Unique trace on a type II$_1$ factor.

Question

Let $M \subseteq B(H)$ be a type $II_1$ factor. Then any two non-zero ultraweakly continuous normalised traces $Tr,tr : \rightarrow \mathbb{C}$ are equal. I'm trying to understand this from what Jones has written about it, but I'm having a hard time understanding it, as it's all rather brief. What I know is, that the equivalence classes of projections on a type $II_1$ factor are abstractly isomorphic to [0,1]. Apparently, that is something I have to use. Could someone explain to me how the proof works? Thank you

Answer 1

This is one way of doing it (a completely different and more general way can be found in Section 8.2 in Kadison-Ringrose). Let $\tau$ be a unital trace.

First note that it is enough to determine the values of the trace on the projections, as all operators are linear combinations of norm limits of linear combinations of projections (i.e. real and imaginary parts, plus the spectral theorem).

The fact we need (and requires some work to do, it is done in detail in Sunder's Invitation book, for instance) is that given any projection $p$, we can find a projection $q\leq p$ with $q\sim p-q$ (it is enough to show this for $p=1$ and then to notice that $pMp$ is also a II$_1$-factor).

So, for any $n$, there exist pairwise orthogonal equivalent projections $p_1,\ldots,p_{2^n}$ with sum $1.$ This forces $\tau(p_j)=2^{-j}$.

Now given any projection $q\in M$ and $n\in\mathbb N$, there is a "Division Algorithm": there exist pairwise orthogonal equivalent projections $p_1,\ldots,p_k$ with trace $2^{-j}$ and another projection $r\prec p_1$ with $q=p_1+\cdots+p_k+r$. This forces $$ k2^{-j}\leq\tau(q)\leq k2^{-j}+\tau(r)\leq(k+1)2^{-j}. $$ This way there is a canonically determined sequence of dyadic numbers converging to $\tau(q)$, and so the trace is completely determined by the equivalence of projections.

Note that ultraweak continuity can be then proven as a consequence, it is not needed to argue the uniqueness of the trace: one only needs that it is a unital positive tracial functional.

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In the case of an arbitrary II$_1$ von Neumann algebra, uniqueness fails when not a factor. For instance, let $N$ be a II$_1$-factor with trace $\tau_N$, and put $M=N\oplus N$. Then the two traces $$ (x,y)\longmapsto \frac13\tau_N(x)+\frac23\tau_N(y),\ \ \ (x,y)\longmapsto \frac23\tau_N(x)+\frac13\tau_N(y), $$ are not equal.