Let $\mathcal M\subseteq B(\mathcal H)$ be a von Neumann algebra. Then, what should be natural definition of a von Neumann subalgebra?
1 Answers
I'd say a von Neumann subalgebra of a von Neumann algebra $\mathcal{M}$ on a Hilbert space $\mathcal{H}$ should at least be a subset $\mathcal{N}$ of $\mathcal{M}$ that is itself a von Neumann algebra—or in other words, a weakly closed $C^*$-subalgebra of $\mathcal{M}$. But one might perhaps require in addition that $\mathcal{N}$ contains the unit of $\mathcal{M}$ (depending on the context.)
This definition of von Neumann subalgebra seems (but doesn't) depend on the choice of the Hilbert space $\mathcal{H}$. Indeed, one can show (using Lemma 1 of this paper of Kadison's) that a $C^*$-subalgebra $\mathcal{N}$ of a von Neuman algebra $\mathcal{M}$ is a von Neumann subalgebra of $\mathcal{M}$ if and only if for every bounded directed set $D$ of self-adjoint elements of $\mathcal{M}$ the supremum $\bigvee D$ in $\mathcal{M}$ of $D$ is in $\mathcal{N}$. Although the supremum $\bigvee D$ is just the strong (or weak) limit of the net $(d)_{d\in D}$, the supremum $\bigvee D$ does not depend not on $\mathcal{H}$ (being defined in terms of the order on $\mathcal{M}$.)

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I agree with your definition. But I have seen some places they insist that $\mathcal N$ has to be an ultraweakly closed subalgebra of $\mathcal M$? Are these two definitions same? – A beginner mathmatician Sep 25 '19 at 05:52
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1@ABeginnerMathmatician Yes, and in fact, for a $C^$-algebra $\mathcal{A}$ of operators on a Hilbert space $\mathcal{H}$, the following are equivalent: $\mathcal{A}$ is a von Neumann algebra; $\mathcal{A}$ is weakly closed; $\mathcal{A}$ is strongly closed; $\mathcal{A}$ is ultraweakly(=$\sigma$-weakly=weak$^$) closed; $\mathcal{A}$ is ultrastrongly closed; $\mathcal{A}''=\mathcal{A}$; $\mathcal{A}$ contains all suprema in $\mathcal{M}$ of bounded directed sets of self-adjoint elements from $\mathcal{A}$. – Bram Westerbaan Sep 25 '19 at 08:20