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Let $B_0(H)$ be the compact operators on the Hilbert space $H$ and let $B \subseteq B_0(H)$ a $C^*$-subalgebra that acts non-degenerately on $H$. Let $\{p_i: i \in I\}$ be a maximal family of pairwise orthogonal minimal projections. The existence of this family is ensured by Zorn's lemma. Is it true that $$H = \bigoplus_i p_i (H)$$

Attempt:

Write $K$ for the direct sum. If $K$ is a proper subspace, we may fix a non-zero $\xi \in K^\perp$. The idea is now to use this vector to construct a minimal projection that will contradict maximality, but I'm not sure how to use the non-degeneracy to do this.

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Let $ p = \sum_{i\in I}p_i, $ where the sum is known to converge strongly. Then, for every $a$ in $B$, notice that $$ pa = \sum_{i\in I}p_ia, $$ where the convergence is now in norm since right-multiplication by compact operators turns strong convergence (of bounded nets) into norm convergence. Therefore $pa\in B$.

Now, assuming by contraction that $p\neq 1$, and since $B$ is non-degenenerate, there must be some element $a$ in $B$ such that $(1-p)a\neq 0$, and hence also that $$ c:= (1-p)aa^*(1-p) \neq 0. $$ Since $c$ is a self-adjoint compact operator in $B$, any one of its spectral projections, say $q$, also lie in $B$. It is also clear that $q\perp p_i$ for every $i$, and since $q$ is necessarily finite dimensional, we may choose a minimal projection among the projections of $B$ dominated by $q$. The existence of such a projection then contradicts the maximality of the originaly chosen family.

Ruy
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  • "since multiplication by compact operators turns strong convergence into norm convergence" Can you clarify what you mean by this? –  Dec 04 '20 at 21:00
  • If ${x_i}_i$ is a strongly converging net of bounded operators, with limit $x$, and if $b$ is a compact operator, then $bx_i\to bx$ in norm. – Ruy Dec 04 '20 at 21:03
  • I would expect that we at least need that ${x_i}$ is a norm-bounded net, because then you can easily approximate with finite rank operators. –  Dec 04 '20 at 21:05
  • Ok, I guess you indeed need to assume boundedness! – Ruy Dec 04 '20 at 21:06
  • And this is satisfied in our case by orthogonality of the projections, I believe? –  Dec 04 '20 at 21:07
  • Yes, it is, since the net is formed by finite sums of the $p_i$, which have norm 1. – Ruy Dec 04 '20 at 21:08
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    Indeed I think it is easier to prove that $x_ib\to xb$. Perhaps I should edit to make the answer depend on this instead. – Ruy Dec 04 '20 at 21:10
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    I am starting to doubt the statement in my second comment above in case the convergence of the $x_i$ is not in the $^*$-strong topology (which it is in my answer). – Ruy Dec 04 '20 at 21:25
  • Do you think the answer contains a possible flaw? –  Dec 04 '20 at 21:26
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    My answer is now based on right multiplication by compacts so it is OK. My doubt is just whether or not my comment above holds. – Ruy Dec 04 '20 at 22:38
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    @user839372, please check this: https://math.stackexchange.com/questions/3935048/does-left-multiplication-by-compact-operators-turn-strong-convergence-into-norm – Ruy Dec 04 '20 at 23:28