Let $N\subseteq M$ be an inclusion of semi-finite factors with normal faithful semi-finite traces $\operatorname{Tr}\_N$ and $\operatorname{Tr}\_M$ respectively. Let $T: M^+\to \widehat{N^+}$ be the unique trace-preserving normal faithful semi-finite operator valued weight.


As for normal weights, we define
\begin{align\*}
\mathfrak{n}\_T 
&= 
\left\{x\in M \,\middle|\, T(x^\*x)\in N^+\right\}
\\
\mathfrak{m}\_T 
&= 
\mathfrak{n}\_T^\*\mathfrak{n}\_T
\\
\mathfrak{p}\_T 
&=
\left\{x\in M^+\,\middle|\,T(x)\in N^+\right\}
\end{align\*}
It is straightforward to show that $\mathfrak{n}\_T$ is a left ideal, $\mathfrak{m}\_T\subset \mathfrak{n}\_T\cap \mathfrak{n}\_T^\*$ is a hereditary $\*$-algebra, both $\mathfrak{n}\_T$ and $\mathfrak{m}\_T$ are $N-N$ bimodules, and $\mathfrak{m}\_T$ is spanned by its positive part $\mathfrak{m}\_T^+$, which in turn is equal to $\mathfrak{p}\_T$.
Moreover, $T$ has a canonical extension to a map $\mathfrak{m}\_T\to N$.


Thus if $x\in \mathfrak{m}\_T$ is self-adjoint, we can write $x=x\_1-x\_2$ with $x\_1,x\_2\in \mathfrak{p}\_T$.
But it is not clear to me if $\mathfrak{m}\_T$ is closed under taking the Hahn-Jordan decomposition of $x$. That is:



> 
> Suppose $x\in \mathfrak{m}\_T$ is self-adjoint and $x=x\_+-x\_-$ is the Hahn-Jordan decomposition of $x$, where $x\_+$ and $x\_-$ are positive and $x\_+x\_-=0$. Does it follow that $x\_\pm \in \mathfrak{p}\_T$?
> 
> 
> 



Is the domain of an operator valued weight closed under Hahn-Jordan decomposition?
Here is a counter-example. Which is sadly rather long. I use [Haagerup's original paper](https://www.sciencedirect.com/science/article/pii/0022123679900533) for the definition of $T$ below.


Let $M$ be the $\ell^\infty$ direct sum of the matrix algebras $\mathbb M\_n$ with $\tau\_M$ being the sum of the un-normalised traces on $\mathbb M\_n$. Let $N$ be the subalgebra of all diagonal matrices, and give $N$ the ``weighted'' trace
$$ \tau\_N(x) = \sum\_n \frac{1}{n} \sum\_i x^{(n)}\_i. $$
where $x = (x\_n)\in N$ and each $x\_n$ has diagonal entries $(x^{(n)}\_i)$.
Below I'll regard $N$ as the direct sum of $\ell^\infty\_n$.


Let $T:M\_+\rightarrow \widehat{N\_+}$ be the operator valued weight, so
$$ \tau\_M(y^{1/2}xy^{1/2}) = \tau\_N(y^{1/2} T(x) y^{1/2}) \qquad (x\in M, y\in N). $$
In general, $T(x)$ is not bounded, so a little care is needed to define the right-hand side. However, it follows from this that $T$ must respect the direct-sum decomposition, and so we obtain maps $T\_n:\mathbb M\_n \rightarrow \ell^\infty\_n$ which satisfy
$$ \sum\_i y\_i x\_{i,i} = \frac{1}{n} \sum\_i y\_i T\_n(x)\_i
\qquad (x=(x\_{i,j})\in \mathbb M\_n^+, y=(y\_i)\in(\ell^\infty\_n)^+) $$
Thus $T\_n(x)\_i = n x\_{i,i}$, that is, project onto the diagonal and multiply by $n$.


Suppose we can find positive matrices $a\_n, b\_n$ in $\mathbb M\_n$ so that $\|a\_n\|\leq 1, \|b\_n\|\leq 1$, with the diagonal entries bounded by $1/n$, but with $n |a\_n - b\_n|$ having diagonal entries which are unbounded, as $n\rightarrow\infty$. Set $a=(a\_n) \in M$ and $b=(b\_n)$. By the assumptions on $(a\_n)$ and $(b\_n)$ we see that $a,b \in \mathfrak{m}\_T^+$ but that $|a-b|\not\in\mathfrak{m}\_T^+$. This is equivalent to $x=a-b \in \mathfrak{m}\_T$ but $x\_+\not\in\mathfrak{m}\_T^+$.


How do we find $a\_n,b\_n$? Here's one construction. Let $(\delta\_i)\_{i=1}^n$ be the usual orthonormal basis of $\mathbb C^n$ and let
$$ \xi = \frac{1}{\sqrt n} \sum\_{i=1}^n \delta\_i, \qquad
\eta = \frac{1}{\sqrt n} \sum\_{i=1}^{n-1} \delta\_i. $$
Let $a\_n = a = \theta\_{\xi,\xi}$ the rank-one positive operator, and let $b\_n = b = \theta\_{\eta,\eta}$. As matrices, $a$ has all entries $1/n$ and $b$ has all entries $1/n$ except for the last row and column which are 0. We claim that $|a-b|$ has diagonal entries of size order $\sqrt n$, to be precise, $|a-b|\_{n,n}$ is about $\sqrt n$.


I don't see an easy way to show this. I spent ages working out a formula for $|\theta\_{\xi,\xi} - \theta\_{\eta,\eta}|$ for arbitrary vectors $\xi,\eta$ in a Hilbert space, and then performed the particular calculation here. If anyone has a nice argument, I'd like to see it.


