Fact 2.1. Fix a structure $\mathcal {M}$ and some subset $A \subseteq M$ . Suppose $Z \subseteq M^n$ is a closed subset. Then Z is a definable set in $\mathcal {M}$ over A if and only if for every $\epsilon> 0$ , there exist $\delta> 0$ and a formula $\varphi (x_1, \ldots , x_n)$ , possibly using parameters from A, such that for any $\mathbf {x} \in M^n$ ,

$$\begin{align*}\varphi^{\mathcal{M}}(\mathbf{x}) < \delta \implies d(\mathbf{x}, Z) \leq \epsilon. \end{align*}$$

Proof. By contrapositive, suppose that $\mathcal {M}$ has a direct summand of the form $M_n(\mathbb {C}) \otimes L^\infty [0,1]$ . In $M_n(\mathbb {C}) \otimes L^\infty [0,1]$ , consider the projections $p = 1 \otimes \mathbf {1}_{[0,1/n]}$ and $q = E_{1,1} \otimes 1$ , where $E_{1,1}$ is the canonical matrix unit in $M_n(\mathbb {C})$ . These two projections have the same trace, hence they have the same $*$ -moments, i.e., the same quantifier-free type. However, they do not have the same type because p is central and q is not central in $M_n(\mathbb {C}) \otimes L^\infty [0,1]$ , hence also in $\mathcal {M}$ . So $\mathcal {M}$ cannot admit quantifier elimination. Furthermore, since p and q do not have the same type, they cannot be conjugate by an automorphism of $\mathcal {M}$ .

Proof. Suppose there is some $j, k \in J$ so that $n_j = n_k \geq 2$ , and $\alpha _j = \alpha _k$ . Let p be a projection of rank $2$ in the $M_{n_j}(\mathbb {C})$ summand, and let q be a projection of rank $1$ in both the $M_{n_j}(\mathbb {C})$ and $M_{n_k}(\mathbb {C})$ summands (and p, q are both 0 in all other summands). Then $\tau (p) = \tau (q) = \frac {2 \alpha _j}{n_j}$ , but p and q are not conjugate by any automorphism.

Proof of Theorem A.

(1) $\implies $ (2). Suppose that $\mathcal {M}$ admits elimination of quantifiers. In order to deal with the diffuse $L^\infty $ term and the atomic terms separately, we first show that the central projection $1_{L^\infty }$ is a definable element (see Section 2.3). Note that for each k, the set

$$\begin{align*}S_k = \{e_1,\dots,e_k \in P(\mathcal{M}) \cap Z(\mathcal{M}): e_i e_j = 0, \tau(e_j) = \alpha_0/k \text{ for } i, j = 1, \dots, k \} \end{align*}$$

is definable using the definability of the center (see [Reference Farah, Hart and Sherman29, Lemma 4.2]) and the stability of projections. Moreover, if x is any element satisfying

$$\begin{align*}\inf_{(e_1,\dots,e_k) \in S_k} d\left(x,\sum_{j=1}^k e_j \right) \leq \epsilon, \end{align*}$$

then x is $\epsilon $ -close to a central projection that is divisible into k central projections of trace $\alpha _0/k$ . If k is large enough, then the sum of the weights of discrete summands that are less than or equal to $\alpha _0/k$ will be less than $\epsilon ^2$ . Hence, $\sum _{j=1}^k e_j$ will be $2\epsilon $ -close to $1_{L^\infty }$ . So $1_{L^\infty }$ is definable.

Let $p, q$ be two projections with the same trace. As noted in the proof of Lemma 3.1, p and q then have the same quantifier-free type and hence they have the same type. Because $1_{L^\infty }$ is definable, every formula over $L^\infty $ and every formula over $\mathcal {N} := \mathcal {M} \ominus L^\infty $ can be expressed as a definable predicate over $\mathcal {M}$ . Thus, $1_{L^\infty } p$ and $1_{L^\infty }q$ have the same type in $L^\infty [0,1]$ and $(1_{\mathcal {N}})p$ and $(1_{\mathcal {N}})q$ have the same type in $\mathcal {N}$ . Then, $1_{L^\infty } p$ and $1_{L^\infty }q$ are two projections of the same trace in $L^{\infty }[0,1]$ and therefore conjugate by an automorphism. Meanwhile, $(1 - 1_{L^\infty })p$ and $(1 - 1_{L^\infty })q$ have the same type in $\mathcal {N}$ , hence they are conjugate by an automorphism in some elementary extension $\tilde {\mathcal {N}}$ of $\mathcal {N}$ . Since $\mathcal {N}$ is type I and atomic, $\tilde {\mathcal {N}}$ must equal $\mathcal {N}$ (see [Reference Farah and Ghasemi27, Proposition 4.3] or [Reference Jekel52, Proposition 3.7(2)]). Thus, $(1_{\mathcal {N}})p$ and $(1_{\mathcal {N}})q$ are conjugate by an automorphism of $\mathcal {N}$ , and so p and q are conjugate by an automorphism of $\mathcal {M}$ .

(2) $\implies $ (1) Let $\mathrm {T} := \operatorname {\mathrm {Th}}(\mathcal {M})$ . We must check that every $\mathrm {T}$ type is determined by its quantifier-free type. First note that all $\mathrm {T}$ types can be realized in $\mathcal {M}$ ; indeed, $\mathcal {M}^{\mathcal {U}}$ is countably saturated (see Section 2.2) and is a direct sum of $L^\infty [0,1]^{\mathcal {U}}$ and $(\mathbb {C}^{n_j})^{\mathcal {U}} = \mathbb {C}^{n_j}$ and ${M_{n_j}(\mathbb {C})^{\mathcal {U}} = M_{n_j}(\mathbb {C})}$ . Any tuple of elements in $L^\infty [0,1]^{\mathcal {U}}$ has the same type as some tuple in $L^\infty [0,1]$ , and swapping out the element in the $L^\infty [0,1]^{\mathcal {U}}$ summand for one of the same type will not change the type of the overall element in $\mathcal {M}^{\mathcal {U}}$ .

Fix some $\mathbf {x} = (x_1,\dots ,x_k)$ and $\mathbf {y} = (y_1,\dots ,y_k)$ in $\mathcal {M}$ with the same quantifier-free type. We shall build a sequence of automorphisms $\sigma _n$ of $\mathcal {M}$ such that $\sigma _n(\mathbf {x}) \rightarrow \mathbf {y}$ , so $\operatorname {\mathrm {tp}}^{\mathcal {M}}(\mathbf {x}) = \operatorname {\mathrm {tp}}^{\mathcal {M}}(\mathbf {y})$ . Since there are no identical matrix summands with the same weight by Lemma 3.2, the only possible automorphisms of $\mathcal {M}$ are those which are a direct sum of automorphisms of each component, possibly composed with swaps of copies of $\mathbb {C}$ which have the same weight. This motivates the following decomposition of $\mathcal {M}$ , where we group together copies of $\mathbb {C}$ which have the same weight:Footnote ³

(3.1) $$ \begin{align} \mathcal{M} = (L^\infty[0,1], \alpha_0) \oplus \left( \bigoplus_{j \in J_1} (\mathbb{C}, \alpha_j)^{\oplus n_j} \right) \oplus \left( \bigoplus_{j \in J_2} (M_{n_j}(\mathbb{C}), \alpha_j) \right). \end{align} $$

We will build the automorphisms on each summand of (3.1) separately.

We start with the matrix summands. Let $p_j$ , $j \in J_2$ , be the central projection onto the jth summand $M_{n_j}(\mathbb {C})$ , where $n_j \geq 2$ . We claim that $p_j\mathbf {x} = (p_j x_1,\dots ,p_j x_k)$ and ${p_j \mathbf {y} = (p_j y_1,\dots ,p_j y_k)}$ have the same quantifier-free type in $M_{n_j}(\mathbb {C})$ . Let f be a self-adjoint non-commutative $*$ -polynomial. For Borel $E \subseteq \mathbb {R}$ , we have $\tau (1_E(f(\mathbf {x}))) = \tau (1_E(f(\mathbf {y})))$ , so by assumption there is some automorphism $\sigma $ conjugating $1_E(f(\mathbf {x}))$ to $1_E(f(\mathbf {y}))$ . As noted above, the automorphism $\sigma $ must fix $p_j$ , so $\sigma (p_j 1_E(f(\mathbf {x}))) = p_j 1_E(f(\mathbf {y}))$ . Hence, $\tau (p_j 1_E(f(\mathbf {x}))) = \tau (p_j 1_E(f(\mathbf {y})))$ , or equivalently ${\operatorname {\mathrm {tr}}_{n_j}(1_E(f(p_j \mathbf {x}))) = \operatorname {\mathrm {tr}}_{n_j}(1_E(f(p_j \mathbf {y})))}$ . Since E was arbitrary, $f(p_j \mathbf {x})$ and $f(p_j \mathbf {y})$ have the same empirical spectral distribution, hence also ${\operatorname {\mathrm {tr}}_n(f(p_j \mathbf {x})) = \operatorname {\mathrm {tr}}_n(f(p_j \mathbf {y}))}$ . This holds for all f, so the multivariate Specht’s theorem [Reference Jing54] implies that ${u_j p_j\mathbf {x}u_j^{*} = p_j \mathbf {y}}$ for some unitary $u \in M_{n_j}(\mathbb {C})$ .

The same argument as in the matrix case shows that when $p_j$ for $j \in J_1$ is the central projection onto some summand of the form $\mathbb {C}^{n_j}$ , $n_j \geq 1$ , with each copy of $\mathbb {C}$ having the same weight $\alpha _j$ , we obtain that $\operatorname {\mathrm {qftp}}^{\mathbb {C}^{n_j}} (p_j \mathbf {x}) = \operatorname {\mathrm {qftp}}^{\mathbb {C}^{n_j}}(p_j \mathbf {y})$ , so some automorphism (i.e., permutation) $\pi _j$ of $\mathbb {C}^{n_j}$ sends $p_j \mathbf {x}$ to $p_j \mathbf {y}$ .

Finally, let $p_0$ be the central projection onto the $L^\infty [0,1]$ summand. Then ${p_0 = 1 - \sum _{j \in J_1 \sqcup J_2} p_j}$ , where $p_j$ is the central projection onto the jth summand of $\mathcal {M}$ . Hence, for any non-commutative $*$ -polynomial f,

$$\begin{align*}\tau(p_0 f(\mathbf{x})) = \tau(f(\mathbf{x})) - \sum_{j \in J_1 \sqcup J_2} \tau(p_j f(\mathbf{x})) = \tau(f(\mathbf{y})) - \sum_{j \in J_1 \sqcup J_2} \tau(p_j f(\mathbf{y})) = \tau(p_0 f(\mathbf{y})), \end{align*}$$

so we again obtain that $\operatorname {\mathrm {qftp}}^{L^\infty }(p_0 \mathbf {x}) = \operatorname {\mathrm {qftp}}^{L^\infty } (p_0 \mathbf {y})$ . By [Reference Jekel and Goldbring51, Lemma 2.16], there is a sequence of automorphisms $\alpha _n$ of $L^\infty [0,1]$ such that $\alpha _n(\mathbf {x}) \to \mathbf {y}$ .

To conclude, let $\sigma _n$ be the direct sum of the automorphisms in each summand of $\mathcal {M}$ given by the arguments above, that is,

$$\begin{align*}\sigma_n = \alpha_n \oplus \bigoplus_{j \in J_1} \pi_j \oplus \bigoplus_{j \in J_2} \operatorname{Ad}_{u_j}. \end{align*}$$

Then $\sigma _n(\mathbf {x}) \to \mathbf {y}$ , so $\operatorname {\mathrm {tp}}^{\mathcal {M}}(\mathbf {x}) = \operatorname {\mathrm {tp}}^{\mathcal {M}}(\mathbf {y})$ . Hence, $\mathcal {M}$ admits elimination of quantifiers by Lemma 2.2.

Proof. Suppose $\mathcal {M}$ admits quantifier elimination. Then [Reference Farah26, Theorem 1] implies (1) and Lemma 3.1 implies (2).

For (3), suppose for contradiction that there are two projections p and q in the atomic part with $0 < |\tau (p) - \tau (q)| \leq \alpha _0$ , and without loss of generality suppose that $\tau (p) < \tau (q)$ . Let $p'$ be a projection in $L^\infty [0,1]$ such that $\tau (p') = \tau (q) - \tau (p)$ . Then q and $p' + p$ have the same trace but are not equivalent by an automorphism, so by Theorem A, $\mathcal {M}$ does not have quantifier elimination.

For (4), let p and q be projections in the atomic part with $\tau (p) = \tau (q)$ . By Theorem A, p and q are conjugate by an automorphism. Hence also $E_{Z(\mathcal {M})}[p]$ and $E_{Z(\mathcal {M})}[q]$ are conjugate by an automorphism. In light of Lemma 3.2, every automorphism must fix the central projections associated with $M_n(\mathbb {C})$ terms for $n \geq 2$ . Thus, $E_{Z(\mathcal {M})}[p]$ and $E_{Z(\mathcal {M})}[q]$ must have equal components in each of the $M_n(\mathbb {C})$ summands for $n \geq 2$ . So they differ by an automorphism that merely permutes the one-dimensional summands.

Conversely, assume (1)–(4). Let p and q be two projections of the same trace. Using (3), the traces of p and q in the $L^\infty [0,1]$ summand must agree, so there is an automorphism of $\mathcal {M}$ such that $\alpha (p) - q$ is in the atomic part of $\mathcal {M}$ . So assume without loss of generality that p and q are in the atomic part. By (4), after applying an automorphism, we can assume that $E_{Z(\mathcal {M})}[p] = E_{Z(\mathcal {M})}[q]$ . Hence, the components of p and q in each direct summand $M_n(\mathbb {C})$ of $\mathcal {M}$ (where $n \geq 1$ ), have the same rank, and hence are unitarily conjugate. Overall, p and q are conjugate by an automorphism. By Theorem A, $\mathcal {M}$ admits quantifier elimination.

Proof. Suppose $\mathcal {M}$ admits quantifier elimination. We already know $\mathcal {M}$ decomposes into an optional $L^\infty [0,1]$ term and an atomic part. By grouping the one-dimensional terms with the same weight, we obtain a direct sum decomposition satisfying conditions (1) and (2). It remains to check condition (3). By contrapositive, suppose that there exist integers $|r_j| \leq n_j$ satisfying

$$\begin{align*}\left| \sum_{j \in J_1} r_j \alpha_j + \sum_{j \in J_2} \frac{r_j \alpha_j}{n_j} \right| \leq \alpha_0. \end{align*}$$

For $j \in J_1$ , let $p_j$ and $q_j$ be projections in $(\mathbb {C},\alpha _j)^{\oplus n_j}$ such that

$$\begin{align*}\operatorname{\mathrm{rank}}(p_j) = \max(r_j,0), \qquad \operatorname{\mathrm{rank}}(q_j) = \max(-r_j,0). \end{align*}$$

Similarly, for $j \in J_2$ , let $p_j$ and $q_j$ be projections in $(M_{n_j}(\mathbb {C}),\alpha _j)$ with the same rank conditions. Thus, $\operatorname {\mathrm {rank}}(p_j) - \operatorname {\mathrm {rank}}(q_j) = r_j$ . Finally, let

$$\begin{align*}t = \sum_{j \in J_1} r_j \alpha_j + \sum_{j \in J_2} \frac{r_j \alpha_j}{n_j}, \end{align*}$$

and let $p_0$ and $q_0$ be projections in $(L^\infty [0,1],\alpha _0)$ such that $\tau (p_0) = \max (-t,0)$ and $\tau (q_0) = \max (t,0)$ , so that $\tau (p_0) - \tau (q_0) = -t$ . Let

$$\begin{align*}p = p_0 \oplus \bigoplus_{j \in J_1} p_j \oplus \bigoplus_{j \in J_2} p_j, \qquad q = q_0 \oplus \bigoplus_{j \in J_1} q_j \oplus \bigoplus_{j \in J_2} q_j. \end{align*}$$

By construction,

$$\begin{align*}\tau(p) - \tau(q) = \tau(p_0) - \tau(q_0) + \sum_{j \in J_1} \alpha_j r_j + \sum_{j \in J_2} \frac{\alpha_j r_j}{n_j} = 0. \end{align*}$$

However, p and q are not automorphically conjugate. Indeed, $r_j$ is nonzero for some j. If $j \in J_1$ , the components of p and q in the central summand $(\mathbb {C},\alpha _j)^{\oplus n_j}$ have different ranks, and $(\mathbb {C},\alpha _j)^{\oplus n_j}$ is invariant under automorphisms because we grouped together all the terms with the same weight. Similarly, if $j \in J_2$ , then the components of p and q in $(M_{n_j}(\mathbb {C}),\alpha _j)$ have different ranks, and by Lemma 3.2, $(M_{n_j}(\mathbb {C}),\alpha _j)$ must be invariant under automorphisms since there is only one summand with a given dimension and weight. Hence, if (3) does not hold, then $\operatorname {\mathrm {Th}}(\mathcal {M})$ does not admit quantifier elimination.

Conversely, suppose $\mathcal {M}$ has a decomposition satisfying (1)–(3). Consider two projections $p = p_0 \oplus \bigoplus _{j \in J_1 \sqcup J_2} p_j$ and $q= q_0 \oplus \bigoplus _{j \in J_1 \sqcup J_2} q_j$ in $\mathcal {M}$ with the same trace. Then

$$\begin{align*}\tau(p_0) - \tau(q_0) = \sum_{j \in J_1 \sqcup J_2} \frac{\alpha_j (\text{rank}(q_j) - \text{rank}(p_j))}{n_j}. \end{align*}$$

Hence,

$$\begin{align*}\left| \sum_{j \in J_1} \alpha_j (\operatorname{rank}(q_j) - \operatorname{rank}(p_j)) + \sum_{j \in J_2} \frac{\alpha_j (\operatorname{rank}(q_j) - \operatorname{rank}(p_j))}{n_j} \right| = |\tau(p_0) - \tau(q_0)| \leq \alpha_0. \end{align*}$$

By condition (3), this forces $\operatorname {\mathrm {rank}}(p_j) = \operatorname {\mathrm {rank}}(q_j)$ for all $j \in J_1 \sqcup J_2$ . In particular, for $j \in J_1$ , $p_j$ and $q_j$ are projections in $(\mathbb {C},\alpha _j)^{\oplus n_j}$ with the same rank and hence conjugate by an automorphism permuting the summands. Moreover, for $j \in J_2$ , $p_j$ and $q_j$ are projections in $M_{n_j}(\mathbb {C})$ with the same rank, hence they are unitarily conjugate. Finally, since $p_j$ and $q_j$ have the same trace for $j \in J_1 \sqcup J_2$ , we deduce that $p_0$ and $q_0$ have the same trace in $L^\infty [0,1]$ and hence they are conjugate by a measure-preserving transformation. Patching the automorphisms on each summand together, p and q are automorphically conjugate. Thus, by Theorem A, $\mathcal {M}$ has quantifier elimination.

Proof. To prove the claims, it suffices to show (4.1) holds almost surely for each $\varphi $ in a countable dense set of formulas (as usual in measure theory, “almost surely” distributes over countable conjunctions).

In fact, the dense set of formulas can be chosen to be Lipschitz. Indeed, a formula will be Lipschitz as long as the atomic formulas and the connectives used are all Lipschitz; the quantifiers do not cause any issue since the supremum of a family of L-Lipschitz functions is L-Lipschitz. The atomic formulas are traces of non-commutative polynomials, and for every non-commutative polynomial p and $R> 0$ , there is some L such that $\tau (p)$ is L-Lipschitz with respect to $\left \lVert \cdot \right \rVert _2$ on each operator norm ball of radius R. The connectives in the language are continuous functions $\mathbb {R}^m \to \mathbb {R}$ , which can all be approximated on compact sets by Lipschitz functions.

So assume that $\varphi $ is an L-Lipschitz formula in three variables. Note that $\mathbf {X}^{(n)}$ depends in a Lipschitz manner upon $\mathbf {U}^{(n)} = (U_1^{(n)},U_2^{(n)},U_3^{(n)})$ ; indeed, the mapping $M_n(\mathbb {C}) \to \mathcal {M}$ given by $u \mapsto u \otimes 1_{\mathcal {M}^{1/n}}$ is $1$ -Lipschitz. In particular, $\varphi ^{\mathcal {M}}(\mathbf {X}^{(n)})$ is an L-Lipschitz function of $\mathbf {U}^{(n)}$ . Therefore, applying Proposition 4.1 with $\delta = 1/n$ ,

$$\begin{align*}\mathbb{P}( |\varphi^{\mathcal{M}}(\mathbf{X}^{(n)}) - \mathbb{E}[\varphi^{\mathcal{M}}(\mathbf{X}^{(n)})])| \geq 1/n) \leq e^{-cn / L^2}. \end{align*}$$

By the Borel–Cantelli lemma, this implies that almost surely

$$\begin{align*}\lim_{n \to \infty} |\varphi^{\mathcal{M}}(\mathbf{X}^{(n)}) - \mathbb{E}[\varphi^{\mathcal{M}}(\mathbf{X}^{(n)})]| = 0, \quad \text{hence} \quad \lim_{n \to \mathcal{U}} \varphi^{\mathcal{M}}(\mathbf{X}^{(n)}) = \lim_{n \to \mathcal{U}} \mathbb{E}[\varphi^{\mathcal{M}}(\mathbf{X}^{(n)})]. \end{align*}$$

Proof. Let $\tilde {\mathbf {X}}^{(n)}$ be defined analogously to $\mathbf {X}^{(n)}$ but with $U_4^{(n)}$ in place of $U_3^{(n)}$ , that is, $\tilde {\mathbf {X}}^{(n)} = (U_1^{(n)} \otimes 1_{\mathcal {M}^{1/n}}, U_2^{(n)} \otimes 1_{\mathcal {M}^{1/n}}, U_4^{(n)} \otimes 1_{\mathcal {M}^{1/n}})$ . Since $\tilde {\mathbf {X}}^{(n)}$ has the same probability distribution as $\mathbf {X}^{(n)}$ , the almost sure limit of $\operatorname {\mathrm {tp}}^{\mathcal {M}}(\tilde {\mathbf {X}}^{(n)})$ agrees with that of $\mathbf {X}^{(n)}$ .

In the following, we fix an outcome $\omega $ in the probability space such that the limit as $n \to \mathcal {U}$ of the type of $\mathbf {X}^{(n)}$ and the type of $\tilde {\mathbf {X}}^{(n)}$ at $\omega $ agree with the almost sure limits given by Lemma 4.2. Let $\varphi $ be an existential formula. Then $\varphi $ can be expressed as

$$\begin{align*}\varphi(x_1,x_2, x_3) = \inf_{z_1, \dots, z_k} \psi(x_1,x_2, x_3, z_1,\dots,z_k), \end{align*}$$

where $\psi $ is a quantifier-free formula and each $z_j$ ranges over the unit ball. Since $\mathcal {M}^{\mathcal {U}}$ is countably saturated (see Section 2.2), there exists some $\mathbf {Z} \in (\mathcal {M}^{\mathcal {U}})_1^k$ such that $\varphi ^{\mathcal {M}^{\mathcal {U}}}(\mathbf {X}) = \psi ^{\mathcal {M}^{\mathcal {U}}}(\mathbf {X},\mathbf {Z})$ . Now because $\mathbf {X}$ and $\tilde {\mathbf {X}}$ have the same type in $\mathcal {M}^{\mathcal {U}}$ , there also exists some $\tilde {\mathbf {Z}} \in (\mathcal {M}^{\mathcal {U}})_1^k$ such that $(\mathbf {X},\mathbf {Z})$ and $(\tilde {\mathbf {X}},\tilde {\mathbf {Z}})$ have the same quantifier-free type.

In the hypotheses of Theorem B, we assumed there is an embedding $i: M_2(\mathcal {M}) \to \mathcal {M}^{\mathcal {V}}$ for some ultrafilter $\mathcal {V}$ .Footnote ⁴ Let $i^{(n)}$ be the corresponding embedding

$$\begin{align*}i^{(n)}: \mathcal{M}^{1/n} = M_2(\mathcal{M})^{1/2n} \to (\mathcal{M}^{\mathcal{V}})^{1/2n} \cong (\mathcal{M}^{1/2n})^{\mathcal{V}}. \end{align*}$$

Then let

$$\begin{align*}i^{\mathcal{U}} = \prod_{n \to \mathcal{U}} (\operatorname{\mathrm{id}}_{M_n(\mathbb{C})} \otimes i^{(n)}): \mathcal{M}^{\mathcal{U}} &= \prod_{n \to \mathcal{U}} (M_n(\mathbb{C}) \otimes \mathcal{M}^{1/n})\\ &\quad \to \prod_{n \to \mathcal{U}} (M_n(\mathbb{C}) \otimes (\mathcal{M}^{1/2n})^{\mathcal{V}}) \cong ((\mathcal{M}^{1/2})^{\mathcal{V}})^{\mathcal{U}}. \end{align*}$$

Consider $i^{\mathcal {U}}(\mathbf {X}) \oplus i^{\mathcal {U}}(\tilde {\mathbf {X}})$ and $i^{\mathcal {U}}(\mathbf {Z}) \oplus i^{\mathcal {U}}(\tilde {\mathbf {Z}})$ as elements of

$$\begin{align*}M_2(((\mathcal{M})^{1/2})^{\mathcal{V}})^{\mathcal{U}}) = (\mathcal{M}^{\mathcal{V}})^{\mathcal{U}} = (\mathcal{M}^{\mathcal{U}})^{\mathcal{V}}. \end{align*}$$

Note that $(i^{\mathcal {U}}(\mathbf {X}) \oplus i^{\mathcal {U}}(\tilde {\mathbf {X}}), \, i^{\mathcal {U}}(\mathbf {Z}) \oplus i^{\mathcal {U}}(\tilde {\mathbf {Z}}))$ has the same quantifier-free type as $(\mathbf {X},\mathbf {Z})$ , and in particular,

$$\begin{align*}\varphi^{(\mathcal{M}^{\mathcal{U}})^{\mathcal{V}}}(i^{\mathcal{U}}(\mathbf{X}) \oplus i^{\mathcal{U}}(\tilde{\mathbf{X}})) \leq \psi^{(\mathcal{M}^{\mathcal{U}})^{\mathcal{V}}}(i^{\mathcal{U}}(\mathbf{X}) \oplus i^{\mathcal{U}}(\tilde{\mathbf{X}}), \, i^{\mathcal{U}}(\mathbf{Z}) \oplus i^{\mathcal{U}}(\tilde{\mathbf{Z}})) = \varphi^{\mathcal{M}^{\mathcal{U}}}(\mathbf{X}). \end{align*}$$

On the other hand,

$$\begin{align*}i^{\mathcal{U}}(\mathbf{X}) \oplus i^{\mathcal{U}}(\tilde{\mathbf{X}}) = j(\mathbf{Y}), \end{align*}$$

where j is the diagonal embedding

$$\begin{align*}j: \mathcal{M}^{\mathcal{U}} \to (\mathcal{M}^{\mathcal{U}})^{\mathcal{V}} \text{ or equivalently } \prod_{n \to \mathcal{U}} (M_{2n}(\mathbb{C}) \otimes \mathcal{M}^{1/2n}) \to \prod_{n \to \mathcal{U}} (M_{2n}(\mathbb{C}) \otimes (\mathcal{M}^{1/2n})^{\mathcal{V}}). \end{align*}$$

Hence,

$$\begin{align*}\varphi^{\mathcal{M}^{\mathcal{U}}}(\mathbf{Y}) = \varphi^{(\mathcal{M}^{\mathcal{U}})^{\mathcal{V}}}(j(\mathbf{Y})) = \varphi^{(\mathcal{M}^{\mathcal{U}})^{\mathcal{V}}}(i^{\mathcal{U}}(\mathbf{X}) \oplus i^{\mathcal{U}}(\tilde{\mathbf{X}})) \leq \varphi^{\mathcal{M}^{\mathcal{U}}}(\mathbf{X}). \end{align*}$$

This proves the asserted inequality (4.2).

Proof. Let $T: L^2(\mathcal {M}) \to L^2(\mathcal {M})^d$ be given by $T(a) = ([u_1,a], \dots , [u_d,a])$ . By elementary computation,

$$\begin{align*}T^{*}T(a) = 2d\,a - \sum_{j=1}^d u_jau_j^{*} - \sum_{j=1}^d u_j^{*}au_j. \end{align*}$$

Let $\mathcal {P} = \mathcal {N}' \cap \mathcal {M}$ . Note that T vanishes on $\mathcal {P}$ and $a - E_{\mathcal {P}}(a)$ is the orthogonal projection of a onto $\mathcal {P}^{\perp }$ . Therefore, condition (1) can be restated as $\epsilon \left \lVert a \right \rVert _2^2 \leq \left \lVert T(a) \right \rVert _2^2 = \left \langle a,T^{*}T(a) \right \rangle $ for $a \in \mathcal {P}^{\perp }$ , which is equivalent to the spectrum of $T^{*}T |_{\mathcal {P}^{\perp }}$ being contained in $[\epsilon ,\infty )$ . Meanwhile, condition (2) can be restated as $\left \lVert (2d - T^{*}T)|_{\mathcal {P}^{\perp }} \right \rVert \leq 2d - \epsilon $ ; since $\left \lVert T^{*}T \right \rVert \leq 2d$ , this is equivalent to the above.

Proof. The situation above is the Hermitian case with $D = 2d$ in Hastings’s terminology. Hastings [Reference Hastings45] at the top of the second page asserts convergence in probability of $\lambda _2^{(n)}$ . Hastings’s arguments in fact yield almost sure convergence. Indeed, $\liminf _{n \to \infty } \lambda _2^{(n)} \geq \sqrt {2d-1} / d$ follows from a deterministic lower bound on $\lambda _2^{(n)}$ in [Reference Hastings45, equation (12)] which gives (using $\lambda _H=2\sqrt {D-1}/D=\sqrt {2d-1}/d$ , see [Reference Hastings45, equation (3)]), $\lambda _2 \geq \frac {2d-1}d(1-O(\ln (\ln (n))/\ln (n))$ .

For the converse inequality, at the end of Section II.F, Hastings shows that for $c>1$ , the probability that $\lambda _2^{(n)}$ is greater than $c \sqrt {2d - 1}/ d$ is bounded by

$$\begin{align*}c^{-(1/4)n^{2/15}} (1 + O(\log(n)n^{-2/15})). \end{align*}$$

Because this is summable, the Borel–Cantelli lemma implies that almost surely we have $\limsup _{n \to \infty } \lambda _2^{(n)} \leq c \lambda _H$ . Since $c> 1$ was arbitrary, this yields almost sure convergence.

Proof. Let $\Phi ^{(n)}$ be as in the previous theorem. Note that $\ker (\operatorname {\mathrm {tr}}_n)$ is the orthogonal complement of $\mathbb {C} 1$ , which is the $\lambda _1^{(n)}$ -eigenspace of $\Phi ^{(n)}$ . Hence,

$$\begin{align*}\left\lVert \sum_{j=1}^d (U_j^{(n)} (A - \operatorname{\mathrm{tr}}_n(A)) (U_j^{(n)})^{*} + (U_j^{(n)})^{*} (A - \operatorname{\mathrm{tr}}_n(A)) U_j^{(n)}) \right\rVert_2 \leq 2d \lambda_2^{(n)} \left\lVert A - \operatorname{\mathrm{tr}}_n(A) \right\rVert_2. \end{align*}$$

This means that $U_1^{(n)}$ , …, $U_d^{(n)}$ satisfy Lemma 4.4 (2) with $\mathcal {N} = \mathbb {C}$ , $\mathcal {M} = M_n(\mathbb {C})$ , and $2d - \epsilon ^{(n)} = 2d \lambda _2^{(n)}$ . Hence, by Lemma 4.4, the $U_j^{(n)}$ witness spectral gap for $\mathbb {C} \subseteq M_n(\mathbb {C})$ with constant $\epsilon ^{(n)} = 2d(1 - \lambda _2^{(n)})$ . By Hastings’s theorem, almost surely,

$$\begin{align*}\frac{1}{\epsilon^{(n)}} = \frac{1}{2d(1 - \lambda_2^{(n)})} \to \frac{1}{2d - 2 \sqrt{2d-1}} = \frac{d + \sqrt{2d-1}}{2(d^2 - 2d + 1)}. \end{align*}$$

We can bound the right-hand side by $d / (d-1)^2$ because $\sqrt {2d - 1} < d$ .

Proof. By Corollary 4.7 with $d = 3$ , for sufficiently large n and $A \in M_n(\mathbb {C})$ , we have almost surely

$$\begin{align*}\left\lVert A - \operatorname{\mathrm{tr}}_n(A) \right\rVert_2^2 \leq \frac{3}{4} \sum_{j=1}^3 \left\lVert [A,U_j^{(n)}] \right\rVert_2^2. \end{align*}$$

Because this is an inequality between linear operators on a Hilbert space, we may tensorize with the identity on $L^2(\mathcal {M}^{1/n})$ (see, e.g., [Reference Goldbring, Jekel, Elayavalli and Pi39, Lemma 4.18]), to obtain for $a \in M_n(\mathbb {C}) \otimes \mathcal {M}^{1/n} = \mathcal {M}$ that

$$\begin{align*}\left\lVert a - E_{\mathcal{A}_n}[a] \right\rVert_2^2 \leq \frac{3}{4} \sum_{j=1}^3 \left\lVert [a,X_j^{(n)}] \right\rVert_2^2 \text{ for } a \in \mathcal{M}. \end{align*}$$

Then in the ultralimit, we obtain

$$\begin{align*}\left\lVert a - E_{\mathcal{A}}[a] \right\rVert_2^2 \leq \frac{3}{4} \sum_{j=1}^3 \left\lVert [a,X_j] \right\rVert_2^2 \text{ for } a \in \mathcal{M}^{\mathcal{U}}, \end{align*}$$

since conditional expectations commute with ultraproducts. This is the desired estimate for $\mathcal {A}$ . For the final claim, $\mathcal {A} \subseteq \{\mathbf {X}\}' \cap \mathcal {M}^{\mathcal {U}}$ is immediate from the construction of $\mathbf {X}$ , and the opposite inclusion follows from the spectral gap estimate that we just proved.

Proof. To prove the estimate for $\mathcal {B}$ , the same tensorization and ultralimit argument as in the proof of Lemma 4.8 apply, and so it suffices to show that for $B = \begin {bmatrix} B_{1,1} & B_{1,2} \\ B_{2,1} & B_{2,2} \end {bmatrix} \in M_{2n}(\mathbb {C})$ , we have

$$ \begin{align*} \left\lVert \begin{bmatrix} B_{1,1} & B_{1,2} \\ B_{2,1} & B_{2,2} \end{bmatrix} - \begin{bmatrix} \operatorname{\mathrm{tr}}_n(B_{1,1}) & 0 \\ 0 & \operatorname{\mathrm{tr}}_n(B_{2,2}) \end{bmatrix} \right\rVert_2^2 &\leq 7 \sum_{j=1}^3 \left\lVert [B,Y_j^{(n)}] \right\rVert_2^2 \\ &= 7 \Bigg( \sum_{j=1}^2 \left\lVert \begin{bmatrix} U_j^{(n)} B_{1,1} - B_{1,1} U_j^{(n)} & U_j^{(n)} B_{1,2} - B_{1,2} U_j^{(n)} \\ U_j^{(n)} B_{2,1} - B_{2,1} U_j^{(n)} & U_j^{(n)} B_{2,2} - B_{2,2} U_j^{(n)} \end{bmatrix} \right\rVert_2^2 \\ &+ \left\lVert \begin{bmatrix} U_3^{(n)} B_{1,1} - B_{1,1} U_3^{(n)} & U_3^{(n)} B_{1,2} - B_{1,2} U_4^{(n)} \\ U_4^{(n)} B_{2,1} - B_{2,1} U_3^{(n)} & U_4^{(n)} B_{2,2} - B_{2,2} U_4^{(n)} \end{bmatrix} \right\rVert_2^2 \Bigg). \end{align*} $$

Equivalently, we want to show that

$$ \begin{align*} \left\lVert B_{1,1} - \operatorname{\mathrm{tr}}_n(B_{1,1}) \right\rVert_2^2 &+ \left\lVert B_{2,2} - \operatorname{\mathrm{tr}}_n(B_{2,2}) \right\rVert_2^2 + \left\lVert B_{1,2} \right\rVert_2^2 + \left\lVert B_{2,1} \right\rVert_2^2 \\ &\leq 7 \Big( \sum_{j=1}^3 \left\lVert [B_{1,1},U_j^{(n)}] \right\rVert_2^2 + \sum_{j=1}^3 \left\lVert [B_{2,2},U_j^{(n)}] \right\rVert_2^2 \\ & \quad + \sum_{j=1}^2 \left\lVert [B_{1,2},U_j^{(n)}] \right\rVert_2^2 + \left\lVert U_3^{(n)} B_{1,2} - B_{1,2} U_4^{(n)} \right\rVert_2^2 \\ & \quad + \sum_{j=1}^2 \left\lVert [B_{2,1},U_j^{(n)}] \right\rVert_2^2 + \left\lVert U_4^{(n)} B_{2,1} - B_{2,1} U_3^{(n)} \right\rVert_2^2 \Big). \end{align*} $$

From Corollary 4.7, we already know

$$\begin{align*}\left\lVert B_{1,1} - \operatorname{\mathrm{tr}}_n(B_{1,1}) \right\rVert_2^2 \leq \frac{3}{4} \sum_{j=1}^3 \left\lVert [B_{1,1},U_j^{(n)}] \right\rVert_2^2, \end{align*}$$

and similarly for the $B_{2,2}$ term. Thus, it remains to estimate the $B_{1,2}$ and $B_{2,1}$ terms. We will handle the $B_{1,2}$ term and show that

(4.6) $$ \begin{align} \left\lVert B_{1,2} \right\rVert_2^2 \leq 7 \left( \sum_{j=1}^2 \left\lVert [B_{1,2},U_j^{(n)}] \right\rVert_2^2 + \left\lVert U_3^{(n)} B_{1,2} - B_{1,2} U_4^{(n)} \right\rVert_2^2 \right); \end{align} $$

the argument for the $B_{2,1}$ term is symmetrical. First, we note that by Corollary 4.7 with $d = 2$ , we have almost surely for sufficiently large n,

(4.7) $$ \begin{align} \left\lVert B_{1,2} - \operatorname{\mathrm{tr}}_n(B_{1,2}) \right\rVert_2^2 \leq 2 \sum_{j=1}^2 \left\lVert [B_{1,2},U_j^{(n)}] \right\rVert_2^2. \end{align} $$

Thus, it remains to estimate $\operatorname {\mathrm {tr}}_n(B_{1,2})$ . We note that

$$ \begin{align*} |\operatorname{\mathrm{tr}}_n(B_{1,2})| \left\lVert U_3^{(n)} - U_4^{(n)} \right\rVert_2 &= \left\lVert U_3^{(n)} \operatorname{\mathrm{tr}}_n(B_{1,2}) - \operatorname{\mathrm{tr}}_n(B_{1,2}) U_4^{(n)} \right\rVert_2 \\ &\leq \left\lVert U_3^{(n)}(B_{1,2} - \operatorname{\mathrm{tr}}_n(B_{1,2})) - (B_{1,2} - \operatorname{\mathrm{tr}}_n(B_{1,2})) U_4^{(n)} \right\rVert_2 \\ & \quad + \left\lVert U_3^{(n)} B_{1,2} - B_{1,2} U_4^{(n)} \right\rVert_2 \\ &\leq 2 \left\lVert B_{1,2} - \operatorname{\mathrm{tr}}_n(B_{1,2}) \right\rVert_2 + \left\lVert U_3^{(n)} B_{1,2} - B_{1,2} U_4^{(n)} \right\rVert_2. \end{align*} $$

Note that $\mathbb {E} \operatorname {\mathrm {tr}}_n((U_3^{(n)})^{*} U_4^{(n)}) = 0$ , and so by Proposition 4.1, we have $\operatorname {\mathrm {tr}}_n((U_3^{(n)})^{*} U_4^{(n)}) \to 0$ almost surely, and thus $\left \lVert U_3^{(n)} - U_4^{(n)} \right \rVert _2^2=\left \lVert 1-(U_3^{(n)})^{*} U_4^{(n)} \right \rVert _2^2 \to 2$ almost surely, and hence is eventually larger than $9/5$ . Hence, we have that for sufficiently large n,

$$\begin{align*}|\operatorname{\mathrm{tr}}_n(B_{1,2})| \leq \sqrt{5/9}\ \left( 2 \left\lVert B_{1,2} - \operatorname{\mathrm{tr}}_n(B_{1,2}) \right\rVert_2 + \left\lVert U_3^{(n)} B_{1,2} - B_{1,2} U_4^{(n)} \right\rVert_2 \right). \end{align*}$$

By the Cauchy–Schwarz inequality and our previous estimate for $\left \lVert B_{1,2} - \operatorname {\mathrm {tr}}_n(B_{1,2}) \right \rVert _2$ ,

$$ \begin{align*} |\operatorname{\mathrm{tr}}_n(B_{1,2})|^2 &\leq \frac{5}{9} (1 + 1/8) \left( 4 \left\lVert B_{1,2} - \operatorname{\mathrm{tr}}_n(B_{1,2}) \right\rVert_2^2 + 8 \left\lVert U_3^{(n)} B_{1,2} - B_{1,2} U_4^{(n)} \right\rVert_2^2 \right) \\ &\leq \frac{5}{8} \left( 4 \cdot 2 \sum_{j=1}^2 \left\lVert [B_{1,2},U_j^{(n)}] \right\rVert_2^2 + 8 \left\lVert U_3^{(n)} B_{1,2} - B_{1,2} U_4^{(n)} \right\rVert_2^2 \right) \\ &= 5 \sum_{j=1}^2 \left\lVert [B_{1,2},U_j^{(n)}] \right\rVert_2^2 + 5 \left\lVert U_3^{(n)} B_{1,2} - B_{1,2} U_4^{(n)} \right\rVert_2^2. \end{align*} $$

Hence, using this and (4.7),

$$ \begin{align*} \left\lVert B_{1,2} \right\rVert_2^2 &= \left\lVert B_{1,2} - \operatorname{\mathrm{tr}}_n(B_{1,2}) \right\rVert_2^2 + \operatorname{\mathrm{tr}}_n(B_{1,2})^2 \\ &\leq \left(2 + 5 \right) \sum_{j=1}^2 \left\lVert [B_{1,2},U_j^{(n)}] \right\rVert_2^2 + 5 \left\lVert U_3^{(n)} B_{1,2} - B_{1,2} U_4^{(n)} \right\rVert_2^2 \\ &\leq 7 \left( \sum_{j=1}^2 \left\lVert [B_{1,2},U_j^{(n)}] \right\rVert_2^2 + \left\lVert U_3^{(n)} B_{1,2} - B_{1,2} U_4^{(n)} \right\rVert_2^2 \right) \end{align*} $$

as desired.

Proof of Theorem B in the $\mathrm {II}_1$ factor case.

Referring to the outline of the proof stated in Section 4.1, we have shown (1) in Lemma 4.3, (2) in Lemma 4.8, and (3) in Lemma 4.9. Item (1) shows that, almost surely, $\varphi ^{\mathcal {M}^{\mathcal {U}}}(\mathbf {X}) \leq \varphi ^{\mathcal {M}^{\mathcal {U}}}(\mathbf {Y})$ for all $\inf $ -formulas. If $\mathcal {M}$ were model complete, then $\mathbf {X}$ and $\mathbf {Y}$ would have the same type by Lemma 2.3. Hence, to finish the argument, it suffices to show that $\mathbf {X}$ and $\mathbf {Y}$ do not have the same type.

In fact, we claim that $\mathbf {X}$ and $\mathbf {Y}$ do not even have the same two-quantifier type. Consider the formula

$$ \begin{align*} \psi(x_1, x_2, x_3)& = \\ &\inf_{z_1} \left[ 1 - \left\lVert z_1 \right\rVert_2^2 + |\operatorname{\mathrm{tr}}(z_1)|^2 + 7 \sum_{j=1}^3 \left\lVert [x_j,z_1] \right\rVert_2^2 + \sup_{z_2} \left[ \left\lVert [z_1,z_2] \right\rVert_2^2 \dot{-} 28 \sum_{j=1}^3 \left\lVert [x_j,z_2] \right\rVert_2^2 \right] \right], \end{align*} $$

where $z_1$ and $z_2$ range over the unit ball. Then the condition $\psi (x_1,x_2, x_3)=0$ attempts to assert the existence of $z_1$ with $\|z_1\|_2=1$ and $\operatorname {\mathrm {tr}}(z_1)=0$ such that $z_1$ commutes with $x_j$ for $j = 1,2,3$ and also commutes with every $z_2$ in the relative commutant of $\{x_1,x_2, x_3\}$ .Footnote ⁵ We will find a self-adjoint unitary $z_1$ that commutes with $Y_j$ for $j = 1,2,3$ , has zero trace, and commutes with everything in the relative commutant of $\{Y_1,Y_2, Y_3\}$ ; this will suffice to show that $\psi ^{\mathcal {M}^{\mathcal {U}}}(\mathbf {Y})=0$ .

Indeed, $\{\mathbf {Y}\}' \cap \mathcal {M}^{\mathcal {U}} = \mathcal {B}$ is a direct sum of two copies of $\prod _{n \to \mathcal {U}} \mathcal {M}^{1/2n}$ so it has a central projection p of trace $1/2$ . Let $z_1 = 2p - 1$ , so that $\left \lVert z_1 \right \rVert _2^2 = 1$ and $\operatorname {\mathrm {tr}}(z_1) = 0$ and $\sum _{j=1}^3 \left \lVert [Y_j,z_1] \right \rVert _2 = 0$ . Also for every $z_2$ , we have

$$\begin{align*}\left\lVert [z_1,z_2] \right\rVert_2^2 \leq \left( \left\lVert [z_1,E_{\mathcal{B}}[z_2]] \right\rVert_2 + 2 \left\lVert z_1 \right\rVert d(z_2,\mathcal{B}) \right)^2 = 4 d(z_2,\mathcal{B})^2 \leq 28 \sum_{j=1}^3 \left\lVert [Y_j,z_2] \right\rVert_2^2, \end{align*}$$

because of Lemma 4.9 and the fact that $[z_1,E_{\mathcal {B}}[z_2]] = 0$ .

On the other hand, we claim that $\psi ^{\mathcal {M}^{\mathcal {U}}}(\mathbf {X}) = 1$ . Because the ultraproduct $\mathcal {M}^{\mathcal {U}}$ is countably saturated (see Section 2.2), there is some $z_1 \in \mathcal {M}^{\mathcal {U}}$ that attains the infimum in the formula. Let $z_1^{\prime } = E_{\mathcal {A}}[z_1]$ . Because $\mathcal {A}$ is a factor, a Dixmier averaging argument (see, e.g., [Reference Farah, Hart and Sherman29, Lemma 4.2]) shows that

$$\begin{align*}\left\lVert z_1^{\prime} \right\rVert_2^2 - |\operatorname{\mathrm{tr}}(z_1^{\prime})|^2 =\|z-\operatorname{\mathrm{tr}}(z)\|_2^2\leq \sup_{z_2 \in \mathcal{A}_1} \left\lVert [z_1^{\prime},z_2] \right\rVert_2^2, \end{align*}$$

where $\mathcal {A}_1$ is the unit ball of $\mathcal {A}$ . Using choices of $z_2 \in \mathcal {A}$ witnessing this inequality as candidates for the supremum in $\psi $ , we conclude

$$\begin{align*}\psi^{\mathcal{M}^{\mathcal{U}}}(\mathbf{X}) \geq 1 - \left\lVert z_1 \right\rVert_2^2 + |\operatorname{\mathrm{tr}}(z_1)|^2 + d(z_1,\mathcal{A})^2 + \left[ \left\lVert z_1^{\prime} \right\rVert_2^2 - |\operatorname{\mathrm{tr}}(z_1^{\prime})|^2 + 0 \right], \end{align*}$$

where we have also applied the spectral gap inequality from Lemma 4.8 to get the $d(z_1, \mathcal {A})^2$ term. Noting that $d(z_1,\mathcal {A})^2 = \left \lVert z_1 - z_1^{\prime } \right \rVert _2^2 = \left \lVert z_1 \right \rVert _2^2 - \left \lVert z_1^{\prime } \right \rVert _2^2$ and that $\operatorname {\mathrm {tr}}(z_1^{\prime }) = \operatorname {\mathrm {tr}}(z_1)$ , the entire expression evaluates to $1$ . For the upper bound ${\psi ^{\mathcal {M}^{\mathcal {U}}}(\mathbf {X}) \leq 1}$ , simply take $z_1 = 0$ .

Proof. Let $\mathcal {M}$ be a tracial von Neumann algebra which decomposes as a direct sum $\mathcal {M}_1 \oplus \mathcal {M}_2$ with weights $\alpha $ and $1 - \alpha $ . Assume the theory of $\mathcal {M}$ is model complete; we will prove that the theory of each one of $\mathcal {M}_1$ and $\mathcal {M}_2$ is model complete.

Let $\mathcal {N}_1 \equiv \mathcal {M}_1$ and $\mathcal {N}_2 \equiv \mathcal {M}_2$ , and $\iota _1: \mathcal {M}_1 \to \mathcal {N}_1$ and $\iota _2: \mathcal {M}_2 \to \mathcal {N}_2$ be trace-preserving $*$ -homomorphisms; we need to show that $\iota _1$ and $\iota _2$ are elementary. Let $\mathcal {N}$ be the direct sum of $\mathcal {N}_1$ and $\mathcal {N}_2$ with weights $\alpha $ and $1 - \alpha $ . Note that by [Reference Farah and Ghasemi27], $\mathcal {N} \equiv \mathcal {M}$ since the theory of $\mathcal {N}$ is uniquely determined by the theories of the direct summands. By model completeness of $\mathcal {M}$ , the map $\iota = \iota _1 \oplus \iota _2: \mathcal {M} \to \mathcal {N}$ is elementary.

Let $\varphi (x_1,\dots ,x_n)$ be an $\mathcal {L}_{\operatorname {\mathrm {tr}}}$ -formula, and we will show that $\varphi ^{\mathcal {N}_1}(\iota _1(\mathbf {a})) = \varphi ^{\mathcal {M}_1}(\mathbf {a})$ for ${\mathbf {a} = (a_1,\dots ,a_n) \in \mathcal {M}_1^n}$ . Because prenex formulas are dense in the space of all formulas [Reference Ben Yaacov, Berenstein, Henson, Usvyatsov, Chatzidakis, Macpherson, Pillay and Wilkie11, Section 6], assume without loss of generality that

$$\begin{align*}\varphi(x_1,\dots,x_n) = \inf_{y_1} \sup_{y_2} \dots \inf_{y_{2m-1}} \sup_{y_{2m}} F(\operatorname{\mathrm{Re}} \operatorname{\mathrm{tr}}(p_1(\mathbf{x},\mathbf{y})),\dots,\operatorname{\mathrm{Re}} \operatorname{\mathrm{tr}}(p_k(\mathbf{x},\mathbf{y}))), \end{align*}$$

where $y_1$ , …, $y_{2m}$ are variables in the unit ball, $F: \mathbb {R}^k \to \mathbb {R}$ is continuous, and $p_1$ , …, $p_k$ are non-commutative $*$ -polynomials. Define

$$\begin{align*}\psi(x_1,\dots,\tilde{x}_n,z) = \inf_{y_1} \sup_{y_2} \dots \inf_{y_{2m-1}} \sup_{y_{2m}} F\left(\frac{1}{\alpha} \operatorname{\mathrm{Re}} \operatorname{\mathrm{tr}}(p_1(\mathbf{x},z\mathbf{y})),\dots, \frac{1}{\alpha} \operatorname{\mathrm{Re}} \operatorname{\mathrm{tr}}(p_k(\mathbf{x},z\mathbf{y})) \right), \end{align*}$$

where $z\mathbf {y} = (zy_1,\dots ,zy_{2m})$ . Observe that

$$\begin{align*}\varphi^{\mathcal{M}_1}(a_1,\dots,a_n) = \psi^{\mathcal{M}}(a_1 \oplus 0,\dots,a_n \oplus 0, 1 \oplus 0), \end{align*}$$

because $(1 \oplus 0)(y \oplus y') = y \oplus 0$ . Similarly,

$$\begin{align*}\varphi^{\mathcal{N}_1}(\iota_1(a_1),\dots,\iota_1(a_n)) = \psi^{\mathcal{N}}(\iota(a_1 \oplus 0),\dots,\iota(a_n \oplus 0), \iota(1 \oplus 0)). \end{align*}$$

The mapping $\iota : \mathcal {M} \to \mathcal {N}$ is elementary, and hence

$$\begin{align*}\psi^{\mathcal{N}}(\iota(a_1 \oplus 0),\dots,\iota(a_n \oplus 0), \iota(1 \oplus 0)) = \psi^{\mathcal{M}}(a_1 \oplus 0,\dots,a_n \oplus 0, 1 \oplus 0). \end{align*}$$

This shows $\varphi ^{\mathcal {N}_1}(\iota _1(\mathbf {a})) = \varphi ^{\mathcal {M}_1}(\mathbf {a})$ , so the mapping $\iota _1$ is elementary as desired. The same argument applies to $\iota _2$ . Therefore, $\mathcal {M}_1$ and $\mathcal {M}_2$ are model complete.

Proof. Let $\mathcal {N} = L^\infty [0,1] \otimes \mathcal {M}$ . Note that

$$\begin{align*}\mathcal{N} = \int_{[0,1]^2} \mathcal{M}_\omega \,d\omega \,d\omega'. \end{align*}$$

Thus, the distribution of $\operatorname {\mathrm {Th}}(\mathcal {M}_\omega )$ over $[0,1]^2$ is the same as the distribution of the $\operatorname {\mathrm {Th}}(\mathcal {M}_\omega )$ over $[0,1]$ . Therefore, it follows from [Reference Farah and Ghasemi27, Theorem 2.3] that $\mathcal {M} \equiv \mathcal {N}$ . Moreover, $\mathcal {N} \oplus \mathcal {N} \cong \mathcal {N}$ . Now fix an ultrafilter $\mathcal {U}$ on $\mathbb {N}$ and note that $M_2(\mathcal {M}_\omega )$ embeds into $\mathcal {M}_\omega ^{\mathcal {U}}$ for all $\omega $ , hence $M_2(\mathcal {N})$ embeds into $\mathcal {N}^{\mathcal {U}}$ . Consider a trace preserving $*$ -homomorphism

$$\begin{align*}\mathcal{N} \to \mathcal{N} \oplus \mathcal{N} \to M_2(\mathcal{N}) \to \mathcal{N}^{\mathcal{U}}, \end{align*}$$

where the first map is an isomorphism and the second map is the block diagonal embedding. Then $1 \oplus 0$ is central in $\mathcal {N} \oplus \mathcal {N}$ but $1 \oplus 0$ is not central in $M_2(\mathcal {N})$ . Hence, our homomorphism does not map $Z(\mathcal {N})$ into $Z(\mathcal {N}^{\mathcal {U}})$ , so it is not elementary.

Proof of Theorem B.

Suppose $\mathcal {M}$ has a direct integral decomposition where $\mathcal {M}_\omega $ is a $\mathrm {II}_1$ factor such that $M_2(\mathcal {M}_\omega )$ embeds $\mathcal {M}_{\omega }^{\mathcal {U}}$ , for $\omega $ in some positive measure set. If the positive measure set has an atom, then $\mathcal {M}$ has a direct summand $\mathcal {N}$ which is a $\mathrm {II}_1$ factor such that $M_2(\mathcal {N})$ embeds into $\mathcal {N}^{\mathcal {U}}$ . The results of the previous section show that $\mathcal {N}$ is not model complete, hence by Lemma 5.1, $\mathcal {M}$ is not model complete.

If there is no atom in our positive measure set, then $\mathcal {M}$ has a direct summand of the form $\mathcal {N} = \int _{[0,1]} \mathcal {N}_\alpha \,d\alpha $ where the integral occurs with respect to Lebesgue measure and $\mathcal {N}_\alpha $ is a $\mathrm {II}_1$ factor such that $M_2(\mathcal {N}_\alpha )$ embeds into $\mathcal {N}_\alpha ^{\mathcal {U}}$ . Hence, by Lemma 5.4, $\mathcal {N}$ is not model complete, and so by Lemma 5.1, $\mathcal {M}$ is not model complete.

Proof. We use the following observation several times: For any two tracial von Neumann algebras $\mathcal {M}_0$ and $\mathcal {M}_1$ , the theory of $\mathcal {M}_\alpha = (\mathcal {M}_0,1-\alpha ) \oplus (\mathcal {M}_1,\alpha )$ depends continuously on $\alpha \in [0,1]$ . This idea was used in [Reference Goldbring and Hart35, Proposition 5.1]. Indeed, one can show by induction that for each formula $\varphi $ , the quantity $\varphi ^{\mathcal {M}_\alpha }(x_1 \oplus x_1^{\prime }, \dots , x_n \oplus x_n^{\prime })$ is continuous in $\alpha $ uniformly over $x_j$ and $x_j^{\prime }$ in the unit ball.

Now we proceed to the main claims:

1. (1) $M_n(\mathbb {C})$ admits quantifier elimination. Fixing an ultrafilter $\mathcal {U}$ on the natural numbers, $\lim _{n \to \mathcal {U}} \operatorname {\mathrm {Th}}(M_n(\mathbb {C})) = \operatorname {\mathrm {Th}}(\prod _{n \to \mathcal {U}} M_n(\mathbb {C}))$ , which does not admit quantifier elimination by [Reference Farah26] since the matrix ultraproduct is a $\mathrm {II}_1$ factor.Footnote ⁶

2. (2) Consider $(M_n(\mathbb {C}),1-\alpha ) \oplus (\mathcal {R},\alpha )$ . This does not admit quantifier elimination when $\alpha> 0$ but does admit quantifier elimination when $\alpha = 0$ .

3. (3) This follows from the same argument as (1).

4. (4) This follows from the same argument as (2) since $(M_n(\mathbb {C}),1-\alpha ) \oplus (\mathcal {R},\alpha )$ is not model complete by Theorem B.

Proof. Consider quantifier elimination. Since the language is separable, choose for each n a countable dense set $\mathcal {F}_n$ of formulas in n variables (if there are multiple sorts, then we choose such a set for each tuple of sorts). For each n and $\varphi \in \mathcal {F}_n$ , for each $k \geq 1$ , let $G_{\varphi ,k}$ be the set of complete theories $\mathrm {T}$ such that there exists a quantifier-free formula $\psi $ such that $\mathrm {T}$ models

$$\begin{align*}\sup_{x_1,\dots,x_n} |\varphi(x_1,\dots,x_n) - \psi(x_1,\dots,x_n)| < \frac{1}{k}. \end{align*}$$

Then $G_{\varphi ,k}$ is open and $\bigcap _{\varphi , k} G_{\varphi ,k}$ is precisely the set of theories that admit quantifier elimination, since it suffices to approximate a dense subset of formulas by quantifier-free formulas. The argument for model completeness works the same way using Lemma 2.3 (3).

Proof. By [Reference Farah, Hart and Sherman29, Lemma 4.2], the center $Z(\mathcal {M})$ is definable relative to the theory of tracial von Neumann algebras. Hence, similar to [Reference Farah, Hart, Lupini, Robert, Tikuisis, Vignati and Winter28, Definition 3.2.3 and Lemma 3.2.5] in the $\mathrm {C}^{*}$ -algebra case, $d(y,Z(\mathcal {M}))^2$ is a definable predicate (or it is a formula in an expanded language with a sort added for $Z(\mathcal {M})$ ). Thus, consider the sentence

$$\begin{align*}\inf_{x_1,\dots,x_d \in B_1^{\mathcal{M}}} \sup_{y \in B_1^{\mathcal{M}}} \left( d(y,Z(\mathcal{M}))^2 \dot{-} C \sum_{j=1}^d \left\lVert [x_j,y] \right\rVert_2^2 \right) = 0. \end{align*}$$

Note that $\mathcal {M}$ has $(C,d)$ -spectral gap, then $\mathcal {M}$ satisfies this sentence. The converse holds when $\mathcal {M}$ is countably saturated because we can choose some $x_1$ , …, $x_d$ that realize the infimum. Since every complete theory had a countably saturated model, the set of theories of von Neumann algebras with $(C,d)$ -spectral gap is equal to the set of theories satisfying this sentence, hence is closed. To see that its complement is dense, note that for every tracial von Neumann algebra $\mathcal {M}$ , the direct sum $(\mathcal {M},1-\alpha ) \oplus (\mathcal {R},\alpha )$ does not have spectral gap, and $\operatorname {\mathrm {Th}}((\mathcal {M},1-\alpha ) \oplus (\mathcal {R},\alpha )) \to \mathcal {M}$ as $\alpha \to 0$ .

Proof. (1) By Lemma 6.3 the $(C,d)$ -spectral gap property defines a closed set whose complement is dense. Taking the union over C and d in $\mathbb {N}$ yields a meager $F_\sigma $ set.

(2) Hastings’s result (see Theorem 4.6 and Corollary 4.7 above) shows that matrix algebras $M_n(\mathbb {C})$ have spectral gap for a fixed C and d (for instance one can take $d = 2$ and $C = 2$ ). It is straightforward to check that a direct integral of tracial von Neumann algebras with $(2,2)$ -spectral gap also has $(2,2)$ -spectral gap. Hence, all separable type $\mathrm {I}$ tracial von Neumann algebras have $(2,2)$ -spectral gap, so their theories are contained in the meager set from (1).

(3) Quantifier elimination can only hold for the theories of type I tracial von Neumann algebras [Reference Farah26, Theorem 1].

Proof. [Reference Farah and Ghasemi27, Theorem 2.3] implies that the theory of $\mathcal {M}_{\vec {\alpha }}$ depends continuously on the weights $\vec {\alpha }$ . The construction in [Reference Farah and Ghasemi27, Lemma 3.2] shows that $\alpha _{m,n} = \rho _{\mathcal {M}_{\vec {\alpha }}}(m,n)$ can be recovered from $\operatorname {\mathrm {Th}}(\mathcal {M}_{\vec {\alpha }})$ . In particular, one can see from this that for each $m, n \geq 1$ , if $\vec {\alpha } \in \Delta $ and the theory of $\mathcal {N}$ is sufficiently close to that of $\mathcal {M}_{\vec {\alpha }}$ , then $\rho _{\mathcal {N}}(m,n)$ will be close to $\alpha _{m,n}$ .

Proof. Let $\mathcal {M}_{\vec {\alpha },k} = \bigoplus _{1 \leq m,n \leq k} (M_m(\mathbb {C}),\alpha _{m,n}) \subseteq \mathcal {M}_{\vec {\alpha }}$ , and let $\mathcal {M}_{\vec {\alpha },k}^{\perp }$ be the direct sum over the complementary indices. Let $\tau _{\vec {\alpha }}$ be the trace on $\mathcal {M}_{\vec {\alpha }}$ . Let

$$\begin{align*}\epsilon_k(\vec{\alpha}) = \min \{ |\tau_{\vec{\alpha}}(p) - \tau_{\vec{\alpha}}(q)|: p, q \text{ projections in } \mathcal{M}_{\vec{\alpha},k} \text{ with } \tau(p) \neq \tau(q) \}. \end{align*}$$

Let

$$\begin{align*}G_k = \{ \vec{\alpha} \in \Delta: 1 - \sum_{1 \leq m,n \leq k} \alpha_{m,n} < \epsilon_k(\vec{\alpha}) \}. \end{align*}$$

Note that $G_k$ is open in $\Delta $ , hence also $\bigcup _{k \geq \ell } G_k$ is open. Moreover, contains the set of $\vec {\alpha }$ such that $\vec {\alpha }$ is supported on $\{1,\dots ,k\}^2$ , and so $\bigcup _{k \geq \ell } G_k$ is dense. Therefore,

$$\begin{align*}G = \bigcap_{\ell \in \mathbb{N}} \bigcup_{k \geq \ell} G_k \end{align*}$$

is comeager. Furthermore,

$$\begin{align*}F = \{\vec{\alpha} \in \Delta: \alpha_{m,n} \text{ are linearly independent over } \mathbb{Q}\} \end{align*}$$

is comeager because non-vanishing of $\mathbb {Q}$ -linear combinations is a countable family of open conditions. Hence, $F \cap G$ is comeager.

We claim that if $\vec {\alpha } \in F \cap G$ , then $\mathcal {M}_{\vec {\alpha }}$ admits quantifier elimination. Let p and q be projections of the same trace in $\mathcal {M}_{\vec {\alpha }}$ . For each k, write $p = p_k \oplus p_k^{\perp }$ and $q = q_k \oplus q_k^{\perp }$ with respect to the decomposition $\mathcal {M}_{\vec {\alpha }} = \mathcal {M}_{\vec {\alpha },k} \oplus \mathcal {M}_{\vec {\alpha },k}^{\perp }$ . If $\vec {\alpha } \in G_k$ , then by construction of $G_k$ , we have

$$\begin{align*}|\tau_{\vec{\alpha}}(p_k) - \tau_{\vec{\alpha}}(q_k)| = |\tau_{\vec{\alpha}}(p_k^{\perp}) - \tau_{\vec{\alpha}}(q_k^{\perp})| < \epsilon_k(\vec{\alpha}), \end{align*}$$

which forces $\tau _{\vec {\alpha }}(p_k) = \tau _{\vec {\alpha }}(q_k)$ by definition of $\epsilon _k(\vec {\alpha })$ . Now let $p_{m,n}$ and $q_{m,n}$ be the components of p and q respectively in the direct summand $(M_m(\mathbb {C}),\alpha _{m,n})$ . Because the $\alpha _{m,n}$ ’s are linearly independent over $\mathbb {Q}$ , the condition that $\tau _{\vec {\alpha }}(p_k) = \tau _{\vec {\alpha }}(q_k)$ forces that $\operatorname {\mathrm {tr}}_m(p_{m,n}) = \operatorname {\mathrm {tr}}_m(q_{m,n})$ for $m, n \leq k$ . Because $\vec {\alpha } \in G$ , we know that $\vec {\alpha } \in G_k$ for infinitely many k, and thus $\operatorname {\mathrm {tr}}_m(p_{m,n}) = \operatorname {\mathrm {tr}}_m(q_{m,n})$ for all $m, n$ , which means that p and q are conjugate by an automorphism. Therefore, by Theorem A, $\operatorname {\mathrm {Th}}(\mathcal {M}_{\vec {\alpha }})$ admits quantifier elimination.

Proof. Consider an existential sentence $\varphi = \inf _{x_1,\dots ,x_n} \psi (x_1,\dots ,x_n)$ where $\psi $ is a quantifier-free formula and $x_j$ ranges over the unit ball. We can express

(6.1) $$ \begin{align} \psi(\mathbf{x}) = F(\operatorname{\mathrm{Re}} \operatorname{\mathrm{tr}}(p_1(\mathbf{x})), \dots, \operatorname{\mathrm{Re}} \operatorname{\mathrm{tr}}(p_k(\mathbf{x}))) \end{align} $$

for some non-commutative $*$ -polynomials $p_j$ and a continuous real-valued function F. By rescaling the input variables to F, assume without loss of generality that $\left \lVert p_j(\mathbf {x}) \right \rVert \leq 1$ when $x_1$ , …, $x_n$ are in the unit ball. Let $\omega _F$ be the modulus of continuity of F with respect to the $\ell ^\infty $ -norm on $[-1,1]^k$ . Suppose that $s < t$ . Write

$$\begin{align*}t = ms + \epsilon, \text{ where } m \in \mathbb{N} \text{ and } \epsilon \in [0,t/s). \end{align*}$$

Let $\iota _{s,t}: \mathcal {M}^s \to \mathcal {M}^t$ be the non-unital $*$ -homomorphism $\iota _{s,t}(x) = x^{\oplus m} \oplus 0_{\mathcal {M}^\epsilon }$ , and note that

$$\begin{align*}|\operatorname{\mathrm{tr}}^{\mathcal{M}^t}(\iota_{s,t}(y)) - \operatorname{\mathrm{tr}}^{\mathcal{M}^s}(y)| \leq \frac{\epsilon}{t} \left\lVert y \right\rVert \leq \frac{s}{t} \left\lVert y \right\rVert. \end{align*}$$

Hence, from (6.1) and the uniform continuity of F,

$$\begin{align*}\psi^{\mathcal{M}^t}(\iota_{s,t}(\mathbf{x})) \leq \psi^{\mathcal{M}^s}(\mathbf{x}) + \omega_F(s/t), \text{ hence } \varphi^{\mathcal{M}^t} \leq \varphi^{\mathcal{M}^s} + \omega_F(s/t). \end{align*}$$

For each $s \in (0,\infty )$ , we have

$$\begin{align*}\limsup_{t \to \infty} \varphi^{\mathcal{M}^t} \leq \liminf_{t \to \infty} \left[ \varphi^{\mathcal{M}^s} + \omega_F(s/t) \right] = \varphi^{\mathcal{M}^s}. \end{align*}$$

Since s was arbitrary, it follows that $\lim _{t \to \infty } \varphi ^{\mathcal {M}^t} = \inf _{t \in (0,\infty )} \varphi ^{\mathcal {M}^t}$ . A similar argument shows that $\lim _{t \to 0^+} \varphi ^{\mathcal {M}^t} = \sup _{t \in (0,\infty )} \varphi ^{\mathcal {M}^t}$ . It remains to show that the limit as $t \to \infty $ agrees with $\operatorname {\mathrm {Th}}_{\exists }(\mathcal {M} \otimes \mathcal {R})$ . First, note that $\mathcal {M}$ embeds into $\mathcal {M} \otimes \mathcal {R}$ , so also $\mathcal {M}^t$ embeds into $(\mathcal {M} \otimes \mathcal {R})^t = \mathcal {M} \otimes \mathcal {R}^t \cong \mathcal {M} \otimes \mathcal {R}$ . Thus, for each $\inf $ -sentence $\varphi $ , we have $\varphi ^{\mathcal {M} \otimes \mathcal {R}} \leq \lim _{t \to \infty } \varphi ^{\mathcal {M}^t}$ . For the opposite inequality, note that $\mathcal {M} \otimes \mathcal {R}$ embeds into $\mathcal {N} = \prod _{n \to \mathcal {U}} \mathcal {M} \otimes M_n(\mathbb {C})$ .

Proof. (1) $\implies $ (2) because $(1/2) \mathcal {M} \oplus (1/2) \mathcal {M}$ is contained in $M_2(\mathcal {M})$ .

(2) $\implies $ (3). Let $\iota : \alpha \mathcal {M} \oplus (1 - \alpha ) \mathcal {M} \to \mathcal {M}^{\mathcal {U}}$ be an embedding where $\mathcal {U}$ is an ultrafilter on index set I. Let $p = \iota (1 \oplus 0)$ . Let $\Delta : \mathcal {M} \to \alpha \mathcal {M} \oplus (1 - \alpha ) \mathcal {M}$ be the diagonal map. Then $\Delta (\mathcal {M})$ commutes with p and hence $\operatorname {Ad}_p \circ \iota \circ \Delta $ gives an embedding $\mathcal {M} \to p(\mathcal {M}^{\mathcal {U}})p$ since $\mathcal {M}$ is a $\mathrm {II}_1$ factor. Now p lifts to a family of projections $(p_i)_{i \in I}$ with $\operatorname {\mathrm {tr}}^{\mathcal {M}}(p_i) = \operatorname {\mathrm {tr}}^{\mathcal {M}^{\mathcal {U}}}(p) = \alpha $ . Since $p_i$ is unitarily conjugate to some fixed projection $p_0 \in \mathcal {M}$ for all i, we have $p \mathcal {M}^{\mathcal {U}} p = \prod _{i \to \mathcal {U}} p_i \mathcal {M} p_i = (p_0 \mathcal {M} p_0)^{\mathcal {U}}$ . In other words, $\mathcal {M}$ embeds into an ultraproduct of $\mathcal {M}^{\alpha }$ . This also implies that $\mathcal {M}^t$ embeds into an ultraproduct of $\mathcal {M}^{t\alpha }$ for each $t \in (0,\infty )$ . Hence, $\mathcal {M}^{1/\alpha ^k}$ embeds into an ultraproduct of $\mathcal {M}$ for each $k \in \mathcal {N}$ . Thus, for an $\inf $ -formula $\varphi $ ,

$$\begin{align*}\varphi^{\mathcal{M} \otimes \mathcal{R}} = \lim_{t \to \infty} \varphi^{\mathcal{M}^t} = \lim_{k \to \infty} \varphi^{\mathcal{M}^{1/\alpha^k}} \leq \varphi^{\mathcal{M}} \leq \varphi^{\mathcal{M} \otimes \mathcal{R}}. \end{align*}$$

Hence, $\operatorname {\mathrm {Th}}_{\exists }(\mathcal {M}) = \operatorname {\mathrm {Th}}_{\exists }(\mathcal {M} \otimes \mathcal {R})$ .

(3) $\implies $ (1). Note $M_2(\mathcal {M})$ embeds into $\mathcal {M} \otimes \mathcal {R}$ , which embeds into $\mathcal {M}^{\mathcal {U}}$ .

(3) $\iff $ (4). When (3) holds, $\mathcal {M}$ and $\mathcal {M} \otimes \mathcal {R}$ are embeddable into each other’s ultrapowers, which implies that $\mathcal {M}^t$ and $(\mathcal {M} \otimes \mathcal {R})^t \cong \mathcal {M} \otimes \mathcal {R}^t \cong \mathcal {M} \otimes \mathcal {R}$ are embeddable into each other’s ultrapowers. Hence, $\operatorname {\mathrm {Th}}_{\exists }(\mathcal {M}^t) = \operatorname {\mathrm {Th}}_{\exists }(\mathcal {M} \otimes \mathcal {R}) = \operatorname {\mathrm {Th}}_{\exists }(\mathcal {M})$ for all $t \in (0,\infty )$ . Conversely, if $\operatorname {\mathrm {Th}}_{\exists }(\mathcal {M}) = \operatorname {\mathrm {Th}}_{\exists }(\mathcal {M}^t)$ for all t, then we have $\operatorname {\mathrm {Th}}_{\exists }(\mathcal {M} \otimes \mathcal {R}) = \lim _{t \to \infty } \operatorname {\mathrm {Th}}_{\exists }(\mathcal {M}^t) = \operatorname {\mathrm {Th}}_{\exists }(\mathcal {M})$ .

(4) $\implies $ (5) is immediate.

(5) $\implies $ (6). As in the proof of Proposition 6.7 or in (2) $\implies $ (3), since $\mathcal {M}^t$ and $\mathcal {M}$ embed into each other’s ultrapowers, the same holds for $\mathcal {M}^{t^k}$ for each $k \in \mathbb {Z}$ , which implies (6).

(6) $\implies $ (4). This follows immediately from the fact that for any $\inf $ -sentence $\varphi $ , we have $\lim _{t \to \infty } \varphi ^{\mathcal {M}^t} = \inf _{t \in (0,\infty )} \varphi ^{\mathcal {M}^t}$ and $\lim _{t \to 0} \varphi ^{\mathcal {M}^t} = \sup _{t \in (0,\infty )} \varphi ^{\mathcal {M}^t}$ , which we showed in the proof of Proposition 6.7.

(3) $\implies $ (7) $\implies $ (8) is immediate by definition.

(8) $\implies $ (2). By assumption $\mathcal {M}$ embeds into $\mathcal {N}^{\mathcal {U}}$ . Since $\mathcal {N}$ has property Gamma, there exists a projection $p \in \mathcal {N}^{\mathcal {U}}$ that commutes with the image of $\mathcal {M}$ (provided that ultrafilter $\mathcal {U}$ is on a sufficiently large index set). Then $\mathcal {M}$ and p generate a copy of $\alpha \mathcal {M} \oplus (1 - \alpha ) \mathcal {M}$ in $\mathcal {N}^{\mathcal {U}}$ , where $\alpha = \operatorname {\mathrm {tr}}^{\mathcal {N}^{\mathcal {U}}}(p)$ . Finally, $\mathcal {N}^{\mathcal {U}}$ embeds into $\mathcal {M}^{\mathcal {V}}$ for some ultrafilter $\mathcal {V}$ , hence (2) holds.