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Holonomy and discrete geometric interfaces — antisymmetric invariant tensor at a bivalent SU(2) vertex

Statement

Assume the full Axiom of Choice. Take the graph with three distinct vertices $u_{1}, u_{2}, v$ , edges $a_{1} : u_{1} \to v$ , $a_{2} : u_{2} \to v$ and their reverses ${\overset{ˉ}{a}}_{1}, {\overset{ˉ}{a}}_{2}$ , and no other edges. Order the edges as $a_{1} < a_{2} < {\overset{ˉ}{a}}_{1} < {\overset{ˉ}{a}}_{2}$ and fix any vertex order. Assign the defining representation $ρ (g) w : = g w$ on $V = C^{2}$ , with its standard inner product, to $a_{1}, a_{2}$ ; assign the trivial representation on $C$ to each reverse edge. Thus $ρ_{{\overset{ˉ}{a}}_{j}} (g) z = z$ for every $g \in S U (2)$ and $z \in C$ ; the Haar projection's hypotheses impose no relation between representations assigned to reverse orientations.

The full vertex space is $W_{v} = V \otimes V \otimes C^{*} \otimes C^{*}$ . Write $1^{*} (z) = z$ and identify $V \otimes V$ with $W_{v}$ by the linear isomorphism $ι (ξ) = ξ \otimes 1^{*} \otimes 1^{*}$ . For any basis $e_{1}, e_{2}$ of $V$ , the nonzero tensor

$ϵ = e_{1} \otimes e_{2} - e_{2} \otimes e_{1}$

is invariant under $ρ (g) \otimes ρ (g)$ . Consequently the normalized Haar projection $P_{v} : W_{v} \to W_{v}$ fixes $ι (ϵ)$ ; equivalently, $ι^{- 1} P_{v} ι (ϵ) = ϵ$ .

Statement references

Finite graph vertex Haar averaging is an orthogonal projection

Finite SU(2) graph holonomy model

The Axiom of Choice

Facts & Assumptions

Given: Full AC, the specified four oriented edges and their representations, the vertex space $W_{v}$ , the map $ι$ , and a basis $e_{1}, e_{2}$ of $V$ .

[F1]

Finite graph vertex Haar averaging is an orthogonal projection: under full AC, the normalized Haar average projects onto vertex-invariant tensors.

[F2]

Finite SU(2) graph holonomy model: the graph includes both edge orientations, and a finite-dimensional unitary representation is a strongly continuous group homomorphism into the unitary operators. The complete vertex factor formula is supplied by [F1].

[F3]

The Axiom of Choice: full AC is the set-theoretic assumption inherited by the averaging supplier.

Proof

technique · direct determinant calculation

1.1givenF1F2algebra

The defining action is a group homomorphism, preserves the standard inner product because $g^{†} g = I_{2}$ , and is strongly continuous because its matrix entries are continuous. The trivial reverse-edge actions have these properties as well. By [F1], their dual factors at $v$ are trivial one-dimensional spaces, so the displayed $W_{v}$ and $ι$ are well-defined and $R_{v} (g) ι = ι (ρ (g) \otimes ρ (g))$ . For $g \in S U (2)$ , write the matrix of $ρ (g)$ in the chosen basis as $(\begin{array}{cc} a & b \\ c & d \end{array})$ ; its determinant is $\det g = 1$ . Expanding the tensor product gives $(g \otimes g) ϵ = (a e_{1} + c e_{2}) \otimes (b e_{1} + d e_{2}) - (b e_{1} + d e_{2}) \otimes (a e_{1} + c e_{2}) = (a d - b c) ϵ = (\det g) ϵ = ϵ$ . The tensor is nonzero because $e_{1} \otimes e_{2}$ and $e_{2} \otimes e_{1}$ are distinct basis tensors.

2.1givenF1F3step 1.1algebra∎

By [F1], $P_{v} ι (ϵ) = \int_{S U (2)} R_{v} (g) ι (ϵ) d μ (g)$ . Step 1.1 makes the integrand the constant vector $ι (ϵ)$ , and the normalized Haar probability has total mass one; hence the integral equals $ι (ϵ)$ . Applying $ι^{- 1}$ gives the equivalent asserted formula.

Remarks

The diagram records the full vertex action under the specified trivial reverse-edge factors; the determinant computation proves invariance independently of the diagram.

Holonomy and discrete geometric interfaces — antisymmetric invariant tensor at a bivalent SU(2) vertex

Statement

Statement references

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

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