5
$\begingroup$

This feels like something I may have asked before (in which case, apologies) and it also might be some kind of "standard counterexample in a book on C* algebras".

Let M be a unital C*-algebra and let A, B be unital, closed star-subalgebras (so they are C*-algebras in their own right). Let q be the quotient map of Banach spaces from M onto M / B. Is q(A) necessarily closed in M / B ?

(Of course if B were an ideal then the answer is yes, because any star-homomorphism between C*-algebras has closed range.)

If it makes any difference, I'm primarily interested in the case where M is a von Neumann algebra and A and B are sub- von Neumann algebras.

$\endgroup$
  • $\begingroup$ (This question brought to you by the department of "could use MathJax, but didn't feel it was necessary") $\endgroup$ – Yemon Choi Sep 20 '14 at 20:01
  • $\begingroup$ Let D be the closed unit disk. Take M=C(D), A=A(D), and let B be the algebra of continuous on D functions that are constant on [-1,1]. Then q is essentially the restriction map (up to factoring out constants, but that won't save the day), so q(A) is dense in M/B but certainly not the whole space... (The answer is formatted for sending to the same department :-)) $\endgroup$ – fedja Sep 20 '14 at 21:41
  • $\begingroup$ @fedja Nice example in general (I'll consider it as an exercise to show my student!) but I am after Cstar subalgebras. I'll edit to make it clearer $\endgroup$ – Yemon Choi Sep 20 '14 at 21:50
6
$\begingroup$

If $q(A)$ is closed in $M/B$, then its preimage $A+B$ is closed in $M$. Here is an example when it is not: Let $M=C([0,1],M_2)$, let $p,q\in M$ be the rank one projections $$ p= \begin{pmatrix} 1 & 0\\ 0&0 \end{pmatrix}, q(t)= \begin{pmatrix} 1-t & t^{1/2}(1-t)^{1/2}\\ t^{1/2}(1-t)^{1/2} & t \end{pmatrix}. $$ Let $A$ be the elements of the form $fp$, with $f\in C[0,1]$ and $B$ the element of the form $gq$, with $g\in C[0,1]$ (i.e., $A=pMp$, $B=qMq$). Then the elements of $A+B$ have the form $$ fp+gq= \begin{pmatrix} f+g\cdot (1-t) & gt^{1/2}(1-t)^{1/2}\\ gt^{1/2}(1-t)^{1/2} & gt \end{pmatrix}. $$ Observe that the rate of decay near zero of the bottom right corner is at least linear. So choosing functions $g_n\in C[0,1]$ such that $g_nt^{1/2}\to t^{1/4}$ and $f_n=-g_n\cdot (1-t)$ we get the matrix $$ \begin{pmatrix} 0 & t^{1/4}(1-t)^{1/2}\\ t^{1/4}(1-t)^{1/2} & t^{3/4}, \end{pmatrix} $$ in the closure but not in the algebraic sum. The subalgebras $A$ and $B$ don't share $M$'s unit but this can be fixed by adding the unit to them: the elements of $A+B+\lambda 1_2$ have $\lambda+gt$ in the bottom right corner, so the same matrix as before does not belong to it.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks! This is especially nice since my underlying question behind the scenes concerned the case where M is a Type I vN algebra and A, B are the ranges of conditional expectations. If I understand your construction correctly it works, with trivial modifications, for M being $L\otimes M_2$ when $L$ is any of $C(\bf N \cup\{\infty\})$, $\ell^\infty$ or $L^\infty([0,1])$ -- in all these cases things in $A+B$ are forced to converge at least as fast as some prescribed rate in the (2,2) position, and you have something in the closure which doesn't. Is that correct? $\endgroup$ – Yemon Choi Sep 21 '14 at 12:59
  • $\begingroup$ Hi Yemon. Yes, those work too. The example can be adapted to this situation: $p$ and $q$ are projections in a C*-algebra $M$ such that 0 is not an isolated point of the spectrum of $(1-p)q(1-p)$. Then $pMp+qMq$ is not closed. $\endgroup$ – Leonel Robert Sep 21 '14 at 13:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.