Finite projection in Von Neumann algebra - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T18:26:38Z http://mathoverflow.net/feeds/question/108357 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/108357/finite-projection-in-von-neumann-algebra Finite projection in Von Neumann algebra Qingyun 2012-09-28T16:26:39Z 2013-01-02T15:52:07Z <p>I had the following question when I am learning von Neumann algebras: Let p be a finite projection in a finite von Neumann algebra $M$, let $p>p_1>p_2>\cdots$ be a decreasing sequence of projections such that $p_i\neq p_{i+1}$. Is such a sequence necessarily of finite length? Any help or recommendations on books/papers are greatly appreciated.</p> http://mathoverflow.net/questions/108357/finite-projection-in-von-neumann-algebra/108360#108360 Answer by Manny Reyes for Finite projection in Von Neumann algebra Manny Reyes 2012-09-28T17:08:16Z 2013-01-02T15:52:07Z <p>The answer to your question is <strong>no</strong>. Abelian von Neumann algebras are finite, but it is easy to find examples of such algebras with infinite decreasing sequences of projections. For instance, one may take $M = L^\infty(\mathbb{R},\lambda)$ where $\lambda$ is the Lebesgue measure, and let $p_i \in M$ be the element corresponding to the characteristic function of $[0,\frac{1}{i}]$.</p> <p><strong>Update:</strong> Now that you've seen some counterexamples, this might be a good follow-up "homework" problem for you (depending on how much you know about <a href="http://en.wikipedia.org/wiki/Maharam%27s_theorem" rel="nofollow">measure theory</a> and <a href="http://en.wikipedia.org/wiki/Abelian_von_Neumann_algebra#Classification" rel="nofollow">abelian von Neumann algebras</a>): prove that a (finite) von Neumann algebra $M$ satisfies your descending chain condition on projections if and only if it is a finite dimensional algebra. </p> <p>(Hint: think about a type decomposition.)</p> http://mathoverflow.net/questions/108357/finite-projection-in-von-neumann-algebra/108363#108363 Answer by Ulrich Pennig for Finite projection in Von Neumann algebra Ulrich Pennig 2012-09-28T17:16:44Z 2012-09-28T22:03:05Z <p>I do not think that this true. Take for example the sequence of matrix algebras with $A_0 = \mathbb{C}$ and $A_n = M_2(\mathbb{C}) \otimes A_{n-1}$ equipped with their normalized traces and the diagonal embeddings $A_{n} \to A_{n+1}$. The direct limit $A$ of this sequence is a $*$-algebra with a trace and the GNS-construction with respect to this trace state yields a (faithful) representation $H$, on which $A$ acts. The weak closure $M$ of $A$ is a von Neumann algebra, which is of type $II_1$ (in particular, it is finite). This is known as the hyperfinite type $II_1$-factor. </p> <p>Now consider the sequence of projections starting with $p_0 = 1 \in A_0$, $$p_1 = \begin{pmatrix} 1 &amp; 0 \\ 0 &amp; 0 \end{pmatrix} \in A_1 = M_2(\mathbb{C})$$</p> <p>$$p_2 = \begin{pmatrix} 1 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 0\end{pmatrix} \in A_2 \cong M_2(\mathbb{C}) \otimes M_2(\mathbb{C})$$</p> <p>etc. (I hope the pattern is clear). We have $tr(p_n) = 2^{-n}$. Therefore $p_i \neq p_j$ if $i \neq j$. But $p_0 > p_1 > p_2 > \dots$</p> <p>In fact, constructions like this are the reason for many of the cool features of type $II$ von Neumann algebras. </p>