedom to define one of the translation vectors as the x-axis of the coordinate system and the remaining freedom to define the xy-plane. In other words, two vectors define a plane so the two translation vectors can be used to define the xy-plane. Thus the two translations $T_i^{E_1}$ can be parametrized most generally as quoted in the text (3.36),

$$ T_1^{E_{11}} = \begin{pmatrix} L_{1x} \\ 0 \\ 0 \end{pmatrix}, \quad T_2^{E_{11}} = \begin{pmatrix} L_{2x} \\ L_{2y} \\ 0 \end{pmatrix}, \quad (A.78) $$

or equivalently

$$ T_1^{E_{11}} = L_1 \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \quad T_2^{E_{11}} = L_2 \begin{pmatrix} \cos \alpha \\ \sin \alpha \\ 0 \end{pmatrix}. \quad (A.79) $$

### A.8 $E_{12}$: Chimney space with half turn

The chimney space with half turn is a root of $E_{11}$ and is a limit of the half-turn space $E_2$ with one non-compact dimension. It has one generator, $g_A$, that is a translation and one generator, $g_B$, that is a corkscrew motion. The rotational freedom to align the coordinate system allows the axis of rotation to be chosen as the z-axis. Freedom to rotate around the z-axis allows the choice $T_{Ay} = 0$. Finally, we can choose a special origin that allows us to set $L_{Bx} = L_{By} = 0$; we will restore an arbitrary origin at the end of the calculation. With all of this, we can choose our general starting case to be

$$ T_A = \begin{pmatrix} L_{Ax} \\ 0 \\ L_{Az} \end{pmatrix}, \quad T_B = \begin{pmatrix} 0 \\ 0 \\ L_{Bz} \end{pmatrix}. \quad (A.80) $$

From these a set of pure translations, an associated $E_{11}$, can be constructed. The action of $g_A$ is already a pure translation, so one of the needed translation vectors can be chosen as $T_1 = T_A$. A second translation vector can be defined based on the fact that $(M_B^{E_{12}})^2 = 1$. Note that

$$ g_2^{E_{12}} = (g_B^{E_{12}})^2 : x \rightarrow x + (1 + M_B^{E_{12}}) T_B^{E_{12}} = x + M_{02}^{E_{12}} T_B^{E_{12}}, \quad (A.81) $$

for $M_{02}^{E_{12}} = \text{diag}(0, 0, 2)$ from (4.35). Thus the second translation vector of the associated $E_{11}$ for $E_{12}$ is as quoted in the main text (3.42),

$$ T_2^{E_{12}} = \begin{pmatrix} 0 \\ 0 \\ 2L_{Bz} \end{pmatrix}. \quad (A.82) $$

For our subsequent development here we choose $T_2 = T_2^{E_{12}}$.
The remaining condition to impose is that

$$g_B^{-1}g_A^{-1}g_Bg_A : \boldsymbol{x} \to \boldsymbol{x} + \boldsymbol{T}_A - \boldsymbol{M}_B\boldsymbol{T}_A = \boldsymbol{x} + \sum_k m_k \boldsymbol{T}_k, \qquad (A.83)$$

for $m_k \in \mathbb{Z}$. Since $\boldsymbol{T}_A$ is already the pure translation $\boldsymbol{T}_1$, this is equivalent to requiring

$$\mathbf{M}_B \boldsymbol{T}_A = \sum_k m_k \boldsymbol{T}_k, \quad \text{for some } m_k \in \mathbb{Z}, \qquad (A.84)$$

leading to the set of conditions

$$\begin{pmatrix} -L_{Ax} \\ 0 \\ L_{Az} \end{pmatrix} = \begin{pmatrix} m_1 L_{Ax} \\ 0 \\ m_1 L_{Az} + 2m_2 L_{Bz} \end{pmatrix}. \qquad (A.85)$$

From the $x$-component of these conditions, we see that $m_1 = -1$. Applying this to the $z$-component we immediately see that $L_{Az} = 0$ (and $m_2 = 0$) is the general solution. Shifting back to an arbitrary origin we have the generic result quoted in the main text (3.39),

$$\boldsymbol{T}_A^{E_{12}} = \begin{pmatrix} L_{Ax} \\ 0 \\ 0 \end{pmatrix}, \quad \boldsymbol{T}_B^{E_{12}} = \begin{pmatrix} L_{Bx} \\ L_{By} \\ L_{Bz} \end{pmatrix}. \qquad (A.86)$$

As stated, we notice that $E_{12}$ is a limit of $E_2$ with $|\boldsymbol{T}_{A_2}^{E_2}| \to \infty$. As shown above, we were able to use the same arguments as for $E_2$ to argue that $L_{Ay} = 0$, or equivalently that $\alpha = 0$.

## A.9 $E_{16}$: slab space including rotations

The slab space is distinct from the other topologies as it allows a corkscrew motion. The standard definition of the slab space does not contain this rotation since the rotated slab is topologically the same as the unrotated slab. However, a rotation does lead to physically observable effects, and thus unrotated and rotated slab manifolds must be treated separately. For this reason, we split the description of $E_{16}$ into two cases labeled $E_{16}^{(h)}$ and $E_{16}^{(i)}$.

### A.9.1 $E_{16}^{(h)}$: conventional unrotated slab space

The general form of the generators for the unrotated slab space follows directly from the fact that it contains only a single generator that is a pure translation. Since $\mathbf{M}_A^{E_{16}^{(h)}} = 1$, $E_{16}^{(h)}$ is homogeneous and we can use the rotational freedom to define the translation vector as the $z$-axis of the coordinate system. Thus we immediately arrive at the general form quoted in the text (3.44),

$$\boldsymbol{T}_A^{E_{16}^{(h)}} = L \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}. \qquad (A.87)$$
### A.9.2 $E_{16}^{(i)}$: general rotated slab space

An arbitrary rotation by an angle $\zeta$ is allowed in the slab space. We can use part of the rotational freedom to choose the $z$-axis of the coordinate system as the rotation axis. We can then use the freedom of rotation about this axis to set the $y$-component of the translation vector to zero. Thus we can start with

$$ M = R_{\hat{z}}(\zeta), \quad \text{with} \quad T = \begin{pmatrix} L_x \\ 0 \\ L_z \end{pmatrix}. \qquad (A.88) $$

In principle $0 < \zeta < 2\pi$ (where the endpoints are excluded so that $E_{16}^{(i)}$ does not include $E_{16}^{(h)}$) and this would be the general action of a group element. However, in this work, we focus on the observable consequences of our universe having a non-trivial topology. To this end, in section 4 the eigenmodes of each topology are determined. In order to use a Fourier basis to expand all functions on the manifold, a required condition is (effectively) that $M^q = 1$ for some finite $q \in \mathbb{Z}^{>0}$. This in turn requires that $q\zeta = 2\pi p$ for $p \in \mathbb{Z}^{\neq 0}$. In other words, $\zeta$ must be a rational multiple of $2\pi$. If $\zeta$ is not a rational multiple of $2\pi$ then only cylindrically symmetric functions can be expanded in a Fourier basis. The translation vector given above is general, it is the rotation matrix that has further constraints. With this, we arrive at the general form given in the text (3.45),

$$ \begin{gathered} \mathbf{M}_B^{E_{16}^{(i)}} = \mathbf{R}_{\hat{z}}(2\pi p/q) = \begin{pmatrix} \cos(2\pi p/q) & -\sin(2\pi p/q) & 0 \\ \sin(2\pi p/q) & \cos(2\pi p/q) & 0 \\ 0 & 0 & 1 \end{pmatrix}, \quad \text{with} \\ \mathbf{T}_B^{E_{16}^{(i)}} = \begin{pmatrix} L_x \\ 0 \\ L_z \end{pmatrix} = L \begin{pmatrix} \cos \beta \\ 0 \\ \sin \beta \end{pmatrix}, \qquad (A.89) \end{gathered} $$

and $p \in \mathbb{Z}^{\neq 0}$, $q \in \mathbb{Z}^{>0}$, and $|p|$ and $q$ relatively prime. From this, a pure translation, an associated $E_{16}^{(h)}$, can be constructed. Based on the fact that $(\mathbf{M}_B^{E_{16}^{(i)}})^q = 1$ we note that

$$ (g_B^{E_{16}^{(i)}})^q : x \to x + \mathbf{M}_{0q}^{E_{16}^{(i)}} \mathbf{T}_B^{E_{16}^{(i)}}, \qquad (A.90) $$

for $\mathbf{M}_{0q}^{E_{16}^{(i)}} = \text{diag}(0, 0, q)$ from (4.35). Thus the translation vector of the associated $E_{16}^{(h)}$ for $E_{16}^{(i)}$ is as quoted in the main text (3.46),

$$ T_1^{E_{16}^{(i)}} = \begin{pmatrix} 0 \\ 0 \\ qL_z \end{pmatrix}. \qquad (A.91) $$

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