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One-flavor U(1) classical F/D-flat vacuum quotient

Physical scope

Exact classical tree-level calculation for a canonical rigid four-dimensional N=1 U(1) gauge-model example with one oppositely charged chiral pair, zero superpotential, and zero Fayet–Iliopoulos parameter; it says nothing about quantum corrections.

Example

Use natural units $c = ℏ = 1$ . Take the finite-dimensional scalar field space $V = C^{2}$ with coordinates $(q, \tilde{q})$ , standard Hermitian norm, compact gauge group $U (1)$ acting by $(q, \tilde{q}) \mapsto (e^{i α} q, e^{- i α} \tilde{q})$ , gauge coupling $g > 0$ , zero superpotential $W = 0$ , canonical scalar kinetic terms, and zero Fayet--Iliopoulos parameter. In this classical model the auxiliary maps are $F_{q} = F_{\tilde{q}} = 0$ and $D = - g (∣ q ∣^{2} - ∣ \tilde{q} ∣^{2})$ . The classical supersymmetric vacuum set is therefore ${(q, \tilde{q}) : ∣ q ∣ = ∣ \tilde{q} ∣} / U (1)$ . The gauge-invariant coordinate $m = q \tilde{q}$ gives a bijection of this orbit set with $C$ . This is a classical tree-level result and is not a claim about the quantum vacuum space.

Example references

Tong, §4.3.1, printed pp. 112–116.

Classical F/D-flat quotient and quantum-moduli boundary: Classical and quantum supersymmetric vacuum-moduli data.

Facts & Assumptions

Given: The U(1) action, canonical field data, $W = 0$ , and zero Fayet--Iliopoulos parameter in the Example.

[A1]

Adopt the finite-dimensional canonical rigid $N = 1$ U(1) classical model stated in the Example; $q$ and $\tilde{q}$ have mass dimension one and $g$ is dimensionless.

[F1]

Classical and quantum supersymmetric vacuum-moduli data: classical vacua are the simultaneous zeros of the F and D maps, modulo the compact gauge group.

Verification

Given: [A1] and the classical vacuum-moduli definition [F1].

1.1A1F1givenalgebra

Since $W = 0$ , both holomorphic derivatives vanish identically, so every $(q, \tilde{q})$ is F-flat. The D equation is $- g (∣ q ∣^{2} - ∣ \tilde{q} ∣^{2}) = 0$ ; because $g > 0$ , it is equivalent to $∣ q ∣ = ∣ \tilde{q} ∣$ . By [F1], the classical vacuum set is exactly that D-flat set modulo the displayed U(1) action.

2.1A1step 1.1algebra

The product $m = q \tilde{q}$ is invariant under the action, since $(e^{i α} q) (e^{- i α} \tilde{q}) = q \tilde{q}$ . It therefore defines a map from the D-flat orbit set to $C$ .

2.2step 1.1constructalgebra

For any $m_{0} \in C$ with $m_{0} \neq 0$ , choose $q = \sqrt{∣ m_{0} ∣} > 0$ and $\tilde{q} = m_{0} / q$ . Then $q \tilde{q} = m_{0}$ and $∣ q ∣ = ∣ \tilde{q} ∣ = \sqrt{∣ m_{0} ∣}$ ; for $m_{0} = 0$ , choose $(q, \tilde{q}) = (0, 0)$ . Thus every $m_{0}$ is attained by a D-flat pair.

3.1step 2.1step 2.2algebra∎

If two D-flat pairs have the same product $m_{0} = 0$ , their equal-modulus condition forces both pairs to be $(0, 0)$ . If $m_{0} \neq 0$ , both $q$ -entries are nonzero and have modulus $\sqrt{∣ m_{0} ∣}$ ; hence their ratio is a phase $e^{i α}$ , and equality of products forces the corresponding $\tilde{q}$ -entries to differ by $e^{- i α}$ . The two pairs are therefore U(1)-gauge-equivalent. The map in step 2.1 is injective as well as surjective, proving the stated bijection.

One-flavor U(1) classical F/D-flat vacuum quotient

Example

Example references

Facts & Assumptions

Verification

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