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An indefinite Coxeter form: infinite, but not of affine type

Example

Let $S = {s_{1}, s_{2}, s_{3}}$ with $m (s_{1}, s_{2}) = m (s_{2}, s_{3}) = 3$ and $m (s_{1}, s_{3}) = 5$ , so that $Γ$ is the triangle with labels $(3, 3, 5)$ , and let $B$ be the cosine matrix $B = (\begin{array}{ccc} 1 & - \frac{1}{2} & - \cos \frac{π}{5} \\ - \frac{1}{2} & 1 & - \frac{1}{2} \\ - \cos \frac{π}{5} & - \frac{1}{2} & 1 \end{array}), \cos \frac{π}{5} = \frac{1 + \sqrt{5}}{4} .$ Then:

(i) $B$ is indefinite: the vector $v = e_{s_{1}} + e_{s_{2}} + e_{s_{3}}$ satisfies $B (v, v) = 3 + 2 (- \frac{1}{2} - \frac{1}{2} - \cos \frac{π}{5}) = 1 - 2 \cos \frac{π}{5} = \frac{1 - \sqrt{5}}{2} < 0$ because $\cos \frac{π}{5} = \frac{1 + \sqrt{5}}{4}$ and $\sqrt{5} > 1$ (the value of $\cos \frac{π}{5}$ is derived in the verification, not cited). Since $B (e_{s_{1}}, e_{s_{1}}) = 1 > 0$ , $B$ has both positive and negative values. Its leading principal minors are $1$ , $\frac{3}{4}$ , and $\det B = \frac{1}{2} - \frac{1}{2} \cos \frac{π}{5} - \cos^{2} \frac{π}{5} = - \frac{\sqrt{5}}{4} < 0$ .

(ii) $W$ is infinite, but $(W, S)$ is not of affine form type: affine form type requires $B$ positive semidefinite of corank one (Irreducible affine Coxeter type: the corank-one form, the radical quotient, and the affine slice (1)), and by (i) $B$ is negative on some vector. It is infinite by the finite-type criterion (Finiteness criterion: W is finite exactly when the Coxeter form is positive definite (1)): an indefinite form is not positive definite.

(iii) The angle sum of the triangle is $1 / 3 + 1 / 3 + 1 / 5 = 13 / 15 < 1$ , and the form is indefinite and nondegenerate: $B (v, v) < 0$ by (i) while $B (e_{s_{1}}, e_{s_{1}}) = 1 > 0$ , and $\det B = - \frac{\sqrt{5}}{4} \neq 0$ ; every nonempty proper principal submatrix is positive definite by Reflections: involutivity, form invariance, fixed hyperplane, and exact rank-two order (3)(i), whose rank-two computations give determinants $\sin^{2} (π / 3) = \frac{3}{4} > 0$ and $\sin^{2} (π / 5) = 1 - \cos^{2} (π / 5) = \frac{5 - \sqrt{5}}{8} > 0$ ; the empty principal matrix is vacuously positive definite. No geometric hyperbolic realization is constructed on this page; only the algebraic definiteness type is asserted. For this connected system, the positive-definite, affine-form, and indefinite cases are mutually exclusive: the first is finite, the second is positive semidefinite of corank one, and this example is the infinite indefinite case (Finiteness criterion: W is finite exactly when the Coxeter form is positive definite (1), Irreducible affine Coxeter type: the corank-one form, the radical quotient, and the affine slice (1), Positive radical, corank one, positive definiteness of proper submatrices, and domination exclusions (1)).

(iv) By contrast the triangle with labels $(3, 3, 3)$ is ${\tilde{A}}_{2}$ , positive semidefinite of corank one (The standard affine diagrams A-tilde, B-tilde, C-tilde, D-tilde, E-tilde, F-tilde and G-tilde (1), Crystallographic alcove diagrams: the affine list realized by Weyl types A–G (3)). For any triangle Coxeter matrix with finite labels $m_{12}, m_{23}, m_{13} \geq 3$ , the cosine-form determinant is zero exactly when $π / m_{12} + π / m_{23} + π / m_{13} = π$ , and is negative when the sum is below $π$ , as calculated in Proof Step 2.2. In particular, $(3, 3, 4)$ is indefinite: for $v = e_{s_{1}} + e_{s_{2}} + e_{s_{3}}$ , $B (v, v) = 1 - 2 \cos (π / 4) = 1 - \sqrt{2} < 0$ , where $\cos (π / 4) = \sqrt{2} / 2$ follows from the half-angle identity at $π / 2$ (Half-angle identities with the sign determined by the quadrant).

(v) Consequently, in the classification statement of Classification of affine Coxeter diagrams and their Euclidean simplex realization (1) the hypothesis "positive semidefinite" cannot be replaced by "infinite", and a consumer testing a diagram for affine type must examine the definiteness of the whole form and not only the finiteness of proper subdiagrams: $Γ$ here has all proper subdiagrams of finite type (the rank-two subdiagrams $I_{2} (3)$ , $I_{2} (3)$ , $I_{2} (5)$ are all finite), yet is not affine.

Facts & Assumptions

Given: The finite labelled triangle $(3, 3, 5)$ and its form.

[F1]

The cosine form is a symmetric bilinear form with diagonal one and off-diagonal $- \cos (π / m)$ (The real Coxeter form, its radical, reflections, and form-preserving maps).

[F2]

Affine form type means connected and positive semidefinite of corank one (Irreducible affine Coxeter type: the corank-one form, the radical quotient, and the affine slice (1)).

[F3]

A Coxeter diagram joins distinct vertices exactly when their label is at least $3$ ; thus this labelled triangle is connected (Coxeter diagrams: edges, labels, components and finite type).

[F4]

A finite-rank Coxeter group is finite exactly when its Coxeter form is positive definite (Finiteness criterion: W is finite exactly when the Coxeter form is positive definite (1)).

[F5]

Every standard parabolic has the restricted Coxeter presentation; for an empty generator set it is the trivial group (Support, intrinsic parabolic presentations, minimal coset representatives and length additivity, with the type-A identification (2)).

[F6]

Fifth roots of unity have the form $e^{2 π i k / 5}$ , Euler's identity is $e^{i x} = \cos x + i \sin x$ , and the double-angle identities hold (The $n$ -th roots of a complex number and the $n$ distinct roots of unity for every $n \geq 1$ , Euler's formula: $\exp (i θ) = \cos θ + i \sin θ$ for every real $θ$ , Double-angle and quadratic power-reduction identities).

[F7]

The defining power series give cosine even, sine odd and $\cos 0 = 1$ (Sine and cosine defined by their real power series).

[F8]

Cosine strictly decreases on $[0, π]$ , $\cos (π / 2) = 0$ , $\cos π = - 1$ , $\sin π = 0$ , and sine is positive on $(0, π)$ (Signs, monotonicity intervals, and ranges of sine and cosine, Quarter-turn values and shifts by pi/2 and pi).

[F9]

Nonnegative square roots exist uniquely and squaring is strictly increasing on nonnegative reals (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ , Squaring is monotone on the nonnegatives).

[F10]

The all- $3$ triangle is ${\tilde{A}}_{2}$ and has a positive-semidefinite corank-one form (The standard affine diagrams A-tilde, B-tilde, C-tilde, D-tilde, E-tilde, F-tilde and G-tilde (1), Crystallographic alcove diagrams: the affine list realized by Weyl types A–G (3)).

[F11]

Sine and cosine satisfy their addition formulas (The addition formulas for sine and cosine).

[F12]

The half-angle identity with its sign determined by the quadrant gives $\cos (π / 4) = \sqrt{2} / 2$ (Half-angle identities with the sign determined by the quadrant).

[F13]

For two distinct generators with finite label $m$ , their rank-two form is positive definite with determinant $\sin^{2} (π / m)$ (Reflections: involutivity, form invariance, fixed hyperplane, and exact rank-two order (3)(i)).

[F14]

For a connected Coxeter diagram with nonpositive off-diagonal entries, a positive-semidefinite Coxeter form with nonzero radical has corank one and a positive radical vector (Positive radical, corank one, positive definiteness of proper submatrices, and domination exclusions (1)).

[F15]

The connected positive-semidefinite corank-one Coxeter diagrams are exactly the standard affine diagrams (Classification of affine Coxeter diagrams and their Euclidean simplex realization (1)).

[F16]

The standard angle constant satisfies $π > 0$ (Pi as twice the smallest positive zero of cosine).

Proof

technique · direct; all finite coordinate constructions use no choice principle

1.1F6F7F8F9F11F16algebra

Put $ζ = e^{2 π i / 5}$ . Since $ζ^{5} = 1$ and $ζ \neq 1$ , multiplication by $ζ - 1$ proves $1 + ζ + ζ^{2} + ζ^{3} + ζ^{4} = 0$ . Divide by $ζ^{2}$ and put $y = ζ + ζ^{- 1}$ to get $y^{2} + y - 1 = 0$ . Euler's identity and parity [F6,F7] give $y = 2 \cos (2 π / 5) > 0$ , as $0 < 2 π / 5 < π / 2$ and cosine is strictly decreasing to $\cos (π / 2) = 0$ [F8]. Hence $(2 y + 1)^{2} = 5$ , so $y = (\sqrt{5} - 1) / 2$ by positivity and uniqueness of square roots [F9]. Double angle [F6] gives $c^{2} : = \cos^{2} (π / 5) = (3 + \sqrt{5}) / 8$ ; $c > 0$ by [F8] and $(1 + \sqrt{5})^{2} / 16 = (3 + \sqrt{5}) / 8$ , so $c = (1 + \sqrt{5}) / 4$ . For $d = \cos (π / 3) > 0$ , the double-angle identity gives $2 d^{2} - 1 = \cos (2 π / 3) = - d$ , since the addition formula, parity, $\cos π = - 1$ and $\sin π = 0$ give $\cos (π - x) = - \cos x$ [F7,F8,F11]; thus $(2 d - 1) (d + 1) = 0$ and $d = 1 / 2$ . Finally, for $h = \cos (π / 4) > 0$ , the double-angle identity gives $2 h^{2} - 1 = \cos (π / 2) = 0$ , so $h = 1 / \sqrt{2} = \sqrt{2} / 2$ by [F9].

2.1F1F2F4F5F9F13step 1.1algebra

On $v = (1, 1, 1)$ the form has value $3 - 2 (1 / 2 + 1 / 2 + c) = 1 - 2 c = (1 - \sqrt{5}) / 2 < 0$ , whereas on $e_{s_{1}}$ it has value one. It is therefore indefinite. Expansion of the determinant gives $1 - 1 / 4 - 1 / 4 - c^{2} - 2 (1 / 2) (1 / 2) c = 1 / 2 - c / 2 - c^{2} = - \sqrt{5} / 4 \neq 0$ . The empty principal matrix is vacuously positive definite; a one-coordinate principal form is $[1]$ ; and a two-coordinate form is $(a - d b)^{2} + (1 - d^{2}) b^{2}$ for $d = 1 / 2$ or $c$ , with positive determinant $3 / 4$ or $(5 - \sqrt{5}) / 8$ since $1 < \sqrt{5} < 3 < 5$ . Thus every proper principal submatrix is positive definite. For $T = \emptyset$ , $W_{T} = {1}$ ; for nonempty proper $T$ , the restricted presentation [F5] and finite-type criterion [F4] make $W_{T}$ finite. The full group is infinite by [F4], and it is not affine by [F2]. The numerical angle sum is $1 / 3 + 1 / 3 + 1 / 5 = 13 / 15 < 1$ ; no hyperbolic realization is needed for these algebraic conclusions.

2.2F1F7F8F10F11F12F13F16step 1.1algebra

The all- $3$ triangle is affine by [F10]. For $(3, 3, 4)$ , the same evaluation on $(1, 1, 1)$ is $1 - \sqrt{2} < 0$ , using [F12] (the value was also computed in Step 1.1); thus it is not positive semidefinite. For any triangle Coxeter matrix with finite labels $m_{12}, m_{23}, m_{13} \geq 3$ , put $θ_{1} = π / m_{12}$ , $θ_{2} = π / m_{23}$ , $θ_{3} = π / m_{13}$ and $A = \cos θ_{1}$ , $B = \cos θ_{2}$ , $C = \cos θ_{3}$ . Its cosine-form determinant is $1 - A^{2} - B^{2} - C^{2} - 2 A B C = (\sin θ_{1} \sin θ_{2})^{2} - (C + A B)^{2}$ , using [F13] for $\sin^{2} θ_{i} = 1 - \cos^{2} θ_{i}$ . Since $0 < θ_{i} \leq π / 3$ , all three cosines are at least $1 / 2 > 0$ and all three sines are positive [F8]; hence the second factor $\sin θ_{1} \sin θ_{2} + C + A B$ is positive. The first factor is $\sin θ_{1} \sin θ_{2} - C - A B = - \cos (θ_{1} + θ_{2}) - \cos θ_{3} = \cos (π - θ_{1} - θ_{2}) - \cos θ_{3}$ by the addition formulas and angle-shift values [F7,F8,F11]. Strict decrease [F8] shows the determinant is zero exactly when $θ_{1} + θ_{2} + θ_{3} = π$ , and negative when the sum is below $π$ . The sum is at most $π$ , with equality only for $(3, 3, 3)$ . If it is below $π$ , at least one $θ_{i} < π / 3$ , so its cosine exceeds $1 / 2$ while the others are at least $1 / 2$ ; therefore the sum-vector has value $3 - 2 (A + B + C) < 0$ , proving indefiniteness. This proves the asserted angle-sum boundary without constructing a hyperbolic triangle.

3.1F2F3F4F14F15step 2.1algebra∎

The Coxeter diagram is connected by [F3]. For this system, the positive-definite case is finite by [F4]. If $B$ is positive semidefinite but not positive definite, choose $0 \neq x$ with $B (x, x) = 0$ . For any $y \in V$ , positive semidefiniteness gives $0 \leq B (x + t y, x + t y) = 2 t B (x, y) + t^{2} B (y, y)$ for every real $t$ . If $B (x, y) \neq 0$ , then either $B (y, y) = 0$ and a $t$ of opposite sign makes the expression negative, or $B (y, y) > 0$ and a sufficiently small $t$ of opposite sign does so. Thus $B (x, y) = 0$ for all $y$ , so $x \in rad (B) ∖ {0}$ ; [F14] gives corank one, and the system is affine by [F2]. Otherwise the form is indefinite and the group is infinite by [F4], but not affine by [F2]. These cases are mutually exclusive. By Step 2.1, the $(3, 3, 5)$ system is in the last case and all its proper parabolics are finite. Thus neither infiniteness nor finiteness of proper subdiagrams can replace positive semidefiniteness in [F15]'s classification. The whole-form calculation, rather than only rank-two tests, is essential. All calculations are explicit and use no Choice.

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Sources

M. W. Davis and G. Moussong, Notes on nonpositively curved polyhedra (Turan Workshop lecture notes, 1998/1999; 65 PDF pages) (standard reference, not scraped)
M. W. Davis, The Geometry and Topology of Coxeter Groups (first-edition author manuscript, Princeton University Press, 2008; 600 PDF pages) (standard reference, not scraped)