Structure of coincidence isometry groups

Definition 2.1

Let V be an n -dimensional Euclidean space.

1. Two vectors α , β ∈ V are orthogonal, written α ⊥ β , if ( α , β ) = 0 ;

2. Two subsets X , Y ⊂ V are orthogonal, written X ⊥ Y , if ( α , β ) = 0 for all α ∈ X and β ∈ Y . We write α ⊥ Y in place of { α } ⊥ Y ;

3. The orthogonal complement of a subset X is the set

   X ⊥ ≔ { β ∈ V : β ⊥ X } .

   We also write α ⊥ instead of { α } ⊥ ;

4. A nonempty subset X of V is called an orthogonal set if 0 ∉ V and ( α , β ) = 0 for pairwise distinct vectors in X .

Definition 2.2

A discrete subset L of R n is a called a lattice of dimension d if it is a free abelian group spanned by a linearly independent set { α 1 , … , α d } of R n . We always write it in the form ⊕ i = 1 d Z α i . A sublattice L ′ of L is also a free abelian group. Let [ L : L ′ ] denote the index of L ′ in L . Two lattices L 1 , L 2 of R n are called commensurate if both the indices [ L 1 : L 1 ∩ L 2 ] and [ L 2 : L 1 ∩ L 2 ] are finite.

Definition 2.3

Let L be a lattice of R n and let V be the Euclidean subspace of R n spanned by the elements in L . An orthogonal transformation A of V is called a coincidence isometry of L if [ L : L ∩ A L ] < ∞ . The set of all coincidence isometries of L is a subgroup of O ( V ) and is denoted by O C ( L ) .

Remark 2.1

Each subspace V of R n has a unique orthogonal complement V ⊥ = { α ∈ R n : α ⊥ V } such that R n = V ⊕ V ⊥ . There exists a canonical way to extend an isometry A of V to an isometry of R n :

Hence, the coincidence isometry group O C ( L ) of a lattice in V can be viewed as a subgroup of O ( R n ) .

Theorem 2.1

[16, Theorem 1] Let L 1 = ⊕ i = 1 n Z α i and L 2 = ⊕ i = 1 n Z β i be two lattices in R n . Then the following are equivalent.

1. L 1 and L 2 are commensurate;

2. The transition matrix from α 1 , … , α n to β 1 , … , β n is a rational matrix;

3. ⊕ i = 1 n Q α i = ⊕ i = 1 n Q β i .

Lemma 2.1

Let L 1 = ⊕ i = 1 n Z α i , L 2 = ⊕ i = 1 n Z β i be two lattices of R n . If ( β 1 , … , β n ) = ( α 1 , … , α n ) P and P is a non-singular rational matrix, then O C ( L 1 ) = O C ( L 2 ) .

Proof

It follows immediately from Theorem 2.1 and Lemma 7.2 of [5].□

Lemma 2.2

Suppose that a lattice L of R n is the orthogonal direct sum of sublattices L 1 and L 2 . Then O C ( L 1 ) ⊕ O C ( L 2 ) is isomorphic to a subgroup of O C ( L ) .

Proof

Without loss of generality, we may assume that L is n -dimensional, L 1 = ⊕ i = 1 r Z α i and L 2 = ⊕ i = r + 1 n Z α i . Let V 1 = ⊕ i = 1 r R α i and V 2 = ⊕ i = r + 1 n R α i be the respective R -subspaces of R n spanned by the basis vectors of L 1 and L 2 . It is routine to check that the following map

is an injective group homomorphism from O C ( L 1 ) ⊕ O C ( L 2 ) to O ( R n ) . It remains to show that A ∈ O C ( L ) . Let A 1 be the matrix of A 1 (viewed as an element of O ( V 1 ) ) with respect to the basis α 1 , … , α r and A 2 be the matrix of A 2 (viewed as an element of O ( V 2 ) ) with respect to the basis α r + 1 , … , α n . Then the matrix of A with respect to the basis α 1 , … , α n is the quasi-diagonal matrix diag ( A 1 , A 2 ) . Since A i is a coincidence isometry, A i is a rational matrix and is also A . Hence, A is a coincidence isometry of L .□

Theorem 2.2

[14, Theorem 3.1 and Theorem 3.2] Let L = ⊕ i = 1 n Z α i be a lattice of R n . Then every non-zero vector of L defines a coincidence reflection of L if and only if ( α i , α j ) ( α k , α k ) are all rational for all 1 ≤ i , j , k ≤ n . Moreover, in this case any coincidence isometry of L is a product of at most n reflections defined by the vectors in L .

Lemma 3.1

Let L = ⊕ i = 1 n Z α i be a lattice of R n . Suppose that ℛ v ∈ O C ( L ) . Then there exists a linearly independent subset { β 1 , … , β n } of L such that β 1 is parallel to v and ( β 1 , β i ) = 0 for i > 1 .

Proof

We put the proof in the appendix.□

Corollary 3.1

Let L = ⊕ i = 1 n Z α i be a lattice of R n and let S = { v ∈ ⊕ i = 1 n Q α i : v ≠ 0 , ℛ v ∈ O C ( L ) } . Then S is an invariant set of any A ∈ O C ( L ) , that is, A ( v ) ∈ S for any A ∈ O C ( L ) and v ∈ S .

Proof

By Lemma 3.1, there exists a linearly independent subset { β 1 , … , β n } of L such that β 1 is parallel to v ∈ S and ( β 1 , β i ) = 0 for i > 1 . It suffices to show that A ( β 1 ) ∈ S . In fact, let γ i = A ( β i ) , L 1 = ⊕ i = 1 n Z β i and L 2 = ⊕ i = 1 n Z γ i . Since A is an isometry, one has ( γ 1 , γ i ) = 0 for i > 1 . It follows that ℛ γ 1 ( γ 1 ) = − γ 1 and ℛ γ 1 ( γ i ) = γ i for i > 1 . Hence, ℛ γ 1 is a coincidence reflection of L 2 . Since L and L 2 are commensurate, we have ℛ γ 1 ∈ O C ( L 2 ) = O C ( L ) by Lemma 2.1. This finishes the proof.□

Lemma 3.2

Let L = ⊕ i = 1 n Z α i be a lattice of R n . Let V 1 = ⊕ i = 1 r R α i and V 2 = ⊕ i = r + 1 n R α i . Suppose that V 1 ⊥ V 2 . Let v = v 1 + v 2 , 0 ≠ v i ∈ V i . If ℛ v ∈ O C ( L ) , then ℛ v i ∈ O C ( L ) .

Proof

We put the proof in the appendix.□

Lemma 3.3

Let L = ⊕ i = 1 n Z α i be a lattice of R n and let V = ⊕ i = 1 n Q α i . Suppose that { v 1 , … , v r } is a linearly independent subset of V such that ℛ v i ∈ O C ( L ) . Applying the Gram-Schmidt orthogonalization process to v 1 , … , v r gives that

Then w i ∈ V and ℛ w i ∈ O C ( L ) for i = 1 , … , r .

Proof

We put the proof in the appendix.□

Theorem 3.1

Let L = ⊕ i = 1 n Z α i be a lattice of R n and let r be dimension of the Q -space spanned by { v ∈ L : ℛ v ∈ O C ( L ) } . Then

1. r ≠ n − 1 .

2. If 1 ≤ r ≤ n − 2 , then there exist two orthogonal sublattices L 1 of dimension r and L 2 of dimension n − r such that O C ( L ) = O C ( L 1 ) ⊕ O C ( L 2 ) , and O C ( L ) is not generated by the reflections.

Proof

Let V = ⊕ i = 1 n Q α i and V 1 be the Q -subspace of V spanned by { v ∈ L : ℛ v ∈ O C ( L ) } . Suppose that w 1 , … , w r is a basis of V 1 . By Lemmas 3.3 and 3.1, we may assume that { w 1 , … , w r } is orthogonal subset of L . By Lemma 3.1 again, W j = { α ∈ V : ( α , w j ) = 0 } is an ( n − 1 ) -dimensional subspace of V . Let V 2 = ∩ j = 1 r W j . Then V 2 is the orthogonal complement of V 1 in the Q -space V and V = V 1 ⊕ V 2 . Choosing a basis { w r + 1 , … , w n } ⊂ L of V 2 . Then L 1 = ⊕ i = 1 r Z w i and L 2 = ⊕ i = r + 1 n Z w i are orthogonal sublattices of L . Since { w 1 , … , w n } is a linearly independent subset of V = ⊕ i = 1 n Q α i , we have V = ⊕ i = 1 n Q w i and L is commensurate to L ′ = L 1 ⊕ L 2 by Theorem 2.1. So we have O C ( L ) = O C ( L ′ ) by Lemma 2.1.

If r = n − 1 , then { w 1 , … , w n − 1 , w n } is an orthogonal basis of L ′ . Thus, ℛ w n ∈ O C ( L ′ ) = O C ( L ) , which is a contradiction. This proves (1).

Let A ∈ O C ( L ) . By Corollary 3.1, the reflection defined by A ( w j ) is also a coincidence isometry of L for j = 1 , … , r . It follows that A ( w j ) ∈ V 1 and V 1 is an A -invariant subspace of V . Since V 1 ⊥ V 2 and A is an isometry, V 2 is also an A -invariant subspace of V . It follows that the group homomorphism defined in Lemma 2.2 is an isomorphism in this case.

If ℛ v is a coincidence isometry of L defined by a vector v ∈ L , then v ∈ V 1 by hypothesis. Since V 1 ⊥ V 2 , it follows that ℛ v ∣ V 2 = I , and is also for any coincidence isometry which is a product of reflections. Hence, − I is not contained in the subgroup of O C ( L ) generated by reflections.□

Corollary 3.2

Let L be an n -dimensional lattice of R n . Suppose that 0 ∉ X ⊂ L and O C ( L ) is generated by the reflections defined by the vectors of X . Then ∣ X ∣ ≥ n .

Proof

Suppose to the contrary that ∣ X ∣ < n . Let { v 1 , … , v r } be a maximal linearly independent subset of X , r ≤ ∣ X ∣ < n . Since O C ( L ) is generated by reflections, by Theorem 3.1 there exist vectors v r + 1 , … , v n ∈ { v ∈ L : ℛ v ∈ O C ( L ) } such that { v 1 , … , v n } is linearly independent. By Lemma 3.3, applying the Gram-Schmidt orthogonalization process to v 1 , … , v n , we obtain an orthogonal set { w 1 , … , w n } such that ℛ w i ∈ O C ( L ) . Since X ⊂ ⊕ i = 1 r Q v i = ⊕ i = 1 r Q w i , so w n ⊥ X . It follows that ℛ v ( w n ) = w n for any v ∈ X , and is also for any coincidence isometry of L . This contradicts ℛ w n ∈ O C ( L ) .□

Lemma 4.1

Let α ∈ R n be a nonzero vector. Then α ⊥ has a basis consisting of vectors in Q n if and only if α is parallel to a nonzero vector of Q n .

Proof

We put the proof in the appendix.□

Theorem 4.1

Let L = ⊕ i = 1 n Z α i be a lattice in R n and let G = ( ( α i , α j ) ) be the Gram matrix of α 1 , … , α n . Let v = ∑ i = 1 n k i α i be a non-zero vector of L . Then ℛ v ∈ O C ( L ) if and only if G ( k 1 , … , k n ) T is parallel to a nonzero vector in Q n .

Proof

Let B = ( k 1 , … , k n ) T . First assume that ℛ v ∈ O C ( L ) . By Lemma 3.1, there exists a linearly independent subset { β 1 , … , β n } of ⊕ i = 1 n Q α i such that β 1 = v and ( β 1 , β i ) = 0 for i > 1 . Suppose β i = ∑ j = 1 n b i j α j . By the property of Gram matrix, we have for i > 1 ,

where B i = ( b i 1 , … , b i n ) . By Lemma 4.1, G B is parallel to a nonzero vector in Q n .

Now suppose that G B = r C and C ∈ Q n . By Lemma 4.1, there exist n − 1 linearly independent row vectors C 1 = ( a 11 , … , a 1 n ) , … , C n − 1 = ( a n − 1 , 1 , … , a n − 1 , n ) such that C i ∈ Q n and C i G B = 0 . Set v i = ∑ j = 1 n a i j α j ∈ ⊕ i = 1 n Q α i . Then { v 1 , … , v n − 1 } is also linearly independent and

by the property of Gram matrix again. We observe that v 1 , … , v n − 1 , v n = v is a basis of ⊕ i = 1 n Q α i , and ℛ v ∈ O C ( ⊕ i = 1 n Z v i ) = O C ( L ) by Lemma 2.1.□

Corollary 4.1

Let L = ⊕ i = 1 n Z α i be a lattice in R n with an orthogonal basis α 1 , … , α n . Suppose that v = ∑ i = 1 n k i α i ∈ L is a nonzero vector and S = { i : k i ≠ 0 } . Then ℛ v ∈ O C ( L ) if and only if ( α i , α i ) ( α j , α j ) ∈ Q for any i , j ∈ S .

Theorem 4.2

Let L = ⊕ i = 1 n Z α i be a lattice in R n and let L ℛ be the sublattice of L spanned by a maximal orthogonal subset of { v ∈ L : ℛ v ∈ O C ( L ) } . Then the following are equivalent.

1. L contains an orthogonal subset of order n ;

2. dim L ℛ = n ;

3. Any coincidence isometry of L can be decomposed as a product of at most n reflections defined by the vectors in L .

Proof

( 1 ) ⇒ ( 2 ) Suppose that { w 1 , … , w n } is an orthogonal subset of L . Then { w 1 , … , w n } is linearly independent and L is commensurate to L ′ = ⊕ i = 1 n Z w i by Theorem 2.1. Hence, ℛ w i ∈ O C ( L ′ ) = O C ( L ) by Lemma 2.1 and we have dim L ℛ = n .

( 2 ) ⇒ ( 1 ) is trivial.

( 3 ) ⇒ ( 1 ) follows from Theorem 3.1 and Lemma 3.3.

( 1 ) ⇒ ( 3 ) Without loss of generality we may assume that α 1 , … , α n is an orthogonal basis of L . Let V = ⊕ i = 1 n Q α i . Suppose that { α 1 , … , α n } = ∪ i = 1 r X i be the partition corresponding to the equivalence relation ∼ Q restricted on { α 1 , … , α n } . Let L i = ⊕ γ ∈ X i Z γ , V i = ⊕ γ ∈ X i Q γ and W i = ⊕ γ ∈ X i R γ be the sublattice of L , the Q -subspace of V and the R -subspace of R n generated by the elements of X i , respectively. Then L = ⊕ i = 1 r L i . Let α , β ∈ V be two nonzero vectors.

We claim that:

1. If α , β ∈ V i for some i ∈ { 1 , … , r } , then ( α , α ) ( β , β ) ∈ Q ;

2. If α ∈ V i , β ∈ V j and i ≠ j , then ( α , α ) ( β , β ) ∉ Q .

where s k = ( α k , α k ) ( α ′ , α ′ ) ∈ Q . This proves claim ( 1 ) . If i ≠ j , then α ′ ≁ Q β ′ by the construction of the partition of { α 1 , … , α n } = ∪ i = 1 r X i . By claim ( 1 ) , we have α ∼ Q α ′ and β ∼ Q β ′ . Since ∼ Q is an equivalence relation, one has α ≁ Q β . This proves claim ( 2 )

By Corollary 4.1, for any nonzero vector w ∈ L , ℛ w ∈ O C ( L ) if and only if w ∈ L i for some i . Now choose a nonzero vector v ∈ L i and A ∈ O C ( L ) . By Corollary 3.1, there exists an index j such that A ( v ) ∈ V j . Since A is an isometry, one has ( v , v ) = ( A ( v ) , A ( v ) ) . By claim ( 2 ) , we have i = j . This shows that A ( V i ) = V i , A ( W i ) = W i and A ∣ W i is a coincidence isometry of the lattice L i considered in the subspace W i . It follows that

and A = A 1 A 2 ⋯ A r , where A i ∣ W i = A ∣ W i and A i ∣ W j = I for i ≠ j .

By Theorem 2.2, each A i is a product of at most ∣ X i ∣ reflections defined by the vectors in L i , and hence A is a product of at most ∑ i = 1 r ∣ X i ∣ = n reflections defined by the vectors in L . The proof is complete.□

Corollary 4.2

With the same notations in the above theorem. The subgroup of O C ( L ) generated by coincidence reflections is isomorphic to O C ( L ℛ ) .

Proof

It follows immediately from Theorems 4.2 and 2.2.□

Remark 4.1

Now we compare our main result and Theorem 2.2 by Zou. Let L be a lattice of rank n in R n . Set

1. ( α , α ) ( β , β ) ∈ Q for any nonzero vectors α , β ∈ L ;

2. Every nonzero vector of L defines a coincidence reflection of L ;

3. Any coincidence isometry of L is a product of at most n reflections defined by the vectors in L ;

4. L contains an orthogonal subset of order n .

and Theorem 4.2 states that

Example 4.1

Let L = Z α 1 ⊕ Z α 2 be a lattice in R 2 . Then one of the following is satisfied.

1. There are no reflections in O C ( L ) and O C ( L ) = { ± I } ≃ Z 2 ;

2. There are exactly two reflections ℛ v 1 , ℛ v 2 in O C ( L ) and O C ( L ) = { ± I , ℛ v 1 , ℛ v 2 } ≃ Z 2 2 ;

3. Each vector in L defines a coincidence reflection of L and O C ( L ) is generated by coincidence reflections.

Example 4.2

Let L = ⊕ i = 1 3 Z α i be a lattice in R 3 . Suppose that X is a maximal orthogonal subset of { v ∈ L : ℛ v ∈ O C ( L ) } . Then exactly one of the following is satisfied.

1. X = ∅ and O C ( L ) = { ± I } ≃ Z 2 ;

2. X = { v } and O C ( L ) = { ± I , ± ℛ v } ≃ Z 2 2 ;

3. X = { v 1 , v 2 , v 3 } and one of the following holds:

   1. Neither ( v 1 , v 1 ) ( v 2 , v 2 ) nor ( v 2 , v 2 ) ( v 3 , v 3 ) is rational, O C ( L ) = { ± I , ± ℛ v 1 , ± ℛ v 2 , ± ℛ v 3 } ≃ Z 2 3 ;

   2. ( v 1 , v 1 ) ( v 2 , v 2 ) ∈ Q and ( v 2 , v 2 ) ( v 3 , v 3 ) ∉ Q , O C ( L ) ≃ O C ( Z v 1 ⊕ Z v 2 ) ⊕ Z 2 , where every nonzero vector of L ′ = Z v 1 ⊕ Z v 2 defines a coincidence reflection of L ′ ;

   3. Both ( v 1 , v 1 ) ( v 2 , v 2 ) and ( v 2 , v 2 ) ( v 3 , v 3 ) are rational, and every nonzero vector of L defines a coincidence reflection of L .