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## Twisted Symmetry Groups

We continue to suppose that is a symmetry group with .We have where H is given in Table 1 . The canonical projection induces a projection .

We say that a symmetry group is twisted if it is not conjugate to an untwisted symmetry group. Equivalently, there exists an such that .

The next lemma states that, without loss of generality, we can always supposethat the element is not responsible for twisting.

Lemma 2.5   Suppose that is a symmetry group and that .Then there is a subgroup of that is conjugate to and contains .The conjugacy leaves unchanged.

Proof: Recall that acts as -I on .By assumption for some .We conjugate by the element .Compute that

as required.

Proposition 2.6   Let be a twisted symmetry group. Then either , or .In addition, is one of the three subgroups , and .

Remark: The possibility will be eliminated in the proof of Theorem  2.7.

Proof: It follows from Lemma  2.5 that we can eliminate the subgroups for which ,that is we can eliminate and .

Next, suppose that contains .As observed in theprevious subsection, contains .If is larger than ,then .It follows from Lemma 2.5 that and hence .In either case, is untwisted. The possibility that contains can be eliminated similarly. This completes the proof that is one of the groups , or .

Recall that is a subgroup of .If ,then .If ,then is conjugate to by Lemma 2.5. Hence, for to be twisted, must be one of the three remaining subgroups of .

Theorem 2.7   Up to conjugacy and scaling, there are seven twisted symmetry groups in .These are as listed in Table 3 .

Proof: By Proposition  2.6, we can assume that , or and that is one of the subgroups , or .We consider the three possibilities for K in turn.

Suppose that .Then for some .Conjugating by ,we can set .Note that

When ,it follows that t=0 in which case ,and there is no twisting. When ,there is the additional possibility that but .Since ,this reduces to the case t=1/2.The argument is more complicated when .Squaring yields the condition .Workingmodulo ,we can choose so that t=1. But still working modulo ,we can replace by .Conjugating once again, we have and there is no twisting.

The case is similar. Conjugation reduces to and squaring yields the condition .Twisting occurs when but onlyfor and . Finally, suppose that .We concentrate attention on the twogenerators

of modulo .Since the reflections are orthogonal, we can simultaneously conjugate so that .Squaring the generators, we obtain that and either t1=0, or dependingon whether , or .The various combinations of generators yieldone untwisted subgroup and one twisted subgroup for ,and one untwisted subgroup and three twisted subgroups for .Once again, there is no twisting when .The arguments are similar to the previous cases of K; we replace by untwisted group elements.

Next: Classification of Columns Up: Symmetries of Columns Previous: Untwisted Symmetry Groups
Marty Golubitsky
2001-01-29