But this implies $ \chi ({X_{1},\dots ,X_{n}}|{Y_{1},\dots ,Y_{m}})\leq \chi ({X_{1},\dots ,X_{n}}|{Z_{1},\dots ,Z_{r}}) $.

$ \chi ({X_{1},\dots ,X_{n}}|{Y_{1},\dots ,Y_{m}})=\chi ({X_{1},\dots ,X_{n}}|W^{*}({Y_{1},\dots ,Y_{m}})) $.

By definition, $$\chi =\chi ({X_{1},\dots ,X_{n}}|W^{*}({Y_{1},\dots ,Y_{m}}))=\inf _{r}\inf _{{Z_{1},\dots ,Z_{r}}\in W^{*}({Y_{1},\dots ,Y_{m}})}\chi ({X_{1},\dots ,X_{n}}|{Z_{1},\dots ,Z_{r}}).$$ In particular, if $ r=m $ and $ ({Z_{1},\dots ,Z_{r}})=({Y_{1},\dots ,Y_{m}}) $, we have that $$\chi \leq \chi ({X_{1},\dots ,X_{n}}|{Y_{1},\dots ,Y_{m}}).$$ But by Proposition \[prop: generation\], we also have that, for $ Z_{1},\dots ,Z_{r}\in W^{*}({Y_{1},\dots ,Y_{m}}) $, $$\chi ({X_{1},\dots ,X_{n}}|{Y_{1},\dots ,Y_{m}})\leq \chi ({X_{1},\dots ,X_{n}}|{Z_{1},\dots ,Z_{r}})\leq \chi ,$$ so that $ \chi =\chi ({X_{1},\dots ,X_{n}}|{Y_{1},\dots ,Y_{m}}) $.

If $ {Y_{1},\dots ,Y_{m}}\in B $, then $ \chi ({X_{1},\dots ,X_{n}}|B)\leq \chi ({X_{1},\dots ,X_{n}}:{Y_{1},\dots ,Y_{m}}) $.

We have $ \chi ({X_{1},\dots ,X_{n}}|B)\leq \chi ({X_{1},\dots ,X_{n}}|{Y_{1},\dots ,Y_{m}})\leq \chi ({X_{1},\dots ,X_{n}}:{Y_{1},\dots ,Y_{m}}) $.

Let $ B=L^{\infty }[0,1]\subset M $ be a diffuse commutative von Neumann subalgebra. For each $ r=1,\dots ,n $ let $ \mu _{r} $ be a measure on $ [0,1]^{2} $ determined by $$\iint f(x)g(y)d\mu _{r}(x,y)=\tau (X_{r}fX_{r}g).$$ Assume that for at least one $ r\in \{1,\dots ,n\} $, Lebesgue measure on $ [0,1]^{2} $ is singular with respect to $ \mu _{r} $. Then $ \chi ({X_{1},\dots ,X_{n}}|B)=-\infty  $.

Let $ Y $ be a self-adjoint 