Manuscript XIV-B@1 · XIV-B.tex · theorem · 53202eaa

Theorem: Free-shell ghost sector of

B2: the free-shell ghost boundary · lines 751–786 · proof included or linked

XIV-B@1#thm:ghost, XIV-B@1#eq:intersection

Back to manuscript
  1. A ghost with is zero. An antighost with is zero.

  2. Consequently

    the bosonic free shell with zero ghost sector.

  3. At every configuration in (eq:intersection) the kinematic rule (eq:kin) takes the values

    Every -dependent term of vanishes there, and the only nonzero component is the frozen free rule . For an antifield-independent , on the intersection; for without antighost argument, in particular for , there.

Proof. (i) , compact in , so and Lemma lem:uniq applies; is compact in the open slab, so and the lemma applies again. (ii) A configuration of lies in exactly when its bosonic part lies in and its ghosts solve the free ghost equations; by (i) the latter forces . (iii) Each -dependent term of (eq:kin) is linear in , as is ; the antighost rule is -independent. The last assertion is (eq:sF) evaluated at .

Displayed formulas in this passage

  • XIV-B@1#eq:intersection · line 758

  • unlabeled · line 765

Nearby in this section