Partial order - Unbounded normal operators affiliated with von Neumann algebra. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T18:07:35Z http://mathoverflow.net/feeds/question/67791 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/67791/partial-order-unbounded-normal-operators-affiliated-with-von-neumann-algebra Partial order - Unbounded normal operators affiliated with von Neumann algebra. Romanov 2011-06-14T18:34:10Z 2011-06-16T16:52:06Z <p>Hello, I have a question which is related to a partial order in a set of self-adjoint operators.</p> <p>Let $\mathcal{M}$ be a semifinite von Neumann algebra with a faithful semi-finite normal trace $\tau$. Let $T$ and $S$ be two self-adjoint operators (possibly unbounded) $\tau$-measurable (here probably the assumption that they are affiliated with $\mathcal{M}$ is enough) such that $0 \leq T \leq S$ i.e. $S-T$ is positive. How to get that $$E_{(s, \infty)}(|T|) \preceq E_{(s, \infty)}(|S|), \ \ s \geq 0,$$ where $E_I(|T|)$ (resp. $E_I(|S|)$) stands for a spectral projection of $T$ (resp. $S$) corresponding to the interval $I$ and $\preceq$ means sub-equivalence relation in Murray-von Neumann sense.</p> <p>I am looking also for some good references which describe the relation between $U|T|$ the elements of the polar decomposition of closed densely defined (possibly unbounded) operator $T$ affiliated with some von Neumann algebra $\mathcal{M}$. I mean that $U$ and each spectral projection of $|T|$ are in this von Neumann algebra. Probably, I can find this in Takesaki vol 2 or vol 3. </p> <p>I will be really grateful for any help.</p> <p>Thank you, VdM</p> http://mathoverflow.net/questions/67791/partial-order-unbounded-normal-operators-affiliated-with-von-neumann-algebra/67974#67974 Answer by Martin Argerami for Partial order - Unbounded normal operators affiliated with von Neumann algebra. Martin Argerami 2011-06-16T16:52:06Z 2011-06-16T16:52:06Z <p>I assume you are following the proof in Fack-Kosaki (if you are not, we are talking here about Proposition 2.2 and 2.5 there). </p> <p>Note that there is no need for absolute value bars since both $T,S$ are positive. </p> <p>The key fact is that $E_{(s,\infty)}(T)\wedge E_{[0,s]}(S)=0$ (to be proven afterwards). Using this, we have (using Kaplansky's formula) <code>$E_{(s,\infty)}(T)=E_{(s,\infty)}(T)-E_{(s,\infty)}(T)\wedge E_{[0,s]}(S)\sim E_{(s,\infty)}(T)\vee E_{[0,s]}(S)-E_{[0,s]}(S)\leq I-E_{[0,s]}(S)=E_{(s,\infty)}(S)$</code></p> <p>So we only need to prove that $E_{(s,\infty)}(T)\wedge E_{[0,s]}(S)=0$. Now, if $\xi\in E_{(s,\infty)}(T)H \cap E_{[0,s]}(S)H$ with $\|\xi\|=1$, the following happens: <code>$\langle T\xi,\xi\rangle=\langle TE_{(s,\infty)}(T)\xi,E_{(s,\infty)}(T)\xi\rangle =\|T^{1/2}E_{(s,\infty)}(T)\xi\|^2&gt;s,$</code> <code>$\langle T\xi,\xi\rangle=\langle E_{[0,s]}(S)TE_{[0,s]}(S)\xi,\xi\rangle \leq\langle E_{[0,s]}(S)SE_{[0,s]}(S)\xi,\xi\rangle=\|S^{1/2}E_{[0,s]}(S)\xi\|^2\leq s$</code> The contradiction implies that $\xi$ cannot exist. </p>