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Let $\mathcal{H}$ be Hilbert space and $\mathfrak{B(}\mathcal{H}\mathcal{)}$ of all bounded linear operators on $\mathcal{H}$. Let $\mathcal{A}$ be a maximal commutative sub-algebra of $\mathfrak{B(}\mathcal{H)}$.

  1. Is there an explicit formula a conditional expectation (a norm one projection) $\pi:\mathfrak{B(}\mathcal{H}\mathcal{)\longrightarrow A}$?
  2. If Yes, I would be greatful if you could please give me it. \end{enumerate}
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frege, maximal abelian subalgebras of von Neumann algebras are von Neumann algebras, hence are of the form $L_\infty(\mu)$ for some measure $\mu$. See this post of Bill Johnson to see a neat way to produce a projection. The problem is that in general masas of $\mathscr{B}(\mathcal{H})$ are elusive, so are the corresponding norm-one projections.

In some particular cases, however, this is not difficult. For instance, when $\mathcal{H}=\ell_2$ and you regard operators on $\mathcal{H}$ as matrices with respect to the canonical basis, you just chop off the diagonal to produce a projection.

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  • $\begingroup$ In their famous 1959 paper where they introduce the Kadison-Singer problem (recently solved), R.V. Kadison and I.M. Singer prove that, in the case of $\ell^2$, taking the diagonal is the only projection onto the corresponding MASA of diagonal operators; and that for $L^2[0,1]$ there are at least two projections on the MASA $L^\infty [0,1]$; see jstor.org/discover/10.2307/… $\endgroup$
    – Alain Valette
    Commented Dec 15, 2013 at 21:50
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    $\begingroup$ If there is a norm one projection, then $\cal A$ has to be self-adjoint, because unital norm one maps are automatically positive. Every self-adoint maximal abelian subalgebra ${\cal A}\subset{\cal B}(H)$ is unitarily equivalent to $L^\infty(X,\mu)\subset{\cal B}(L^2(X,\mu))$. $\endgroup$
    – Narutaka OZAWA
    Commented Dec 15, 2013 at 23:25

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