I have a number of confusions that I think a good reference might be able to sort out.

Let $A$ be a von Neumann Algebra. At the moment I am mostly concerned with $\dim A<\infty$, but am interested in more general results.

Consider these three ideas:

  1. Let $\nu\in S(A)$ be a state on $A$. Define the following null space:

$$N_\nu=\left\{b\in A\,|\,\nu(|b|^2)=0\right\}.$$

This set is a $\sigma$-weakly closed left ideal. Therefore there exists a projection $Q_\nu$ such that $N_\nu=AQ_\nu$. Some properties include the fact that $b\in N_\nu$ if and only if $bQ_\nu=b$. Also, for all $a\in A$ we have $$\nu(Q_\nu)=\nu(aQ_\nu)=\nu(Q_\nu a)=0.$$

Question 1: Is the map $a\mapsto aQ_\nu$ the projection onto $N_\nu$?

Define the projection $P_\nu:=1_A-Q_\nu$. We have $$\nu(a)=\nu(aP_\nu)=\nu(P_\nu a)=\nu(P_\nu aP_\nu),$$ and $\nu(P_\nu)=1$.

Question 2: Is there a projection $P\leq P_\nu$ such that $\nu(P)=1$?

  1. Following this answer, consider the projection $$\operatorname{supp}\,\nu=1_A-\bigvee\left\{p\in A,\,\text{a projection such that }\nu(p)=0\right\}.$$

Connecting with the above:

Question 3: Does $\operatorname{supp}\,\nu$ coincide with $P_\nu$ (from above)?

  1. I cannot find the definition of the support projection of $\nu$. My naive definition is that the support projection of $\nu$ would be the smallest projection $p$ such that $\nu(p)=1$. However my understanding is that the support projection would usually live in the same space as $\nu$ (and that is not what I am trying to model).

Question 4: Does the definition of $\operatorname{supp}\,\nu$ (above) coincide with "the smallest projection $p$ such that $\nu(p)=1$".

An overarching question

Question: Is there a good reference (possibly lecture notes) which can discuss the (non?-)relationships between these concepts?

Context: let $F(X)$ be the algebra of functions on a finite set $X$. Let $\nu\in M_p(X)\subset \mathbb{C}X$ be a probability on $X$. There are a number of ways that we can think about the support of $\nu$:

  • it is the subset $S\subset X$ comprised of elements $s\in X$ such that $\nu(\delta_s)>0$. Fix $S$ in the remaining.
  • Consider the set $S_\nu=\{g\in F(X)\,|\,\nu(|g|^2)>0\}$. This a subspace of $F(X)$. Let $Q$ be the projection in $L(F(X))$ onto $S_\nu$. The projection $P=1_X-Q$ coincides with the map $f\mapsto \mathbf{1}_Sf$.
  • Note that there is no projection $P$ in $F(X)$, $P\leq \mathbf{1}_S$ such that $\nu(P)=1$. In this sense $\mathbf{1}_S$ is the smallest projection $p\in F(X)$ such that $\nu(p)=1$.
  • 1
    $\begingroup$ Q1. What do you mean with "the" projection onto? A von Neumann algebra isn't necessarily a Hilbert space. Q2. No. Q3. Yes. Q4. Yes. Q5. arxiv.org/pdf/1804.02203.pdf See "carrier of a map" (63) and "corner" (94). (Edit: note that a state is automatically completely positive.) $\endgroup$ – westerbaan May 31 at 11:05
  • $\begingroup$ @westerbaan this is an answer. Submit it as such and I will accept. I needed Q.1 for intuition more than anything. Is there anyway to interpret that map as a projection onto the subspace $N_\nu$? $\endgroup$ – JP McCarthy May 31 at 11:32
  • $\begingroup$ I guess you are working in the GNS space ? $\endgroup$ – Epsilon May 31 at 14:01
  • $\begingroup$ @Epsilon I know it seems as if it is set up to quotient out and go there I don't think I am. $\endgroup$ – JP McCarthy May 31 at 19:45
  1. There is no "the" projection (and sometimes not even "a" projection) onto a von Neumann algebra. Your map is an idempotent that maps onto $N_\nu$, but there is nothing interesting about it, as far as I can tell.

  2. No. If $\nu(P)=1$, then $\nu(P_\nu-P)=0$, which implies $P_\nu-P\in N_\nu$, so $$ 0=Q_\nu(P_\nu-P)=P_\nu-P. $$

  3. Yes, you can prove this using 2. As for "I cannot find the definition..." I have no idea what you mean, as you wrote the definition three lines above.

  4. Yes. That's 2.

I think this is in every classic text on von Neumann algebras. The first one that comes to mind is Chapter 7 in Kadison-Ringrose.

  • $\begingroup$ Thank you. As commented upon above, I asked Q.1 out of intuition more than anything. If I have an algebra with a faithful trace, and I define the inner product in the usual way using it, is the map above a projection in that case (it seems to me to me)? Just curious, not a massive deal $\endgroup$ – JP McCarthy Jun 1 at 5:26
  • $\begingroup$ Just trying to finish off 2. Is it the case that $P_\nu-P\in N_\nu$ implies that $(P_\nu-P)Q_\nu=P_\nu-P$. Then, because $P\leq P_\nu$ we have $P=PP_\nu$. Then $(P_\nu-P)Q_\nu=(P_\nu-PP_\nu)Q_\nu=(1_A-P)P_\nu Q_\nu=0$? $\endgroup$ – JP McCarthy Jun 1 at 7:43
  • 1
    $\begingroup$ The map is indeed a projection in that sense (you can check that it is selfadjoint with respect to that product). What you don't necessarily have is that the range is closed (in the same sense that in general a von Neumann algebra with a trace is not closed with respect to the inner product). $\endgroup$ – Martin Argerami Jun 2 at 23:23

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