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Euclidean Yang–Mills self-dual action bound

Statement

Assume full Axiom of Choice. Let $X$ be a nonempty, connected, compact, oriented, finite-dimensional Hausdorff second-countable smooth four-manifold without boundary, with Riemannian metric $g_{X}$ . Let $G \subseteq U (V, h)$ be a compact non-Abelian matrix Lie group in its specified global form, let $P \to X$ be a smooth right principal $G$ -bundle, and let $ω$ be a supplied smooth principal connection. Let $B$ be a positive-definite, symmetric, $Ad (G)$ -invariant bilinear form on $g$ . For any local section $s_{i}$ , set $A_{i} = s_{i}^{*} ω$ and $F_{i} = d A_{i} + \frac{1}{2} [A_{i} \land A_{i}]$ . Invariance of $B$ makes the local four-forms $β_{B} ∣_{U_{i}} = \frac{1}{8 π^{2}} B (F_{i} \land F_{i})$ and $γ_{B} ∣_{U_{i}} = B (F_{i} \land *_{g_{X}} F_{i})$ restrictions of global forms. Define $q_{B} = \int_{X} β_{B}$ . For a real scalar $g > 0$ , define the Euclidean Yang–Mills functional by

S_{B} (P, ω) : = \frac{1}{g^{2}} \int_{X} γ_{B},

where the Hodge operator is characterized by $α \land *_{g_{X}} β = ⟨ α, β ⟩_{g_{X}} d {vol}_{g_{X}}$ and acts componentwise on $g$ -valued forms. Then

S_{B} (P, ω) \geq \frac{8 π^{2}}{g^{2}} ∣ q_{B} (P, ω) ∣ .

Equality holds exactly in these cases: if $q_{B} > 0$ , then $F_{i} = *_{g_{X}} F_{i}$ ; if $q_{B} < 0$ , then $F_{i} = - *_{g_{X}} F_{i}$ ; if $q_{B} = 0$ , then $F_{i} = 0$ . Every self-dual or anti-self-dual supplied connection satisfies the source-free Euclidean Yang–Mills equation $D_{A_{i}} (*_{g_{X}} F_{i}) = 0$ , where $D_{A_{i}} η = d η + [A_{i} \land η]$ . The statement asserts neither existence nor uniqueness of a connection in any charge sector.

Statement references

Belavin–Polyakov–Schwartz–Tyupkin, printed pp. 85–86

Tong, Lecture 1 §1.1.1

Charge, orientation, Hodge star, and invariant-form conventions

Curvature Bianchi identity and covariant exterior derivative

Full Axiom of Choice used by the topological charge interface

Facts & Assumptions

Given: Full AC, the nonempty compact oriented Riemannian four-manifold and bundle of the instanton-number definition, a supplied connection, invariant positive-definite $B$ , and a real $g > 0$ .

[A1]

Full AC means every family of nonempty sets has a choice function. The Axiom of Choice

[F1]

The global charge form is $β_{B} = \frac{1}{8 π^{2}} B (F \land F)$ , and $q_{B} = \int_{X} β_{B}$ ; the Hodge star and the self-dual/ASD decomposition are those fixed by the orientation and metric. Trace-normalized instanton number and self-dual sectors

[F2]

The bilinear form $B$ is positive definite. Trace-normalized instanton number and self-dual sectors

[F3]

The Hodge star on two-forms is an orthogonal involution whose $+ 1$ and $- 1$ eigenspaces are orthogonal for the inner product induced by $B$ . Hodge decomposition of two-forms in four dimensions

[F4]

The local curvature satisfies $D_{A_{i}} F_{i} = d F_{i} + [A_{i} \land F_{i}] = 0$ . Principal curvature, Cartan structure equation, and Bianchi identity

Proof

1.1givenA1F1F2F3algebra

Write $F_{i} = F_{i}^{+} + F_{i}^{-}$ with $*_{g_{X}} F_{i}^{+} = F_{i}^{+}$ and $*_{g_{X}} F_{i}^{-} = - F_{i}^{-}$ . Let $a = \int_{X} B (F_{i}^{+} \land *_{g_{X}} F_{i}^{+})$ and $b = \int_{X} B (F_{i}^{-} \land *_{g_{X}} F_{i}^{-})$ . Orthogonality of the two Hodge eigenspaces [F3] makes the cross terms vanish; positivity of $B$ [F2] gives $a, b \geq 0$ . Consequently $\int_{X} B (F_{i} \land *_{g_{X}} F_{i}) = a + b$ , while $\int_{X} B (F_{i} \land F_{i}) = a - b$ . By [F1], $S_{B} = (a + b) / g^{2}$ and $q_{B} = (a - b) / (8 π^{2})$ .

2.1F1F2step 1.1algebra

For nonnegative $a, b$ , $a + b \geq ∣ a - b ∣$ , so the claimed bound follows after division by $g^{2} > 0$ . If $q_{B} > 0$ , then $a > b$ and equality is equivalent to $b = 0$ . Its integrand is the continuous nonnegative squared norm of $F_{i}^{-}$ ; if $F_{i}^{-}$ were nonzero at a point, it would be bounded away from zero on a neighborhood of positive Riemannian volume, contradicting $b = 0$ . Thus $F_{i}^{-} = 0$ . If $q_{B} < 0$ , equality in $a + b \geq b - a$ is equivalent to $a = 0$ ; the same continuity and positivity argument gives $F_{i}^{+} = 0$ . If $q_{B} = 0$ , equality forces $a + b = 0$ , so nonnegativity gives $a = b = 0$ and both components vanish, hence $F_{i} = 0$ . Conversely each of the three stated conditions makes its corresponding norm vanish and gives equality.

3.1givenF4∎

By [F4], $D_{A_{i}} F_{i} = 0$ . For a self-dual connection, $D_{A_{i}} (*_{g_{X}} F_{i}) = D_{A_{i}} F_{i} = 0$ ; for an anti-self-dual connection, $D_{A_{i}} (*_{g_{X}} F_{i}) = - D_{A_{i}} F_{i} = 0$ . This is the stated source-free Euclidean Yang–Mills equation. No boundary term or variational argument is used.

Euclidean Yang–Mills self-dual action bound

Statement

Statement references

Facts & Assumptions

Proof

Depends on

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Sources