Let $B=\int_{Y}^{\oplus}B_yd_y$ be the direct integral decomposition of vNa $B$ into factors and if $P=\int_{Y}^{\oplus} p_ydy$ is a projection in $B$ and $p_y$ is equivalent to a projection $q_y$ in $B_y$ for all or a.e. $y\in Y$. Is there a good choice of $q_y$ up to equivalence such that $Q=\int_{Y}^{\oplus} q_ydy$ makes sense and $P$ and $Q$ are equivalent in $B$?
$\begingroup$
$\endgroup$
6
-
1$\begingroup$ I don't think the question makes sense as is. Why not choose $q_y = p_y$? $\endgroup$– Jesse PetersonCommented Oct 17, 2011 at 21:40
-
$\begingroup$ @Jesse: I presume that when the OP says “good choice of $q_y$ up to equivalence” he means the choice of a representative of the equivalence class of a section with respect to sets of measure 0. $\endgroup$– Dmitri PavlovCommented Oct 18, 2011 at 7:47
-
$\begingroup$ That's certainly one interpretation. I wonder if the OP would mind clarifying what is meant by "good", "choice" (since the $q_y$ are already defined), and "up to equivalence"? $\endgroup$– Jesse PetersonCommented Oct 18, 2011 at 12:46
-
$\begingroup$ For each $y\in Y$ we can find a $q_y$ which is equivalent to $p_y$ (we don't know whether $q_y$ equals $p_y$ or not, let's say they are different), but $y\mapsto q_y$ is not necessarily measurable in the sense that $Q=\int_Y^{\oplus}q_ydy$ makes sense. So I want to find $q'_y$ which is equivalent to $q_y$ in $B_y$ and $y\mapsto q'_y$ is measurable. I need a kind of measurable selection theorem. Thanks for your attention :) $\endgroup$– m07klCommented Oct 18, 2011 at 15:18
-
$\begingroup$ @m07kl: So what is wrong with choosing $q_y' = p_y$? $\endgroup$– Jesse PetersonCommented Oct 18, 2011 at 15:35
|
Show 1 more comment
1 Answer
$\begingroup$
$\endgroup$
Suppose B is a trivial bundle whose fibers are type I2 factors and p is a constant section of B corresponding to some projection with 1-dimensional image. Projections with 1-dimensional image in a type I2 factor can be identified with angles, i.e., elements of iR/πiZ. If q is given by some non-measurable section of B, then there is no way to choose its equivalence class in such a way that Q exists.
answered Oct 17, 2011 at 19:58