Subfactor

Remarks

Example: let $N$ be a $\mathrm{II}_1$ factor and $k \ge 1$. Then $N \otimes I_k$, the matrices $\operatorname{diag}(x, \dots, x)$ with $x \in N$, is a subfactor of $M = N \otimes M_k(\C)$, the $k \times k$ matrices over $N$. Both are factors with the same unit, and $L^2(M)$ is the direct sum of $k^2$ copies of $L^2(N)$ as an $N$-module, so $[M:N] = k^2$.

Non-example: the diagonal matrices $D_2 \subseteq M_2(\C)$ form a unital von Neumann subalgebra of the factor $M_2(\C)$. But $D_2$ is commutative and two-dimensional, so its center is $D_2 \ne \C I$, and $D_2$ is not a factor.

Another example: if a finite group $G$ acts on a $\mathrm{II}_1$ factor $M$ by outer automorphisms, the fixed-point algebra $M^G$ is a subfactor with $[M : M^G] = \abs{G}$. The index is $1$ exactly when $N = M$.