von Neumann algebra contains the hyperfinite factor $R$

Question

We know the fact that any type II$_1$ factor contains the hyperfinite II$_1$ factor $R$.

My question: If we let $M$ be an arbitrary type II$_1$ von Neumann algebra. can we have $R\subseteq M$?

Answer 1

Yes.

In an arbitrary von Neumann algebra, the centre could be insanely big. So the first step is to work in a reasonably sized area. Fix a nonzero normal state $\varphi\in M_*$, and let $s$ be its support. Then $sMs$ is a type II$_1$ algebra with faithful state $\varphi$ (if $M$ acts on a separable Hilbert space, one can get $\varphi$ faithful from the start). If $\tau$ denotes the unique centre-valued trace on $M$, then $\psi=\varphi\circ\tau$ is a faithful normal tracial state on $sMs$. So we will assume from now on that $M$ has a faithful normal tracial state $\psi$.

We will need on the following lemma:

Lemma.(Halving) Let $M$ be a von Neumann algebra with no type I part, and let $p\in M$ be a projection. Then there exists a projection $q\in M$ with $q\leq p$ and $q\sim p-q$.

Using the Lemma, one starts with $p_0=1\in M$ and finds $e_{11}^1\leq p_0$ with $e_{11}^1\sim p_0-e_{11}^1=e_{22}^1$. So there exists a partial isometry $e_{12}^1\in M$ with $(e_{12}^1)^*e_{12}^1=e_{22}^1$ and $e_{12}^1(e_{12}^1)^*=e_{11}^1$. Letting $e_{21}^1=(e_{12}^1)^*$ the four elements $e_{11}^1,e_{12}^1,e_{21}^1,e_{22}^1$ behave like matrix units. That is, $\def\espan{\operatorname{span}}$ $\espan\{e_{kj}^1,\ k,j=1,2\}\simeq M_2(\mathbb C)$. We also have, since $\psi$ is a trace, $\psi(e_{11}^1)=\psi(e_{22}^1)=\frac12$, $\psi(e_{12}^1)=\psi(e_{21}^1)=0$.

Now for induction suppose that we have matrix units $\{e_{kj}^\ell\}\subset M$. Via the Halving Lemma there exist equivalent projections $e_{11}^{\ell+1}$ and $e_{22}^{\ell+1}$ with $e_{11}^{\ell+1}+e_{22}^{\ell+1}=e_{11}^{\ell}$, and a partial isometry $e_{12}^{\ell+1}$ with $(e_{12}^{\ell+1})^*e_{12}^{\ell+1}=e_{11}^{\ell+1}$ and $e_{12}^{\ell+1}(e_{12}^{\ell+1})^*=e_{11}^{\ell+1}$. Now we define, for $k,j=1,\ldots,2^\ell$,$\def\abajo{\\[0.2cm]}$ \begin{align} e_{2k-1,2j-1}^{\ell+1}&=e_{k1}^\ell e_{11}^{\ell+1} e_{1j}^\ell\abajo e_{2k,2j-1}^{\ell+1}&=e_{k2}^\ell e_{21}^{\ell+1} e_{1j}^\ell\abajo e_{2k,2j}^{\ell+1}&=e_{k2}^\ell e_{22}^{\ell+1} e_{2j}^\ell\abajo e_{2k-1,2j}^{\ell+1}&=e_{k1}^\ell e_{12}^{\ell+1} e_{2j}^\ell\abajo \end{align} Then one checks that these are also matrix units, so $\espan\{e_{kj}^{\ell+1}:\ k,j\}\simeq M_{2^{\ell}}(\mathbb C)$. This way, we have constructed embeddings $$ M_2(\mathbb C)\subset M_4(\mathbb C)\subset M_8(\mathbb C)\subset\cdots\subset M, $$ where the restriction of $\psi$ to any of these subalgebras gives the normalized trace. Then $\Big(\bigcup_\ell M_{2^\ell}(\mathbb C)\Big)''\subset M$ is isomorphic to $R$.

It remains to prove the lemma.

Proof of the Lemma. By working on $pMp$, we may assume without loss of generality that $p=1$. Consider the family of collections $\{(q_j,r_j)\}$ such that $q_j,r_j$ are projections in $M$ with $q_j\sim r_j$, $q_jr_j=0$, and with $\{q_j\}$ pairwise orthogonal and $\{r_j\}$ pairwise orthogonal. This is nonempty because we can fix a non-trivial non-central projection $q$ and apply comparison to $q$ and $1-q$; so there exists a central projection $z$ such that $zq\preceq z(1-q)$ and $(1-z)(1-q)\preceq (1-z)q$ and at least one of $zq$, $(1-z)(1-q)$ is nonzero; in either case we find projections $q_0,r_0$ with $q_0\sim r_0$ and $q_0r_0=0$. We order the family by inclusion. If we have a chain of such collections, the union is also an acceptable collection, so it is an upper bound. Then Zorn's Lemma gives us a maximal collection $\{(q_j,r_j)\}$. Let $q=\sum_jq_j$, $r=\sum_jr_j$. We have $q\sim r$ and $qr=0$.

It remains to show that $q+r=1$. It it is not, let $p_0=1-(q+r)$. As $p_0Mp_0$ is not abelian (there are no abelian projections in a II$_1$ algebra), there is a projection $p_1\in p_0$ that is not central in $p_0Mp_0$. Then we can reason as in the previous paragraph: we apply comparison to $p_1$ and $p_0-p_1$, so there exists $z\in Z(p_0Mp_0)$ with $zp_1\preceq z(1-p_1)$ and $(1-z)(1-p_1)\preceq (1-z)p_1$. If $zp_1\ne0$, then there exists $p_2$ with $zp_1\sim p_2\leq z(1-p_1)\leq 1-p_1$, so we get a pair $zp_1,p_2\leq p_0$ with $zp_1\sim p_2$ and $(zp_1)p_2=0$, and this contradicts the maximality. And if $zp_1=0$ then $z(1-p_1)=z\ne0$, so we can repeat the reasoning and contradict the maximality. It follows that $p+r=1$.