Weyl Group of π΅π as Automorphisms of π-Cube: Isomorphism and Conjugacy David Chen University of California, Los Angeles Abstract The Weyl groups are important for Lie algebras. Lie algebras arise in the study of Lie groups, coming from symmetries of diο¬erential equations, and of diο¬erentiable manifolds. The Weyl groups have been used to classify Lie algebras up to isomorphism. The Weyl group associated to a Lie algebra of type π΅π and the group of graph automorphisms of the π-cube π΄π’π‘(ππ ) are known to be isomorphic to β€π 2 β ππ . We provide a direct isomorphism between them via correspondence of generators. Geck and Pfeiο¬er have provided a parametrization of conjugacy classes and an algorithm to compute standard representatives. We believe we have a more transparent account of conjugacy in the Weyl group by looking at π΄π’π‘(ππ ). We give a complete description of conjugacy in the automorphism group. We also give an algorithm to recover a canonical minimal length (in the Weyl group sense) representative from each conjugacy class, and an algorithm to recover that same representative from any other in the same conjugacy class. Under the correspondence with the Weyl group, this representative coincides precisely with the minimal length representative given by Geck and Pfeiο¬er, leading to an easier derivation of their result. 1 Introduction The Weyl groups are important for Lie algebras. A gentle undergraduate introduction to Lie algebras is given in [3]. Lie algebras arise in the study of Lie groups, coming from symmetries of diο¬erential equations, and of diο¬erentiable manifolds. The Weyl groups and their associated root systems have been used to classify Lie algebras up to isomorphism. That is, we associate a Weyl group to a Lie algebra and that group is the same as a reο¬ection group we associate to a ο¬nite set of vectors in βπ , called a root system. Most Lie algebras fall into the types π΄π , π΅π , πΆπ , or π·π , which arise from diο¬erent root systems. We’ll deο¬ne everything in the next section. The Weyl group π of the root system of π΅π has already been determined to be the semidirect product of the ο¬nite group of sign changes on a basis of an π- 1 dimensional Euclidean space and the group of permutations on π letters ππ . (see details in [5]) The graph automorphisms of the π-cube ππ , π΄π’π‘(ππ ) (or simply π΄π’π‘), have also been determined to be the semidirect product of transpositions and of the coordinate permutations. Each are isomorphic abstractly to β€π2 β ππ . We will provide a canonical isomorphism directly between the two, via generators. Geck and Pfeiο¬er [4] provide an account of conjugacy in all Weyl groups of the classical types. In doing so, they develop a lot of machinery: cuspidal classes, Coxeter elements, signed and unsigned blocks, etc.. The upside is that it generalizes to types π΄π , π΅π , and π·π , with each as a particular case. The downside is that it lacks transparency and exhibits a general diο¬culty in determining conjugacy of two elements. We believe we can provide an easier account of conjugacy in π of type π΅π by seeing π as π΄π’π‘. We can provide a complete description of conjugacy. In particular, we can computationally determine the conjugacy of two elements. Geck and Pfeiο¬er also are able to parametrize the classes by certain pairs of partitions of π, and provide an algorithm using blocks to compute a complete system of canonical minimal length representatives. We recover the same parametrization, provide algorithms in π΄π’π‘ to recover the same minimal length representatives and to recover, given any automorphism, its standard minimal length representative. By length in the Weyl group, we mean the minimal number of occurrences of generators required to write a member of the Weyl group. Length in the Weyl sense is tricky when viewed in π΄π’π‘ as it is not mentioned in general for graph automorphism groups. We provide a deο¬nition for length in the graph automorphism group of the cube that is equivalent to length in the Weyl group. 1.1 One Application to π-cube One graph construction from a graph πΊ and its automorphisms π΄ is the quotient graph πΊ/π΄. In [1], Brouwer details this construction and its use. The quotient is the set of πΊ-orbits under π΄ with adjacency when two orbits contain two adjacent vertices. We often look at quotient graphs under a group generated by a single automorphism. It is an exercise to show that two conjugate automorphisms induce isomorphic quotient graphs. In particular, if we look at the π-cube, we use our parametrization to compute representatives from each conjugacy class. We know their quotient graphs make up all possible quotient graphs. 2 Preliminaries In this section we deο¬ne the vocabulary used above. The details about reο¬ection groups and Weyl groups can be found in [5]. Fix a real inner product space π with basis π1 , . . . , ππ . Deο¬nition 2.1. A πππ ππππ‘πππ π πΌ associated to πΌ ∈ π is a linear map from π to itself sending πΌ to −πΌ and ο¬xing everything orthogonal to πΌ. 2 Deο¬nition 2.2. A ππππ‘ π π¦π π‘ππ Φ is a ο¬nite subset of π such that, if the line generated by πΌ, βπΌ, intersects Φ nontrivially, the intersection is {πΌ, −πΌ}, and all the reο¬ections associated to πΌ ∈ Φ permute Φ. An example of a root system is (1, 1), (1, −1), (−1, −1), (−1, 1) in β2 . Deο¬nition 2.3. A πππ ππππ‘πππ ππππ’π associated to a root system Φ is the group < π πΌ >πΌ∈Φ generated by reο¬ections associated to πΌ ∈ Φ. Deο¬nition 2.4. The π πππππ ππππ‘π Φ′ of root system Φ are a subset of Φ that generates all of Φ. It is well known that every root system Φ admits a subset of simple roots. It is also well known that < π πΌ >πΌ∈Φ =< π πΌ >πΌ∈Φ′ (see [3] for both results). We thus only need to look at the simple roots. Deο¬nition 2.5. Fix simple roots Φ′ = {π1 , π1 − π2 , . . . , ππ−1 − ππ }. π ππ¦π ππππ’π of π΅π is < π πΌ >πΌ∈Φ′ . We denote it π . The Deο¬nition 2.6. Let π€ ∈ π . The πππππ‘β of π€ is the minimal number of occurrences of generators to write π€ as a product of generators. The π-cube is the generalization of the line segment (1-cube), square (2cube), and cube (3-cube) to higher dimensions. Formally, we have: Deο¬nition 2.7. The π − ππ’ππ, denoted ππ , is a simple graph with vertices β€π2 and adjacency when exactly one component diο¬ers. Deο¬nition 2.8. Let πΊ be a graph, with ∼ as the edge relation. Then π : πΊ → πΊ is a ππππβ ππ’π‘πππππβππ π if it is bijective and for all π, π in πΊ, π ∼ π iο¬ π(π) ∼ π(π). We denote the group of graph automorhpisms of the π − ππ’ππ as π΄π’π‘(ππ ), or simply π΄π’π‘. We look at two particular types of automorphisms on the π − ππ’ππ: translations and coordinate permutations. Deο¬nition 2.9. A π‘ππππ πππ‘πππ π‘ ∈ (β€π2 ; +) acts on vertex π£ by π‘π£ = π‘ + π£. We write π‘ = +π1 . . . ππ , with ππ ∈ β€π2 . Deο¬nition 2.10. A πππππ’π‘ππ‘πππ π ∈ ππ acts on vertex π£ = π1 . . . ππ by ππ£ = ππ1 . . . πππ It is well known that permutations can be written in cycles and are unique products (up to order) of disjoint cycles (see [2] for details). We will use this fact again and again. Deο¬nition 2.11. Let π» and πΎ be ο¬nite groups. The set of ordered pairs π» × πΎ together with the following multiplication is the π πππππππππ‘ πππππ’ππ‘ of π» and πΎ, denoted π» β πΎ. Let βπ ∈ π», ππ ∈ πΎ. Deο¬ne (β1 , π1 )(β2 , π2 ) = (β1 π1 β2 π1−1 , π1 π2 ). We identify π» with∩{(β, 1) : β ∈ π»} and πΎ with {(1, π) : π ∈ πΎ}. π» is normal in π» β πΎ, and π» πΎ = 1. β£π» β πΎβ£ = β£π»β£β£πΎβ£. 3 Much more on semidirect product groups can be found in [2]. We are interested in a paritcular semidirect product group β€π2 β ππ . It is comprised of ordered pairs of a translation (under +) and a permutation. Multiplication is deο¬ned as: (π‘′ , π ′ )(π‘, π) = (π‘′ + π ′ ⋅ π‘, π ′ π), with π ′ ⋅ π‘ as +ππ′ (1) . . . ππ′ (π) for π‘ = +π1 . . . ππ . (π , π)−1 = (π −1 ⋅ π , π −1 ). For reference, (π , π)(π‘, π)(π , π)−1 = (π + π ⋅ π‘ + πππ −1 ⋅ π , πππ −1 ). 3 A Canonical Isomorphism between π and π΄π’π‘ We’ll start with determining the automorphism group of the π-cube. The details are provided in [6]. One can check that the coordinate permutations normalize the translations. We thus have: Proposition 3.1. The translations and coordinate permutations each form subgroups of π΄π’π‘, and intersect trivially, with the translations making a normal subgroup. Together, the two subgroups induce β€π2 β ππ ⊆ π΄π’π‘. Proof. To see that two subgroups π» and πΎ of group πΊ induce a semidirect product π»βπΎ, it follows from the deο¬nition of semidirect product that it suο¬ces to show π» ∩ πΎ = 1 and that for β ∈ π», π ∈ πΎ, we have πβπ −1 ∈ π». One can check that the translations and coordinate permutations form subgroups of π΄π’π‘. Here, (β€π2 ; +) is π», and ππ is πΎ. Clearly β€π2 ∩ ππ = 1. Take π‘ = +π1 . . . ππ ∈ β€π2 and π ∈ ππ . We have ππ‘π −1 = +ππ1 . . . πππ ∈ β€π2 . Remark. To establish β€π2 β ππ ∼ = π΄π’π‘, we need only show β£π΄π’π‘β£ ≤ 2π π!. We use a counting argument. Any automorphism will map a vertex to one of all 2π vertices. Then it must permute the neighbors, of which there are π! choices. Once the permutation of neighbors is chosen, all other points are determined because the π-cube is an (0, 2)-graph. In fact, we do not need the above isomorphism. From now, we proceed only knowing that all transpositions and coordinate permutations are automorphisms. For a real inner product space π with basis π1 , . . . , ππ , π arises as the group generated by the reο¬ections π πΌπ where πΌπ is a simple root. Canonically, πΌ1 = π1 − π2 , . . . , πΌπ−1 = ππ−1 − ππ , πΌπ = π1 . The reο¬ections corresponding to the simple roots generate π . The goal of this section is to prove the following theorem given by a remarkable correspondence between the simple root ππ − ππ+1 and the transposition (π π + 1), and between π1 and the translation +10 . . . 0. In our particular root system, the roots have one of two Euclidean lengths. We note that translations correspond to reο¬ections in short roots, while coordinate permutations correspond to those in long roots. So for example, we get other translations +000...1...000 (1 in πth position) by reο¬ection in ππ . Theorem 3.2. There is a one-to-one correspondence between the reο¬ections induced by the simple roots and the automorphisms of the π-cube induced by 4 these reο¬ections. These automorphisms generate π΄π’π‘ and satisfy the relations of the Weyl group π of type π΅π . This establishes the canonical isomorphism π ∼ = π΄π’π‘. The idea behind the proof is essentially to ο¬rst embed the graph into the unit cube in π , translate it to the origin, apply a reο¬ection given by a simple root, translate the image back to the unit cube, then recover a new graph from images of the embedded graph. This process induces an automorphism. From there we ο¬nd that each reο¬ection induces a unique automorphism, and each automorphism recovers a unique reο¬ection. We then show that the images of the generators of π preserve the relations, thereby achieving our isomorphism. Lemma 3.3. 1. There is a map π embedding the vertices of an π-cube, β€π2 , into π . This map π is bijective to the set of vectors π ∈ π with coordinates 0 or 1. 2. The translation π‘ in π by −0.5(1π1 + . . . + 1ππ ) is injective. 3. The reο¬ection π πΌ on π moving root πΌ, when restricted to π‘(π) is a permutation. 4. The map ππΌ = π −1 β π‘−1 β π πΌ β π‘ β π on π is a well deο¬ned graph automorphism. Proof. Deο¬ne the natural map π : β€π2 → βπ taking π1 . . . ππ to π1 π1 + . . . + ππ ππ , with ππ = 1 or 0. The idea is to have, for example in β2 , the graph’s vertex 01 rest at 0π1 +1π2 , 11 rest at 1π1 +1π2 , etc., so the graph rests naturally inside β2 . The translation π‘ as above centers the graph. The image of π under π‘, π‘(π), is clearly a root system. As such, π πΌ restricted to π‘(π) is a permutation. Applying the reο¬ection should ο¬x the embedded graph, as we are reο¬ecting roots. Undoing π‘ translates the vectors of π‘(π) back to π, so we apply π −1 to π to recover a graph, with incidence when two vertices diο¬er by exactly one entry. The map ππΌ = π −1 β π‘−1 β π πΌ β π‘ β π on π is a well deο¬ned graph permutation, as each map in the composition is bijective. The map ππΌ preserves the edge relation because π πΌ permutes and preserves orthogonality, in particular, relative distance in V by exactly 1. Using π and π −1 will preserve the diο¬erence in exactly one entry between two vertices. Example. Take π = β2 and consider the 2-cube with vertices 00, 10, 01, 11. The map π puts 00 at 0π1 + 0π2 , 10 at 1π1 + 0π2 , 01 at 0π1 + 1π2 , and 11 at 1π1 + 1π2 . If we apply the reο¬ection now, we wouldn’t stay in π = {vectors with coordinates 0 or 1}. We translate via π‘ our four vectors about the origin and apply the reο¬ection. The idea is to make π into a root system and closed under reο¬ection. Let’s apply the ο¬ip across the x-axis π π2 . A quick check tells us the image of π2 under π π2 is π‘(π). Undoing π‘, we see 0π1 + 0π2 is sent to 0π1 + 1π2 , 1π1 + 0π2 is sent to 1π1 + 1π2 , 0π1 + 1π2 is sent to 0π1 + 0π2 , and 1π1 + 1π2 is sent to 1π1 + 0π2 . Applying π −1 to each image and establishing adjacency in the usual way recovers the 2-cube. Thus our ππ2 is an automorphism. 5 Proposition 3.4. There is a one-to-one correspondence between simple root ππ − ππ+1 and the automorphism given by the coordinate permutation (π π + 1), and the simple root π1 and the automorphism given by translation by +10 . . . 0, via π πΌ ↔ ππΌ . Proof. Reο¬ections in π are of the form π π (π£) = π£ − 2(π£, π)/(π, π)π, π ∈ π . A routine check of π πΌ β π‘(π1 . . . ππ ), where πΌ is a simple root and π1 . . . ππ ∈ β€π2 , will demonstrate π πΌ induces the desired ππΌ . Going backwards from ππΌ , we must have π πΌ because each map is bijective by the Lemma and reο¬ections are unique up to scalars. The Weyl group π of type π΅π and indeed all ο¬nite Weyl groups have been determined (see [5]). π is < π πΌ1 , . . . , π πΌπ > satisfying: 1. πΌπ simple 2. π 2πΌπ = 1 3. π πΌπ π πΌπ+1 π πΌπ = π πΌπ+1 π πΌπ π πΌπ+1 for π + 1 < π 4. π πΌ1 π πΌπ π πΌ1 π πΌπ = π πΌπ π πΌ1 π πΌπ π πΌ1 5. π πΌπ π πΌπ = π πΌπ π πΌπ for β£π − πβ£ > 1. Proposition 3.5. The images of π πΌπ under the correspondence in the previous proposition satisfy the same relations as above. Proof. The images are transpositions (π π + 1) following the usual rules. The only thing tricky is the relation with +10 . . . 0. Keeping in mind the semidirect product multiplication, one can check the relation holds. 4 4.1 Conjugacy in π΄π’π‘ Conjugacy Classes of π΄π’π‘ We have a simple description of the conjugacy classes of the Weyl group of π΅π which can be seen by looking at π΄π’π‘ as ordered pairs of translations normalized by permutations, following the usual rules of semidirect product multiplication. The conjugacy classes behave in a nice way, and deciding conjugacy is simple. We prove necessary and suο¬cient conditions for deciding conjugacy. π΄π’π‘ ∼ = β€π2 β ππ is comprised of ordered pairs of a translation (under +) and a permutation. Multiplication works as in semidirect products: (π‘′ , π ′ )(π‘, π) = (π‘′ + π ′ ⋅ π‘, π ′ π), with π ′ ⋅ π‘ as +ππ′ (1) . . . ππ′ (π) for π‘ = +π1 . . . ππ . We also have (π , π)−1 = (π −1 ⋅ π , π −1 ). For reference, (π , π)(π‘, π)(π , π)−1 = (π + π ⋅ π‘ + πππ −1 ⋅ π , πππ −1 ). Clearly if (π‘′ , π ′ ) ∼ (π‘, π) (conjugacy) then π ′ ∼ π. We need ∏π to ο¬rst introduce some deο¬nitions and notation. Let (π‘, π) ∈ π΄π’π‘. Let π = π=1 ππ , a product of disjoint nontrivial cycles, with the shorter length cycles indexed lower. Explicitly, ππ = (ππ,1 . . . ππ,ππ ), where ππ is the length of ππ and ππ > 1. A standard result tells us the product is unique up to order and 6 cyclically permuting the terms inside each cycle (see [2]). For example, we have π = (1 3)(4 7 8) = (8 4 7)(3 1). Deο¬nition 4.1. The ππ¦πππ π π‘ππ’ππ‘π’ππ of π is the sequence (π1 , . . . , ππ ). For example, if π = (1 2)(4 3 7), the cycle structure of π is (2, 3). Deο¬nition 4.2. Consider the πππ£ππ πππ ππ‘ππππ of π‘ under a cycle ππ those indices in the cycle (as π‘ ∈ β€π2 ). Deο¬nition 4.3. Consider the π ππ₯ππ πππ ππ‘ππππ of π‘ those indices moved by no cycle. We ο¬nish the section by proving the following theorem. Theorem 4.4. Let (π‘′ , π ′ ), (π‘, π) ∈ π΄π’π‘. Denote ππ‘π the number of 1’s among the moved positions of π‘ under cycle ππ , and ππ‘ the number of 1’s among the ο¬xed positions. Then (π‘′ , π ′ ) ∼ (π‘, π) (conjugacy) iο¬ 1. π ∼ π ′ 2. ππ‘′π ≡ ππ‘π (mod 2), all π, though the indices of cycles of the same length may be permuted. 3. ππ‘′ = ππ‘ Denote these conditions by π». By the second condition in the theorem above, we mean that for example, ππ‘1 ≡ ππ‘′2 and ππ‘′1 ≡ ππ‘2 , with π = π1 π2 = (1 2)(3 4) and π ′ = π′1 π′2 = (1 2)(3 4). That is, ππ‘1 and ππ‘′1 don’t exactly match up, but there is exactly one other cycle that matches it in parity, and for π2 , there is also exactly one cycle matching it in parity. We use the following commonly known lemma and theorem (see details in [2]): ∏ Lemma 4.5. If permutation π = ππ , where ππ = (ππ,1 . . . ππ,ππ ), then we have ∏ πππ −1 = (πππ,1 . . . πππ,ππ ). Proof. To see this, we consider permutations π, π = (π1 . . . ππ ). We can show that ππ π −1 = (ππ1 . . . πππ ). To achieve the lemma, we simply extend the result for all permutations by inserting π −1 π between all the ππ and ππ+1 . Theorem 4.6. Two permutations are conjugate iο¬ they have the same cycle structure. ∏π ∏π That is, consider π = π=1 ππ and π = π=1 ππ , where ππ = (ππ,1 . . . ππ,ππ ), and ππ = (ππ,1 . . . ππ,ππ ), and ππ , ππ > 1. Then π ∼ π iο¬ π = π and (π1 , . . . , ππ ) = (π1 , . . . , ππ ). One can easily prove the following lemma: 7 Lemma 4.7. Any permutation with a given cycle structure is conjugate to the permutation with 1, 2, . . . running through the cycle structure. For example (π1 π2 )(π3 π4 π5 ) ∼ (1 2)(3 4 5), and so (π‘, (π1 π2 )(π3 π4 π5 )) ∼ (π‘, (1 2)(3 4 5)). Proposition 4.8. (π‘′ , π ′ ) ∼ (π‘, π) via (π , π) satisfy π». Proof. Reduce to the case of π = π ′ , with 1, . . . , π running through the cycle structure of π. By assumption, π‘′ = π +π⋅π‘+πππ −1 ⋅π . Because π = π ′ = πππ −1 , π permutes the ο¬rst π positions, and πππ −1 by construction permutes the ο¬rst π positions and ο¬xes those after π. π permutes cyclically within each cycle, and so πππ −1 as well, so the stronger ππ‘′π = ππ‘π + 2ππ ≡ ππ‘′π (mod 2) for the number of 1’s ππ under a cycle, each πth cycle. Note that when there are two π-cycles, their indices may be permuted for matching up the ππ‘π ’s, as on a pair of π-cycles, π may permute not only cyclically within a cycle but also permute the π-cycles themselves (see the remark under the above theorem). After π, πππ −1 ⋅ π and π coincide. Adding them gives 0 after π, and we satisfy π». Proposition 4.9. (π‘′ , π ′ ) and (π‘, π) satisfying π» are conjugate. Proof. We prove it for base case of π as a π-cycle, then proceed by induction by assuming it holds for all π-cycles and prove it for a product of π-cycles. Reduce to the case of π = π ′ = (1 2 . . . π). We want π‘′ = π + π ⋅ π‘ + πππ −1 ⋅ π with appropriate choice of (π , π). First set π = 1 on moved positions and have it change π‘ to π‘′ on ο¬xed positions. We can impose the latter requirement because ππ‘ = ππ‘′ . By construction, we reduce the task to proving π‘′ = π + π ⋅ π‘ + π ⋅ π . Looking at the ο¬xed positions, we see π‘′ = π ⋅ π‘. By construction, π = π ⋅ π . Thus π‘′ = π + π‘′ + π = π + π ⋅ π‘ + π ⋅ π , for any choice of π . Looking at moved positions, we see π ⋅ π‘ = π‘. We’ll pick π satisfying π‘′ + π‘ = π + π ⋅ π . Let π = π1 . . . ππ . π ⋅ π = ππ π1 . . . ππ−1 . For notational simplicity, let π‘ + π‘′ = 1 . . . 10 . . . 0, with an even number π of 1’s by hypothesis. Then π1 . . . ππ satisfy: π1 +ππ = 1, π2 +π1 = 1, . . . , ππ +ππ−1 = 1, ππ+1 +ππ = 0, . . . , ππ +ππ−1 = 0. Setting π1 = 0, we have ππ = 1 for π even, ππ = 0 if odd, π ≤ π, and ππ = 1 for π ≤ π ≤ π. This π works. Induction Step: we lift the π ’s found in the previous proposition for each π-cycle to a solution for π. Again let π = 1 on moved positions and let it morph π‘ to π‘′ on ο¬xed positions. For π , we concatenate the fragments of π ’s corresponding to each cycle, as they were only computed for their own moved positions and could behave∏regardless of what’s chosen in ο¬xed positions. That is, π = π 1 . . . π π . . . for π = ππ , π cycles, and π π is the part of π computed for ππ as above stripped of its ο¬xed positions and adjusted for reindexing. Anything after π π in π is whatever is desired. Example. Take (1111001, (1 2)(3 4 5)) = (π‘′ , π) and (0001110, (1 2)(3 4 5)) = (π‘, π) in π΄π’π‘(7). We’ll ο¬nd (π , π) establishing conjugacy. Note ππ‘′(1 2) ≡ ππ‘(1 2) (mod 2) and ππ‘′(3 4 5) ≡ ππ‘(3 4 5) (mod 2). Note ππ‘′ = ππ‘ . As in the proof above, 8 we reduce the task to proving π‘′ = π + π ⋅ π‘ + π ⋅ π , as π and π are chosen to be disjoint. We ο¬rst look at (11 . . . , (1 2)) and (00 . . . , (1 2)). We can ignore the ο¬xed positions even when they don’t have exact same number of 1’s. On moved positions 1, 2, set π = 1, so π ⋅ 00 . . . = 00 . . .. We want π (1 2) so that 11 . . . + 00 . . . = π (1 2) + (1 2) ⋅ π . For π (1 2) = π1 π2 . . ., (1 2) ⋅ π (1 2) = π2 π1 . . .. Set 11 = π1 π2 + π2 π1 and set π1 = 0, so π2 = 1. Thus our π (1 2) = 01 . . .. Now look at (. . . 110 . . . , (3 4 5)) and (. . . 011 . . . , (3 4 5)). Ignore the ο¬xed positions again. On moved positions 3, 4, 5 set π = 1, so π ⋅ . . . 011 . . . = . . . 011 . . .. We want π (3 4 5) so that . . . 110 . . . + . . . 011 . . . = π (3 4 5) + (3 4 5) ⋅ π (3 4 5) . For π (3 4 5) = . . . π3 π4 π5 . . ., (3 4 5) ⋅ π (3 4 5) = . . . π5 π3 π4 . . .. Set 101 = π3 π4 π5 + π5 π3 π4 and set π3 = 0, so π4 = 0 and π5 = 1. Thus our π (3 4 5) = . . . 001 . . .. Now lift the π (1 2) and π (3 4 5) to π = π (1 2) π (3 4 5) 00 = 0100100. Set π = (6 7). Then π‘′ = π + π ⋅ π‘ + π ⋅ π = 0100100 + 0001101 + 1010000. We establish conjugacy. 4.2 Parametrization of Conjugacy Classes and Algorithm for Minimal Length Representative for Each We give algorithms to compute conjugacy class representatives from elements of a class, and from a parametrization, with the latter’s conditions given by Geck and Pfeiο¬er. We deο¬ne 3 diο¬erent notions of a representative using the same name, but we also prove they all coincide. We ο¬rst note that a permutation may be written as a unique product of nontrivial cycles, and that trivial cycles may be inserted anywhere in the product, changing neither the cycle structure nor the permutation. For example (1 2)(5 8 6) = (3)(1 2)(9)(5 8 6). We will need to consider the trivial cycles in this section. 4.2.1 Representatives and Algorithms Deο¬nition 4.10. For (π‘, π) when π has 1, . . . , π running through its cycle structure, the πππππ of π‘ moved by the cycle ππ is the subtuple of π‘ with indices the same as in ππ . For example, let π‘ = +1001 and π = (1 2)(3 4). The block of π‘ moved by cycle (1 2) is 10. Deο¬nition 4.11. For (π‘, π), the (π, π) computed by the algorithm below is called the πΏ − ππππππ πππ‘ππ‘ππ£π (πΏπππ) computed from (π‘, π). We call it an πΏ − ππππππ πππ‘ππ‘ππ£π because it will be shown to have minimal πππππ‘β. Algorithm 4.12. Compute ππ‘ , ππ‘π . Initially form π to consist of, from left to right, ππ‘ trivial cycles, then the cycle structure of π, with 1, . . . , π running through. Initially form π, put a 1 on each ο¬xed position, then for cycles ππ where 9 ππ‘π ≡ 0, put 0’s on the positions it moves, for cycles where ≡ 1, put 1 on the least moved position and 0’s on all other moved positions. Permute the blocks of π moved by the cycles so that the those of ≡ 1 are to the left, with shorter cycles more left. This is π. Then permute the cycles of π to reο¬ect π; Order the cycles where ≡ 0 so the longer cycles are more left. Lastly have 1, . . . , π running through again. This is π. By all the work above, the πΏπππ computed from (π‘, π) is conjugate to (π‘, π). Because conjugate automorphisms meet all the conditions π», an inspection of the algorithm will show that the construction only depended on the cycle structure and the numbers in π», which are the same to the algorithm. Also notice if even one number changes, or the cycle structures diο¬er, we get a diο¬erent automorphism. We thus have: Lemma 4.13. Automorphisms are conjugate iο¬ they compute the same πΏ − ππππππ πππ‘ππ‘ππ£π. With this uniqueness, we can associate an πΏπππ to a conjugacy class. Deο¬nition 4.14. Given a conjugacy class πΆ of π΄π’π‘, let the πΏ − ππππππ πππ‘ππ‘ππ£π of πΆ, π€πΆ , be the πΏ − ππππππ πππ‘ππ‘ππ£π of any element in πΆ. Clearly then, we have: Proposition 4.15. The set of πΏ − ππππππ πππ‘ππ‘ππ£ππ from all conjugacy classes form a complete system of representatives of the conjugacy classes. Example. Care needs to be taken when computing an πΏπππ. Consider π‘ = 10100111, π = (1 2)(6 7)(3 4 5). π = 1, π(1 2) = 1, π(6 7) = 2, π(3 4 5) = 1. Applying the algorithm, we have π = (1)(2 3)(4 5)(6 7 8) and one would guess π = 11000100. That is not exactly right: we need to have π reο¬ect the cycle structure (1)(2 3)(6 7 8)(4 5), so that all the chunks of 1’s and 0’s corresponding to cycles with a 1 are to the left. We can do this because cycle structure is unique only in counting the number of π-cycles. Then the real π = 11010000. Deο¬nition 4.16. Consider a pair of partial partitions of π, (πΌ, π½) with πΌ + π½ = π. The πΏ − ππππππ πππ‘ππ‘ππ£π computed from (πΌ, π½), denoted π€(πΌ,π½) is the automorphism computed using the following algorithm:. Algorithm 4.17. Let π½ = π1 +. . .+ππ and πΌ = ππ+1 +. . .+ππ . For 1 ≤ π ≤ π, set ππ = 10 . . . 0, ππ − 1 0’s. For π + 1 ≤ π ≤ π, ∏set ππ = 0 . . . 0, ππ 0’s. Then set π = π1 . . . ππ . Set ππ as an ππ -cycle. Set π = ππ with 1 . . . π running through the cycle structure including the trivial cycles. π€(πΌ,π½) = (π, π). Remark. The algorithm to compute the πΏπππ from (πΌ, π½) computes the same representative, under isomorphism, computed by the algorithm given by Geck and Pfeiο¬er that takes as argument (πΌ, π½). Indeed, we chose notation π€(πΌ,π½) to coincide with theirs. 10 Example. We’ll compute an πΏπππ from πΌ = 2 + 1, π½ = 1 + 3. π1 = 1, π2 = 100, π3 = 00, π4 = 0, so π = 1100000. π1 = 1, π2 = (1 2 3), π3 = (1 2), π4 = 1, so π = (2 3 4)(5 6). It’s easy to see that with πΌ decreasing and π½ increasing, π€(πΌ,π½) is trivially reduced to the πΏπππ computed from it. Thus: Lemma 4.18. The πΏ − ππππππ πππ‘ππ‘ππ£π computed from (πΌ, π½) with πΌ decreasing and π½ increasing, in conjugacy class πΆ, is exactly the πΏ − ππππππ πππ‘ππ‘ππ£π of the same conjugacy class. 4.2.2 Parametrization and Proof With the deο¬nitions in place (except length: see next section), we present the main theorem of the section: Theorem 4.19. (Geck and Pfeiο¬er) There is a one-to-one correspondence between the pairs of partitions of π, (πΌ, π½), with πΌ decreasing and π½ increasing, πΌ + π½ = π, and the conjugacy classes of π΄π’π‘. Furthermore, there is a way to compute a canonical representative of minimal length from these (πΌ, π½), and all of them are representatives from all conjugacy class of π΄π’π‘. Namely, the πΏπππ from (πΌ, π½) is that canonical representative. To prove the correspondence exists, we prove the following: Proposition 4.20. The πΏ − ππππππ πππ‘ππ‘ππ£ππ computed from pairs of partial partitions of π, (πΌ, π½), πΌ decreasing, π½ increasing, πΌ+π½ = π, are representatives from all conjugacy class of π΄π’π‘, with each pair picking out a distinct conjugacy class. Lemma 4.21. Let (πΌ, π½) and (πΌ′ , π½ ′ ) be distinct pairs of partial partitions of π, with πΌ + π½ = πΌ′ + π½ ′ = π, with πΌ and πΌ′ decreasing, π½ and π½ ′ increasing. Then the πΏπππ’s they compute are not conjugate. ∑π ∑π ′ Proof. Let πΌ = π=1 π∑ π, π½ = 1 +1+ π=π+1 ππ , and without ∑π ∏π loss let πΌ = π∏ π π ′ ′ ′ π½′ = π , π½ = π −1+ π . We have π = π and π π π+1 π πΌπ½ π πΌ π=2 π=π+2 π=1 π=1 ππ , with ππ an ππ -cycle, even a 1-cycle. For simplicity, suppose we have conjugacy and that ππ corresponds to π′π , in terms of number of 1’s. Imposing the restriction of the same cycle structure, we have that π′π1 +1 corresponds to πππ+1 and ππ1 corresponds to π′ππ+1 −1 , as ππ1 cannot correspond to π′π1 +1 as their lengths diο¬er. We can’t however impose this correspondence as each in each pair has diο¬erent numbers of 1’s. Thus we can’t have conjugacy. With the previous lemma, we provide injectivity: Proposition 4.22. Distinct pairs of partitions of π, (πΌ, π½), with πΌ decreasing and π½ increasing, πΌ + π½ = π, correspond to distinct conjugacy classes of π΄π’π‘ by computing πΏ − ππππππ πππ‘ππ‘ππ£ππ of distinct conjugacy classes. 11 Now for surjectivity. We start by noting that when we lift the restriction of πΌ decreasing, π½ increasing, we get repeats, as we state in the following lemma. Lemma 4.23. The πΏ − ππππππ πππ‘ππ‘ππ£π computed from (πΌ, π½) with πΌ increasing or π½ decreasing is conjugate to some πΏ − ππππππ πππ‘ππ‘ππ£π computed from (πΌ, π½) with πΌ decreasing and π½ increasing. Proof. If π½ is not increasing, start the reduction by permuting the blocks with 1’s until the shortest are more to the left. If π½ is increasing and πΌ is not decreasing, start by changing the cycle structure to have the cycles where ≡ 0 of greater length are more to the left after the cycles ≡ 1. In either case we have nontrivial reductions to πΏππππ of the class. Every conjugacy class gets hit by the (πΌ, π½) parametrization. Proposition 4.24. Let πΆ be any conjugacy class. From the πΏπππ of πΆ, π€πΆ , we recover a pair of partial partitions (πΌ, π½), πΌ decreasing, π½ increasing, πΌ + π½ = π, so that π€πΆ = π€(πΌ,π½) . ∏π Proof. Compute π€πΆ = (π, π = π=1 ππ ), including trivial cycles. The product can be naturally split between the last π with ππ has ππ ≡ 1, and the ο¬rst π where ππ ≡ 0. Denote ∏πthe ο¬rst ∏ππ where ππ ≡ 0 with π = 1: it goes along for π cycles. Then π = π π π=π+1 π=1 ππ . Letting ππ be the length of ππ , we see ∑π ∑π π = π=1 ππ + π=π+1 ππ = πΌ + π½. By construction of π€πΆ , πΌ is decreasing and π½ is increasing. A review of the algorithm for computing π€(πΌ,π½) tells us π€πΆ = π€(πΌ,π½) . We thus establish the surjectivity and the correspondence. That leaves minimal length. 4.2.3 Length in π΄π’π‘ and Proof of Minimal Length We’ll deο¬ne a length in π΄π’π‘ to be compatible to that in the Weyl group, where it is the least number of simple reο¬ections needed to write an element. We’ll deο¬ne length individually on β€π2 and ππ , then let length in π΄π’π‘ be their sum. Deο¬nition 4.25. For 0 . . . 010 . . . 0 ∈ β€π2 , with 1 in the πth place, let the πππππ‘β π be 2π − 1. Extend by additivity. Deο¬nition 4.26. For π ∈ ππ , let the πππππ‘β π be the least number of (π π + 1) needed to write π. Deο¬nition 4.27. For (π‘, π) ∈ π΄π’π‘, let the πππππ‘β π = π(π‘) + π(π). This deο¬nition makes sense because of the Weyl group relations under isomorphism. The above notion of length in π΄π’π‘ seems deο¬ned arbitrarily, but one can check it is exactly length in the Weyl group under isomorphism. Length in β€π2 is obtained by examining the conjugation action on 10 . . . 0. Proposition 4.28. In conjugacy class πΆ of π΄π’π‘, π€πΆ has minimal length in πΆ. 12 Proof. Let π€πΆ = (π, π). The length π(π) is minimal because we pushed all the 1’s as left as possible while preserving the number of 1’s and 1’s mod 2 among cycles, with minimal 1’s. Then let’s prove that π(π) is minimal among all (π‘, π) ∈ πΆ with π(π‘) = π(π). But even this easily follows by construction. The cycle structure of π has 1 . . . π running through it. This guarantees that each π-cycle has minimal length among all π-cycles. Any conjugate to π with strictly lower length would need to have the same cycle structure and not have 1 . . . π running through each cycle, which isn’t possible. We thus establish that the representative computed from our parametrization is of minimal length. We thus recover the entire picture, up to isomorphism, of conjugacy of the Weyl group of type π΅π given by Geck and Pfeiο¬er. We also provide a quick check to see if any two elements are conjugate. Acknowledgements I completed the work for this paper at the Summer 2009 Math Research Experience for Undergraduates at the University of Georgia under Professor Leonard Chastkofsky. I’d like to thank Professor Chastkofsky for all the guidance he gave during and after the program. He helped me start the paper and has continued to help me until publication. I’d also like to thank the University of Georgia for hosting the REU, and to the staο¬ and faculty who made it run smoothly. A special thank you to the referees from the Rose-Hulman Undergraduate Math Journal for their reviews and suggestions. I greatly appreciate their comments on improving the paper. References [1] Andries E. Brouwer. Classiο¬cation of small (0,2)-graphs. J. Combinatorial Theory, 2006. [2] David Steven Dummit and Richard M. Foote. Abstract Algebra. Wiley, 3rd edition, 2004. [3] Karin Erdmann and Mark J. Wildon. Birkhauser, 2006. Introduction to Lie Algebras. [4] Meinolf Geck and Gotz Pfeiο¬er. Characters of ο¬nite Coxeter groups and Iwahori-Hecke algebras. Oxford University Press, 2000. [5] James E. Humphreys. Reο¬ection Groups and Coxeter Groups. Cambridge University Press, 1992. [6] Jennifer Muskovin. On (0,2)-graphs and semibiplanes. Master’s thesis, University of Georgia, 2009. 13