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Nilpotence and contraction of nonminimal doublets

Statement

The Abelian jet derivation $s$ defined in the local BRST algebra satisfies $s^{2} = 0$ . More generally, take any finite set of jet doublets $u^{j}, v^{j}$ with $∣ v^{j} ∣ = ∣ u^{j} ∣ + 1 (m o d 2)$ , $gh (v^{j}) = gh (u^{j}) + 1$ , $s u^{j} = v^{j}$ , and $s v^{j} = 0$ , and suppose the transformations of all other generators are independent of these doublets. Define the doublet-number operator $N = \sum_{j} (u^{j} \frac{\partial_{L}}{\partial u^{j}} + v^{j} \frac{\partial_{L}}{\partial v^{j}})$ and the odd homotopy $h = \sum_{j} u^{j} \frac{\partial_{L}}{\partial v^{j}}$ , using left graded derivatives. Then $[N, s] = 0$ and $s h + h s = N$ . Consequently the inclusion of the doublet-independent polynomial subalgebra induces an isomorphism on polynomial BRST cohomology. In particular, if $s a_{k} = 0$ and $N a_{k} = k a_{k}$ for a positive integer $k$ , then $a_{k} = s (h a_{k} / k)$ .

For a ghost-number grading, write this cohomology as $H^{r} (s, P) = \ker (s : P^{r} \to P^{r + 1}) / im (s : P^{r - 1} \to P^{r}) .$

Statement references

Abelian local jet BRST differential

Doublet diagram

For one fixed jet index, the arrows record the differential on the pair: $s (u^{j}) = v^{j}$ and $s (v^{j}) = 0$ . The zero at the right is the zero element. The proof verifies nilpotence and the homotopy identity algebraically; the diagram is only a record of the generator maps.

Facts & Assumptions

Given: The algebra and odd graded derivation in Abelian local jet BRST differential.

[F1]

Abelian local jet BRST differential: $P$ is a graded-commutative finite-jet polynomial algebra; $s$ is odd, raises ghost number by one, obeys the graded Leibniz rule, and commutes with spacetime derivatives. Its generator values are $s a_{μ} = \partial_{μ} c_{g h}$ , $s c_{g h} = 0$ , $s \overset{ˉ}{c} = B$ , $s B = 0$ , and $s ψ = i g c_{g h} ψ$ .

[A1]

For a finite chosen family of additional jet pairs, $s u^{j} = v^{j}$ and $s v^{j} = 0$ , with opposite parity and ghost number increased by one, and the $s$ -transformations of all other generators do not involve $u^{j}$ or $v^{j}$ .

Proof

Given: [F1] and [A1].

1.1givenF1algebra

On the gauge and nonminimal generators, $s^{2} a_{μ} = s (\partial_{μ} c_{g h}) = \partial_{μ} (s c_{g h}) = 0$ , $s^{2} c_{g h} = 0$ , $s^{2} \overset{ˉ}{c} = s B = 0$ , and $s^{2} B = 0$ . Commutation of $s$ with spacetime derivatives gives the same equality on every jet of these generators.

1.2givenF1algebra

Since $c_{g h}$ is odd in a graded-commutative algebra over $C$ , $c_{g h}^{2} = - c_{g h}^{2}$ and hence $c_{g h}^{2} = 0$ . The graded Leibniz rule therefore gives $s^{2} ψ = s (i g c_{g h} ψ) = - i g c_{g h} (i g c_{g h} ψ) = g^{2} c_{g h}^{2} ψ = 0.$

2.1step 1.1step 1.2algebra

The square of an odd derivation is an even derivation: the two mixed terms in $s^{2} (X Y)$ cancel by the graded Leibniz rule. It vanishes on every generator and every jet by steps 1.1–1.2, so it vanishes on each finite polynomial in $P$ .

3.1A1step 2.1algebra

Now $s^{2} = 0$ by step 2.1, so $s$ is a differential. It preserves the number of the selected doublet variables: its action on the other generators is independent of them, and each term $v^{j} \partial_{L} / \partial u^{j}$ replaces one $u^{j}$ by one $v^{j}$ . Thus $[N, s] = 0$ . On each pair, $h u^{j} = 0$ , $h v^{j} = u^{j}$ , $s u^{j} = v^{j}$ , and $s v^{j} = 0$ , so $(s h + h s) u^{j} = u^{j}$ and $(s h + h s) v^{j} = v^{j}$ . Both $s h + h s$ and $N$ are even derivations and vanish on the doublet-independent generators; equality on generators proves $s h + h s = N$ on the polynomial algebra.

4.1step 3.1algebra

Let $a_{k}$ be $s$ -closed and homogeneous of doublet number $k > 0$ . Since $[N, s] = 0$ , the homogeneous components of any $s$ -closed polynomial are separately $s$ -closed. Applying $s h + h s = N$ to $a_{k}$ gives $s (h a_{k}) = k a_{k}$ because $s a_{k} = 0$ ; division by the nonzero integer $k$ yields $a_{k} = s (h a_{k} / k)$ .

5.1step 3.1step 4.1algebra

Every polynomial has a finite decomposition $a = \sum_{k \geq 0} a_{k}$ . For an $s$ -closed $a$ , step 4.1 makes every $a_{k}$ with $k > 0$ exact, while $a_{0}$ is closed by $[N, s] = 0$ . Therefore every extended cohomology class has a representative in the doublet-independent subalgebra.

6.1step 3.1step 5.1algebra∎

If a doublet-independent closed polynomial $a_{0}$ is $s b$ in the extended algebra, take the doublet-number-zero component $b_{0}$ of the finite decomposition of $b$ . Because $s$ preserves $N$ , the number-zero component of $s b = a_{0}$ is $s b_{0} = a_{0}$ . Thus a class from the minimal subalgebra becomes exact in the extension only if it was already exact there, proving injectivity as well as the surjectivity in step 5.1.

Depends on

Abelian local jet BRST differential