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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 $x_\pm \in \mathfrak{p}_T$?

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Here is a counter-example. Which is sadly rather long. I use Haagerup's original paper 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.

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Edit: I had a simply incorrect attempt to show that $\mathfrak{m}_T$ is closed under taking absolute values.

This is equivalent to showing that $\mathfrak{m}_T$ is closed under absolute values. If $a \in \mathfrak{m}_T$ is self-adjoint, then again by the continuous functional calculus, $|a| = a_+ + a_-$, with $0 \leq a_+ \leq |a|$ and $0 \leq a_- \leq |a|$.

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  • $\begingroup$ there is a problem with the first inequality. If $a\geq 0$ with $\|a\|\leq 1$, then $a^2\leq a$. $\endgroup$ – Dave Penneys Nov 10 '16 at 15:49

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