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I think this should not be too difficult, but I am not an expert. I did not get an answer on stackexchange.

Let $A$ be a $C$*-algebra and let $p,q\in A^{**}$ be two commuting projections. Then there exist self-adjoint nets $(x_i)_i$ and $(y_j)_j$ in $A$ such that $x_i\to p$ and $y_j\to q$ in the weak$^*$-toplogy. Can these nets be chosen to be commutative, that is, $x_iy_j=y_jx_i$ for all $i,j$?

Extra question: Is this true if $A$ is a $JB$-algebra?

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I think the following provides a counterexample, though the bidual of a $C^*$-algebra always makes me nervous.

Let $A=M_2\otimes C[0,1]$. Any bounded, Borel measurable, $M_2$ valued function on $[0,1]$ will give an element of $A^{**}$; for the projection $p\in A^{**}$ we take the function \begin{equation} p(t)=\begin{cases} \begin{pmatrix} 1 & 0\\ 0 & 0\end{pmatrix} & 0\leq t\leq \frac12, \\ \begin{pmatrix} \frac12 & \frac12 \\ \frac12 & \frac12 \end{pmatrix} & \frac12 <t \leq 1\end{cases} \end{equation} and for $q$ we take the complementary projection $1-p$. Suppose by way of contradiction there are nets of selfadjoints $(x_i), (y_i)$ in $A$ with $x_i\to p$ and $y_i\to 1-p$ weak-$*$, and $x_iy_j=y_jx_i$ for all $i,j$. Since the $y_j$ converge to $1-p$ and commute with all the $x_i$, it follows that all the $x_i$ commute with $1-p$, and hence also with $p$. This now means that there are continuous $a_i, b_i, c_i, d_i$ so that $x_i$ has the form \begin{equation} x_i(t) = \begin{cases} \begin{pmatrix} a_i(t) & 0\\ 0 & b_i(t)\end{pmatrix} & 0\leq t\leq \frac12, \\ \begin{pmatrix} c_i(t) & d_i(t)\\ d_i(t) & c_i(t) \end{pmatrix} & \frac12 <t \leq 1\end{cases}. \end{equation} Continuity at $t=\frac12$ forces $x_i(\frac12 )$ to be a scalar multiple of the identity for all $i$, but then $x_i(\frac12 )$ cannot converge to $p(\frac12)$, which is a contradiction since weak-$*$ convergence controls pointwise congvergence.

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  • $\begingroup$ I agree you can map $A^{**}$ to $L^\infty([0,1], M_2)$ but I don't see how you are going back the other way $\endgroup$ – Yemon Choi Sep 20 '14 at 2:50
  • $\begingroup$ Not $L^\infty$, Borel; I am thinking $A^*$ is given by $M_2$-valued Radon measures $\mu$, and an $M_2$-valued Borel function $f$ acts on these by $\langle f, \mu \rangle= tr(\int f\, d\mu)$. (Am I all wet? This is why the bidual makes me nervous.) $\endgroup$ – Mike Jury Sep 20 '14 at 2:54
  • $\begingroup$ I'm not claiming that every element of the bidual looks like this, only that IF I have a bounded Borel function, I get an element of $C[0,1]^{**}$ out of it via $\langle f,\mu\rangle =\int f\, d\mu$. (And I guess I also need that for this part of the bidual, weak-$*$ convergence controls pointwise convergence, which is supposed to follow by testing against point masses.) $\endgroup$ – Mike Jury Sep 20 '14 at 3:03
  • $\begingroup$ OK, I misunderstood earlier. This looks like it works - nice! Do you think we could run the same argument on $A=M_2 \otimes C({\bf N}\cup\{\infty\})$, sticking your all-entries-the-same projection at the point at infinity? $\endgroup$ – Yemon Choi Sep 20 '14 at 14:53
  • $\begingroup$ @YemonChoi - Yes, that's a nice simplification and I think it still works; continuity at infinity would still force the $x_i(\infty)$ to be scalar multiples of $I_2$ which would prevent convergence. Also, I'm not sure why I wanted $q=1-p$ orginally; it seems like $q=p$ works fine. $\endgroup$ – Mike Jury Sep 20 '14 at 15:02

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