On continuity of the index of subfactors of a finite factor (Q1100694)

The authors consider the behaviour of the index of N in L as N varies, where the von Neumann algebra L is a type \(II_ 1\) factor with normalized trace \(tr_ L\) and N is subfactor of L. First the authors set up their notations. According to \textit{V. F. R. Jones} [Invent. Math. 72, 1-25 (1983; Zbl 0508.46040)] the index of N in L is defined as follows: if M is a finite factor acting on a Hilbert space H with M' also finite and \(tr_ M\) resp. \(tr_{M'}\) is the normalized trace on M resp. M', then the coupling constant \(\dim_ M(H)\) is \(tr_ M(E_ f^{M'})/tr_{M'}(E_ f^ M)\), where f is a nonzero vector in H and \(E_ f^ M\) resp. \(E_ f^{M'}\) is the orthogonal projection of H onto \(\overline{Mf}\) resp. \(\overline{M'f}\). If L is acting on a Hilbert space H then the index of N in L is \[ [L:N]:= \begin{cases} \dim_ N(H)/\dim_ L(H) &\text{ if \(N'\) is finite, } \\ \infty &\text{ if \(N'\) is infinite.}\end{cases} \] If M, N are two subfactors of L, then the distance d(M,N) between M and N is defined as the Hausdorff distance between the unit balls of M and N using the norm \(\| \cdot \|_ 2\) on L induced by \(tr_ L\). Let SF(L) denote the set of all subfactors of L with same identity and let SFT(L) denote the subset of SF(L) consisting of subfactors with trivial relative commutant in L, then (SF(L),d) is a complete metric space and SFT(L) is a closed subset of SF(L). Next the authors define a map index by \[ \text{index}(\cdot):SF(L)\to [1,\infty],\quad M\mapsto \text{index}(M):=[L:M]. \] After these prelimaries the authors prove their main results, which can be stated as follows: 1. The map index is a lower semicontinuous function on SF(L), but 2. in general it is not continuous on SF(L), because for any hyperfinite type \(II_ 1\) factor \(M\in SF(L)\) with index\((M)<\infty\), there exists a sequence \(\{M_ n\}_{n\in {\mathbb{N}}}\) in SF(L) with \(M_ n\subseteq M\), (n\(\in {\mathbb{N}})\), and \(d(M_ n,M)\to 0\) but index \((M_ n)\to \infty\), as \(n\to \infty.\) 3. The restriction of the map index to the closed subset SFT(L) is a continuous map from SFT(L) into \([1,\infty]\), and the index is constant on a suitable neighbourhood of any point in SFT(L). 4. The discontinuity described above is only the one that can occur. For \(1\leq C<\infty\), let \(SF_ C(L):=\{M\in SF(L)| index(M)\leq C\}\). For a fixed C the restriction of the map index to \(SF_ C(L)\) is a continuous map into [1,C].