Jones index theorem

Remarks

The proof uses the basic construction: $M$ acts on $L^2(M)$, and adding the orthogonal projection $e_N$ onto $L^2(N)$ gives a larger factor $M_1 = \langle M, e_N \rangle$ with $[M_1 : M] = [M : N]$. Iterating gives a tower $N \subseteq M \subseteq M_1 \subseteq M_2 \subseteq \cdots$ with projections $e_1, e_2, \dots$ that satisfy $e_i e_{i \pm 1} e_i = \lambda e_i$ with $\lambda = [M:N]^{-1}$ and $e_i e_j = e_j e_i$ for $\abs{i - j} \ge 2$. The trace restricted to the algebras generated by $e_1, \dots, e_k$ must be positive, and this forces $\lambda^{-1}$ into the set above. For the values below $4$, the construction of the converse uses these same projections inside $R$.

The first allowed values are $1$, $2$, $(3 + \sqrt 5)/2 \approx 2.618$ and $3$, for $n = 3, 4, 5, 6$. So no subfactor has index strictly between $1$ and $2$, although the index need not be an integer.

Going deeper

For a student

The relations among the projections $e_i$ are those of the Temperley–Lieb algebra. Jones used representations of the braid group built from them to discover the Jones polynomial of knots. Finer invariants of a subfactor, such as its principal graph and its standard invariant, lead to the classification of subfactors of small index. For the fixed points $M^G$ of an outer action of a finite group $G$ on $M$, one has $[M : M^G] = \abs{G}$, so the index generalizes the index of a subgroup.