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LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)verified 2026-11-04 (gpt-6.1-sol)
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Trace pairing and a complete resolution of a split projector

Statement

Let C be a complex modular ribbon fusion category with its spherical pivotal trace, let Y be an object, and let P:Y→Y satisfy P2=P. For each simple label i, put Ei={e:Y→Ui∣eP=e},Fi={f:Ui→Y∣Pf=f}. The bilinear pairing Ei×Fi→C, (e,f)↦tr⁡j(ef), is nondegenerate. For arbitrary bases (ei,a) and (fi,b) of these two spaces, its matrix Cab(i)=tr⁡j(ei,afi,b) is invertible and P=∑i,a,bdi(C(i)−1)ba fi,bei,a,di=tr⁡j(id⁡Ui). Zero-dimensional blocks contribute an empty sum; the inverse of their empty matrix denotes the unique automorphism of the zero vector space. This identity does not divide by tr⁡j(P) or by dY, either of which may vanish.

Facts & Assumptions

Given: The object, idempotent, trace, fixed Hom spaces, and bases in the Statement.

[F1]

Modular ribbon fusion category and Hopf-link data gives finite semisimplicity, End⁡(Ui)=Cid⁡Ui, and the pivotal trace. An idempotent on a finite direct sum of simple objects splits by the ordinary image/kernel decomposition of its finite multiplicity matrices.

[F2]

Simple ribbon fusion dimensions are nonzero gives di≠0. Linearity of the pivotal trace gives tr⁡j(tid⁡Ui)=tdi for t∈C.

Proof

1.1givenF1algebra

Split P as Y→rZ→hY with rh=id⁡Z and hr=P, using the image/kernel decompositions in [F1]. The maps e↦eh and f↦rf identify Ei with Hom⁡(Z,Ui) and Fi with Hom⁡(Ui,Z). Their inverses are a↦ar and b↦hb: for example (eh)r=eP=e and r(hb)=b. Under these maps ef=(eh)(rf), so the pairing is unchanged.

2.1F1F2step 1.1algebra

Write Z=⨁i,tUi with inclusions hi,t:Ui→Z and projections ri,t:Z→Ui. They satisfy ri,thk,s=δikδtsid⁡Ui when i=k, and zero when i≠k, and ∑i,thi,tri,t=id⁡Z. Thus the pairing between Hom⁡(Z,Ui) and Hom⁡(Ui,Z) has matrix diI in these projection/inclusion bases. By [F2] it is nondegenerate, and this property persists under every invertible change of bases and the isomorphisms of step 1.1. Consequently every C(i) in the Statement is square and invertible.

3.1givenF1F2step 2.1algebra

Define R to be the displayed sum in the Statement. It satisfies PR=R=RP, because every ei,a and fi,b is fixed on its stated side. For fk,c in a chosen basis, Schur's lemma gives ei,afk,c=0 if i≠k, and ek,afk,c=(Cac(k)/dk)id⁡Uk if i=k. Therefore Rfk,c=∑a,bdk(C(k)−1)bafk,bCac(k)dk=fk,c. This calculation divides only by the nonzero simple dimension dk.

4.1step 1.1step 2.1step 3.1algebra∎

The maps hhi,t:Ui→Y from steps 1.1 and 2.1 belong to Fi and span its simple inclusions, so step 3.1 gives Rhhi,t=hhi,t for all i,t. Summing against ri,t gives Rh=h. Since R=RP=Rhr, it follows that R=hr=P. If Z=0, all these sums are empty and P=0, so the same conclusion holds. This proves both nondegeneracy and the resolution, without a nonzero total-dimension assumption.

Remarks

This lemma supplies the finite algebraic resolution after a boundary projector and its field-space identifications have been constructed. It does not identify a disk RT correlator with this trace pairing and does not construct the FRS connecting ribbon graph or the marked-torus projector.

Depends on

Dependency tree · 2 levels

Sources