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The finite dihedral rotation and the infinite unipotent rank-two product

Example

Let S={s,t}, m=m(s,t), V=RS, B the Coxeter form, P=Res+Ret and c as in The real Coxeter form, its radical, reflections, and form-preserving maps (c=1 for m=∞). By Reflections: involutivity, form invariance, fixed hyperplane, and exact rank-two order the product A:=rsrt acts on P with matrix A=(4c2−1−2c2c−1) in the basis (es,et) (Coordinate columns [v]B and matrices [T]BC of linear maps relative to ordered bases).

(i) Finite dihedral rotation. For m=3 one has c=cos⁡(π/3)=12 and A=(0−11−1),A2=(−11−10),A3=I2, so A has order 3=m; its trace is −1=2cos⁡(2π/3), consistent with a rotation through 2π/3 of the positive definite plane (P,B∣P), and A≠I2≠A2. For m=2 the same formula gives c=0 and A=−I2, of order 2=m.

(ii) Infinite unipotent product. For m=∞ one has c=1 and A=(3−22−1)=I2+N,N=(2−22−2)≠0,N2=0. Hence Ak=I2+kN for every k∈Z, so Ak(es)=es+2k(es+et) and no nonzero power of A is the identity: the product rsrt has infinite order, in contrast to the finite cases where A has order m. In the abstract group W of Coxeter matrices, the presented Coxeter group, reduced words, length, and standard parabolic subgroups, the element st likewise has infinite order when m=∞, and has order m in the displayed finite cases m=2,3.

Facts & Assumptions

Given: S={s,t} with s≠t, a Coxeter matrix value m=m(s,t)∈{2,3,… }∪{∞}, the space V=RS, the Coxeter form B, the plane P=Res+Ret and the number c of The real Coxeter form, its radical, reflections, and form-preserving maps (c=cos⁡(π/m) for finite m, and c=1 for m=∞).

[F1]

In the ordered basis (es,et) of P the product A=rsrt has matrix [A]=(4c2−1−2c2c−1) of determinant 1, and A acts on P by that matrix (Reflections: involutivity, form invariance, fixed hyperplane, and exact rank-two order, clause (3)(iii)); the same item's clause (3)(iv) records the order conclusions for finite m and the unipotent shape for m=∞, which the computations below verify directly.

[F2]

B(es,es)=B(et,et)=1 and B(es,et)=−c; rs and rt are B-preserving involutions (The real Coxeter form, its radical, reflections, and form-preserving maps, Reflections: involutivity, form invariance, fixed hyperplane, and exact rank-two order, clause (2)).

[F3]

Trigonometric facts: the addition formulas and the resulting triple-angle identity cos⁡3x=4cos⁡3x−3cos⁡x; sin⁡2x+cos⁡2x=1; cos⁡π=−1 and cos⁡(π/2)=0; sin⁡x=0 if and only if x∈πZ (The addition formulas for sine and cosine, Parity and the Pythagorean identity for sine and cosine, The zero sets of sine and cosine and the least positive common period 2 pi, Quarter-turn values and shifts by pi/2 and pi).

[F4]

Matrices of linear maps in an ordered basis, products of matrices and the identity matrix are as defined entrywise; [T2]=[T]2 for a linear endomorphism T and an ordered basis (S finite, V finite-dimensional) (Coordinate columns [v]B and matrices [T]BC of linear maps relative to ordered bases, [S∘T]BD=[S]CD[T]BC, Rectangular matrix multiplication and the identity matrix In, including zero-sized shapes).

[F5]

The group W is presented by s2=t2=1 and, when m<∞, (st)m=1. Any assignment of s,t to involutions in a group satisfying the finite relator extends to a homomorphism from W (Coxeter matrices, the presented Coxeter group, reduced words, length, and standard parabolic subgroups, Universal property).

Verification

1.1givenF1F2F3F4algebra

The case m=3. Here c=cos⁡(π/3). The triple-angle identity of [F3] at x=π/3 gives cos⁡π=4c3−3c; with cos⁡π=−1 this is 4c3−3c+1=0, that is (c+1)(2c−1)2=0. The factor c+1 is nonzero: cos⁡x=−1 forces sin⁡2x=1−cos⁡2x=0, hence sin⁡x=0 and x∈πZ, whereas π/3 is not an integer multiple of π. Hence 2c−1=0 and c=12. Substituting into [F1], [A]=(0−11−1),[A]2=(−11−10),[A]3=[A]2[A]=I2, and [A]≠I2≠[A]2, so A has order exactly 3=m. Its trace is −1, and 2cos⁡(2π/3)=2(2c2−1)=2(12−1)=−1, consistent with the rotation through 2π/3 of the positive definite plane: A preserves B∣P and has determinant 1.

1.2givenF1F3F4algebra

The case m=2. Here c=cos⁡(π/2)=0, so [F1] gives [A]=(−100−1)=−I2, and A2=idP while A≠idP; thus A has order 2=m, again a rotation through 2π/2=π of the positive definite plane.

1.3givenF1F2F4algebra

The case m=∞. Here c=1, so [F1] gives [A]=(3−22−1)=I2+N with N=(2−22−2)≠0, and direct multiplication gives N2=0. The binomial theorem in a ring with N2=0 gives (I2+N)k=I2+kN for every k≥0, and A−1=I2−N because (I2+N)(I2−N)=I2−N2=I2, so (I2−N)j=I2+(−j)N for j≥0 and Ak=I2+kNfor every k∈Z. Therefore Ak(es)=es+kNes=es+2k(es+et), and since kN has first entry 2k, which is nonzero for k≠0, no nonzero power of A is the identity: the product rsrt has infinite order in GL(V).

2.1givenF1F2F5step 1.1step 1.2step 1.3∎

Conclusion in the abstract group. By [F2], rs,rt are invertible involutions. If m<∞, [F1] gives (rsrt)m=idV; the reversed finite relator holds too, since rtrs=(rsrt)−1. If m=∞, there is no finite pair relator to check. Thus [F5] supplies a homomorphism ρ:W→GL(V) with ρ(s)=rs, ρ(t)=rt, and ρ(st)=A. For m=2,3, the relator gives (st)m=1, and steps 1.1–1.2 show that no smaller positive power can be 1: it would map to the corresponding nonidentity power of A. Hence st has order exactly m in these finite cases. For m=∞, if (st)k=1 for any nonzero integer k, applying ρ would give Ak=idV, contradicting step 1.3. Hence st has infinite order. This conclusion uses the homomorphism and the computed nonidentity powers, rather than inferring element order from the absence of a relator.

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