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$I_{2} (5)$ admits no crystallographic scaling and no reduced crystallographic root system with that base pairing

Statement refuted

(a) Every finite Coxeter matrix $(S, m)$ admits a crystallographic scaling (Crystallographic scalings: scaled simple roots, coroots and the root, coroot and weight lattices); in particular the rank-two geometry with $S = {s, t}$ and $m (s, t) = 5$ does.

(b) There is a reduced crystallographic Euclidean root system $Ψ$ with a base ${α, β}$ whose simple roots satisfy $\frac{(α, β)}{∣ α ∣ ∣ β ∣} = - \cos \frac{π}{5},$ the normalized pairing of the two basis vectors of the $I_{2} (5)$ Coxeter form.

Facts & Assumptions

Given: the rank-two Coxeter matrix on $S = {s, t}$ with $m (s, t) = 5$ , the space $V = R^{S}$ with basis $e_{s}, e_{t}$ and the Coxeter form $B$ , a scaling $c = (c_{s}, c_{t})$ with scaled simple roots $a_{s}, a_{t}$ and Cartan numbers $a_{s t}$ , and, in the second refutation, a reduced crystallographic Euclidean root system $Ψ$ with a base ${α, β}$ .

[F1]

$m (s, s) = 1$ , while $m (s, t) = m (t, s) \in {2, 3, \dots} \cup {\infty}$ for $s \neq t$ ; in particular $m (s, t) = 5$ is finite (Coxeter matrices, the presented Coxeter group, reduced words, length, and standard parabolic subgroups).

[F2]

$B$ is the unique symmetric bilinear form on $V$ with $B (e_{s}, e_{s}) = 1$ , $B (e_{t}, e_{t}) = 1$ and $B (e_{s}, e_{t}) = - \cos (π / 5)$ (The real Coxeter form, its radical, reflections, and form-preserving maps).

[F3]

For distinct $s, t$ with finite $m = m (s, t)$ and $c : = \cos (π / m)$ , the plane $P = R e_{s} + R e_{t}$ has $B (x_{s} e_{s} + x_{t} e_{t}, x_{s} e_{s} + x_{t} e_{t}) = (x_{s} - c x_{t})^{2} + \sin^{2} (π / m) x_{t}^{2}$ , so $B ∣_{P}$ is positive definite (Reflections: involutivity, form invariance, fixed hyperplane, and exact rank-two order (3)(i)).

[F4]

$W$ is finite if and only if $B$ is positive definite (Finiteness criterion: W is finite exactly when the Coxeter form is positive definite (1)).

[F5]

$I_{2} (m)$ with $3 \leq m < \infty$ is the diagram of two vertices joined by one edge labelled $m$ , $H_{2} = I_{2} (5)$ , and $I_{2} (m)$ with $m \notin {2, 3, 4, 6}$ admits no crystallographic scaling (Classification of finite Coxeter systems, including the H and dihedral families (1), (4); Crystallographic finite type: the Weyl types, reduced realizations and lattice stability (1)).

[F6]

The scaling data are $a_{s} = c_{s} e_{s}$ , $a_{s}^{\lor} = 2 a_{s} / B (a_{s}, a_{s})$ , $a_{s t} = B (a_{s}, a_{t}^{\lor}) = 2 B (a_{s}, a_{t}) / B (a_{t}, a_{t})$ , and $c$ is crystallographic exactly when $a_{s t} \in Z$ for all $s, t$ (Crystallographic scalings: scaled simple roots, coroots and the root, coroot and weight lattices).

[F7]

For every scaling, $a_{s t} a_{t s} = 4 \cos^{2} (π / m (s, t))$ and $a_{s t} = - 2 \frac{c_{s}}{c_{t}} \cos (π / m (s, t)) \leq 0$ for distinct $s, t$ with finite $m (s, t)$ (Cartan-number products, allowed edge labels, tree scalings and reflection stability (1)).

[F8]

If $B$ is positive definite and $c$ is crystallographic, then for all distinct $s, t$ one has $a_{s t} a_{t s} \in {0, 1, 2, 3}$ and $m (s, t) \in {2, 3, 4, 6}$ (Cartan-number products, allowed edge labels, tree scalings and reflection stability (2)).

[F9]

A reduced crystallographic Euclidean root system is a finite spanning set $Ψ \subseteq E ∖ {0}$ closed under its root reflections, with integral Cartan integers $2 (β, α) / (α, α)$ and $R α \cap Ψ = {α, - α}$ (Reduced crystallographic Euclidean root system).

[F10]

For nonproportional $α, β \in Ψ$ with angle $θ$ one has $n_{α β} n_{β α} = 4 \cos^{2} θ \in {0, 1, 2, 3}$ , where $n_{α β} = 2 (β, α) / (α, α)$ and $n_{β α} = 2 (α, β) / (β, β)$ ; if ${α, β}$ is a base of a rank-two system, then $(α, β) \leq 0$ and $θ$ is one of $90^{\circ}$ , $120^{\circ}$ , $135^{\circ}$ , $150^{\circ}$ (Rank-two root-system classification (i), (iv)).

[F11]

Distinct simple roots of a reduced crystallographic root system satisfy $(α, β) \leq 0$ (Distinct simple roots have nonpositive inner product).

[F12]

$\cos (2 x) = 2 \cos^{2} x - 1$ and $\cos^{2} x = (1 + \cos 2 x) / 2$ for all real $x$ (Double-angle and quadratic power-reduction identities).

[F13]

$\cos (π - x) = - \cos x$ and $\cos (π / 2) = 0$ (Cofunction, supplementary, quarter-turn, and reflection identities for the six trigonometric functions, Quarter-turn values and shifts by pi/2 and pi).

[F14]

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

[F15]

$π > 0$ (Pi as twice the smallest positive zero of cosine).

[F16]

A base is the set of simple roots of a positive system, and the simple roots form a basis of the ambient space (Positive systems and simple roots, Simple roots form a signed integral basis).

Counterexample

1.1F1F2F3F4F5

The geometry: by [F2] the form $B$ has $B (e_{s}, e_{s}) = B (e_{t}, e_{t}) = 1$ and $B (e_{s}, e_{t}) = - \cos (π / 5)$ , and since $V = R e_{s} + R e_{t}$ is the plane $P$ of [F3] with $m = 5$ , [F3] makes $B$ positive definite; hence $W$ is finite by [F4], and the diagram is $I_{2} (5)$ , conventionally also named $H_{2}$ , by the classifier clauses (1), (4) in [F5]. The normalized pairing of the two basis vectors is $\frac{B (e_{s}, e_{t})}{\sqrt{B (e_{s}, e_{s}) B (e_{t}, e_{t})}} = - \cos (π / 5)$ .

1.2F12F13F14F15algebra

The value $\cos (π / 3) = \frac{1}{2}$ : put $q : = \cos (π / 3)$ . By [F13] at $x = π / 3$ one has $\cos (2 π / 3) = \cos (π - π / 3) = - q$ , while [F12] gives $\cos (2 π / 3) = 2 q^{2} - 1$ ; hence $2 q^{2} - 1 = - q$ , that is $(2 q - 1) (q + 1) = 0$ . Since $0 < π / 3 < π$ (as $π > 0$ by [F15]) and cosine is strictly decreasing on $[0, π]$ with $\cos π = - 1$ by [F14], one has $q > - 1$ , so $(2 q - 1) (q + 1) = 0$ forces $q = \frac{1}{2}$ .

2.1F7F12F13F14step 1.2algebra

The product is strictly between $2$ and $3$ : for every scaling $c$ , [F7] gives $a_{s t} a_{t s} = 4 \cos^{2} (π / 5)$ , and [F12] at $x = π / 5$ rewrites this as $2 + 2 \cos (2 π / 5)$ . Since $π > 0$ we have $π / 3 < 2 π / 5 < π / 2$ (because $2 π / 5 - π / 3 = π / 15 > 0$ and $π / 2 - 2 π / 5 = π / 10 > 0$ ), and cosine is strictly decreasing on $[0, π]$ with $\cos (π / 2) = 0$ and $\cos (π / 3) = \frac{1}{2}$ by [F13, F14] and step 1.2; therefore $0 < \cos (2 π / 5) < \frac{1}{2}$ , and hence $2 < a_{s t} a_{t s} < 3$ .

3.1F5F6F8step 2.1

No crystallographic scaling exists: if $c$ were crystallographic, then $a_{s t}$ and $a_{t s}$ would both be integers by [F6], so their product $a_{s t} a_{t s}$ would be an integer; but step 2.1 places that product strictly between the consecutive integers $2$ and $3$ . This contradiction refutes (a) for the geometry $S = {s, t}$ , $m (s, t) = 5$ : at least one of $a_{s t}, a_{t s}$ is non-integral for every scaling. Equivalently, [F8] would force $m (s, t) \in {2, 3, 4, 6}$ , which $m (s, t) = 5$ contradicts, and [F5] records the resulting exclusion of $I_{2} (5)$ from the crystallographic finite types.

3.2F9F10F11F16step 2.1algebra

No root system realizes (b): suppose $Ψ$ were a reduced crystallographic Euclidean root system with base ${α, β}$ and $u : = \frac{(α, β)}{∣ α ∣ ∣ β ∣} = - \cos (π / 5)$ . By [F16], ${α, β}$ is linearly independent, so the roots are nonproportional and [F10] applies. Unfolding the two Cartan integers, $n_{α β} n_{β α} = \frac{2 (β, α)}{(α, α)} \cdot \frac{2 (α, β)}{(β, β)} = \frac{4 (α, β)^{2}}{(α, α) (β, β)} = 4 u^{2} = 4 \cos^{2} \frac{π}{5},$ and by step 2.1 this number lies in $(2, 3)$ ; but [F10] states $n_{α β} n_{β α} = 4 \cos^{2} θ \in {0, 1, 2, 3}$ , a contradiction. Hence no reduced crystallographic root system has a base with the normalized pairing $- \cos (π / 5)$ . This is consistent with [F10] (iv) read together with [F11]: a rank-two base has nonacute angle $θ$ among $90^{\circ}, 120^{\circ}, 135^{\circ}, 150^{\circ}$ , and each of those angles gives $4 \cos^{2} θ \in {0, 1, 2, 3}$ , never the value $4 \cos^{2} (π / 5) \in (2, 3)$ .

4.1F5F6F8F10given∎

The failure and its range: the dropped hypothesis identified by this counterexample is that a crystallographic scaling requires the cross product $a_{s t} a_{t s} = 4 \cos^{2} (π / m (s, t))$ to be an integer, hence (for positive definite $B$ ) equal to one of $0, 1, 2, 3$ , equivalently a label $m \in {2, 3, 4, 6}$ ; the value $m = 5$ gives the non-integral number $4 \cos^{2} (π / 5) = 2 + 2 \cos (2 π / 5) \in (2, 3)$ that is strictly between the admissible integer values of that product. Both refutations are independent of each other: (a) is a statement about scalings of one Coxeter geometry, (b) about bases of reduced crystallographic root systems, and their common obstruction is the same interval $(2, 3)$ for $4 \cos^{2} (π / 5)$ ; the full exclusion of $I_{2} (5)$ from the Weyl types is the criterion (1) of Crystallographic finite type: the Weyl types, reduced realizations and lattice stability. No choice principle is used, and the computations are finite real arithmetic in a two-dimensional space.

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Sources

Jean Michel, Lectures on Coxeter groups (Beijing lecture notes, April-May 2014) (standard reference, not scraped)
Anthony W. Knapp, Lie Groups Beyond an Introduction, 2nd ed. (author-hosted digital edition) (standard reference, not scraped)
J. S. Milne, Lie Algebras, Algebraic Groups, and Lie Groups (course notes, version 2.00) (standard reference, not scraped)

I2(5)I2​(5) admits no crystallographic scaling and no reduced crystallographic root system with that base pairing