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Explicit $N = 2$ shortened rest fiber at central-charge saturation

Example

Let $M > 0$ and $z \in C$ satisfy $∣ z ∣ = M$ , and set $ω = z / ∣ z ∣$ . On $H = Λ^{*} C^{2}$ with the orthonormal occupation basis and fermionic operators $a_{1}, a_{2}$ defined by $a_{1} ∣ n_{1}, n_{2} ⟩ = n_{1} ∣ 0, n_{2} ⟩$ , $a_{2} ∣ n_{1}, n_{2} ⟩ = (- 1)^{n_{1}} n_{2} ∣ n_{1}, 0 ⟩$ , set $P^{0} = M I$ , $P^{j} = 0$ , $Z^{12} = z$ , $Z^{21} = - z$ , and

$Q_{1}^{1} = \sqrt{2 M} a_{1}, Q_{2}^{1} = \sqrt{2 M} a_{2}, Q_{2}^{2} = ω \sqrt{2 M} a_{1}^{*}, Q_{1}^{2} = - ω \sqrt{2 M} a_{2}^{*},$

with every barred charge the actual Hilbert adjoint and parity $(- 1)^{n_{1} + n_{2}}$ . These operators give a four-dimensional positive rest fiber with $M = ∣ z ∣$ . The two combinations $Q_{1}^{1} - ω (Q_{2}^{2})^{*}$ and $Q_{2}^{1} + ω (Q_{1}^{2})^{*}$ vanish identically. Use natural units $ℏ = c = 1$ , so $M$ and $z$ have mass dimension one and each displayed supercharge has mass dimension one-half. This is a finite-dimensional algebra representation, not a claim about a normalizable sharp-momentum subspace in a full spacetime representation.

Example references

Gates et al., §3.2.

Martin, §1.

Four-dimensional real extended super-Poincare algebra: Four-dimensional real extended super-Poincare algebra.

Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ : Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ .

Hilbert space: Hilbert space.

Real and complex inner-product spaces and their induced length: Real and complex inner-product spaces and their induced length.

Facts & Assumptions

Given: $M > 0$ , $z \in C$ with $∣ z ∣ = M$ , $ω = z / ∣ z ∣$ , the finite two-mode Fock space, its CAR operators, and the $N = 2$ mostly-plus rest algebra.

[F1]

Four-dimensional real extended super-Poincare algebra: at rest, ${Q_{α}^{I}, (Q_{β}^{J})^{*}} = 2 M δ_{α β} δ^{I J} I$ , ${Q_{α}^{I}, Q_{β}^{J}} = 2 ϵ_{α β} Z^{I J}$ , and $ϵ_{12} = 1$ .

[F4]

Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ : $2 M > 0$ has a positive square root.

Verification

Given: [F1], [F4] and the four basis vectors in the Example.

1.1givenF1F4algebra

The displayed occupation rules and their adjoints give the two-mode CAR on each of the four basis vectors: for equal mode labels, creation followed by annihilation and annihilation followed by creation are the projections onto occupied and empty states, whose sum is the identity; for distinct labels, the two orders of changing the different occupations have opposite signs from the factor $(- 1)^{n_{1}}$ . Repeated creation or annihilation is zero. Therefore ${a_{i}, a_{j}^{*}} = δ_{i j} I$ and ${a_{i}, a_{j}} = {a_{i}^{*}, a_{j}^{*}} = 0$ . The same-mode mixed charges in the rest relation now give $2 M I$ , while every mismatched label is either a distinct-mode anticommutator or a repeated annihilator/creator and is zero. This verifies the mixed relation in [F1].

2.1F1step 1.1algebra

Since $∣ z ∣ = M > 0$ , the definition $ω = z / ∣ z ∣$ gives $z = M ω$ and direct scalar arithmetic gives $ω \overset{ˉ}{ω} = z \overset{ˉ}{z} / M^{2} = ∣ z ∣^{2} / M^{2} = 1$ and $\overset{ˉ}{ω} z = M$ . For the same-flavor pairs, two annihilators or two creators anticommute by step 1.1, so their central bracket is zero. The two nonzero distinct-flavor brackets are ${Q_{1}^{1}, Q_{2}^{2}} = 2 M ω {a_{1}, a_{1}^{*}} = 2 z I$ and ${Q_{2}^{1}, Q_{1}^{2}} = - 2 M ω {a_{2}, a_{2}^{*}} = - 2 z I$ . Every other distinct-flavor pair is either a distinct-mode anticommutator or a square of one creation/annihilation operator, and so has zero anticommutator. These are exactly $2 ϵ_{α β} Z^{I J}$ with $ϵ_{12} = 1$ and antisymmetric $Z$ ; taking Hilbert adjoints gives the conjugate barred relations.

3.1step 1.1step 2.1algebra∎

Since $(Q_{2}^{2})^{*} = \overset{ˉ}{ω} \sqrt{2 M} a_{1}$ and $(Q_{1}^{2})^{*} = - \overset{ˉ}{ω} \sqrt{2 M} a_{2}$ , direct substitution and $ω \overset{ˉ}{ω} = 1$ give $Q_{1}^{1} - ω (Q_{2}^{2})^{*} = 0$ and $Q_{2}^{1} + ω (Q_{1}^{2})^{*} = 0$ . Thus $M = ∣ z ∣$ and both shortening combinations vanish by direct finite matrix calculation; no general central-charge theorem or Choice hypothesis is used. The Hilbert space has two even and two odd basis vectors, and all domains equal the full finite-dimensional space.

Explicit $N = 2$ shortened rest fiber at central-charge saturation

Example

Example references

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources

Explicit N=2N=2 shortened rest fiber at central-charge saturation

Example

Example references

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources

Explicit $N = 2$ shortened rest fiber at central-charge saturation