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First, let me preface this by saying that I am fairly new to the wide field of (finite) von Neumann algebras. In my studies of $L^2$-invariants, I am mostly concerned with Group von Neumann algebras, but it will be very helpful to gain some more insight into general finite von Neumann algebras.

Let $\mathcal A$ be a von Neumann algebra, i.e a unital, weakly closed $*$-subalgebra of $B(\mathcal H)$, the space of bounded operators over some Hilbert space $\mathcal H$, and assume additionally that $\mathcal A$ is equipped with a finite, positive, faithful and normal trace $\phi: \mathcal A \to \mathbb C$, such that, wlog, $\phi(1) = 1$. Denote by $||.||_1$ the operator norm on $B(\mathcal H)$.

Then, $\phi$ determines an inner product on $\mathcal A$ given by $\langle a,b \rangle := \phi(ab^*)$ for any pair $a,b \in \mathcal A$. Denote the induced tracial norm on $\mathcal A$ by $||.||_\phi$ and by $l^2(\mathcal A)$ the Hilbert space completion of $\mathcal A$ with respect to $\langle \;,\; \rangle$. Assume that $l^2(\mathcal A)$ is infinite-dimensional and seperable, and denote by $||.||_2$ the operator norm on $B(l^2(\mathcal A))$. Then left-multiplication by elements in $\mathcal A$ identifies elements of $\mathcal A$ with certain bounded operators over $l^2(\mathcal A)$, giving rise to the well-known left-regular representation $L: \mathcal A \hookrightarrow B(l^2(\mathcal A))$, an embedding satisfying $\|a\|_\phi \leq \|L(a)\|_2 \leq \|a\|_1$ for all $a \in \mathcal A$.

Of course, in most circumstances, one will not have an inequality of the form $\|L(a)\|_2 \leq C\|a\|_\phi$ for all $a \in \mathcal A$ and for some $C > 0$, but there are certain relations between the tracial norm and the $\|\cdot\|_2$-Norm that do occur regularly and are of very special use to me. Let me list three properties.

(1) There exists an orthonormal unitary system of $\mathcal A$, i.e, an orthonormal base $(b_i)_{i = 1}^\infty \subset \mathcal A$ of $l^2(\mathcal A)$, such that $L(b_i) \in B(l^2(\mathcal A))$ is a unitary operator.

(2) There exists an orthonormal base $(b_i)_{i = 1}^\infty \subset \mathcal A$ of $l^2(\mathcal A)$ such that $\|L(b_i)\|_2 = \|b_i\|_\phi = 1$.

(3) There exists an orthonormal base $(b_i)_{i = 1}^\infty \subset \mathcal A$ of $l^2(\mathcal A)$, such that $\sup_{i \in \mathbb N}\|L(b_i)\|_2 < \infty$.

Of course $(1) \Rightarrow (2) \Rightarrow (3)$ and all von Group von Neumann algebras $\mathcal A = \mathcal N(G)$ with $G$ a countably infinite group already trivially satisfy $(1)$ (with the obvious choice for $\phi$ and orthonormal base).

Do there exist finite von Neumann algebras that satisfy (2), but not (1), and/or (3), but not (2) ? Do there exist finite von Neumann algebras that do not even satisfy (3). If so, are they classified (up to some equivalence) ?

Do there exist finite von Neumann algebras that satisfy (1), but are not Group von Neumann algebras?

Any help is appreciated.

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    $\begingroup$ An idea for possible examples of von Neumann algebras that satisfy (1): consider measure preserving actions on measure spaces and form the crossed product algebra. This is a finite von Neumann algebra and the associated Hilbert space is of the form $L^{2}(X,\mu)\otimes \ell^2(G)$, so if $L^{2}(X,\mu)$ admits an ONB formed by function with absolute value $1$, then this von Neumann algebra satisfies (1). There should be examples, when these algebras are not group von Neumann algebras. $\endgroup$ – Mateusz Wasilewski Jan 25 '18 at 18:42
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    $\begingroup$ Actually, any non-atomic measure space $(X,\mu)$ is isomorphic to the unit circle endowed with the Lebesgue measure, so we can transfer the trigonometric basis in this case. It means that any measure-preserving action on a non-atomic measure space gives rise to a von Neumann algebra that satisfies (1); many of these are not group von Neumann algebras. $\endgroup$ – Mateusz Wasilewski Jan 25 '18 at 21:53
  • $\begingroup$ Thanks for your detailed answer and various hints. As a beginner in this field, I wonder how to show that a given finite von Neumann algebra is not (isomorphic to) a group von Neumann algebra ... $\endgroup$ – Berni Waterman Jan 25 '18 at 22:13
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    $\begingroup$ Well, that's one of the reasons that I put this remark in a comment rather than in an answer; it's not that easy to prove that something is not a group von Neumann algebra. One property of group von Neumann algebras is that they are isomorphic to their opposite algebras (with reversed multiplication), which one can achieve using the inverse map on the group. But examples that do not have this property are, as far as I can tell, hard to obtain. I am sure that experts in the field know way more about von Neumann algebras that are not group algebras. $\endgroup$ – Mateusz Wasilewski Jan 26 '18 at 9:15
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    $\begingroup$ The question of whether every II$_1$ factor has an orthonormal basis consisting of unitaries is a well known open problem which was first asked by Kadison. See: On “Problems on von Neumann Algebras by R. Kadison, 1967”, by Ge. $\endgroup$ – Jesse Peterson Feb 6 '18 at 21:51
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This is only a partial answer.

I will show that (1) and (2) are equivalent. Since the left regular representation is isometric, I will forget about $L$ and I will simply denote the operator norm by $\|a\|$. It suffices to show that if an element satisfies $\|a\|=\|a\|_{\phi}=1$ then it is unitary. In order to prove that, note that $\|a\|=1$ implies that $\|a^{\ast}a\|=1$, hence $a^{\ast}a \leqslant \text{Id}$. By positivity of the trace, we get $\phi(\text{Id}-a^{\ast}a)\geqslant 0$. On the other hand, it is equal to $0$, hence by faithfulness we obtain $a^{\ast}a=\text{Id}$. As $\phi$ is a trace, this works equally well for $aa^{\ast}$, so $aa^{\ast}=a^{\ast}a=\text{Id}$.

I don't have yet a clear idea, where to look for possible counterexamples to other statements, but maybe the following paper would help: Adrian Ioana, Sorin Popa, and Stefaan Vaes, A class of superrigid group von Neumann algebras, Ann. of Math. (2), vol. 178 (2013), no. 1, 231--286; doi: 10.4007/annals.2013.178.1.4, arXiv:1007.1412. The authors show, among other things, that corners (i.e. algebras of the form $pMp$, where $p\in M$ is a projection) of certain group von Neumann algebras are never group von Neumann algebras. It is conceivable that these corners should satisfy condition (3).

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I think this observation may lead to a solution of the first question. Notice that $\mathcal{A} \subset L^2(\mathcal{A})$ is dense and that

  1. If it holds that, given $S \subset \mathcal{A} \subset L^2(\mathcal{A})$ a finite dimensional subspace, its orthogonal complement $$S^\perp = \{ \xi \in L_2(\mathcal{A}) : \phi(a^\ast \, \xi) = 0, \forall a \in S \}$$ has dense intersection with $\mathcal{A}$. Then (3) holds.

  2. If for all $S \subset \mathcal{A}$, $S^\perp$ has dense intersection with the span of $U(\mathcal{A})$, the unitary group of $\mathcal{A}$, then (1) holds.

Observations 1 and 2 hold because they allow us to iteratively choose an orthogonal base with the desired properties.

I do not know whether 1 or 2 hold in general. Maybe you can try to use the fact that $\mathcal{A}$ is closed by functional calculus to define the functional $T_{a,b}: C_b(\mathbb{C}) \to \mathbb{C}$ by $$ f \mapsto \phi(a^\ast \, f(b)). $$ At least for normal $b$ the formula above makes sense. Then, the kernel of $T_{a,b}$ will have codimension 1 and picking an element $f$ in $$ \bigcap_{a \in S} \ker(T_{a,b}) $$ Will give you $f(b) \in S^\perp \cap \mathcal{A}$. Probably refining such type of argument you can prove observation 1 above.

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  • $\begingroup$ I don't understand how $S^\perp \cap \mathcal A$ can ever be dense in $\mathcal A$ if we have a non-trivial set $S \subset \mathcal A$. What topology are you using? $\endgroup$ – Jesse Peterson Feb 6 '18 at 21:42
  • $\begingroup$ I meant that $S^\perp \cap \mathcal{A}$ should be dense inside $S^\perp$ not in $\mathcal{A}$. In any case the argument just need that $S^\perp \cap \mathcal{A}$ is not empty for every finite dimensional $S$. If that is the case, we can choose $x \in S^\perp \cap \mathcal{A}$ to be of $\| \cdot \|_{\mathcal{A}}$-norm one at every step. That will give an orthogonal base $(b_i)_i$ with $\| b_i \|_{\mathcal{A}} = 1$ but it may not be true that $\| b_i \|_{\phi} = 1$. $\endgroup$ – Adrián González-Pérez Feb 7 '18 at 13:06

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