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The six exceptional Coxeter spectra: characteristic polynomials, orders and spectral exponents from exact matrices

Statement

Let $Γ$ be one of the six labelled diagrams $E_{6} : (0, 1), (1, 2), (2, 3), (3, 4), (2, 5), E_{7} : (0, 1), (1, 2), (2, 3), (3, 4), (4, 5), (2, 6),$ $E_{8} : (0, 1), (1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (2, 7), F_{4} : (0, 1), (1, 2), (2, 3),$ $H_{3} : (0, 1), (1, 2), H_{4} : (0, 1), (1, 2), (2, 3),$ all edges labelled $3$ except the single edges $(1, 2)$ of $F_{4}$ labelled $4$ and $(0, 1)$ of $H_{3}, H_{4}$ labelled $5$ ; let $(W, S)$ be the corresponding irreducible Coxeter system (Coxeter diagrams: edges, labels, components and finite type, Classification of finite Coxeter systems, including the H and dihedral families (1), Finiteness criterion: W is finite exactly when the Coxeter form is positive definite); let $V = R^{S}$ with $B (e_{s}, e_{t}) = - \cos (π / m_{s t})$ , $ρ$ the canonical reflection representation, and $c$ the bipartite Coxeter element of The bipartite Coxeter element, its ordered prefix roots, and the conditional vector map mu(a) = -2(c-1)^{-1}a with the colour classes of the tree $Γ$ and the recorded application orders $E_{6} : [0, 2, 4, 1, 3, 5], E_{7} : [0, 2, 4, 1, 3, 5, 6], E_{8} : [0, 2, 4, 6, 1, 3, 5, 7],$ $F_{4} : [0, 2, 1, 3], H_{3} : [0, 2, 1], H_{4} : [0, 2, 1, 3] .$ Then $ρ (c)$ has exact order $h$ and characteristic polynomial as follows, with spectral exponents $r_{1}, \dots, r_{n} \in {1, \dots, h - 1}$ defined by the eigenvalues $ζ_{h}^{r_{i}}$ :

$E_{6} : h = 12, \det (X - ρ (c)) = X^{6} + X^{5} - X^{3} + X + 1 = Φ_{3} Φ_{12}, {r_{i}} = {1, 4, 5, 7, 8, 11};$

$E_{7} : h = 18, \det (X - ρ (c)) = X^{7} + X^{6} - X^{4} - X^{3} + X + 1 = Φ_{2} Φ_{18}, {r_{i}} = {1, 5, 7, 9, 11, 13, 17};$

$E_{8} : h = 30, \det (X - ρ (c)) = X^{8} + X^{7} - X^{5} - X^{4} - X^{3} + X + 1 = Φ_{30}, {r_{i}} = {1, 7, 11, 13, 17, 19, 23, 29};$

$F_{4} : h = 12, \det (X - ρ (c)) = X^{4} - X^{2} + 1 = Φ_{12}, {r_{i}} = {1, 5, 7, 11};$

$H_{3} : h = 10, \det (X - ρ (c)) = X^{3} + (1 - φ) X^{2} + (1 - φ) X + 1 = (X + 1) (X^{2} - φ X + 1), {r_{i}} = {1, 5, 9};$

$H_{4} : h = 30, \det (X - ρ (c)) = X^{4} + (1 - φ) X^{3} + (1 - φ) X^{2} + (1 - φ) X + 1, {r_{i}} = {1, 11, 19, 29},$

where $φ = 2 \cos (π / 5)$ , $φ^{2} = φ + 1$ , embedded in $Q (ζ_{h})$ for $H_{3}, H_{4}$ by $φ = 1 + ζ_{h}^{h / 5} + ζ_{h}^{- h / 5}$ . In each case $\sum_{i} r_{i} = n h / 2 = ∣ Φ_{+} ∣$ and each polynomial divides $X^{h} - 1$ ; the matrices themselves (in the simple-root basis, with entries in $Z$ , $Z [\sqrt{2}]$ or $Z [φ]$ ) are recorded case by case in the proof, and the F4 characteristic polynomial is rational so no embedding of $\sqrt{2}$ into $Q (ζ_{12})$ is used.

Facts & Assumptions

Given: One of the six labelled diagrams of the statement with its Coxeter system $(W, S)$ , the space $V = R^{S}$ with Coxeter form $B$ , the canonical reflection representation $ρ$ , the bipartite Coxeter element $c$ and the recorded application orders, and the exact certificate file research/coxeter-scaffold/math-checks/finite-degree-poincare-certificates.json (generator finite-degree-poincare.py), whose records for the six types contain the matrices of the dual action and the same characteristic polynomials.

[F1]

$B (e_{s}, e_{s}) = 1$ and $B (e_{s}, e_{t}) = - \cos (π / m (s, t))$ for $s \neq t$ ; for $a$ with $B (a, a) \neq 0$ the reflection $r_{a} (v) = v - \frac{2 B (v, a)}{B (a, a)} a$ is linear, involutive, fixes $\ker B (-, a)$ pointwise and preserves $B$ ; the canonical homomorphism $ρ$ satisfies $ρ (s) = r_{e_{s}}$ and is faithful, every root has $B$ -norm one, and $ρ (w s w^{- 1}) = r_{ρ (w) e_{s}}$ (The real Coxeter form, its radical, reflections, and form-preserving maps (2),(3), Reflections: involutivity, form invariance, fixed hyperplane, and exact rank-two order (1)-(2), The canonical reflection homomorphism, roots, reflections, and the positive cone, Descent of the reflection representation, unit root norms, and conjugation of reflections (2)-(4), Complexifying a finite Coxeter reflection representation: faithfulness, complex reflections, and the hypotheses of the invariant-theory suppliers (1)).

[F2]

The Coxeter diagram of an irreducible finite type is a tree with bipartition $J ⊔ K$ ; the colour-class products $a, b$ , their product $c = a b$ and $h = ord (c)$ are well defined and listing the classes in the other order replaces $c$ by a conjugate element, so order and characteristic polynomial are unaffected; the Coxeter plane $P = span (u, v)$ is $ρ (c)$ -invariant with $ρ (c) ∣_{P}$ a rotation by $2 π / h$ ; the prefix roots $ρ_{i}$ enumerate $Φ_{+}$ with $∣ Φ_{+} ∣ = n h / 2$ , and $c - i d$ is invertible (The bipartite Coxeter element, its ordered prefix roots, and the conditional vector map mu(a) = -2(c-1)^{-1}a (1)-(4), The Coxeter plane, ordered-root enumeration, and invertibility of rho(c) - id (2)-(4)).

[F3]

The six diagrams of the statement are the standard diagrams $E_{6}, E_{7}, E_{8}, F_{4}, H_{3}, H_{4}$ of the classification, and for each of them $B$ is positive definite, so $W$ is finite (Classification of finite Coxeter systems, including the H and dihedral families (1),(3), Finiteness criterion: W is finite exactly when the Coxeter form is positive definite (1)-(2), Coxeter diagrams: edges, labels, components and finite type (1)-(2)).

[F4]

Trigonometric and square-root facts: $\cos (π / 4) = \sqrt{2} / 2$ (The finite Viete cosine product and its positive nested-radical factors); for all real $x$ , $\cos (2 x) = 2 \cos^{2} x - 1$ , $\sin (2 x) = 2 \sin x \cos x$ , $\cos (x + y) = \cos x \cos y - \sin x \sin y$ , $\cos (π - x) = - \cos x$ , $\cos (π / 2 - x) = \sin x$ , $\sin^{2} x + \cos^{2} x = 1$ (Double-angle and quadratic power-reduction identities, The addition formulas for sine and cosine, Cofunction, supplementary, quarter-turn, and reflection identities for the six trigonometric functions, Parity and the Pythagorean identity for sine and cosine); $\cos (π / 2) = 0$ and cosine is strictly decreasing on $[0, π]$ (Quarter-turn values and shifts by pi/2 and pi, Signs, monotonicity intervals, and ranges of sine and cosine); $2 \cos u \cos v = \cos (u + v) + \cos (u - v)$ and $\cos u - \cos v = - 2 \sin \frac{u + v}{2} \sin \frac{u - v}{2}$ (Product-to-sum and sum-to-product identities); every $a \geq 0$ has a unique $\sqrt{a} \geq 0$ with ${\sqrt{a}}^{2} = a$ (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ) and $0 \leq a < b$ implies $a^{2} < b^{2}$ (Squaring is monotone on the nonnegatives); sine and cosine are the power-series functions of Sine and cosine defined by their real power series.

[F5]

For $M \in M_{n} (C)$ : $χ_{M} (X) = \det (X I - M)$ is the characteristic polynomial, whose coefficients are the determinants of the expansion of $X I - M$ (For $A \in M_{n} (F)$ , the characteristic polynomial is $χ_{A} (x) = \det (x I_{n} - A)$ when $n \geq 1$ , with $χ_{A} (x) = 1$ for the unique $0 \times 0$ matrix, For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix); $tr (M)$ is the sum of the diagonal entries (The trace of a square matrix over a commutative ring); linearity follows termwise, and $tr (A B) = \sum_{i, j} a_{i j} b_{j i} = \sum_{j, i} b_{j i} a_{i j} = tr (B A)$ , which also gives cyclicity for longer products; $A adj (A) = adj (A) A = \det (A) I_{n}$ (For every positive-sized square matrix over a commutative ring, $A adj (A) = adj (A) A = \det (A) I$ , Deleted-row-and-column minors, cofactors, the cofactor matrix and the adjugate over a commutative ring); $\frac{d}{d t} \det (I - t M) = - tr (adj (I - t M) M)$ for the formal derivative ( $\frac{d}{d x} \det (I - x A) = - tr (adj (I - x A) A)$ ); the formal derivative of polynomials is linear, satisfies the product rule and has $\frac{d}{d t} t^{k} = k t^{k - 1}$ (The formal derivative of a polynomial, Linearity, power rule, Leibniz rule and the degree bound for the formal derivative). Determinants are invariant under similarity (Similar matrices over a commutative ring have the same determinant), so an eigenbasis computes the determinant factors of a diagonalisable operator. Over a field, a positive-sized square matrix is invertible exactly when its determinant is nonzero (A positive-sized square matrix over a commutative ring is invertible if and only if its determinant is a unit); hence $\det (λ I - M) = 0$ exactly when $λ$ is an eigenvalue.

[F6]

Cyclotomic and root-of-unity facts: the $Φ_{m} \in Z [t]$ satisfy $t^{m} - 1 = \prod_{d ∣ m} Φ_{d}$ and $Φ_{1} = t - 1$ (The cyclotomic polynomials $Φ_{n} \in Z [t]$ , defined by $\prod_{d ∣ n} Φ_{d} = t^{n} - 1$ ); over $C$ the roots of $Φ_{m}$ are exactly the primitive $m$ -th roots of unity and the $m$ -th roots of unity are the distinct numbers $e^{2 π i k / m}$ , $k = 0, \dots, m - 1$ (Over a field whose characteristic does not divide $n$ , the roots of $Φ_{n}$ are exactly the primitive roots of unity, The $n$ -th roots of a complex number and the $n$ distinct roots of unity for every $n \geq 1$ , The group $μ_{n} (K)$ of $n$ -th roots of unity in a field, and primitive $n$ -th roots of unity, The complex numbers as $R [x] / (x^{2} + 1)$ , with the real embedding and imaginary unit $i$ ); $e^{i θ} = \cos θ + i \sin θ$ (Euler's formula: $\exp (i θ) = \cos θ + i \sin θ$ for every real $θ$ ); a finite-order operator on a finite-dimensional complex vector space is diagonalisable (Over an algebraically closed field of characteristic $0$ , every element of finite order acts diagonalisably in a finite-dimensional representation), and the finite induction principle is The principle of mathematical induction with The natural numbers $N$ (von Neumann).

[F7]

Complexifying the real representation $ρ$ gives $ρ_{C}$ on $V_{C} = C \otimes_{R} V$ with the same operator matrices and characteristic polynomials in the basis $1 \otimes e_{s}$ ; the eigenvalues of a real matrix acting on $V_{C}$ are the roots of its characteristic polynomial (Complexifying a finite Coxeter reflection representation: faithfulness, complex reflections, and the hypotheses of the invariant-theory suppliers (1), For $A \in M_{n} (F)$ , the characteristic polynomial is $χ_{A} (x) = \det (x I_{n} - A)$ when $n \geq 1$ , with $χ_{A} (x) = 1$ for the unique $0 \times 0$ matrix).

Proof

1.1F4algebra

(The trigonometric constants behind the edge labels.) Put $c : = \cos (π / 3)$ . Since $0 < π / 3 < π / 2$ and cosine is strictly decreasing on $[0, π]$ with $\cos (π / 2) = 0$ , we have $c > 0$ ; the supplementary and double-angle identities give $- c = \cos (π - π / 3) = \cos (2 π / 3) = 2 c^{2} - 1$ , so $2 c^{2} + c - 1 = 0$ , that is $(2 c - 1) (c + 1) = 0$ , and $c + 1 > 0$ forces $c = 1 / 2$ . Next put $d : = \cos (π / 5)$ . Since $0 < π / 5 < π / 2$ we have $d > 0$ , and $\cos (π - 2 π / 5) = - \cos (2 π / 5)$ with $π - 2 π / 5 = 3 π / 5$ gives via the addition, double-angle and Pythagorean identities $\cos (3 π / 5) = \cos (2 π / 5) \cos (π / 5) - \sin (2 π / 5) \sin (π / 5) = (2 d^{2} - 1) d - 2 (1 - d^{2}) d = 4 d^{3} - 3 d,$ while $\cos (3 π / 5) = - \cos (2 π / 5) = - (2 d^{2} - 1)$ ; hence $4 d^{3} + 2 d^{2} - 3 d - 1 = 0$ , that is $(d + 1) (4 d^{2} - 2 d - 1) = 0$ , and $d + 1 > 0$ leaves $4 d^{2} - 2 d - 1 = 0$ . Completing the square, $(d - \frac{1}{4})^{2} = d^{2} - \frac{d}{2} + \frac{1}{16} = \frac{1}{4} + \frac{1}{16} = \frac{5}{16}$ , and $(\frac{\sqrt{5}}{4})^{2} = \frac{5}{16}$ with $\frac{\sqrt{5}}{4} \geq 0$ (here $\sqrt{5}$ is the unique nonnegative square root of $5$ and $\sqrt{5} > 0$ because $5 > 0$ is a nonzero square), so by uniqueness of nonnegative square roots $d - \frac{1}{4} = \pm \frac{\sqrt{5}}{4}$ and $d = \frac{1 \pm \sqrt{5}}{4}$ ; since $\sqrt{5} > 1$ by monotonicity of squaring on $[0, \infty)$ and $1 < 5$ , the root $\frac{1 - \sqrt{5}}{4}$ is negative while $d > 0$ , so $d = \frac{1 + \sqrt{5}}{4}$ . Define $φ : = \frac{1 + \sqrt{5}}{2} = 2 d = 2 \cos (π / 5)$ and $p : = 2 \cos (π / 4)$ ; then $\cos (π / 4) = \sqrt{2} / 2$ gives $p = \sqrt{2}$ and $p^{2} = 2$ , while $φ^{2} = \frac{6 + 2 \sqrt{5}}{4} = \frac{3 + \sqrt{5}}{2} = φ + 1$ ; finally the double-angle identity gives $\cos (2 π / 5) = 2 d^{2} - 1 = \frac{φ + 1}{2} - 1 = \frac{φ - 1}{2}$ , so $1 + 2 \cos (2 π / 5) = φ$ .

2.1F1F2F3F7step 1.1

(The six matrices.) For $t \in S$ let $R_{t}$ be the matrix of $ρ (s_{t}) = r_{e_{t}}$ in the basis $(e_{s})$ : $r_{e_{t}} (e_{j}) = e_{j} - 2 B (e_{j}, e_{t}) e_{t}$ , so the $j$ -th column of $R_{t}$ is $e_{j} - 2 B (e_{j}, e_{t}) e_{t}$ , and in particular the only nonzero off-diagonal entries of $R_{t}$ lie in row $t$ . Order the reflections as in the recorded application order $i_{1}, \dots, i_{n}$ and let $M : = R_{i_{n}} \dots R_{i_{1}}$ be the matrix obtained by applying $r_{i_{1}}$ first and then successively $r_{i_{2}}, \dots, r_{i_{n}}$ to a vector; equivalently $M$ is the matrix of $ρ (c)$ for $c : = s_{i_{n}} \dots s_{i_{1}}$ , the product of the two colour-class products in the order $K$ then $J$ (the classes commute internally), which is one of the two admissible orders of the bipartite definition and is conjugate to the element $a b$ of [F2] since $b a = a^{- 1} (a b) a$ . With $\cos (π / 3) = 1 / 2$ , $\cos (π / 4) = \sqrt{2} / 2$ , $\cos (π / 5) = φ / 2$ from 1.1, so that $B (e_{j}, e_{t})$ takes the values $- \frac{1}{2}$ , $- \frac{p}{2} = - \frac{\sqrt{2}}{2}$ (which we write with the symbol $p = \sqrt{2}$ ), $- \frac{φ}{2}$ , the resulting matrices are $E_{6} : (\begin{array}{cccccc} - 1 & 1 & 0 & 0 & 0 & 0 \\ - 1 & 1 & - 1 & 1 & 0 & 1 \\ 0 & 1 & - 1 & 1 & 0 & 1 \\ 0 & 1 & - 1 & 1 & - 1 & 1 \\ 0 & 0 & 0 & 1 & - 1 & 0 \\ 0 & 1 & - 1 & 1 & 0 & 0 \end{array}), E_{7} : (\begin{array}{ccccccc} - 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ - 1 & 1 & - 1 & 1 & 0 & 0 & 1 \\ 0 & 1 & - 1 & 1 & 0 & 0 & 1 \\ 0 & 1 & - 1 & 1 & - 1 & 1 & 1 \\ 0 & 0 & 0 & 1 & - 1 & 1 & 0 \\ 0 & 0 & 0 & 1 & - 1 & 0 & 0 \\ 0 & 1 & - 1 & 1 & 0 & 0 & 0 \end{array}),$ $E_{8} : (\begin{array}{cccccccc} - 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ - 1 & 1 & - 1 & 1 & 0 & 0 & 0 & 1 \\ 0 & 1 & - 1 & 1 & 0 & 0 & 0 & 1 \\ 0 & 1 & - 1 & 1 & - 1 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 & - 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & - 1 & 1 & - 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & - 1 & 0 \\ 0 & 1 & - 1 & 1 & 0 & 0 & 0 & 0 \end{array}), F_{4} : (\begin{array}{cccc} - 1 & 1 & 0 & 0 \\ - 1 & 2 & - p & p \\ 0 & p & - 1 & 1 \\ 0 & p & - 1 & 0 \end{array}),$ $H_{3} : (\begin{array}{ccc} - 1 & p & 0 \\ - p & 1 + p & - 1 \\ 0 & 1 & - 1 \end{array}), H_{4} : (\begin{array}{cccc} - 1 & p & 0 & 0 \\ - p & 1 + p & - 1 & 1 \\ 0 & 1 & - 1 & 1 \\ 0 & 1 & - 1 & 0 \end{array}),$ with $p = \sqrt{2}$ in the F4 matrix and $p = φ$ in the H3 and H4 matrices; the entries lie in $Z$ , $Z [\sqrt{2}]$ or $Z [φ]$ , and $M$ is obtained from these entries over $C$ through the embeddings $\sqrt{2} \mapsto \sqrt{2}$ , $φ \mapsto φ$ . Since each $R_{t}$ preserves $B$ and is an involution [F1], each $R_{t}$ satisfies $R_{t}^{T} G R_{t} = G$ for the Gram matrix $G = (B (e_{s}, e_{t}))$ , and so does the product $M$ ; in particular $M$ has finite order because $W$ is finite [F3] and $ρ$ is a homomorphism, and $M$ is the matrix of a Coxeter element of the type.

3.1F5step 2.1

(Newton's identities from the cofactor identity.) Let $M \in M_{n} (C)$ be any of the six matrices of step 2.1 and write the expansion $χ (t) : = \det (I - t M) = \sum_{k = 0}^{n} (- 1)^{k} e_{k} t^{k}$ with $e_{0} : = 1$ , so that $\det (X I - M) = \sum_{k} (- 1)^{k} e_{k} X^{n - k}$ . Put $A (t) : = adj (I - t M)$ and $p_{i} : = tr (M^{i})$ for $i \geq 1$ . Then for every $1 \leq k \leq n$ , $k e_{k} = \sum_{i = 1}^{k} (- 1)^{i - 1} e_{k - i} p_{i} .$ Indeed the entries of $A (t)$ are cofactors of $I - t M$ , hence determinants of $(n - 1) \times (n - 1)$ matrices whose entries are affine in $t$ , so $A (t) = \sum_{k = 0}^{n - 1} C_{k} t^{k}$ with $C_{k} \in M_{n} (C)$ ; the adjugate identity $(I - t M) A (t) = χ (t) I_{n}$ gives, on comparing coefficients of $t^{k}$ for $0 \leq k \leq n$ , the recursion $C_{0} = I_{n}$ and $C_{k} = M C_{k - 1} + (- 1)^{k} e_{k} I_{n}$ for $1 \leq k \leq n - 1$ (the coefficient of $t^{n}$ only records $0 = C_{n} = M C_{n - 1} + (- 1)^{n} e_{n} I_{n}$ and is not used); induction on $k$ gives $C_{k} = \sum_{j = 0}^{k} (- 1)^{j} e_{j} M^{k - j}$ for $0 \leq k \leq n - 1$ . Differentiating the adjugate identity gives $χ^{'} (t) = - tr (A (t) M)$ by [F5]; the left side is $\sum_{k = 1}^{n} k (- 1)^{k} e_{k} t^{k - 1}$ and the right side $- \sum_{k = 0}^{n - 1} tr (C_{k} M) t^{k}$ , so comparing coefficients of $t^{k}$ for $0 \leq k \leq n - 1$ gives $(k + 1) (- 1)^{k + 1} e_{k + 1} = - tr (C_{k} M)$ . Taking traces in the displayed expansion of $C_{k}$ and using linearity, cyclicity and $p_{i} = tr (M^{i})$ gives $tr (C_{k} M) = \sum_{j = 0}^{k} (- 1)^{j} e_{j} p_{k + 1 - j}$ ; substituting and multiplying by $(- 1)^{k + 1}$ yields $(k + 1) e_{k + 1} = \sum_{i = 1}^{k + 1} (- 1)^{i - 1} e_{k + 1 - i} p_{i}$ , which is the displayed identity with $k$ replaced by $k + 1$ .

4.1step 2.1step 3.1algebra

(Traces and characteristic polynomials of the six matrices.) Computing the powers $M^{i}$ by exact matrix multiplication of the matrices of step 2.1 gives the following power sums, with $p = \sqrt{2}$ in the F4 case and $p = φ$ in the H3, H4 cases: $E_{6} : (p_{1}, \dots, p_{6}) = (- 1, 1, 2, - 3, - 1, - 2); E_{7} : (p_{1}, \dots, p_{7}) = (- 1, 1, 2, 1, - 1, - 2, - 1);$ $E_{8} : (p_{1}, \dots, p_{8}) = (- 1, 1, 2, 1, 4, - 2, - 1, 1); F_{4} : (p_{1}, \dots, p_{4}) = (0, 2, 0, - 2);$ $H_{3} : (p_{1}, p_{2}, p_{3}) = (φ - 1, φ, - φ); H_{4} : (p_{1}, \dots, p_{4}) = (φ - 1, φ, 2 φ, 1 - φ) .$ (For instance the E6 matrix has diagonal entries $- 1, 1, - 1, 1, - 1, 0$ , whose sum is $- 1 = p_{1}$ ; the higher power sums are the same finite computation.) Applying the Newton identities of step 3.1 successively for $k = 1, 2, \dots, n$ determines the coefficients $e_{k}$ in the order $k = 1, 2, \dots$ alone, and gives: $E_{6} : (e_{1}, \dots, e_{6}) = (- 1, 0, 1, 0, - 1, 1), \det (X I - M) = X^{6} + X^{5} - X^{3} + X + 1;$ $E_{7} : (e_{1}, \dots, e_{7}) = (- 1, 0, 1, - 1, 0, 1, - 1), \det (X I - M) = X^{7} + X^{6} - X^{4} - X^{3} + X + 1;$ $E_{8} : (e_{1}, \dots, e_{8}) = (- 1, 0, 1, - 1, 1, 0, - 1, 1), \det (X I - M) = X^{8} + X^{7} - X^{5} - X^{4} - X^{3} + X + 1;$ $F_{4} : (e_{1}, \dots, e_{4}) = (0, - 1, 0, 1), \det (X I - M) = X^{4} - X^{2} + 1;$ $H_{3} : (e_{1}, e_{2}, e_{3}) = (φ - 1, 1 - φ, - 1), \det (X I - M) = X^{3} + (1 - φ) X^{2} + (1 - φ) X + 1;$ $H_{4} : (e_{1}, \dots, e_{4}) = (φ - 1, 1 - φ, φ - 1, 1), \det (X I - M) = X^{4} + (1 - φ) X^{3} + (1 - φ) X^{2} + (1 - φ) X + 1,$ where each coefficient is obtained from the displayed Newton identity by the finite exact arithmetic in $Z$ , $Z [\sqrt{2}]$ or $Z [φ]$ , using $φ^{2} = φ + 1$ for the H3 and H4 cases.

5.1F2F6F7step 2.1step 4.1algebra

(Cyclotomic factorisations, E6, E7, E8, F4.) The computable cyclotomic polynomials are $Φ_{2} = X + 1$ , $Φ_{3} = X^{2} + X + 1$ , $Φ_{12} = X^{4} - X^{2} + 1$ , $Φ_{18} = X^{6} - X^{3} + 1$ and $Φ_{30} = X^{8} + X^{7} - X^{5} - X^{4} - X^{3} + X + 1$ ; expanding the products coefficientwise gives $Φ_{3} Φ_{12} = X^{6} + X^{5} - X^{3} + X + 1$ , $Φ_{2} Φ_{18} = X^{7} + X^{6} - X^{4} - X^{3} + X + 1$ and $Φ_{12} = X^{4} - X^{2} + 1$ , so by step 4.1 the E6, E7, E8 and F4 characteristic polynomials equal respectively $Φ_{3} Φ_{12}$ , $Φ_{2} Φ_{18}$ , $Φ_{30}$ and $Φ_{12}$ . Since $M$ is the matrix of the finite-order operator $ρ_{C} (c)$ [F7], $M$ is diagonalisable and its eigenvalue multiset is the root multiset of its characteristic polynomial; by [F6] the roots of $Φ_{m}$ are the primitive $m$ -th roots of unity, namely $ζ_{12}^{r}$ with $r \in {1, 5, 7, 11}$ for $Φ_{12}$ , $ζ_{12}^{4}, ζ_{12}^{8}$ for $Φ_{3}$ , $ζ_{18}^{9} = - 1$ for $Φ_{2}$ , $ζ_{18}^{r}$ with $r \in {1, 5, 7, 11, 13, 17}$ for $Φ_{18}$ , and $ζ_{30}^{r}$ with $r \in {1, 7, 11, 13, 17, 19, 23, 29}$ for $Φ_{30}$ . Hence the eigenvalue multisets are ${ζ_{12}^{r} : r \in {1, 4, 5, 7, 8, 11}}$ for E6, ${ζ_{18}^{r} : r \in {1, 5, 7, 9, 11, 13, 17}}$ for E7, ${ζ_{30}^{r} : r \in {1, 7, 11, 13, 17, 19, 23, 29}}$ for E8 and ${ζ_{12}^{r} : r \in {1, 5, 7, 11}}$ for F4; in particular each characteristic polynomial divides the corresponding $X^{h} - 1$ (because $Φ_{m} ∣ X^{m} - 1 ∣ X^{h} - 1$ when $m ∣ h$ ), and in each case $ζ_{h}$ is a root, so the order of $M$ is a multiple of $h$ ; since every eigenvalue is an $h$ -th root of unity and $M$ is diagonalisable, $M^{h} = I$ , so the order of $M$ is exactly the displayed integer $h$ . Faithfulness from [F7] implies $ρ_{C} (c)^{k} = I$ if and only if $c^{k} = 1$ , hence this is also the group order of $c$ used in [F2].

5.2F2F4F6F7step 4.1algebra

(H3 and H4.) For H3 the right side of step 4.1 expands as $(X + 1) (X^{2} - φ X + 1) = X^{3} + (1 - φ) X^{2} + (1 - φ) X + 1$ ; by [F6] the roots of $X^{2} - φ X + 1$ are the two complex numbers with sum $φ$ and product $1$ , and $e^{i π / 5}$ and $e^{- i π / 5}$ have sum $2 \cos (π / 5) = φ$ and product $1$ , so they are the roots; with $ζ_{10} = e^{2 π i / 10} = e^{i π / 5}$ and $- 1 = ζ_{10}^{5}$ the eigenvalue multiset of $M$ is ${ζ_{10}^{1}, ζ_{10}^{5}, ζ_{10}^{9}}$ . For H4 the four numbers $ζ_{30}^{1}, ζ_{30}^{11}, ζ_{30}^{19}, ζ_{30}^{29}$ lie in inverse pairs, so the monic quartic with these roots is $(X^{2} - a X + 1) (X^{2} - b X + 1) = X^{4} - (a + b) X^{3} + (a b + 2) X^{2} - (a + b) X + 1$ with $a = ζ_{30} + ζ_{30}^{- 1} = 2 \cos (π / 15)$ and $b = ζ_{30}^{11} + ζ_{30}^{- 11} = 2 \cos (11 π / 15) = - 2 \cos (4 π / 15)$ ; then $a b = - 4 \cos (π / 15) \cos (4 π / 15) = - 2 (\cos (π / 3) + \cos (- π / 5)) = - (1 + φ) = - φ^{2}$ and $a + b = 2 (\cos (π / 15) - \cos (4 π / 15)) = 2 \cdot 2 \sin (π / 6) \sin (π / 10) = 2 \sin (π / 10) = 2 \cos (2 π / 5) = φ - 1$ , using the product-to-sum and sum-to-product identities, $\cos (- π / 5) = \cos (π / 5) = φ / 2$ , $\cos (π / 3) = 1 / 2$ , $\sin (π / 6) = \cos (π / 3) = 1 / 2$ and $\sin (π / 10) = \cos (2 π / 5)$ ; hence the monic quartic with roots $ζ_{30}^{\pm 1}, ζ_{30}^{\pm 11}$ equals $X^{4} - (φ - 1) X^{3} + (- φ^{2} + 2) X^{2} - (φ - 1) X + 1 = X^{4} + (1 - φ) X^{3} + (1 - φ) X^{2} + (1 - φ) X + 1$ , because $- φ^{2} + 2 = 1 - φ$ and $- (φ - 1) = 1 - φ$ ; this is exactly the H4 polynomial computed in step 4.1, so the eigenvalue multiset of $M$ is ${ζ_{30}^{1}, ζ_{30}^{11}, ζ_{30}^{19}, ζ_{30}^{29}}$ , and the relation $φ = 1 + ζ_{30}^{6} + ζ_{30}^{- 6} = 1 + 2 \cos (2 π / 5)$ is the embedding recorded in the statement (for H3 the corresponding relation is $φ = 1 + ζ_{10}^{2} + ζ_{10}^{- 2}$ ). In both cases all eigenvalues are $h$ -th roots of unity, $M$ is diagonalisable and $ζ_{h}$ occurs among them, so the order of $M$ is exactly the displayed integer $h$ (the same argument as in 5.1); faithfulness from [F7] again identifies it with $ord (c)$ . Finally the displayed residue lists have sums $36, 63, 120, 24, 15, 60$ for $E_{6}, E_{7}, E_{8}, F_{4}, H_{3}, H_{4}$ , which are exactly $n h / 2$ and hence equal $∣ Φ_{+} ∣$ by [F2].

6.1F5step 1.1step 2.1step 3.1step 4.1step 5.1step 5.2∎

(Summary and conventions.) Steps 1.1-5.2 prove all assertions of the statement: for each of the six types the displayed matrix is the matrix of a Coxeter element of that type in the simple-root basis over $Z$ , $Z [\sqrt{2}]$ or $Z [φ]$ , its characteristic polynomial is computed from the exact power sums by Newton's identities and equals the displayed polynomial, the cyclotomic and quadratic factorisations identify the eigenvalue multisets with the displayed residue lists, each polynomial divides $X^{h} - 1$ and has $ζ_{h}$ as a root, the orders are exactly $h$ , and the residue sums equal $n h / 2 = ∣ Φ_{+} ∣$ . The computation uses no floating-point approximation, no enumeration of a Coxeter group and no invariant-degree table; the F4 characteristic polynomial is rational, so no embedding of $\sqrt{2}$ into $Q (ζ_{12})$ is used; the element whose matrix is displayed is the product of the colour-class products in the recorded order, conjugate to the element $a b$ of the bipartite definition [F2], so the conclusions hold for $c$ as well; and the dual-action matrices in the local certificate research/coxeter-scaffold/math-checks/finite-degree-poincare-certificates.json are $M^{- T}$ , not $M^{T}$ . Indeed the dual of each involutive reflection has matrix $R_{t}^{T}$ , and the same application order gives $M^{- T}$ . From $M^{T} G M = G$ we obtain $M^{- T} = G M G^{- 1}$ ; hence its characteristic polynomial agrees with that of $M$ .

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Sources

Bill Casselman, Essays on Coxeter groups: Coxeter elements in finite Coxeter groups (author-hosted PDF, 12 pages) (standard reference, not scraped)
Vivien Ripoll (Strobl seminar notes, joint with Reiner and Stump), Coxeter elements in well-generated reflection groups (57-page PDF) (standard reference, not scraped)