Quantifier elimination in II₁ factors

Let $\Gamma$ be a property (T) group with infinitely many inequivalent irreducible representations on finite-dimensional Hilbert spaces. For the ‘moreover’ part, fix $k\geq 1$ and an $R^{\mathcal{U}}$-embeddable tracial von Neumann algebra $P$ with separable predual, let $\Gamma=\operatorname{SL}(2k+1,{\mathbb{Z}})$ for any $k\geq 1$, let $M_{0}=L(\Gamma)$, $M_{1}=L(\Gamma)*P$, and $\alpha=\operatorname{id}_{M_{0}}*\alpha_{0}$, for some nontrivial automorphism $\alpha_{0}$ of $L({\mathbb{Z}})$. Thus (1) holds. Since $M_{1}$ is $R^{{\mathcal{U}}}$-embeddable, (3) follows by [2, Proposition 3.4(2)].

It remains to prove (2). For $n\geq 2$ let $\rho_{n}\colon\Gamma\curvearrowright\ell_{2}(n)$ be an irreducible action of $\Gamma$ on $\ell_{2}(n)$, if such action exists, and trivial action otherwise. (The choice of $\rho_{n}$ in the latter case will be completely irrelevant.) Then $\rho_{n}$ defines a unital *-homomorphism of the group algebra ${\mathbb{C}}\Gamma$ into $M_{n}({\mathbb{C}})$, also denoted $\rho_{n}$. Let $\rho=\bigoplus_{n}\rho_{n}\colon{\mathbb{C}}\Gamma\to\prod_{n}M_{n}({\mathbb{C}})$.

Fix a nonprincipal ultrafilter ${\mathcal{U}}$ on ${\mathbb{N}}$ that concentrates on the set $\{n\mid\rho_{n}$ is irreducible$\}$. If $q\colon\prod_{n}M_{n}({\mathbb{C}})\to\prod^{\mathcal{U}}M_{n}({\mathbb{C}})$ is the quotient map, then $q\circ\rho\colon{\mathbb{C}}\Gamma\to\prod^{\mathcal{U}}M_{n}({\mathbb{C}})$ is a unital *-homomorphism. Let $M_{0}$ be the ultraweak closure of $q\circ\rho[{\mathbb{C}}\Gamma]$. If $\Gamma=\operatorname{SL}(2k+1,{\mathbb{Z}})$, then [6, Theorem 1] implies that $M_{0}$ isomorphic to the group factor $L(\Gamma)$; this nice fact however does not affect the remaining part of our proof.

For $g\in\Gamma$ we have a representation $(u(g)_{n})/{\mathcal{U}}$, where $u(g)_{n}=\rho_{n}(g)$ is a unitary in $M_{n}({\mathbb{C}})$ for every $n$. Since $P$ is $R^{\mathcal{U}}$-embeddable, it is also embeddable into $\prod^{\mathcal{U}}M_{n}({\mathbb{C}})$. Let $K\subseteq\prod^{\mathcal{U}}M_{n}({\mathbb{C}})$ be a countable set that that generates an isomorphic copy of $P$.

Our copy of $M_{0}$ in $\prod^{\mathcal{U}}M_{n}({\mathbb{C}})$ and the copy of $P$ generated by $K$ need not generate an isomorphic copy of $M_{1}=M_{0}*P$. In order to ‘correct’ this, we invoke some standard results.

At this point we need terminology from free probability. A unitary $u$ in a tracial von Neumann algebra $(M,\tau)$ is a Haar unitary if $\tau(u^{m})=0$ whenever $m\neq 0$. Such Haar unitary is free from some $X\subseteq M$ if for every $m\geq 1$, if $a_{j}$ is such that $\tau(a_{j})=0$ and in the linear span of $X$ and if $k(j)\neq 0$ for $0\leq j<m$, then (note that $\tau(u^{k(j)})=0$ since $u$ is a Haar unitary)

More generally, if $X$ and $Y$ are subsets of $M$ then $X$ is free from $Y$ if for every $m\geq 1$, if $a_{j}$ is such that $\tau(a_{j})=0$ and in the linear span of $X$ and $b_{j}$ is such that $\tau(b_{j})=0$ and in the linear span of $Y$ for $j<m$, then

Since $\Gamma$ is a Kazhdan group ([7]), we can fix a Kazhdan pair $F,\varepsilon$ for $\Gamma$. By [39, Theorem 2.2], there is a Haar unitary $w\in\prod^{\mathcal{U}}M_{n}({\mathbb{C}})$ free from $X=\{(u(g)_{n})/{\mathcal{U}}\mid g\in F\}\cup\{K\}$.

A routine calculation shows that $K^{\prime}=\{waw^{*}\mid a\in K\}$ is free from $X$ and that $K^{\prime}$ generates an isomorphic copy of $P$. Therefore the $\mathrm{W}^{*}$-subalgebra of $\prod^{\mathcal{U}}M_{n}({\mathbb{C}})$ generated by $\{(u(g)_{n})/{\mathcal{U}}\mid g\in F\}\cup\{K^{\prime}\}$ is isomorphic to $M_{1}\cong M_{0}*P$. This defines an embedding $\Phi_{0}\colon M_{1}\to\prod^{\mathcal{U}}M_{n}({\mathbb{C}})$.

Fix a type II₁ tracial von Neumann algebra $N$. In order to find an embedding of $M_{1}$ into $N^{\mathcal{U}}$ as required in (2), for every $n\geq 1$ write $N=M_{n}({\mathbb{C}})\bar{\otimes}N^{1/n}$ (where $N^{1/n}$ is the corner of $N$ associated to a projection whose center-valued trace is $1/n$). Define an embedding of $\prod^{\mathcal{U}}M_{n}({\mathbb{C}})$ into $N^{\mathcal{U}}$ by (the far right side is $N^{\mathcal{U}}$ in disguise)

Let $\Phi\colon M_{1}\to N^{\mathcal{U}}$ be the composition of the embedding $\Phi_{0}$ of $M_{1}$ into $\prod^{\mathcal{U}}M_{n}({\mathbb{C}})$ with this embedding.

It remains to prove that $N^{\mathcal{U}}\cap\Phi[M_{0}]^{\prime}=N^{\mathcal{U}}\cap\Phi[M_{1}]^{\prime}$. Since $M_{0}\subseteq M_{1}$, only the forward inclusion requires a proof. Suppose that $b\in N^{\mathcal{U}}\cap\Phi[M_{0}]^{\prime}$ and fix a representing sequence $(b_{n})$ for $b$. Fix for a moment $\delta>0$. Then the set ($F,\varepsilon$ is a Kazhdan pair for $\Gamma$)

belongs to ${\mathcal{U}}$.⁶⁶6The set $Z_{\delta}$ would have belonged to ${\mathcal{U}}$ regardless of whether the maximum had been taken over $F$ or some other finite subset of $\Gamma$. The maximum has been taken over $F$ because we need to bound the norm of the commutator of $b$ with all elements of $F$, and not elements of some other finite subset of $\Gamma$.

For each $n$ we define an action $\sigma_{n}\colon\Gamma\curvearrowright L^{2}(N,\tau)$ as follows. For $f\in\Gamma$ let $\tilde{\rho}_{n}(f)=\rho_{n}(f)\otimes 1_{N^{1/n}}$. For $c\in N$, let

This gives an action by isometries of $\Gamma$ on $(N,\|\cdot\|_{2,\tau})$, which continuously extends to action $\sigma_{n}$ of $\Gamma$ on $L^{2}(N,\tau)$.

For $g\in F$ and $n\in Z_{\delta}$ we have $\|g.b_{n}-b_{n}\|_{2}=\|u^{g}_{n}b_{n}-b_{n}u^{g}_{n}\|_{2}<\varepsilon\delta$. Since $\rho_{n}$ is an irreducible representation, the space of $\sigma_{n}$-invariant vectors in $L^{2}(N,\tau)$ is equal to $1_{M_{n}({\mathbb{C}})}\otimes L^{2}(N^{1/n},\tau)$. Let $P_{n}$ be the projection onto this space. Since $F,\varepsilon$ is a Kazhdan pair for $\Gamma$, by [13, Proposition 12.1.6] (with $\Gamma=\Lambda$) or [7, Proposition 1.1.9] (the case when $F$ is compact, and modulo rescaling) we have $\|b_{n}-P_{n}(b_{n})\|_{2}<\varepsilon$.

Therefore $(P_{n}(b_{n}))/{\mathcal{U}}=(b_{n})/{\mathcal{U}}$ belongs to $\prod^{\mathcal{U}}(1_{M_{n}({\mathbb{C}})}\bar{\otimes}N^{1/n})$, which is equal to $N^{\mathcal{U}}\cap(\prod^{{\mathcal{U}}}M_{n}({\mathbb{C}}))^{\prime}$. Recall that $\Phi[M_{1}]\subseteq\prod^{\mathcal{U}}M_{n}({\mathbb{C}})$. Since $b$ in $N^{\mathcal{U}}\cap\Phi[M_{0}]^{\prime}$ was arbitrary, $N^{\mathcal{U}}\cap\Phi[M_{0}]’$ is included in $N^{\mathcal{U}}\cap\Phi[M_{1}]^{\prime}$ (and is therefore equal to it), as required. ∎

Suppose for a moment that $N$ is of type II₁. Let $M_{0}$ and $M_{1}$ be as in Lemma 2.1. The embeddings of $M_{1}$ and $M_{1}\rtimes_{\alpha}{\mathbb{Z}}$ into $N^{\mathcal{U}}$ provided by (2) and (3) of this lemma satisfy (2) of Proposition 1.2, and therefore the theory of $N$ does not admit elimination of quantifiers.

In the general case, when $N$ has a type II₁ summand, we can write it as $N=N_{0}\oplus N_{1}$, where $N_{0}$ is type I and $N_{1}$ is type II₁ (e.g., [12, §III.1.4.7]). Then $N^{\mathcal{U}}=N_{0}^{\mathcal{U}}\oplus N_{1}^{\mathcal{U}}$. Let $r=\tau(1_{N_{1}})$, $P_{0}={\mathbb{C}}\oplus M_{1}$, and $P_{1}={\mathbb{C}}\oplus M_{1}\rtimes_{\alpha}{\mathbb{Z}}$, with the tracial states $\tau_{0}$ and $\tau_{1}$ such that $\tau_{0}(1_{M_{1}})=\tau_{1}(1_{M_{1}\rtimes_{\alpha}{\mathbb{Z}}})=r$. Since type II₁ algebra cannot be embedded into one of type I, this choice of tracial states forces that every embedding of $M_{1}$ into $N^{\mathcal{U}}$ sends $1_{M_{1}}$ to the image of $1_{N_{1}}$ under the diagonal map. An analogous fact holds for $M_{1}\rtimes_{\alpha}{\mathbb{Z}}$. Therefore, as in the factorial case, Proposition 1.2 implies that the theory of $N$ does not admit elimination of quantifiers. ∎

The equivalence of (2) and (4) is a well-known consequence of saturation of ultraproducts (see [18, §16]), and (4) will not be used in this paper (it is included only for completeness).

For simplicity and without loss of generality we assume that the language ${\mathcal{L}}$ is single-sorted. Assume (1) and fix finitely generated ${\mathcal{L}}$-substructures $F\leq G\leq N^{\mathcal{U}}$ and an isometric embedding $\Phi\colon F\to N^{\mathcal{U}}$. By Lemma 3.1, $\bar{a}$ and $\Phi(\bar{a})$ have the same quantifier-free type (over the empty set). It suffices to prove that $\Phi$ extends to an isometric embedding of $G$ in case when $G$ is generated by $b$ and $F$ for a single element $b$.

Let $r_{\varphi}=\varphi(\bar{a},b)$ for every $\varphi(\bar{x},y)\in\operatorname{\mathfrak{F}_{QF}^{\bar{x},y}}$. Consider the following quantifier-free type over $\Phi(\bar{a})$,⁹⁹9Since the formulas are quantifier-free, we do not need to specify the algebra in which they are being evaluated.

This type is uncountable, but separable in $\|\cdot\|_{N}$ since the space $\operatorname{\mathfrak{F}^{\bar{x}}}$ is separable in $\|\cdot\|_{N}$.

In order to prove that $\mathsf{t}(y)$ is satisfiable, fix a finite set of conditions in $\mathsf{t}(x)$, say $\varphi_{j}(\bar{a},y)=r_{j}$ for $j<m$, $m\geq 1$. Consider the formula

Then $\psi^{N^{\mathcal{U}}}(\bar{a})=0$, as witnessed by $b$. Since the theory of $N$ admits elimination of quantifiers, there are quantifier-free formulas $\psi_{k}(\bar{x})$, for $k\geq 1$, such that $\|\psi(\bar{x})-\psi_{k}(\bar{x})\|_{N}<1/k$ for all $n$. Therefore, since $\Phi$ is an isometry between the structures generated by $\bar{a}$ and $\Phi(\bar{a})$, we have

Thus $\mathsf{t}(y)$ is consistent, and by countable saturation ([18, §16])some $c\in N^{\mathcal{U}}$ realizes it. Thus $\bar{a},b$ and $\Phi(\bar{a}),c$ have the same quantifier-free type. By mapping $b$ to $c$ and Lemma 3.1, one finds an isometric extension of $\Phi$ to $G$ as required.

To prove that (2) implies (5), assume that (5) fails. Fix a quantifier-free formula $\varphi(\bar{x},y)$ in $n+1$ variables and $\varepsilon>0$ such that every $\psi\in\operatorname{\mathfrak{F}_{QF}^{\bar{x}}}$ satisfies

Let $\mathsf{t}(\bar{x}(0),\bar{x}(1))$ be the type in $2n$ variables with the following conditions.

This type is consistent by our assumptions, and by countable saturation it is realized in $N^{\mathcal{U}}$ by some $\bar{a}(0),\bar{a}(1)$. With $F$ denoting the ${\mathcal{L}}$-structure generated by $\bar{a}(0)$, we have an isometric embedding $\Phi$ of $F$ into $N^{\mathcal{U}}$ that sends $\bar{a}(0)$ to $\bar{a}(1)$.

By using countable saturation again, we can find $b\in N^{\mathcal{U}}$ such that $\inf_{y}\varphi(\bar{a}(1),y)-\varphi(\bar{a}(0),b)\geq\varepsilon$. Let $G$ be the ${\mathcal{L}}$-structure generated by $\bar{a}(0)$ and $b$. Then $\Phi$ cannot be extended to an isometric embedding of $G$ into $N^{\mathcal{U}}$, and (2) fails.

Since (5) implies that for every quantifier-free formula $\varphi(\bar{x},y)$, the formula $\sup_{y}\varphi(\bar{x},y)$ is a uniform $\|\cdot\|_{N}$-limit of quantifier-free formulas (by replacing $\varphi$ with $-\varphi$), the proof that (5) implies (1) follows by induction on complexity of formulas (see [9]). ∎

Suppose that $\tau$ satisfies the assumption, $F$ is a $\mathrm{C}^{*}$-subalgebra of $M_{m}({\mathbb{C}})\oplus M_{n}({\mathbb{C}})$, and $\Phi\colon F\to M_{m}({\mathbb{C}})\oplus M_{n}({\mathbb{C}})$ is a trace-preserving embedding. Then $F$ is a direct sum of matrix algebras. Let $p_{j}$, for $j<k$, be the units of these matrix algebras. Then $\tau(p_{j})=\tau(\Phi(p_{j}))$ for all $j$. By the assumption on $\tau$ there is a partial isometry $v_{j}$ such that $v_{j}^{*}v_{j}=p_{j}$ and $v_{j}v_{j}^{*}=\Phi(p_{j})$. Therefore $u=\sum_{j<k}v_{j}$ is a unitary such that $up_{j}u^{*}=\Phi(p_{j})$ for $j<k$. Since every automorphism of a matrix algebra is implemented by a unitary, for every $j<k$ there is $w_{j}$ such that $w_{j}^{*}w_{j}=w_{j}w_{j}^{*}=p_{j}$ and for $a\in p_{j}F$ we have $w_{j}v_{j}av_{j}^{*}w_{j}^{*}=\Phi(a)$. Therefore $\Phi$ coincides with conjugation by the unitary $u^{\prime}=\sum_{j<k}w_{j}v_{j}$. This implies that $\Phi$ automatically extends to an embedding of any $G$ such that $F\subseteq G\subseteq M_{m}({\mathbb{C}})\oplus M_{n}({\mathbb{C}})$ into $M_{m}({\mathbb{C}})\oplus M_{n}({\mathbb{C}})$. ∎

Assume (1). Then $\Phi[M]$ is a separable elementary submodel of  $N^{\mathcal{U}}$, and the Continuum Hypothesis implies that $M^{\mathcal{U}}\cong N^{\mathcal{U}}$, via an isomorphism $\Psi$ that sends the diagonal copy of $M$ onto $\Phi[M]$. The $\mathrm{C}^{*}$-algebra case is proven in [18, Theorem 16.7.5], and the proof of the general case is virtually identical.

Assume (2) and let $\Phi\colon M\to N$ be an embedding between models of $T$. We need to prove that $\Phi$ is elementary. By replacing $M$ and $N$ with separable elementary submodels large enough to detect a given failure of elementarity, we may assume they are separable. With $\Psi$ as guaranteed by (2), the embedding of $M$ into $N^{\mathcal{U}}$ is, being the composition of an elementary embedding and an isomorphism, elementary. Since the diagonal image of $N$ in $N^{\mathcal{U}}$ is elementary, it follows that $\Phi$ is elementary. ∎

This is an immediate consequence of the results of [20]. Fix a type I tracial von Neumann algebra $M$.

If $M$ is abelian, then $M\cong L^{\infty}(\mu)$ for a probability measure $\mu$. Then [20, Lemma 3.2] implies that theory of $M$ determines the measures of the atoms of $\mu$ (these are the values of $\rho(1,n)$ for $n\geq 1$, using the notation from this lemma and taken in the decreasing order; $\rho(1,0)$ is the measure of the diffuse part).

Therefore if $N\equiv M$ then $N\cong L^{\infty}(\nu)$, and the measures of the atoms of $\nu$ (if any) are exactly the same as the measures of the atoms of $\mu$. Every trace-preserving *-homomorphism $\Phi$ from $M$ into $N$ has to send the diffuse part to the diffuse part. Since these parts are of the same measure, it also sends atomic part to the atomic part. Finally, since the atoms in $M$ and $N$ have the same measures (with multiplicities), and since we are dealing with a probability measure, the restriction of $\Phi$ to the atomic part is an isomorphism. The theory of $L^{\infty}$ space of a diffuse measure admits elimination of quantifiers, hence the restriction of $\Phi$ to the diffuse part is elementary. Thus $\Phi$ can be decomposed as a direct sum of two elementary embeddings, and is therefore elementary by the second part of [20, Corollary 2.6].

This proves Proposition in case when $M$ is abelian.

In general, when $M$ is not necessarily abelian, it is isomorphic to $\bigoplus_{n\in X}A^{M}_{n}$, where $X\subseteq[1,\infty)$, and for every $n\in X$ we have $A^{M}_{n}\cong M_{n}(B^{M}_{n})$, for a commutative von Neumann algebra $B^{M}_{n}$. Suppose that $M\equiv N$. Then [20, Lemma 3.2] implies that $N\cong\bigoplus_{n\in X}A^{N}_{n}$ (with the same set $X$) and that $\tau(1_{A^{M}_{n}})=\tau(1_{A^{N}_{n}})$ for all $n\in X$. (Using the notation from [20, Lemma 3.2], these quantities are both equal to $\sum_{k}\rho_{M}(n,k)$, for every $n\in X$.)

Now assume that $\Phi\colon M\to N$ is a trace-preserving *-homomorphism. For $n^{\prime}<n$ in $X$ we have that $\Phi(1_{A^{M}_{n}})$ and $1_{A^{N}_{n^{\prime}}}$ are orthogonal. By induction on $n\in X$, this implies that $\Phi(1_{A^{M}_{n}})=1_{A^{N}_{n}}$ for all $n$.

By the abelian case, the restriction of $\Phi$ to $A^{M}_{n}$ is elementary for every $n$. By the second part of [20, Corollary 2.6], $\Phi$ is elementary. ∎