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A circle representation with an averaged orthogonal weight form

Example

Assume the Axiom of Choice (The Axiom of Choice). Let $S^{1} = {z \in C : ∣ z ∣ = 1}$ be the unit circle with the subspace topology inherited from $C ≅ R^{2}$ , made a group by complex multiplication; let $V = C^{2}$ , and for $z \in S^{1}$ let $ρ (z) \in GL (V)$ be given by $ρ (z) (v_{1}, v_{2}) = (v_{1}, z v_{2})$ , so that $ρ (z) = diag (1, z)$ is a continuous finite-dimensional complex representation of $S^{1}$ (A finite-dimensional representation $ρ : G \to GL (V)$ over a field, and its degree, Topological group: multiplication and inversion are continuous). Let $h_{0} (v, w) : = 2 v_{1} \overline{w_{1}} + v_{1} \overline{w_{2}} + v_{2} \overline{w_{1}} + 3 v_{2} \overline{w_{2}}$ (Real and complex inner-product spaces and their induced length), a Hermitian inner product on $V$ whose matrix is $(\begin{array}{cc} 2 & 1 \\ 1 & 3 \end{array})$ , and let $h$ be its average over the normalized Haar probability measure $μ$ of $S^{1}$ , $h (v, w) : = \int_{S^{1}} h_{0} (ρ (z) v, ρ (z) w) d μ (z)$ (Averaged Hermitian form for a compact group, Normalized Haar probability on a compact group). Then:

$h (v, w) = 2 v_{1} \overline{w_{1}} + 3 v_{2} \overline{w_{2}}$ for all $v, w \in V$ ;
the coordinate (weight) lines $C e_{1}$ and $C e_{2}$ , on which $ρ$ acts by the characters $z \mapsto 1$ and $z \mapsto z$ , are orthogonal for the averaged form $h$ , since $h (e_{1}, e_{2}) = 0$ , but not for $h_{0}$ , since $h_{0} (e_{1}, e_{2}) = 1$ ;
$h_{0}$ is not $S^{1}$ -invariant, because $h_{0} (ρ (- 1) e_{1}, ρ (- 1) e_{2}) = - 1 \neq 1 = h_{0} (e_{1}, e_{2})$ , whereas $h$ is $S^{1}$ -invariant and positive definite.

Facts & Assumptions

Given: AC; the unit circle $S^{1} = {z \in C : ∣ z ∣ = 1}$ with the subspace topology of $C ≅ R^{2}$ and complex multiplication; the representation $ρ (z) = diag (1, z)$ on $V = C^{2}$ ; the form $h_{0}$ above; the normalized Haar probability $μ$ of $S^{1}$ ; and its averaged form $h$ .

[F1]

$C$ is a field with the usual coordinate-plane model: the map $Φ (a + b i) = (a, b)$ is a bijection carrying products to $(a u - b v, a v + b u)$ , and conjugation is an involutive field automorphism with $z \overline{z} = ∣ z ∣^{2}$ and $∣ z w ∣ = ∣ z ∣ ∣ w ∣$ , so $z^{- 1} = \overline{z}$ whenever $∣ z ∣ = 1$ ( $C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ , $C$ is the real coordinate plane, with coordinate arithmetic, Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

[F2]

Topology toolkit: a map into a product is continuous exactly when its coordinates are; real addition and multiplication are jointly continuous (by specializing vector operations to the real normed space $R$ ), and restrictions and composites of continuous maps are continuous for the subspace and product topologies (A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice, The product set $\prod_{i \in I} X_{i}$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space, Vector addition and scalar multiplication are continuous in a normed space, Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous, Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace, Continuity of a map of topological spaces at a point and globally, For a map of spaces the following agree: continuity at every point, preimages of open sets open, preimages of closed sets closed, preimages of subbasic open sets open, and $f (\overline{A}) \subseteq \overline{f (A)}$ ). A subset of $R^{2}$ with the Euclidean metric is compact exactly when it is closed and bounded, with no choice principle (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies), and metric spaces are Hausdorff (Distinct points of a metric space have disjoint balls around them, Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not).

[F3]

Normalized Haar measure: a compact Hausdorff group has a unique left Haar probability $μ$ , which is right invariant and inversion invariant (Normalized Haar probability on a compact group), positive on every nonempty open set and finite on compact sets (Haar measure is positive on nonempty open sets and finite on compact sets, Measure spaces).

[F4]

Averaged forms: for a continuous finite-dimensional complex representation of a compact Hausdorff group and a Hermitian inner product $h_{0}$ linear in the first variable, the averaged form is well defined, sesquilinear and Hermitian, and its integrand is continuous and integrable (Averaged Hermitian form for a compact group).

[F5]

Integral tools: the Lebesgue integral is linear on integrable functions (The Lebesgue integral is linear on $L^{1} (μ)$ , Integrable real and complex functions, and their integrals), and a for integrable real or complex $f$ , a measure-preserving self-map $T$ satisfies $\int f \circ T d μ = \int f d μ$ (Integral invariance under measure-preserving maps, Measure-preserving transformations and systems).

[F6]

The averaged form of [F4] is positive definite and invariant under the representation, so in the present example $h$ is an inner product on $V$ with $h (ρ (z) v, ρ (z) w) = h (v, w)$ for all $z \in S^{1}$ (Averaging a Hermitian form unitarizes a finite-dimensional compact-group representation, Linear isometries, and orthogonal or unitary operators on finite-dimensional inner product spaces).

Proof

technique · direct

1.1F1F2

The set $S^{1}$ is a compact Hausdorff topological group. It contains $1$ and is closed under multiplication and inversion because $∣ z w ∣ = ∣ z ∣ ∣ w ∣$ and $z^{- 1} = \overline{z}$ with $∣ \overline{z} ∣ = ∣ z ∣$ by [F1]; associativity and the remaining group axioms are inherited from the field $C$ . In the coordinate plane, multiplication has the polynomial formula $(a, b, u, v) \mapsto (a u - b v, a v + b u)$ and inversion the formula $(a, b) \mapsto (a, - b)$ on $∣ z ∣ = 1$ , so both operations are continuous on the product $S^{1} \times S^{1}$ respectively on $S^{1}$ by [F2]. Moreover $S^{1}$ is the preimage of ${1}$ under the continuous map $(a, b) \mapsto a^{2} + b^{2}$ , hence closed in $R^{2}$ , and it is bounded because $a^{2} + b^{2} = 1$ ; by Heine–Borel [F2] it is compact, and it is Hausdorff as a subspace of a metric space.

2.1F1step 1.1

The map $ρ$ is a continuous finite-dimensional complex representation of $S^{1}$ on $V = C^{2}$ , and $h_{0}$ is a Hermitian inner product on $V$ . Indeed $ρ (z) ρ (w) = diag (1, z w) = ρ (z w)$ and $ρ (1) = I$ , each $ρ (z) = diag (1, z)$ is invertible because $z \neq 0$ , and $z \mapsto ρ (z)$ is continuous as a map into the finite-dimensional space $End (V)$ because its matrix entries are continuous and all norms on that space are equivalent (All norms on a finite-dimensional complex normed space are equivalent). The form $h_{0}$ has the real symmetric matrix $(\begin{array}{cc} 2 & 1 \\ 1 & 3 \end{array})$ , hence is conjugate-symmetric, and $h_{0} (v, v) = 2 ∣ v_{1} ∣^{2} + 2 Re (v_{1} \overline{v_{2}}) + 3 ∣ v_{2} ∣^{2} \geq ∣ v_{1} ∣^{2} + 2 ∣ v_{2} ∣^{2} > 0$ whenever $v \neq 0$ , because $2 Re (v_{1} \overline{v_{2}}) \geq - (∣ v_{1} ∣^{2} + ∣ v_{2} ∣^{2})$ ; in particular $h_{0} (e_{1}, e_{2}) = 1$ .

2.2F3F5step 1.1

The integrals of the characters vanish: $\int_{S^{1}} z d μ (z) = 0$ and $\int_{S^{1}} \overline{z} d μ (z) = 0$ . Both characters are continuous and have modulus one, hence are integrable against the probability $μ$ . The map $j (z) : = - z = (- 1) \cdot z$ is a continuous self-map of $S^{1}$ with $j^{- 1} (E) = (- 1) E$ , so $μ (j^{- 1} E) = μ (E)$ by left invariance of [F3]: it is measure preserving, and [F5] gives $\int z d μ = \int (- z) d μ = - \int z d μ$ , hence $\int z d μ = 0$ ; replacing $z$ by $\overline{z}$ , whose composite with $j$ is $- \overline{z}$ , gives $\int \overline{z} d μ = - \int \overline{z} d μ = 0$ in the same way.

3.1F1F4step 2.1

For $z \in S^{1}$ one has $ρ (z) v = (v_{1}, z v_{2})$ and $∣ z ∣^{2} = 1$ , so expanding $h_{0}$ in [F4] gives $h_{0} (ρ (z) v, ρ (z) w) = 2 v_{1} \overline{w_{1}} + \overline{z} v_{1} \overline{w_{2}} + z v_{2} \overline{w_{1}} + 3 ∣ z ∣^{2} v_{2} \overline{w_{2}} = 2 v_{1} \overline{w_{1}} + 3 v_{2} \overline{w_{2}} + \overline{z} (v_{1} \overline{w_{2}}) + z (v_{2} \overline{w_{1}})$ for all $v, w \in V$ .

4.1F4F5step 3.1step 2.2

Integrating the expansion of step 3.1 and pulling out the constants $v_{i} \overline{w_{j}}$ by linearity of the integral [F5] yields $h (v, w) = \int_{S^{1}} (2 v_{1} \overline{w_{1}} + 3 v_{2} \overline{w_{2}} + \overline{z} v_{1} \overline{w_{2}} + z v_{2} \overline{w_{1}}) d μ (z) = 2 v_{1} \overline{w_{1}} + 3 v_{2} \overline{w_{2}} + (\int \overline{z} d μ) v_{1} \overline{w_{2}} + (\int z d μ) v_{2} \overline{w_{1}} = 2 v_{1} \overline{w_{1}} + 3 v_{2} \overline{w_{2}}$ by step 2.2.

5.1F6step 2.1step 4.1∎

Consequences. By step 4.1, $h (e_{1}, e_{2}) = 0$ , while $h_{0} (e_{1}, e_{2}) = 1$ by step 2.1: the two weight lines are orthogonal for $h$ but not for $h_{0}$ . Since $ρ (z) e_{1} = e_{1}$ and $ρ (z) e_{2} = z e_{2}$ , the invariance failure is visible at $z = - 1$ : $h_{0} (ρ (- 1) e_{1}, ρ (- 1) e_{2}) = h_{0} (e_{1}, - e_{2}) = - h_{0} (e_{1}, e_{2}) = - 1 \neq 1 = h_{0} (e_{1}, e_{2})$ , conjugate-linearity in the second variable producing the sign. The averaged form $h$ is positive definite and $S^{1}$ -invariant by [F6], in agreement with the explicit formula of step 4.1.

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Sources

Emmanuel Kowalski, An Introduction to the Representation Theory of Groups, §§5.2–5.6 (standard reference, not scraped)
Vera Serganova, Representation Theory, Chapter III §§1.6–2.1 (standard reference, not scraped)