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Classical SU(2) monopole energy bound and Bogomolny equations

Physical scope

Conditional classical energy identity and lower bound for smooth finite-energy fields in the specified static SU(2) adjoint-Higgs Prasad–Sommerfield sector. It asserts neither existence of a saturating solution nor any quantum particle or supersymmetry state.

Statement

Assume ACω. Use natural units on Minkowski R1,3 with η=diag⁡(−1,1,1,1) and oriented Euclidean spatial slice with ϵ123=1. Fix the global group SU(2), basis Ta=−iσa/2, and positive Ad⁡-invariant form κ(X,Y)=−2Tr⁡2(XY) with κ(Ta,Tb)=δab. Let g>0 be dimensionless and let Ai,Φ be smooth static fields in the zero-potential adjoint-Higgs model, with [Ai]=[Φ]=length−1, Fij=∂iAj−∂jAi+[Ai,Aj], DiΦ=∂iΦ+[Ai,Φ], and Bi=12ϵijkFjk. Require the model's smooth finite-energy boundary reduction: ∣Φ∣κ→v>0 for a fixed v of dimension length−1, with n∞(ω)=lim⁡R→∞Φ(Rω)/v in C∞(S2,su(2)). Use the adapted limiting connections Aα,∞ of the static-sector Definition: the tangential condition is [Aα,∞,T3]=0 in each adapted frame. The scaled radial limit B∞(ω):=lim⁡R→∞R2∑iωiBi(Rω) exists uniformly on each closed hemisphere and satisfies [B∞(ω),n∞(ω)]=0. Suppose the energy

E=12g2∫R3∑i(κ(Bi,Bi)+κ(DiΦ,DiΦ)) d3x

is finite and the model's asymptotic reduction and surface limit

m=lim⁡R→∞∫SR2κ(Φ,Bi)x^i dS

exist. Put T=m/g2. Then both identities hold:

E=12g2∫R3∑iκ(Bi−DiΦ,Bi−DiΦ) d3x+T,

E=12g2∫R3∑iκ(Bi+DiΦ,Bi+DiΦ) d3x−T.

Consequently E≥∣T∣. If T>0, equality holds exactly when Bi=DiΦ for every i and every x; if T<0, equality holds exactly when Bi=−DiΦ everywhere. If T=0, equality holds exactly when Bi=0 and DiΦ=0 everywhere. These are conditional classical deductions for the adopted energy functional; the word BPS here refers only to classical energy saturation and asserts no quantum state or supersymmetry central-charge relation.

Facts & Assumptions

[A1]

ACω is the Axiom of Countable Choice, required by the exact Bianchi and divergence suppliers. The Axiom of Countable Choice (ACω)

[F1]

The sector fixes Fij=∂iAj−∂jAi+[Ai,Aj], DiΦ=∂iΦ+[Ai,Φ], Bi=12ϵijkFjk, the positive coupling g, the energy functional, and the finite surface limit. Static SU(2) adjoint-Higgs Prasad–Sommerfield monopole sector

[F2]

The form κ is a positive-definite symmetric real bilinear form and is Ad⁡-invariant, so κ([X,Y],Z)+κ(Y,[X,Z])=0. Compactness and normalized invariant form for SU(2)

[F3]

In every local section, F=dA+12[A∧A] and DAF=0; contracting the latter with ϵijk gives DiBi=0. Principal curvature, Cartan structure equation, and Bianchi identity

[F4]

For a bounded C1 domain and a C1 vector field on its closure, the integral of the divergence equals the outward boundary flux; this theorem assumes ACω. Divergence on a bounded C1 Euclidean domain

Proof

Given: [A1], [F1]–[F4], smooth fields, and the finite-energy and surface-limit hypotheses.

1.1A1F1F2F3algebra

Define Ji=κ(Φ,Bi). The product rule gives ∂iJi=κ(∂iΦ,Bi)+κ(Φ,∂iBi). By [F1]–[F3], the sum is κ(DiΦ,Bi)+κ(Φ,DiBi)=κ(DiΦ,Bi): the two connection terms cancel by Ad-invariance and DiBi=0 by Bianchi.

2.1A1F1F2F4step 1.1algebra

On each closed ball BR, J is C1, so [F4] and step 1.1 give ∫BR∑iκ(DiΦ,Bi) d3x=∫SR2κ(Φ,Bi)x^i dS. Positive definiteness in [F2] gives 2∣κ(DiΦ,Bi)∣≤κ(DiΦ,DiΦ)+κ(Bi,Bi), so finite energy makes the cross term integrable. Taking R→∞ and using the declared surface limit yields ∫R3∑iκ(Bi,DiΦ) d3x=m.

3.1F1F2step 2.1algebra

Expand the first square: its integral is ∫∑i(κ(Bi,Bi)+κ(DiΦ,DiΦ)−2κ(Bi,DiΦ)) d3x. By [F1] the first two terms give 2g2E, and by step 2.1 the cross term integrates to 2m; multiplying by 1/(2g2) and adding T=m/g2 gives the first identity. Expanding with the plus sign changes the cross term to +2m, so subtracting T gives the second identity. All integrals converge by [F1], [F2], and step 2.1.

4.1F1F2step 3.1algebra∎

Each square integral in step 3.1 is nonnegative by [F2], giving E≥T and E≥−T, hence E≥∣T∣. For T>0, equality in E≥T holds exactly when the integral of the continuous nonnegative function ∑iκ(Bi−DiΦ,Bi−DiΦ) is zero, which is equivalent to Bi=DiΦ everywhere; the converse follows from the first identity. For T<0, the second identity gives equivalently Bi=−DiΦ. For T=0, equality means E=0; positivity and continuity of the energy density imply Bi=DiΦ=0 everywhere, and these conditions also make E=0.

Remarks

For anti-Hermitian X,Y, put XH=iX and YH=iY. Then κ(X,Y)=2Tr⁡2(XHYH). Consequently the energy coefficient κ/(2g2) becomes Tr⁡2/g2, and the surface coefficient κ/g2 becomes 2Tr⁡2/g2. Both match Tong's equation (2.16) at g2=e2. The proof is a classical model deduction and establishes no existence result for any specified topological sector.

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