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  "body": "\n## Statement\n\nLet $N \\subseteq M$ be a [[subfactor|9RWS7TZHXH]] of type $\\mathrm{II}_1$ factors with Jones index $[M:N] \u003c \\infty$.\nThen\n$$[M : N] \\in \\set{4 \\cos^2(\\pi / n) : n = 3, 4, 5, \\dots} \\cup [4, \\infty).$$\nConversely, every number in this set is the index of a subfactor of the [[hyperfinite type II1 factor|NZ4NY8SK1C]] $R$.\n\n## Remarks\n\nThe 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 $[M_1 : M] = [M : N]$.\nIterating 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 $\\abs{i - j} \\ge 2$.\nThe trace restricted to the algebras generated by $e_1, \\dots, e_k$ must be positive, and this forces $\\lambda^{-1}$ into the set above.\nFor the values below $4$, the construction of the converse uses these same projections inside $R$.\n\nThe first allowed values are $1$, $2$, $(3 + \\sqrt 5)/2 \\approx 2.618$ and $3$, for $n = 3, 4, 5, 6$.\nSo no subfactor has index strictly between $1$ and $2$, although the index need not be an integer.\n\n## Explanations\n\n### L1 human\n\nThe index of a subfactor measures how much bigger the large algebra is than the small one.\nOne might expect any number at least $1$ to be possible.\nJones found that below $4$ only a discrete list of values $4\\cos^2(\\pi/n)$ can occur, while every number from $4$ on can occur.\n\n### L2 human\n\nThe relations among the projections $e_i$ are those of the Temperley–Lieb algebra.\nJones used representations of the braid group built from them to discover the Jones polynomial of knots.\nFiner invariants of a subfactor, such as its principal graph and its standard invariant, lead to the classification of subfactors of small index.\nFor 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.\n",
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