Strong Additivity, Haag Duality, and the μ-Index

Haag duality identifies the algebra of one interval with the commutant of its complementary interval; strong additivity says that deleting one interior point does not delete observables; and the $\mu$ -index measures the failure of duality for two disjoint intervals. These are distinct properties. Only their precise combination supports the finite-sector conclusions used later.

Required background. Conformal Nets and Covariance Axioms supplies interval locality, and Circle and Interval Nets: Möbius and Diffeomorphism Covariance fixes circle complements and geometric transport. Helpful background. Von Neumann Factors and Type-III Local Algebras explains why index replaces Hilbert-space dimension, while The Split Property and Approximate Tensor Products explains separated-interval tensor products.

One interval, one point, and a complement

For a proper interval $I\subset S^1$ , write $I'$ for the interior of its complement. Locality always gives

\mathcal A(I)\subset\mathcal A(I')'.

Haag duality on the circle is equality. In the usual irreducible Möbius-covariant vacuum formulation it follows from the Bisognano–Wichmann modular action and is therefore a theorem, not an additional finiteness assumption. The corresponding assertion on the punctured circle, identified with $\mathbb R$ , needs care because complements can be disconnected. Passing between circle and line duality without naming the point at infinity can conceal a dual completion. The basic definitions and this distinction appear in Kawahigashi, Longo, and Müger 2001, §2, pp. 635–638.

Now choose $p\in I$ and let $I_1,I_2$ be the two connected components of $I\setminus\{p\}$ . Strong additivity is

\mathcal A(I_1)\vee\mathcal A(I_2)=\mathcal A(I).

Ordinary additivity does not imply this formula: an ordinary interval cover may overlap near every point, whereas $I_1$ and $I_2$ are separated precisely by the deleted point. Strong additivity asserts that no additional bounded observable is localized only by retaining that point. It is independent of circle Haag duality and must be checked or obtained from another theorem.

The two-interval inclusion

Take four intervals $I_1,I_2,I_3,I_4$ in cyclic order and set $E=I_1\cup I_3$ , so $E'=I_2\cup I_4$ . Define

\mathcal A(E)=\mathcal A(I_1)\vee\mathcal A(I_3), \qquad \widehat{\mathcal A}(E)=\mathcal A(E')'.

Locality gives the subfactor inclusion $\mathcal A(E)\subset\widehat{\mathcal A}(E)$ . When the split property and strong additivity hold, the Jones index is independent of the chosen four-interval configuration; the common value is the $\mu$ -index,

\mu_{\mathcal A}= \bigl[\mathcal A(E')':\mathcal A(E)\bigr].

Finite $\mu_{\mathcal A}$ is a quantitative defect of two-interval Haag duality. The value $1$ means equality for this disconnected region; a value larger than $1$ records charged intertwiners visible in the dual algebra but absent from the algebra generated componentwise. Proposition 5 and Definition 8 of Kawahigashi, Longo, and Müger 2001, §§2–3, pp. 638–641 establish configuration independence under the stated hypotheses and place finite index inside complete rationality.

The proof mechanism is subfactor-theoretic. Split identifies the two component algebra with a spatial tensor product, strong additivity controls changes of endpoints, and Möbius covariance moves configurations. Jones-index invariance under the resulting isomorphisms then makes $\mu$ geometric rather than coordinate-dependent. None of those steps says that every local net has finite index.

Current-net application and checks

For the level-one $SU(2)$ current net, the vacuum and fundamental sectors have statistical dimensions $1$ and $1$ . Complete rationality therefore gives

\mu_{SU(2)_1}=1^2+1^2=2.

This is the exact two-interval statement returned to Affine Current Algebras and WZW Models: remove a point, verify the two remaining component algebras generate the original interval algebra, and then form the alternating two-interval inclusion. The identity $\mu=\sum_\rho d(\rho)^2$ is Theorem 33 of Kawahigashi, Longo, and Müger 2001, pp. 654–656; using it here is licensed because the current net is completely rational, not as a definition for an arbitrary net.

An independent check uses the sector fusion rule $\frac12\otimes\frac12=0$ : both sectors are invertible, so both dimensions must be $1$ , and their squared sum is $2$ . A second check is order-theoretic: Jones index is at least $1$ , so any proposed $\mu<1$ is impossible before one examines the model.

What finite index controls

Finite index is stronger than finite-dimensionality of a selected relative commutant. It supplies a normal conditional expectation from $\widehat{\mathcal A}(E)$ onto $\mathcal A(E)$ with a Pimsner–Popa lower bound governed by $\mu_{\mathcal A}^{-1}$ . That bound controls the canonical endomorphism and hence the total statistical dimension of localized sectors. Conversely, observing that one relative commutant is finite-dimensional does not provide the expectation, configuration independence, or sector completeness. This is why the two-interval inclusion—not a character count—is the quantitative input to complete rationality.

Adversarial failure

If ordinary additivity is substituted for strong additivity, the point-removal equality has not been proved. If circle Haag duality is written with a disconnected real-line complement as though it were one interval, the wrong commutant is used. Finally, a finite list of familiar sectors does not prove finite $\mu$ : completeness of the list is part of the theorem’s input or conclusion. The strongest surviving claim in each case is a local inclusion, not complete rationality or modularity.

Exercises

Assume a completely rational net has irreducible sectors of dimensions $1,1,\sqrt2$ . Compute its $\mu$ -index and explain why the answer cannot be inferred if further sectors may exist.

Solution

Completeness of the list and the global-dimension theorem give $\mu=1+1+2=4$ . If the list is only partial, positivity gives merely $\mu\ge4$ when the global-dimension formula applies; without complete rationality even that formula is not automatically licensed.