Decomposition of Sohncke space groups into products of Bieberbach and symmorphic parts

A.1 24, I2₁2₁2₁

As another example, consider Γ=I2₁2₁2₁. To determine T, we use the Bilbao function HERMANN which gives

α = (−1/2−1/200−1/2−1/211/4−1/21/200)

(where the perfunctory bottom row in the affine transformation as a 4×4 matrix has been removed to save space.)

The coset representatives in the decomposition of Γ with respect to T (in the basis of T) are given below:

ℱΓ/T = {(x, y, z); (−y, −x, −z);(−x + 1/2, −y + 1/2, −x − y + z + 1/2);(y + 1/2, x + 1/2, x + y − z + 1/2)}.

Here S={(x, y, z); (−y, −x, −z)} is identical with ℱC2/TC, which is consistent with the fact that MAXSUB gives [I2₁2₁2₁:C2]=2. If ℱ_Γ/T can be semi-decomposed, then it should be possible to find a Bieberbach subgroup with [Γ_B :P1]=2. But the only Bieberbach group that could satisfy this is P2₁, and MAXSUB does not list P2₁ as a maximal subgroup. Therefore, there is no need to investigate the case when [I2₁2₁2₁:T_I ]=4 further, and so we investigate [I2₁2₁2₁:Σ]=8, which is the next smallest candidate.

When Σ=P1<T the Bilbao COSET function (left and right are the same so Σ◁Γ) gives

ℱI212121/P1 = {(x, y, z); (−x + 1/2, −y, z + 1/2);(−x, y + 1/2, −z + 1/2); (x + 1/2, −y + 1/2, −z);(x + 1/2, y + 1/2, z + 1/2); (−x, −y + 1/2, z);(−x + 1/2, y, −z); (x, −y, −z + 1/2)}

Here (x+1/2, y+1/2, z+1/2) is a pure translation. Such will only occur when Σ<T. In contrast, (–x+1/2, –y, z+1/2); (–x, y+1/2, –z+1/2); (x+1/2, –y+1/2, –z) are all screw motions with d=1/2, and all of the rotation axes are mutually orthogonal with an angle of rotation of π. So this is a clue that it may be related to P2₁2₁2₁/P1. We verify this by using the Bilbao COSET function to give

ℱP212121/P1 = {(x, y, z); (−x + 1/2, −y, z + 1/2);(−x, y + 1/2, −z + 1/2); (x + 1/2, −y + 1/2, −z)}

This matches with four entries in ℱI212121/P1. In more complicated cases we might need to search for an affine transformation to relate a subgroup of ℱ_Γ/Σ with a candidate Bieberbach group quotient by an appropriate translation group.

In contrast, looking at the elements (–x, –y+1/2, z); (–x+1/2, y, –z); (x, –y, –z+1/2), we observe that they are all rotations because d=0. But if we seek 𝔹 and 𝕊 such that their product reproduces ℱI212121/P1, then both cannot have order 4. And because 𝔹=2₁2₁2₁/P1 is of order 2 in ℱI212121/P1 when evaluated modulo P1, it is normal, and so we do not need to check that the conjugation action of 𝕊_i on ℱI212121/P1.

We see that we can define a two-element point group (which fixes a point other than the origin) from the identity and any one of (–x, –y+1/2, z); (–x+1/2, y, –z); (x, –y, –z+1/2). For example,

S1: = {(x, y, z); (−x, −y + 1/2, z)},S2: = {(x, y, z); (−x + 1/2, y, −z)},S3: = {(x, y, z); (x, −y, −z + 1/2)}.

Abstractly, every two-element group is isomorphic, and so candidates for these are P2/P1 or C2/T_C . But concretely, we can only find affine transformations such that αiℱP2/P1αi−1 = Si. Again examining left and right COSETS using Bilbao (or doing finite calculations in the factor group) we find that

I212121/P1 = (P212121/P1) ⋊ (P2/P1).

There is no guarantee that this decomposition is unique.

Using the bottom-up approach employing the Bilbao MAXSUB function, we see that [I2₁2₁2₁:P2₁2₁2₁]=2. and as [P2₁2₁2₁:P1]=4 and [P2:P1]=2 we conclude that it is not possible to find a decomposition with smaller values than this, and so our search terminates.

A.2 90, P42₁2

Consider group no. 90, and use Γ=P42₁2 and T=P1. When using the identity transformation, the Bilbao COSET function gives

ℱP4212/P1 = {(x, y, z); (−x, −y, z);(−y + 1/2, x + 1/2, z); (y + 1/2, −x + 1/2, z);(−x + 1/2, y + 1/2, −z); (x + 1/2, −y + 1/2, −z);(y, x, −z); (−y, −x, −z)}

This is convenient because we can immediately identify

S1: = {(x, y, z); (−x, −y, z); (y, x, −z); (−y, −x, −z)}

with a point group of order 4 (and index 2 in Γ). Therefore it is normal. Searching for every symmorphic space group with point group of order 4, we find that this corresponds to C222/T_C . The complementing subgroup can be chosen as

B: = {(x, y, z); (−x + 1/2, y + 1/2, −z)}.

(In fact, there are four possible choices, but they all act in the same way as described below.) It is easy to check that this is a group.

For the nontrivial transformation in this group, R is a rotation around n=e₂ by π. And v(R)=[1/2, 1/2, 0]^t. And so v·n=1/2. We conclude that 𝔹 is isomorphic with P2₁/P1. To find affine transformations that will make the relationship between 𝔹 and P2₁/P1, we solve the equation αg=g′α where g and g′ are respectively representatives taken from 𝔹 and P2₁/P1.

Clearly ℱP4212/P1 = S1B. Next we ask if 𝔹 is normal. To do this, we use the Bilbao COSET function with the information about α that has been computed. We find that left and right cosets do not match modulo P1, and 𝔹 is not normal, and

ℱP4212/P1 = S1 ⋊ B.

The benefit of this top-down approach is that we need not worry about whether the two components correspond to space subgroups with a common lattice, since that is guaranteed a priori. But this does not provide a complete list of all possible decompositions. We therefore go to the Bilbao MAXSUB function, and look for other low-index symmorphic and Bieberbach subgroups. We find that [P42₁2:P4]=2. Hence P4◁P42₁2. The same subroutine provides a transformation, and we find (after moding out extraneous 1’s returned by Bilbao) that

ℱ(αP4212α−1)/P1 = {(x, y, z); (−x, −y, z); (−y, x, z); (y, −x, z);(−x + 1/2, y + 1/2, −z); (x + 1/2, −y − 1/2, −z);(y + 1/2, x − 1/2, −z); (−y − 1/2, −x − 1/2, −z)}.

Here we see that the first four elements correspond to 𝕊₂:=P4/P1, and with (–x+1/2, y+1/2, –z) clearly visible, we can define the complement 𝔹 as before. And it is not normal here either. Therefore, we conclude that

ℱ(αP4212α−1)/P1 = S2 ⋊ B.

Though the quotient is “reverse-semi-decomposable” the Bieberbach subgroup is not normal (and no other Bieberbach subgroup with [Γ:Γ_B ]<[Γ:T] is normal either.

There is a copy of P2₁2₁2₁ that has index 4, which we can identify using the HERMANN subroutine in the Bilbao server, and gradually increasing the index of candidate Bieberbach subgroups. But left and right cosets do not match, and hence it is not normal.

Another route to index-4 P2₁2₁2₁ is by observing that none of the coset representatives in ℱ(αP4212α−1)/P1 given above have translations in the z component. Then, using the affine transformation α with linear part diag[1, 1, 2] and zero translation in the COSET function in Bilbao can double the size of the quotient group, giving enough room to fit both a copy of a conjugated ℱP212121/P1 and the same 𝕊 as before, which is left unaffected by this affine conjugation. Moreover, 𝕊 is normal in the resulting 16-element quotient group as well. And so we can write

Γ/Σ=(C222/T_C ) ⋊ (P2₁2₁2₁/P1). But P2₁2₁2₁ is not normal in P42₁2, and because this quotient group has 16 elements as opposed to the 8 of the original, we do not include this in the table.

There is an affine-conjugated copy of P2₁ with α=diag[1, 1, 2] that is normal in P42₁2 with index 8, but this index does not match the order of the largest point subgroup, and so the compatibility conditions fail, and so it cannot be used in a semi-decomposition.

We therefore conclude that 90 cannot be efficiently semi-decomposed, but the quotient group can be “reverse-semi-decomposed” in multiple ways, each with the pure rotation part being normal rather than the Bieberbach part. As a consequence of this, and of Theorem 3, it is possible to write

P4212 = P21P4 = P21C222

where it is understood that in the above different copies of P2₁ are used. Moroever, from Lemma 2.1, this is a case where even though Γ_B is not normal, it is nevertheless possible to perform decompositions of the form Γ=Γ_BS.

A.3 93,P4₂22

Using the Bilbao COSET function, and taking the top-down approach, [P4₂22:P1]=8 and

ℱP4222/P1 = {(x, y, z); (−x, −y, z);(−y, x, z + 1/2); (y, −x, z + 1/2);(−x, y, −z); (x, −y, −z);(y, x, −z + 1/2); (−y, −x, −z + 1/2)}.

Of these,

S = {(x, y, z); (−x, −y, z);(−x, y, −z); (x, −y, −z)} = ℱP222/P1

is a subgroup with [P4₂22:P222]=2 and hence it is normal. But there is no way to construct a two-element subgroup from the remaining elements, which appear to be P4₁ or P4₃ transformations. And so we look for ways to conjugate so that all elements of these subgroups are present as representatives in the coset decomposition.

Using HERMANN routine in Bilbao with G=93 and H=76 with index 4, we find that with

α = (100001000021/2)ℱ(αP4222α−1)/P1 = {(x, y, z); (−x, −y, z);(−y, x, z + 1/4); (y, −x, z + 1/4);(−x, y, −z − 1/2); (x, −y, −z − 1/2);(y, x, −z − 1/4); (−y, −x, −z − 1/4);(x, y, z + 1/2); (−x, −y, z + 1/2);(−y, x, z + 3/4); (y, −x, z + 3/4);(−x, y, −z); (x, −y, −z);(y, x, −z − 3/4); (−y, −x, −z − 3/4)}.

Several subgroups can be visually identified:

S = {(x, y, z); (−x, −y, z);(−x, y, −z); (x, −y, −z)} = ℱP222/P1,B1 = {(x, y, z); (y, −x, z + 1/4);(−x, −y, z + 1/2); (−y, x, z + 3/4)} = ℱP41/P1,B2 = {(x, y, z); (−y, x, z + 1/4);(−x, −y, z + 1/2); (y, −x, z + 3/4)} = ℱP41/P1.

We find that 𝔹₁𝕊=𝔹₂𝕊=F_P222/_P1 mod P1. And by conjugating each element of 𝔹₁ by all elements of 𝕊 and evaluating mod P1, we see that 𝔹₁ is closed under conjugation. The same is true for 𝔹₂. Hence, we find that 𝔹₁ and 𝔹₂ are normal in ℱP4222/P1, and so

P4222 = P41 ⋊ (P222/P1)

and

P4222 = P43 ⋊ (P222/P1)

A.4 94,P4₂2₁2

[P4₂2₁2, P1]=8. When computing ℱP42212/P1 using the Bilbao COSET function, we find four coset reps that are pure rotations.

But also present are elements such as (–x+1/2, y+1/2, –z+1/2) and (x+1/2, –y+1/2, –z+1/2), each of which are conjugated versions of elements of P2₁. But, computing all right coset reps for P4₂2₁2/P2₁ in HERMANN, and computing the corresponding left cosets in COSET, we see that the P2₁ elements corresponding to this are not normal.

In contrast, (–y+1/2, x+1/2, z+1/2) and (y+1/2, –x+1/2, z+1/2) which are in P4₁, and neither of which forms a 2-element group mod P1 with the identity. But when using

α = (100001000020)

in Bilbao, [(P4₂2₁2)^α,P1]=16.

P4₁◁(P4₂2₁2)^α; [(P4₂2₁2)^α, P4₁]=4 and [(P4₂2₁2)^α, C222]=4.

In particular,

ℱ(P42212)α/P1 = {(x, y, z); (−x, −y, z); (y, x, −z); (−y, −x, −z);(−y + 1/2, x + 1/2, z + 1/4); (y + 1/2, −x + 1/2, z + 1/4);(−x + 1/2, y + 1/2, −z + 1/4);(x + 1/2, −y + 1/2, −z + 1/4);(x, y, z + 1/2); (−x, −y, z + 1/2);(−y + 1/2, x + 1/2, z + 3/4); (y + 1/2, −x + 1/2, z + 3/4);(−x + 1/2, y + 1/2, −z − 1/4); (x + 1/2, −y + 1/2, −z − 1/4);(y, x, −z − 1/2); (−y, −x, −z − 1/2)}

And

S = {(x, y, z); (−x, −y, z); (y, x, −z); (−y, −x, −z)}

looks like ℱC222/TC

Of the remainder of elements, we can construct

B: = {(x, y, z); (−y + 1/2, x + 1/2, z + 1/4);(−x, −y, z + 1/2); (y + 1/2, −x + 1/2, z + 3/4)}.

𝔹 is normal in ℱ(P42212)α/P1, and so

P42212 = P41 ⋊ (C222/TC)

And the same is true for P4₃:

P42212 = P43 ⋊ (C222/TC)

We note also that there are normal affine-conjugated copies of P222 and P2₁2₁2₁ inside of P4₂2₁2 with [(P4₂2₁2)^α:P222]=[(P4₂2₁2)^β:P2₁2₁2₁]=4. The difficulty with P222 is that it has a different lattice than the Bieberbach subgroups with compatible orders, and so it cannot be used to semi-decompose. P2₁2₁2₁ does have a complement that is C222/T_C (about a point that is not the origin when expressed in the basis of P2₁2₁2₁ in the standard setting) giving

P42212 = P212121 ⋊ (C222/TC).

A.5 98,I4₁22

P4₁ with index 4 is normal. [I4₁22:P1]=16 in standard setting. Subgroup of pure rotations is order 4. So should be decomposable.

ℱI4122/P1 = {(x, y, z); (−x + 1/2, −y + 1/2, z + 1/2);(−y, x + 1/2, z + 1/4); (y + 1/2, −x, z + 3/4);(−x + 1/2, y, −z + 3/4); (x, −y + 1/2, −z + 1/4);(y + 1/2, x + 1/2, −z + 1/2); (−y, −x, −z);(x + 1/2, y + 1/2, z + 1/2); (−x, −y, z);(−y + 1/2, x, z + 3/4); (y, −x + 1/2, z + 1/4);(−x, y + 1/2, −z + 1/4); (x + 1/2, −y, −z + 3/4);(y, x, −z); (−y + 1/2, −x + 1/2, −z + 1/2)}

Of these,

S = {(x, y, z); (−y, −x, −z); (−x, −y, z); (y, x, −z);

This is a conjugated version of ℱC222/TC.

B = {(x, y, z); (y, −x + 1/2, z + 1/4);(−x + 1/2, −y + 1/2, z + 1/2); (−y + 1/2, x, z + 3/4)}

This is of the form αℱP41/P1α−1 where α∈SE(3), and 𝔹 is normal in ℱI4122/P1, and so

I4122 = P41 ⋊ (C222/TC).

Similarly, by choosing different coset representatives to define 𝔹 we find

I4122 = P43 ⋊ (C222/TC).

It is also possible to identify elements of P2₁2₁2₁ in ℱI4122/P1 and to write

I4122 = P212121 ⋊ (C222/TC).

A.6 171P6₂– A case that can be alternatively decomposed with larger [Γ:Σ]

It is obvious from examining ℱP62P1 that it is possible to write P6₂=P3₂⋊ (P2/P1), and in this case [Γ:Γ_B ]=[P6₂:P3₂]=2 and [Γ_B :T]=[P3₂:P₁]=3.

Here we show that it is possible to find other decompositions where [Γ:Γ_B′] is the same but [Γ_B′:Σ] is larger. In particular, [P6₂:P6₁]=2 and [P6₁:P1]=6 with

α = (100001000020).

The coset representatives from the two cosets are

{(x, y, z); (−y, x−y, z + 1/3);(−x + y, −x, z + 2/3); (−x, −y, z + 1/2);(y, −x + y, z + 5/6); (x − y, x, z + 1/6)}

and

{(−x + y, −x, z + 1/6); (x, y, z + 1/2);(−y, x − y, z + 5/6); (x − y, x, z + 2/3);(−x, −y, z); (y, −x + y, z + 1/3)}.

From these we can construct

ℱP62/P61 = {(x, y, z); (−x, −y, z)},

which is a pure rotation group.

This is a case where we could decompose Γ/Σ (which has 12 elements)

Γ/Σ = B ⋊ S

where 𝔹=P6₁/Σ and

S: = {(x, y, z); (−x, −y, z)} = P2/P1.

Similar less-efficient decompositions are possible for P222₁, C222₁, P6₃, P6₄, P6₂22, P6₄22, P6₃22. These are listed in the table after the | symbol. In most cases they have the same type of complementing S as the Γ_B in the more efficient decomposition of the same group, and the answer to the question of whether 𝕊 is normal in Σ\Γ is the same as well. When the answers differ, the answer for the less efficient decomposition is written to the right of the | symbol.

A.7 208 and 210

These two groups each have large symmorphic subgroups of index 2, indicating normality of Γ_S . Since in a semi-decomposition all of the torsion must reside in 𝕊, if a compatible Bieberbach subgroup exists it must be the case that [Γ:Γ_B ]=|𝕊|. Each of the groups 208 and 210 have a symmorphic subgroup with point group of order 12 (195 in 208, and 196 in 210). Moreover, in each of these two groups are Bieberbach groups of index 12 (76 and 78 in 208, and 76, 78 and 19 in 210). Checking these (e.g. by conjugating generators of each candidate Γ_B by all generators of Γ) none of these Bieberbach subgroups are normal. In some cases there is a faster way to rule out Γ_B ’s that are not normal, based on the discussion at the end of Section 6, which leads to the following theorem.

Theorem 4: If Σ=Γ_B ∩T◁Γ_B <Γ and Σ is not normal in Γ, then Γ_B is not normal in Γ.

Proof: Every element of Γ_B can be written as γ_B =σb for some σ∈Σ and b ∈ ℱΓBΣ. By construction, ℱΓBΣ ∩ T = {e} where T is the minimal-index translation subgroup of Γ. Conjugating by an arbitrary γ∈Γ gives

γγBγ−1 = γσbγ−1 = (γσγ−1)(γbγ−1).

Conjugation does not change the nature of a rigid-body displacement, i.e. pure translations remain pure translations, pure rotations remain pure rotations, and screw displacements remain screw displacements with the same θ and d. Therefore, γσγ^–1 must be a translation, but since Σ is not normal in Γ it must be that γσγ^–1∉Σ for at least one γ∈Γ. Therefore, γγ_Bγ^–1 cannot be decomposed as σ′b′. But since every element of Γ_B can be decomposed in this way, it must be that γγ_Bγ^–1∉Γ_B and therefore Γ_B cannot be normal in Γ. □

Note that whereas the proof of this theorem is specific to space groups, it can be viewed as an example of the contrapositive of the statement

Σ, ΓB ⊲ Γ ⇒ Σ ∩ ΓB ⊲ Γ

which is true in more general (abstract) settings. That is, the intersection of normal subgroups is also normal.

As a consequence of this theorem, if Σ=Γ_B ∩T is the primitive translation group for Γ_B , and Γ_B <Γ, then if Σ is not normal in Γ we need not even check if Γ_B is normal, because it is guaranteed not to be. Therefore, in such cases it cannot be the case that Γ=Γ_B ⋊𝕊.

Moreover, even if there were normal Bieberbach subgroups further down the subgroup tree, these would be irrelevant for such decompositions because there would be no way to match [Γ:Γ_B ] and |𝕊|. In searching for Γ_B and S such that Γ=Γ_BS, the indices and orders of coset spaces must match even if Γ_B is not normal in Γ because Γ_B ∩S=e. The above mentioned groups have compatible indices and orders for this, and these are explored in the subsections that follow.

A.8 213,P4₁32 and 212,P4₃32

Taking the top-down approach, we use COSET to compute representatives of the cosets in P4₁32/P1. The result is [P4₁32:P1]=24. Of these, we identify four elements to construct ℱP212121/P1, and we find that this is normal in ℱP4132/P1.

We also find three representatives (x, y, z); (z, x, y); (y, z, x) that are clearly rotations that preserve the origin. However, these rotations all have the axis of rotation nt = [1, 1, 1]/3, and so any point on the line passing through the origin with this direction will be preserved by them. Since the goal is to clearly separate point and Bieberbach transformations, we seek the translation to the point along this line that maximizes the number of entries that have v=0. In particular, we find that when using

α = (1003/80103/80013/8)