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Aug 7, 2015 at 4:13 comment added Sebastien Palcoux @JessePeterson: can you weaken the assumption " $\tilde{T} \in M$ " ?
Aug 6, 2015 at 18:59 comment added Jesse Peterson You still need that $T$ is contained in $M$. In Michael's example the operator $T$ is not contained in $M$.
Aug 6, 2015 at 18:18 comment added Sebastien Palcoux @JessePeterson: I don't understand your last sentence about the abelian case, it seems in contradiction with the comment of Michael.
Aug 5, 2015 at 21:23 comment added Jesse Peterson You just need a dimension function which assigns equal dimensions to equivalent projections. You don't need factoriality. In the abelian case $M = L^\infty(X, \mu)$ this just says $\mu( E \cup F ) = \mu(E) + \mu(F)$ for disjoint measurable sets $E$ and $F$.
Aug 5, 2015 at 4:57 comment added Sebastien Palcoux @JessePeterson: is it still true if we remove one of the two extra assumptions (a priori "$M$ is a factor" is useless in your argument)? What could be your ultimate form for the rank-nullity theorem for finite von Neumann algebras (at least conjecturally)?
Aug 4, 2015 at 17:00 comment added Jesse Peterson With the extra assumption this is just as easily seen to be true since then the projections onto the ranges of $\tilde Tp$ and $(\tilde T p)^*$ are equivalent in $M$ (just consider the partial isometry in the polar decomposition of $\tilde T p$).
Aug 4, 2015 at 7:34 comment added Michael Without the extra assumption in your comment, it seems to me you could just take something like $H = L^2([0,1])$, $M = L^\infty([0,1]) \subset B(H)$, $\mathrm{tr} = \int dx$, $p=1$, $q_1 = \chi_{[0.5]}$, $q_2=0$ and $T \in B(H)$ given by $(T \xi)(x) = \begin{cases} \xi(2x) & \text{ if } 0 \leq x \leq .5 \\ 0 & \text{ if } .5 \leq x \leq 1 \end{cases}$.
Aug 4, 2015 at 6:41 comment added Sebastien Palcoux If necessary we can assume that $M$ is a factor and $T = \tilde{T}p\vert_{pH}$ with $\tilde{T} \in M$.
Aug 4, 2015 at 6:24 history asked Sebastien Palcoux CC BY-SA 3.0