Relation between tracial norm and operator norm on a von Neumann algebra

Question

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.

Answer 1

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).

Answer 2

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.