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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.

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  • $\begingroup$ (This question brought to you by the department of "could use MathJax, but didn't feel it was necessary") $\endgroup$
    – Yemon Choi
    Commented Sep 20, 2014 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
    Commented Sep 20, 2014 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
    Commented Sep 20, 2014 at 21:50

1 Answer 1

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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.

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  • $\begingroup$ 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
    Commented Sep 21, 2014 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
    Commented Sep 21, 2014 at 13:55

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