Operator Valued Weights - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T20:19:33Z http://mathoverflow.net/feeds/question/946 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/946/operator-valued-weights Operator Valued Weights Dave Penneys 2009-10-17T21:54:29Z 2010-05-31T22:23:13Z <p>One of the basic tools in subfactors is the conditional expectation. If $N\subset M$ is a $II_1$-subfactor (or an inclusion of finite factors), then there is a unique trace-preserving conditional expectation of $M$ onto $N$. This should be thought of as a (Banach space) projection of norm 1. In fact, it is the restriction of the Jones projection $e_N$ on $L^2(M)$ to $M$. In the finite index case, we get another conditional expectation (Jones projection...) from the basic construction $M_1=\langle M, e_N\rangle$ onto $M$.</p> <p>In his thesis, Michael Burns showed that if we iterate the basic construction in the infinite index case, we only get half the conditional expectations (we only get the odd Jones projections). The other half of the time, we get a generalization of the conditional expectation called an operator valued weight, originally defined by Haagerup.</p> <p>Given an inclusion of semifinite von Neumann algebras $(N, tr_N)\subset (M, tr_M)$, there is a unique normal, faithful, semi-finite trace-preserving operator valued weight $T\colon M_+\to \widehat{N_+}$, where we must take the "extended part" of the positive cone $N_+$ of $N$. </p> <p>Edit as per @Dmitri's answer: Let $$n_T=\{x\in M| T(x^\ast x)\in N_+\}$$ and set $$m_T=n_T^\ast n_T=span\{x^\ast y| x,y\in n_T\}.$$ There is a natural extension of $T$ to $m_T$. Is there an example of a normal, faithful, semifinite operator valued weight such that </p> <ul> <li>$N$ is not contained in $T(m_T)$, and/or</li> <li>$1\notin T(M_+)$?</li> </ul> <p>What about when $M$ and $N$ are factors ($M$ is $II_\infty$)?</p> http://mathoverflow.net/questions/946/operator-valued-weights/1250#1250 Answer by Dmitri Pavlov for Operator Valued Weights Dmitri Pavlov 2009-10-19T18:11:44Z 2009-10-19T18:11:44Z <p>If T is a faithful semifinite operator valued weight from M to N, then the image of the extended positive positive cone of M coincides with the extended positive cone of N. See Proposition 1.13 in the paper by Falcone and Takesaki. So the answer to your answer is yes if we extend T from the positive cone of M to the extended positive cone of M (such an extension is unique).</p> http://mathoverflow.net/questions/946/operator-valued-weights/26632#26632 Answer by Jesse Peterson for Operator Valued Weights Jesse Peterson 2010-05-31T22:23:13Z 2010-05-31T22:23:13Z <p>Since $T(m_T)$ is a non-trivial two sided ideal in $N$, it follows from spectral calculus that $T(m_T)$ contains a non-zero projection. $T(m_T)$ then contains all subprojections, hence there exists $x \in m_T$ such that $p = T(x)$ is a projection with trace Tr$(p) = 1/n$ for some natural number $n$. </p> <p>We may assume that $pxp = x$, and by considering $(x + x^*)/2$ we may assume that $x$ is self-adjoint. Since $m_T$ is spanned by its positive part it follows that we can write $x$ as $x_1 - x_2$ where $x_j \in m_T$ are positive. Hence $T(x_1) \geq p$ and so by considering the element $(pT(x_1)^{-1/2}p)x_1(pT(x_1)^{-1/2}p) \in m_T$ we may assume that $x \geq 0$.</p> <p>If $N$ is a factor, then since Tr$(p) = 1/n$ we may find a (finite, if $N$ is finite) sequence of partial isometries $v_k$ such that $v_k^*v_k = p$ for each $k$ and $\Sigma_k v_kv_k^* = 1$. Then $x_m = \Sigma_{k = 1}^m v_k x v_k^*$ is a bounded increasing sequence in $m_T$ and we have that $T(x_m)$ increases to $1$. Since the weight is normal we have that $x_\infty = \Sigma_k v_k x v_k^* \in m_T \cap M_+$ and $T(x_\infty) = 1$. In particular, since $T(m_T)$ is an ideal we have $N \subset T(m_T)$.</p> <p>If $N$ has infinite dimensional center then as Dmitri explained there are many examples (even bounded ones) for which $1 \not\in T(M_+)$. I imagine that for finite dimensional center the argument above should still work.</p>