A Multiplicative Ergodic Theorem for von Neumann Algebra Valued Cocycles

Abstract

The classical Multiplicative Ergodic Theorem of Oseledets is generalized here to cocycles taking values in a semi-finite von Neumann algebra. This allows for a continuous Lyapunov distribution.

Appendices

Diffuse Finite von Neumann Algebras

We recall some definitions from the introduction.

Definition 13

Let $M\subseteq B(\mathcal {H}),$ $N\subseteq B(\mathcal {K})$ be von Neumann algebras. A linear map $\theta :M\rightarrow N$ is normal if $\theta \big |_{\{x\in M:\Vert x\Vert \le 1\}}$ is weak operator topology-weak operator topology continuous.

For linear functionals, this agrees with our previous notion of normality introduced in Sect. 1.3. See [KR97, Theorem 7.1.12].

It can be shown that a $C^{*}$-algebra is a von Neumann algebra if and only if M has a predual (i.e. is isometrically isomorphic to the dual of a Banach space), and that moreover this predual is unique [Tak02, Theorem III.3.5 and Corollary III.3.9]. It is also known that a linear map is normal if and only if it is weak$^{*}$-weak$^{*}$ continuous. This explains why normality is the correct continuity condition for maps between von Neumann algebras: it is intrinsic to the algebra and does not depend upon how it is represented.

Suppose that $(M,\tau )$ is a finite tracial von Neumann algebra. We leave it as an exercise to verify that a sequence (or more generally a net) $x_{n}\in M$ with $\Vert x_{n}\Vert \le 1$ tends to x in the weak operator topology acting on $\text {L}^{2}(M,\tau )$ if and only if

$$\begin{aligned} \tau (x_{n}a)\rightarrow _{n\rightarrow \infty }\tau (xa)\text { for all }a\in M. \end{aligned}$$

(17)

In particular, normality of the trace implies that the regular representation of M on $\text {L}^{2}(M,\tau )$ is normal. It follows from the fact that (17) characterizes convergence in the WOT on the unit ball of M that a trace-preserving $*$-homomorphism between tracial finite von Neumann algebras is normal.

Definition 14

Let M be a von Neumann algebra. A projection $p\in M$ is a minimal projection if whenever $q\in M$ is a projection with $q\le p$ we have either $q=p$ or $q=0$. We say that M is diffuse if it has no nonzero minimal projections.

Proposition A.1

Let $(M,\tau )$ be a finite tracial von Neumann algebra. The following are equivalent:

• 1. M is diffuse,

• 2. there is an atomless standard probability space $(X,\mu )$ and an injective, trace-preserving $*$-homomorphism $\iota :\text {L}^{\infty }(X,\mu )\rightarrow M$, (here $\text {L}^{\infty }(X,\mu )$ is equipped with the trace $\int \cdot \,d\mu$),

• 3. there is a sequence $u_{n}$ of unitaries in M with $u_{n}\rightarrow 0$ in the weak operator topology.

Proof

(1) implies (2): Let $A\subseteq M$ be a unital, abelian von Neumann algebra which is maximal under inclusion among all abelian von Neumann subalgebras of M. The existence of such an A follows from Zorn’s Lemma. We start with the following claim.

Claim: A is diffuse.

To prove the claim suppose, for contradiction, that $p\in A$ is a nonzero minimal projection. Since every von Neumann algebra is the norm closed linear span of its projections by [Con90, Proposition IX.4.8], it follows that the von Neumann algebra Ap is 1-dimensional. Since M is diffuse, there exists a nonzero projection $q\in M$ so that $q\le p$ and $q\ne p$. We claim that q commutes with A. To see this, let $a\in A$. Since Ap is 1-dimensional, there is a $\lambda \in {{\mathbb {C}}}$ so that $ap=\lambda p.$ So

$$\begin{aligned}aq=(ap+a(1-p))q=(\lambda p+a(1-p))q=\lambda q,\end{aligned}$$

the last equation following as $q\le p$ implies that $pq=q$ and $(1-p)q=0$. Similarly,

$$\begin{aligned}qa=q(ap+a(1-p))=q(\lambda p+(1-p)a)=\lambda q,\end{aligned}$$

where in the second equality we are using that A is abelian and that $p\in A$. So q commutes with A. Now set

$$\begin{aligned}B=\overline{{\text {span}}(A\cup \{qa:a\in A\})}^{WOT}.\end{aligned}$$

Because q commutes with A, we know that ${\text {span}}(A\cup \{qa:a\in A\})$ is an abelian $*$-algebra. So B is an abelian von Neumann algebra containing A. By maximality, this forces $B=A$ and so $q\in A$. But then p is not a minimal projection in A, which gives a contradiction. This proves the claim.

Having shown the claim, note that by [Tak02, Theorem III.1.18] we have that $(A,\tau )\cong (\text {L}^{\infty }(Y,\nu ),\int \cdot \,d\nu )$ where Y is a compact Hausdorff space and $\nu$ is a probability measure on Y (we can choose Y to be metrizable if and only if $\text {L}^{2}(Y,\nu )$ is separable). Saying that $\text {L}^{\infty }(Y,\nu )$ is diffuse is equivalent to saying that for every measurable $E\subseteq Y$ with $\nu (E)>0$ there is a measurable $F\subseteq E$ with $\nu (F)<\nu (E)$. By a standard measure theory exercise this forces $\{\nu (F):F\subseteq E \text { is measurable}\}=[0,\nu (E)]$ for every measurable $E\subseteq Y$. By a recursive construction, this implies that for every $n\in {{\mathbb {N}}}$ and for every $\sigma \in \{0,1\}^{n}$ there is a measurable $E_{\sigma }\subseteq Y$ which satisfy the following properties:

• $\nu (E_{\sigma })=2^{-n}$ for every $n\in {{\mathbb {N}}}$ and every $\sigma \in \{0,1\}^{n}$,

• for every $n\in {{\mathbb {N}}}$ and every $\sigma ,\omega \in \{0,1\}^{n}$ with $\sigma \ne \omega$ we have $E_{\sigma }\cap E_{\omega }=\varnothing$,

• $X=E_{0}\cup E_{1},$

• for every $\sigma \in \{0,1\}^{n}$ we have $E_{\sigma }=E_{\sigma 0}\cup E_{\sigma 1}$ where $\sigma 0=(\sigma _{1},\ldots ,\sigma _{n},0)$, $\sigma 1=(\sigma _{1},\ldots ,\sigma _{n},1)\in \{0,1\}^{n+1}$.

It follows from the above properties that there is a unique map $\pi :Y\rightarrow \{0,1\}^{{{\mathbb {N}}}}$ with the property that for all $y\in Y,n\in {{\mathbb {N}}}$ we have that $(\pi (y)_{1},\cdots ,\pi (y)_{n})=\sigma$ if and only if $y\in E_{\sigma }$. Moreover, the above properties imply that $\mu =\pi _{*}\nu$ is the infinite power of the uniform measure on $\{0,1\}$. The map $\pi$ induces a trace-preserving $*$-homomorphism

$$\begin{aligned}\theta :\text {L}^{\infty }(\{0,1\}^{{{\mathbb {N}}}},\mu )\rightarrow \text {L}^{\infty }(Y,\nu )\cong A\end{aligned}$$

by $\theta (f)=f\circ \pi$. Combining with the inclusion of A into M we have a trace-preserving $*$-homomorphism $\text {L}^{\infty }(\{0,1\}^{{{\mathbb {N}}}},\mu )\rightarrow M$. Since $(\{0,1\}^{{{\mathbb {N}}}},\mu )$ is an atomless standard probability space, we are done.

(2) implies (3): By [Tak02, Theorem V.1.22] we may, and will, assume that $(X,\mu )=([0,1],m)$ where m is Lebesgue measure. Define unitaries $v_{n}$ in $\text {L}^{\infty }([0,1],m)$ by $v_{n}(x)=e^{2\pi i nx}$. Let $\iota :\text {L}^{\infty }([0,1],m) \rightarrow B(\text {L}^{2}([0,1],m))$ be the inclusion map. So for $\theta \in \text {L}^{\infty }([0,1],m)$ and $f \in \text {L}^{2}([0,1],m)$, $\iota (\theta )(f)(x) = \theta (x)f(x).$

It follows from the Riemann-Lebesgue Lemma that $\iota (v_{n})\rightarrow 0$ in the weak operator topology (as $n\rightarrow \infty$). By normality, we have that $v_{n}\rightarrow 0$ in the weak operator topology as well.

(3) implies (1): Suppose that $p\in M$ is a minimal projection. Let z be the central support of M, namely the smallest projection in the center of M which dominates p. Since M is a minimal projection, it follows by [KR97, Proposition 6.4.3 and Corollary 6.5.3] that we have a normal isomorphism of von Neumann algebras $Mz\cong B(\mathcal {K})$ for some Hilbert space $\mathcal {K}$. Since Mz (and thus $B(\mathcal {K})$) has a faithful, finite, normal tracial state it follows that $\mathcal {K}$ is finite-dimensional. So Mz is finite-dimensional. Note that $u_{n}z\rightarrow 0$ in the weak operator topology. Since there is only one Hausdorff vector space topology on a finite-dimensional space, we know that $\Vert u_{n}z\Vert \rightarrow 0$. But by unitarity we know that $\Vert z\Vert =\Vert u_{n}z\Vert$. So $z=0$, and the fact that $p\le z$ implies that $p=0$. So M has no nonzero minimal projection, and thus M is diffuse. $\quad \square$

As we remarked in the introduction, for most of the tracial von Neumann algebras we are interested in, the limiting operator in our Multiplicative Ergodic Theorem cannot be compact (unless it is zero). This is because of the following result.

Proposition A.2

Let $M\subseteq B(\mathcal {H})$ be a diffuse von Neumann algebra. Then M does not contain any nonzero compact operators.

Proof

Suppose that $x\in M$ is a nonzero compact operator. Then $x^{*}x$ is also a nonzero compact operator, and is self-adjoint. By the spectral theorem for compact normal operators, there is a $\lambda \in (0,\infty )$ which is a eigenvalue for $x^{*}x$ with finite dimensional eigenspace. Let p be the projection onto the kernel of $\lambda I-x^{*}x$. By functional calculus, we know that $p=1_{\{\lambda \}}(x^{*}x)$. So it follows by [Con90, Proposition IX.8.1] that $p\in M.$ Since $p\mathcal {H}$ is finite-dimensional, we may choose a nonzero projection $q\le p$ so that $\dim (q\mathcal {H})\le \dim (e\mathcal {H})$ whenever $e\in M$ is a nonzero projection with $e\le p.$ It is direct to see that q is a minimal projection in M, and this contradicts our assumption that M is diffuse. $\quad \square$

We close with some facts about $\text {L}^{2}(M,\tau )$ and ${{\text {GL}}}^{2}(M,\tau )$ when M is diffuse. We need a few preliminaries. Suppose that $(N,\widetilde{\tau })$, $(M,\tau )$ are tracial-von Neumann algebras and that $\iota :N\rightarrow M$ is a trace-preserving $*$-homomorphism. Then $\mu _{|\iota (x)|}=\mu _{|x|}$ for all $x\in N$ and thus $\iota$ is uniformly continuous for the measure topology (with respect to the unique translation-invariant uniform structure on a topological vector space). By [Tak03, Lemma 2.3 and Theorem 2.5], we may regard $\text {L}^{0}(M,\tau )$ and $\text {L}^{0}(N,\widetilde{\tau })$ as the measure topology completions of M and N, and so there is a unique measure topology continuous extension $\text {L}^{0}(N,\widetilde{\tau })\rightarrow \text {L}^{0}(M,\tau )$ of $\iota$. We will still use $\iota$ for this extension. Since $\text {L}^{0}(M,\tau )$, $\text {L}^{0}(N,\widetilde{\tau })$ are $*$-algebras extending M, N, the map $\iota :\text {L}^{0}(N,\widetilde{\tau })\rightarrow \text {L}^{0}(M,\tau )$ is a $*$-homomorphism. By measure topology continuity of $\iota$, density of N in $\text {L}^{0}(N,\widetilde{\tau })$, and Proposition 5.7 we have $\iota (f(|x|))=f(\iota (|x|))$ for all continuous, compactly supported $f:[0,\infty )\rightarrow {{\mathbb {R}}}$. In particular

$$\begin{aligned}\int f\,d\mu _{|\iota (x)|}=\tau (f(|\iota (x)|))=\widetilde{\tau }(f(|x|))=\int f\,d\mu _{|x|},\end{aligned}$$

for all continuous, compactly supported $f:[0,\infty )\rightarrow {{\mathbb {R}}}$. Thus $\mu _{|\iota (x)|}=\mu _{|x|}$.

Proposition A.3

Let $(M,\tau )$ be a semifinite tracial von Neumann algebra and suppose that M is diffuse. Then:

• 1. $\text {L}^{2}(M,\tau )$ is not closed under products,

• 2. there is an unbounded operator in $\text {L}^{2}(M,\tau )$,

• 3. $\mathcal {P}^{\infty }(M,\tau )$ is not complete in the metric $d_{\mathcal {P}}$.

Proof

By Proposition A.1 and [Tak02, Proposition V.1.40] there is a nonempty set J and a trace-preserving map $\iota$ of $(\text {L}^{\infty }([0,1]\times J),\int \cdot \,d(m\otimes \eta ))$ into $(M,\tau )$ where m is Lebesgue measure and $\eta$ is counting measure on the set J.

(1) Let $g:[0,1]\rightarrow {{\mathbb {C}}}$ be a measurable function so that $g\in \text {L}^{2}([0,1])$ but $g\notin \text {L}^{4}([0,1])$. E.g. we can take $g(x)=1_{(0,1]}(x)x^{-1/4}$. Fix $j_{0}\in J$ and define $f:X\times J\rightarrow {{\mathbb {C}}}$ by $f(x,j)=1_{\{j_{0}\}}(j)g(x)$. Since $\iota$ is trace-preserving, we have that

$$\begin{aligned}\mu _{|\iota (f)|}=\mu _{|f|}=|g|_{*}(m).\end{aligned}$$

So $\Vert \iota (f)\Vert _{2}=\Vert g\Vert _{2}<\infty$ and similarly $\Vert \iota (f)^{2}\Vert _{2}=\Vert \iota (f^{2})\Vert _{2}=\Vert g\Vert _{4}=\infty$. So $\iota (f)\in \text {L}^{2}(M,\tau )$ and $\iota (f)^{2}\notin \text {L}^{2}(M,\tau )$. So $\text {L}^{2}(M,\tau )$ is not closed under products.

(2) Let $f\in \text {L}^{2}([0,1]\times J)$ with $f\notin \text {L}^{\infty }([0,1]\times J)$. As in (1) we have that $\iota (f)\in \text {L}^{2}(M,\tau )$ and $\Vert \iota (f)\Vert _{\infty }=\Vert f\Vert _{\infty }=\infty$, so $\iota (f)$ is not bounded.

(3) Let $g\in \text {L}^{2}([0,1]\times J)$ with $g\notin \text {L}^{\infty }([0,1]\times J)$. Let $f=\exp (|g|)$. Since $\mu _{|\iota (f)|}=\mu _{|f|}$, we have that $\log |\iota (f)|\in \text {L}^{2}(M,\tau )$ but $\log |\iota (f)|\notin M$. Thus $\mathcal {P}(M,\tau )\ne \mathcal {P}^{\infty }(M,\tau )$. So $\mathcal {P}^{\infty }(M,\tau )$ is a proper dense subspace of a complete metric space, and is thus not complete. $\quad \square$

More Examples

1.1 Example: the hyperfinite factor

Let $M_n({{\mathbb {C}}})$ denote the algebra of $n\times n$ square matrices with entries in ${{\mathbb {C}}}$. Let $\phi _n:M_n({{\mathbb {C}}}) \rightarrow M_{2n}({{\mathbb {C}}})$ be the homomorphism

$$\begin{aligned} A \mapsto \left( \begin{array}{cc} A &{} 0 \\ 0 &{} A \end{array}\right) . \end{aligned}$$

Let $\tau _n:M_n({{\mathbb {C}}}) \rightarrow {{\mathbb {C}}}$ be the normalized trace defined by $\tau _n(A) = \frac{1}{n} \sum _{i=1}^n A_{ii}$. Note that $\phi _n$ preserves normalized traces in the sense that $\tau _n(A) = \tau _{2n}(\phi _n(A))$.

Let ${{\mathcal {A}}}:=\cup _n M_{2^n}({{\mathbb {C}}})$ be the direct limit of the matrix algebras $M_{2^n}({{\mathbb {C}}})$ under the family of maps $\phi _n$ (in other words, ${{\mathcal {A}}}$ is the disjoint union of $M_{2^n}({{\mathbb {C}}})$ after quotienting out by the equivalence relation generated by $A \sim \phi _{2^n}(A)$ for all $A \in M_{2^n}({{\mathbb {C}}})$). Because the maps $\phi _n$ preserve normalized traces, there is a trace $\tau :{{\mathcal {A}}}\rightarrow {{\mathbb {C}}}$ satisfying $\tau (A) = \tau _{2^n}(A)$ for all $A \in M_{2^n}({{\mathbb {C}}})$.

Define the inner product $\langle A, B \rangle := \tau (A^*B)$ for all $A,B \in {{\mathcal {A}}}$. The Hilbert space completion of this inner product is a Hilbert space, denoted by $\text {L}^2({{\mathcal {A}}},\tau )$. Moreover, each operator $A \in {{\mathcal {A}}}$ acts on $\text {L}^2({{\mathcal {A}}},\tau )$ by left-composition. Thus we have embedded ${{\mathcal {A}}}$ into the algebra $B(\text {L}^2({{\mathcal {A}}},\tau ))$ of bounded operators. Let R denote the weak operator closure of ${{\mathcal {A}}}$ in $B(\text {L}^2({{\mathcal {A}}},\tau ))$. The trace $\tau$ admits a unique normal extension to R. The pair $(R,\tau )$ is called the hyperfinite II$_1$-factor. For more details, see [AP16, Theorem 11.2.2].

1.2 Example: infinite tensor powers

Let $(M,\tau )$ be a finite tracial von Neumann algebra with $M \subset B({\mathcal {H}})$ SOT-closed. Then there is a natural embedding of $B(\ell ^2({{\mathbb {N}}}))\otimes M \rightarrow B(\ell ^2({{\mathbb {N}}}) \otimes {\mathcal {H}})$. Let $M_\infty :=B(\ell ^2({{\mathbb {N}}})) {\bar{\otimes }}M$ be the SOT-closure of $B(\ell ^2({{\mathbb {N}}}))\otimes M$ in $B(\ell ^2({{\mathbb {N}}}) \otimes {\mathcal {H}})$. We can think of elements of $M_\infty$ as ${{\mathbb {N}}}\times {{\mathbb {N}}}$ matrices with entries in M. Define a trace $\tau _\infty$ on $M_\infty$ by

$$\begin{aligned} \tau _\infty (x)=\sum _{i\in {{\mathbb {N}}}} \tau (x_{ii}). \end{aligned}$$

Equipped with this trace, $M_\infty$ is a semi-finite tracial von Neumann algebra. Moreover, this is the typical example of a non-finite semi-finite tracial von Neumann algebra. See [AP16, Chapter 8] for details.

Glossary

Let ${\mathcal {H}}$ be a separable Hilbert space and $B({\mathcal {H}})$ be the algebra of all bounded linear operators on ${\mathcal {H}}$.

• A von Neumann algebra M is a sub-algebra of $B({\mathcal {H}})$ satisfying: $1 \in M$, M is WOT-closed and $*$-closed.

• A tracial von Neumann algebra is a pair $(M,\tau )$ where M is a von Neumann algebra and $\tau$ is a faithful, normal, semi-finite trace. If $\tau ({\text {I}})<\infty$ then $(M,\tau )$ is a finite tracial von Neumann algebra.

• An operator $x \in B({\mathcal {H}})$ is positive if $\langle x\xi ,\xi \rangle \ge 0$ $\forall \xi \in {\mathcal {H}}$.

• $x \ge y$ iff $x-y$ is positive.

• $M_+ =\{x \in M:~x \ge 0\}$.

• $M^\times = {{\text {GL}}}^\infty (M,\tau )$ is the group of operators $x\in M$ with $x^{-1}\in M$.

• $M_{sa} = \{x \in M:~ x=x^*\}$.

• ${\mathcal {P}}^\infty = M_+ \cap M^\times =\exp (M_{sa})={\mathcal {P}}\cap M^\times$.

• ${\mathcal {N}}=\{x \in M:~ \tau (x^*x)<\infty \}$.

• $\text {L}^2(M,\tau )$ is the Hilbert space completion of ${\mathcal {N}}$ with respect to the inner product $\langle x,y \rangle = \tau (x^*y)$. Also $\text {L}^2(M,\tau ) = \{x \in \text {L}^0(M,\tau ):~ \tau (x^*x)<\infty \}$.

• $\text {L}^2(M,\tau )_{sa} = \{x \in \text {L}^2(M,\tau ):~x =x^*\}$.

• $\text {L}^0(M,\tau )$ is the algebra of $\tau$-measurable operators affiliated with M.

• $\text {L}^0(M,\tau )_+ = \{x \in \text {L}^0(M,\tau ):~x \ge 0\}$.

• $\text {L}^0(M,\tau )^\times$ is the group of operators $x\in \text {L}^0(M,\tau )$ with $x^{-1}\in \text {L}^0(M,\tau )$.

• $\text {L}^0(M,\tau )_{sa} = \{x \in \text {L}^0(M,\tau ):~x =x^*\}$.

• ${{\text {GL}}}^2(M,\tau ) = \{x\in \text {L}^0(M,\tau )^\times :~ \log |x| \in \text {L}^2(M,\tau )\}$.

• ${\mathcal {P}}= \text {L}^0(M,\tau )_+ \cap {{\text {GL}}}^2(M,\tau ) = \exp (\text {L}^2(M,\tau )_{sa})$.