Fact 2.1. Let T be a normal operator on H and let V be a T-invariant subspace of H. Then $T\! \upharpoonright V$ is normal if and only if V is reducing for T.

Proof Let E be the linear closed subspace spanned by all eigenvectors of T and let $E^\bot $ be its orthogonal complement, so $H=E\oplus E^\bot $ . We wish to show that $H=E$ . Since $\sigma (T)$ is a non-empty, countable closed subset of $\mathbb {C}$ , it is Cantor–Bendixson analyzable and has an isolated point, whence $\dim (E)>0$ . Note that E is reducing for T. Assume, towards a contradiction, that $E^\perp \neq \{0\}$ . Since $\sigma (T\! \upharpoonright {E^\perp }) \subseteq \sigma (T)$ , it is at most countable. As above, since $\sigma (T\! \upharpoonright {E^\perp })$ is a non-empty countable closed subset of $\mathbb {C}$ , it has an isolated point and thus $(E^\perp ,T\! \upharpoonright {E^\perp })$ has an eigenvector, a contradiction to the definition of E.

Proof The assumption implies that $\|(T-\lambda I)^{-1}\|>\epsilon $ . On the other hand, by continuous functional calculus (see [Reference Conway11, Theorem VIII.2.6]), setting $f(z):=1/(z-\lambda )$ for $z\in \sigma (T)$ , one has $\|(T-\lambda I)^{-1}\|=\|f\|_\infty =1/d(\lambda ,\sigma (T))$ .

Fact 2.4. Suppose that E is a PVM defined on $(X,\mathcal {B})$ with values in the Hilbert space H. Then:

1. (1) For all $v,w\in H$ , the function $\mu _{v,w}$ on $\mathcal {B}$ given by $\mu _{v,w}(A):=\langle E(A)v,w\rangle $ is a countably additive measure on $(X,\mathcal {B})$ and $\|\mu _{v,w}\|\leq \|v\|\|w\|$ . Setting $\mu _v:=\mu _{v,v}$ , we have that $\mu _{v}$ is a positive measure with $\mu _{v}(X)=\|v\|^2$ .

2. (2) If $\phi :X\to \mathbb {C}$ is a bounded $\mathcal {B}$ -measurable function, then there is a unique bounded operator on H, denoted $\int _X \phi dE$ , satisfying the following property: whenever $\{A_1,\ldots ,A_n\}$ is a finite partition of X into elements of $\mathcal {B}$ for which $\sup \{|\phi (x)-\phi (x')| \ : \ x,x'\in A_i\}<\epsilon $ for all $i=1,\ldots ,n$ , then for any choice of $a_i\in A_i$ , $i=1,\ldots ,n$ , we have $\|\int _X \phi dE-\sum _{i=1}^n \phi (a_i)E(A_i)\|<\epsilon $ .

3. (3) For $\mu _{v,w}$ as in (1) and $\int _X \phi dE$ as in (2), we have that

   $$ \begin{align*}\left\langle \left(\int_X \phi dE\right) v,w\right\rangle=\int_X \phi d\mu_{v,w}.\end{align*} $$

Fact 2.5 (Spectral theorem for normal operators).

If T is a normal operator on H, then there is a PVM E defined on the Borel subsets of $\sigma (T)$ with values in H such $\int _{\sigma (T)}zdE=T$

Proof We claim that $\delta :=\epsilon \cdot (\eta /2)$ satisfies the conclusion of the lemma. Write $x=y+z$ , with $y\in E(U)(H)$ and $z\in E(U^c)(H)$ . Set $S:=T\upharpoonright E(U)(H)^\perp $ , so $\sigma (S)\subseteq \{\lambda '\in K \ : \ |\lambda -\lambda '|\geq \eta \}$ . Take a partition $I_1,\ldots ,I_t$ of $\sigma (S)$ such that each $I_i$ has diameter at most $\eta /2$ and choose $\lambda _i\in I_i$ for all $i=1,\ldots ,t$ . Then $\|Tz-\sum _i \lambda _i E(I_i)(z)\|<(\eta /2)\|z\|$ . On the other hand,

$$ \begin{align*}\|\lambda z-\sum_i \lambda_i E(I_i)(z)\|^2=\|\sum_i \lambda E(I_i)(z)-\sum_i \lambda_i E(I_i)(z)\|^2\geq \eta^2\|z\|^2.\end{align*} $$

Finally, note that, since $E(U)(H)$ and $E(U^c)(H)$ are T-invariant, we have

$$ \begin{align*}\|Tx-\lambda x\|^2=\|Ty-\lambda y\|^2+\|Tz-\lambda z\|^2,\end{align*} $$

and thus $\|Tz-\lambda z\|^2\leq \|Tx-\lambda x\|^2<\delta ^2$ . Putting this all together, we arrive at

$$ \begin{align*}\eta \|z\| &\leq \|\lambda z-\sum_i \lambda_iE(I_i)(z)\| \\ &\leq \|\lambda z-Tz\|+\|Tz-\sum_i \lambda_i E(I_i)(z)\|<\delta+(\eta/2)\|z\|.\end{align*} $$

It follows that $\|z\|<\delta /(\eta /2)=\epsilon $ .

Fact 2.8. Suppose that $T_i:H_i\to H_i$ is a normal operator for $i=1,2$ with associated spectral measures $E_1$ and $E_2$ , viewed as defined on all Borel subsets of $\mathbb {C}$ . Then the following are equivalent:

1. (1) $T_1$ and $T_2$ are approximately unitarily equivalent, meaning that there is a sequence of unitary operators $U_n:H_1\to H_2$ such that

   $$ \begin{align*}\lim_{n\to \infty}\|U_nT_1U_n^*-T_2\|=0.\end{align*} $$

2. (2) For each open set $\mathcal O\subseteq \mathbb C$ , the spectral projections $E_1(\mathcal O)$ and $E_2(\mathcal O)$ have the same rank.

If, in addition, the spaces $H_1$ and $H_2$ are separable, (1) and (2) are equivalent to:

1. (3) $T_1$ and $T_2$ are spectrally equivalent.

Proof For each $n\geq 1$ , consider the L-sentence $\sigma _n$ given by

$$ \begin{align*}\sup_x\sup_y \min(1/n \dot{-} \sup_z \left| \langle Tz,x\rangle - \langle z,y\rangle \right|, | {\|Tx\|-\|y\|} | \dot{-} 1/n).\end{align*} $$

We claim that $(H,T)\models \sigma _n$ for all $n\geq 1$ if and only if T is normal. Indeed, first suppose that $(H,T)\models \sigma _n$ for all $n\geq 1$ and let $x\in H$ be such that $\|x\|\leq 1$ . Since $\sup _z|\langle Tz,x\rangle -\langle z,T^*x\rangle |=0$ , letting $y=T^*x$ in the above axiom, we have that $|\|Tx\|-\|T^*x\||\leq 1/n$ for all $n\geq 1$ , whence $\|Tx\|=\|T^*x\|$ ; since x was arbitrary, we see that T is normal. Conversely, suppose that T is normal and let $x,y\in H$ be such that $\|x\|,\|y\|\leq 1$ and $\sup _z |\langle Tz,x\rangle -\langle z,y\rangle |<1/n$ . It follows that $\sup _z|\langle z,T^*x-y\rangle |<1/n$ , whence $\|T^*x-y\|<1/n$ and thus $|\|T^*x\|-\|y\||<1/n$ . Since T is normal, $\|T^*x\|=\|Tx\|$ , whence $|\|Tx\|-\|y\||<1/n$ , as desired.

Proof Let $\varphi (x,y)$ be the formula $\sup _z |\langle Tz,x\rangle -\langle z,y\rangle |$ . The zeroset of $\varphi $ is the graph of $T^*$ . We must show that $\varphi $ is an almost-near formula. Towards that end, suppose that $\sup _z |\langle Tz,x\rangle -\langle z,y\rangle |<\epsilon $ . Then, as in the proof of the previous proposition, we have $\|T^*x-y\|<\epsilon $ and thus $d((x,y),(x,T^*x))<\epsilon $ .

Proof Let $(H',T')\succeq (H,T)$ be sufficiently saturated. By Proposition 3.5, we have $\dim (H^{\prime }_\lambda )=n$ . Since the unit ball in a finite dimensional Hilbert space is compact, the orbit of any $x\in H_\lambda $ is a compact subset in $(H',T')$ , and thus $x\in \mathrm {acl}(\emptyset )$ . For the second statement, if $\{\lambda _i:i\in I\}$ are different isolated points in $\sigma (T)$ with $\dim (H_{\lambda _i})\in \mathbb N$ for each $i\in I$ , then $P_{H_{\lambda _i}}(x)\in \mathrm {acl}(\emptyset )$ for all $x\in H$ and $i\in I$ . If $x=\sum _{i\in I}P_{H_{\lambda _i}}(x)$ , then $x\in \mathrm {dcl}(P_{H_{\lambda _i}}(x) :i\in I)\subseteq \mathrm {acl}(\emptyset )$ .

Proof Recall that every element of the spectrum of a normal operator is an approximate eigenvalue. Given $\lambda \in \mathbb C$ , to say that $\lambda $ is an approximate eigenvalue of the normal operator T can be captured by an existential condition, namely by $\inf _x \max (1\dot{-} \|x\|,\|Tx-\lambda x\|)=0$ . It follows that $\sigma (T_1)=\sigma (T_2)$ .

Now suppose that $\lambda \in \mathrm {Isol}(\sigma (T_1))$ . Suppose also that $\ker (T_1-\lambda )$ has dimension at least n. Then,

$$ \begin{align*}(H_1,T_1)\models \inf_{x_1}\ldots\inf_{x_n} \max\left(\max_{1\leq i\leq n}\|Tx_i-\lambda x_i\|,\max_{1\leq i\leq j\leq n}|\langle x_i,x_j\rangle-\delta_{ij}|\right)=0.\end{align*} $$

By assumption, the same condition is satisfied in $(H_2,T_2)$ , meaning that there are n vectors in $H_2$ which are approximate eigenvectors corresponding to $\lambda $ that are almost orthonormal. By Proposition 3.5, there are n eigenvectors for $T_2$ with eigenvalue $\lambda $ that are almost orthonormal; this is enough to conclude that the dimension of $\ker (T_2-\lambda )$ is at least n. Switching the roles of $T_1$ and $T_2$ yields the desired equality.

Proof The only direction needing some justification is the implication (3) implies (1); but this follows from the previous fact by passing to separable elementary substructures and noting that approximately unitarily equivalent operators have isomorphic ultrapowers, and are thus elementarily equivalent.

Proof For (1), we add existential axioms stating that each $\lambda \in K$ has approximate eigenvectors, which is enough to conclude that K is contained in the spectrum of all models. To ensure that all models have spectrum exactly K, fix $\lambda \notin K$ and set $\delta :=\mathrm {dist}(\lambda ,K)$ . We then add the universal axioms $\sup _x (\delta \dot{-} \|Tx-\lambda x\|)=0$ ; by Lemma 2.3, these axioms do indeed ensure that $\sigma (T)=K$ .

For (2), first suppose that $\lambda \in \mathrm {Isol}(K))$ and $m(\lambda )=n\in \mathbb N$ . Then by Proposition 3.5 the eigenspace $H_{\lambda }$ is definable, and can thus be quantified over. We then add the following axioms to ensure that all $m(\lambda )=n$ in all models:

$$ \begin{align*}\inf_{x_1\in H_{\lambda}}\ldots \inf_{x_n\in H_\lambda }\max\left(\max_{1\leq i \leq j\leq n}|\langle x_i,x_j\rangle-\delta_{ij}|,\sup_{y\in H_{\lambda}} \|y-\sum_{k=1}^n \langle y,x_k\rangle x_k\|\right)=0.\end{align*} $$

If $\lambda \in \mathrm {Isol}(K)$ and $m(\lambda )=\infty $ , then add the following scheme of axioms indexed by $n\geq 1$ to ensure that $m(\lambda )=\infty $ in all models:

$$ \begin{align*}\inf_{x_1\in H_{\lambda}}\ldots \inf_{x_n\in H_\lambda }\left(\max_{1\leq i \leq j\leq n}|\langle x_i,x_j\rangle-\delta_{ij}|\right)=0.\end{align*} $$

Perhaps it is not entirely obvious that these axioms are $\forall \exists $ -axioms, but it is clear that a union of a chain of models of $\Sigma _{K,m}$ is once again a model of $\Sigma _{K,m}$ , whence the theory is indeed $\forall \exists $ -axiomatizable.

To see that $\Sigma _{K,m}$ is satisfiable, extend m to a function defined on all of K by setting $m(\lambda )=\infty $ for all $\lambda \in K\setminus \mathrm {Isol}(K)$ . For each $\lambda \in K$ , let $H_\lambda $ be a Hilbert space of dimension $m(\lambda )$ and define $T_\lambda (x)=\lambda x$ for all $x\in H_\lambda $ . Then $(H,T):=\bigoplus (H_\lambda ,T_\lambda )$ is a model of $\Sigma _{K,m}$ .

Proof Without loss of generality, we may assume that $\mathcal U$ is a non-principal ultrafilter. For each $i\in I$ , take models $(H_i,T_i)$ of $\Sigma _{K_i,m_i}$ . It suffices to show that $(H,T):=\prod _{\mathcal {U}} (H_i,T_i)$ is a model of $\Sigma _{K,m}$ , where K and m are defined by the formulas (1) and (2) above. We first show that $K=\lim _{\mathcal {U}} K_i$ . Fix $\epsilon>0$ ; we need to show that $d(\sigma (T),K_i)< \epsilon $ for $\mathcal U$ -almost all $i\in I$ . Fix that $\lambda \in \sigma (T)$ . Since the structure $(H,T)$ is $\aleph _1$ -saturated, there is $v\in H$ with $\|v\|=1$ and $Tv=\lambda v$ . Writing $v=(v_i)_{\mathcal {U}}$ , we see that $\|T_iv_i-\lambda v_i\|<\epsilon /2$ for $\mathcal U$ -almost all $i\in I$ . We may assume that $\|v_i\|=1$ for all i and thus for those indices satisfying $\|T_iv_i-\lambda v_i\|<\epsilon /2$ , by Lemma 2.3, it follows that $d(\lambda ,K_i)<\epsilon /2$ . Fixing a finite $\epsilon /2$ -net $\lambda _1,\ldots ,\lambda _t\in \sigma (T)$ , we have, for $\mathcal U$ -almost all $i\in I$ and all $j=1,\ldots ,n$ , that $d(\lambda _j,K_i)<\epsilon /2$ and thus $\sup \{d(\lambda ,K_i) \ : \ \lambda \in \sigma (T)\}<\epsilon $ for $\mathcal U$ -almost all $i\in I$ . Now suppose, towards a contradiction, that $A_{\epsilon }:=\{i\in I : d(\lambda _i,\sigma (T))\geq \epsilon \text { for some }\lambda _i\in K_i\}\in \mathcal {U}$ . Fix positive real numbers $(r_i:i\in I)$ such that $\lim _{\mathcal {U}}r_i=0$ . For $i\in A_{\epsilon }$ , take $v_i\in H_i$ with $\|v_i\|=1$ such that $\|T_iv_i-\lambda _i v_i\|<r_i$ ; for $i\notin A_\epsilon $ choose any $v_i\in H_i$ with $\|v_i\|=1$ . Setting $\lambda :=\lim _{\mathcal {U}} \lambda _i$ , we have that $Tv=\lambda v$ , where $v=(v_i)_{\mathcal {U}}$ . Then $\lambda \in \sigma (T)$ , contradicting the fact that $d(\lambda ,\sigma (T))\geq \epsilon $ . It follows that $d(\sigma (T),K_i)< \epsilon $ for $\mathcal U$ -many $i\in I$ , as desired.

For (2), take $\lambda \in \mathrm {Isol}(K)$ and $\eta>0$ such that $d(\lambda ,K\setminus \{\lambda \})>\eta $ . First assume that $m(\lambda )\geq N$ and take orthonormal eigenvectors $x_1,\ldots ,x_N\in H$ with eigenvalue $\lambda $ . Write each $x_j=(x_j^i)_{\mathcal {U}}$ , where, without loss of generality, each set $\{x^i_1,\ldots ,x^i_n\}$ forms an orthonormal set in $H_i$ . Take $k\in \mathbb N$ such that $1/k<\eta /3$ . For $i\in I$ , consider the open set

$$ \begin{align*}U_i=\{B(\rho;1/k)\ : \ \rho \in K_i\setminus B(\lambda;1/k)\}.\end{align*} $$

Setting $A_k:=\{i\in I: \ B(\lambda ;1/k)\cap U_i=\emptyset \}$ , we claim that, $A_k\in \mathcal U$ . Indeed, for $\mathcal U$ -almost all $i\in I$ , we have that $d(K,K_i)<\frac {1}{k}$ . Suppose $B(\lambda ;1/k)\cap B(\rho ;1/k)\not =\emptyset $ , where $d(\rho ,\lambda )\geq 1/k$ . Take $\rho '\in K$ such that $d(\rho ,\rho ')<1/k$ ; it follows that $\rho '\not =\lambda $ and $d(\lambda ,\rho ')<3/k<\eta $ , contradicting the choice of $\eta $ . For any $i\in A_k$ and any $y^i\in H_i$ , we have

$$ \begin{align*}\|y^i\|^2=\|E(B(\lambda;1/k))(y^i)\|^2+\|E(U_i)(y^i)\|^2.\end{align*} $$

Take $\epsilon>0$ sufficiently small such that whenever $\{x_1,\dots ,x_N\}$ and $\{y_1,\dots ,y_N\}$ are vectors in a Hilbert space satisfy $\langle x_i,x_j \rangle =\delta _{ij}$ and $\|x_i-y_i\|<\epsilon $ , then $\dim (\{y_1,\dots ,y_N\})=N$ . (Note that $\epsilon $ only depends on N.) Fix $j\in \{1,\ldots N\}$ ; since $\lim _{\mathcal {U}} T_ix^i_j=\lambda x^i_j$ , Lemma 2.6 implies that $\lim _{\mathcal {U}}\|E(B(\lambda ;1/k))(x^i_j)\|^2=1$ . Setting $B_{\epsilon }:=\{i\in I: \|E_i(B(\lambda ;1/k)(x^i_j)\|>1-\epsilon \text { for all }j=1,\ldots ,N\}$ , we have $B_\epsilon \in \mathcal U$ . For $j=1,\ldots ,N$ , set $y^i_j=E_i(B(\lambda ;1/k)(x^i_j)$ . For $i\in B_\epsilon $ and $j=1,\ldots ,N$ , we have $d(x^i_j,y^i_j)<\epsilon $ . By the way we chose $\epsilon $ , it follows that $\dim (\{y^i_1,\dots ,y^i_N\})=N$ . As a result, we have $\lim _{\mathcal {U}}(m_i(B(\lambda ;1/k)))\geq N$ ; since the only condition on k was that $1/k<\eta /3$ , we also conclude $\inf _k \lim _{\mathcal {U}}(m_i(B(\lambda ;1/k)))\geq N$ .

Assume now that $\inf _{k}\lim _{\mathcal {U}}(m_i(B(\lambda ;1/k)))\geq N$ . Let $\epsilon $ be as in the previous paragraph. Let $\delta =\delta (\eta ,\epsilon )>0$ be as in Lemma 2.6. Let $k>0$ be such that $1/k<\delta $ and $\lim _{\mathcal {U}}(m_i(B(\lambda ;1/k)))\geq N$ . For $\mathcal U$ -almost all $i\in I$ , take orthonormal $x_1^i,\ldots ,x_N^i\in E_i(B(\lambda ;1/k))(H_i)$ . For each of these indices we have $\|T_ix^i_j-\lambda x^i_j\|\leq 1/k<\delta $ ; for the other indices, choose any $x_1^i,\ldots ,x_N^i\in H_i$ . Set $x_j=(x^i_j)_{\mathcal U}$ . Then $\{x_1,\ldots ,x_N\}$ form an orthonormal set satisfying $\|Tx_j-\lambda x_j\|\leq 1/k<\delta $ for each $j=1,\ldots ,N$ . By Lemma 2.6, there are $y_1,\dots ,y_N\in H_{\lambda }$ such that $\|x_j-y_j\|<\epsilon $ . By the way we chose $\epsilon $ , we have $\dim (\{y_1,\dots ,y_N\})=N$ , and thus $m(\lambda )\geq N$ .

Proof Let $(\lambda _n)_{n\in \alpha }$ enumerate $\mathrm {Isol}(K)$ , where $\alpha \in \omega \cup \{\omega \}$ , $\alpha $ being chosen depending on whether the family of isolated points is finite or countably infinite. For each $k\geq 1$ , let $(\rho _i^k)_{i=1,\ldots ,\ell (k)}$ enumerate a finite $1/k$ -net in $K'$ . For each $k\geq 1$ and $n\leq \min (k,\alpha )$ , let $H_{n,k}$ have dimension $\min \{k,m(\lambda _n)\}$ and let $T_{n,k}(x)=\lambda _nx$ for all $x\in H_{n,k}$ . For each $i=1,\ldots ,\ell (k)$ , let $H^{\prime }_{i,k}$ be an one-dimensional Hilbert space with $T^{\prime }_{i,k}(x)=\rho _i^kx$ for all $x\in H^{\prime }_{i,k}$ . Let $H_k=(\bigoplus _{m=1}^{\min k,\alpha } H_{m,k})\oplus (\bigoplus _{i=1}^{\ell (k)}H^{\prime }_{i,k})$ and $T_k=(\bigoplus _{m=1}^{\min (k,\alpha )} T_{m,k})\oplus (\bigoplus _{i=1}^{m(k)}T^{\prime }_{i,k})$ . Let $\mathcal U$ be a nonprincipal ultrafilter on $\mathbb {N}$ . It follows that $(H,T):=\prod _{\mathcal U}(H_k,T_k)$ is a model of $\Sigma _{K,m}$ . Since each of the spaces $H_{m,k}$ and $H^{\prime }_{i,k}$ is finite-dimensional, it follows that $\Sigma _{K,m}$ is pseudocompact.

Proof It suffices to note that, if $(p_n)$ is a sequence of $*$ -polynomials that converge uniformly to f, then $d(f(T)(x),y)=\lim _n \|p_n(T)(x)-y\|$ .

Proof Since $K_1$ and $K_2$ are uniformly separated by $\epsilon $ , there is a continuous function $f:\mathbb D\to \mathbb D$ satisfying $f(\lambda )=1$ for all $\lambda \in K_1$ and $f(\lambda )=0$ for all $\lambda \in K_2$ . By Lemma 3.14, $f(T)$ is $\Sigma _{\mathrm {normal}}$ -definable. But in models of $\Sigma _K$ , $f(T)=E(K_1)$ . An analogous argument shows that $E(K_2)$ is $\Sigma _K$ -definable.

Proof Since $T_1$ and $T_2$ are spectrally equivalent, we may take a sequence $(U_n)$ of unitary maps such that $\|T_2-U_nT_1U_n^*\|\to 0$ . Then,

$$ \begin{align*}\|f(T_2)-U_nf(T_1)U_n^*\| &=\|f(T_2)-f(U_nT_1U_n^*)\| \\ &\leq \|f\|_\infty \|T_2-U_nT_1U_n^*\| \to 0,\end{align*} $$

showing that $f(T_1)$ and $f(T_2)$ are approximately unitarily equivalent.

Proof Let $((H_i,T_i):i\in I)$ be a collection of models of $\Sigma _{\mathrm {normal}}$ , $\mathcal U$ an ultrafilter on I, and $(H,T):=\prod _{\mathcal U}(H_i,T_i)$ ; it suffices to show that $\prod _{\mathcal U}(H_i,f(T_i))=(H,f(T))$ . However, this follows immediately from Lemma 3.14.

Proof We only treat the case of the theory $\Sigma _{K,m}$ , the other cases being simpler. We first establish the joint embedding property. Given models $(H_1,T_1),(H_2,T_2)$ of $ \Sigma _{K,m}$ , we can write

$$ \begin{align*}(H_1,T_1)=(H_{\mathrm{fin}},T_{\mathrm{fin}}) \oplus [(H_1,T_1)\ominus (H_{\mathrm{fin}},T_{\mathrm{fin}})]\end{align*} $$

and

$$ \begin{align*}(H_2,T_2)=(H_{\mathrm{fin}},T_{\mathrm{fin}}) \oplus [(H_2,T_2)\ominus (H_{\mathrm{fin}},T_{\mathrm{fin}})].\end{align*} $$

Set $(H_3,T_3)=(H_{\mathrm {fin}},T_{\mathrm {fin}})\oplus [(H_1,T_1)\ominus (H_{\mathrm {fin}},T_{\mathrm {fin}})] \oplus [((H_2,T_2)\ominus (H_{\mathrm {fin}}, T_{\mathrm {fin}})]$ . Clearly $\sigma (T_3)=K$ . Since the eigenspaces associated with isolated eigenvalues of finite multiplicity are contained in $H_{\mathrm {fin}}$ , we have $(H_3,T_3)\models \Sigma _{K,m}$ . It is clear that $(H_i,T_i)$ embeds into $(H_3,T_3)$ for $i=1,2$ .

Given three models $(H_0,T_0),(H_1,T_1),(H_2,T_2)$ of $\Sigma _{K,m}$ with $(H_0,T_0)\subseteq (H_i,T_i)$ for $i=1,2$ , we can amalgamate the structures $(H_1,T_1)$ and $(H_2,T_2)$ over $(H_0,T_0)$ by repeating the construction above over $(H_0,T_0)$ . In this case, we can define $(H_3,T_3)=(H_0,T_0)\oplus [(H_1,T_1)\ominus (H_0,T_0)] \oplus [((H_2,T_2)\ominus (H_0,T_0)]$ . As in the case of the joint embedding property, we have $(H_3,T_3)\models \Sigma _{K,m}$ and is an amalgam of $(H_1,T_2)$ and $(H_2,T_2)$ over $(H_0,T_0)$ .

Proof Fix models $(H_1,T_1,T_1^*),(H_2,T_2,T_2^*)\models \Sigma _{K}^*$ with $(H_2,T_2,T_2^*)$ $\kappa ^+$ -saturated, where $\kappa $ is the density character of $H_1$ . Further suppose that $(H_0,T_0,T_0^*)$ is a substructure of $(H_1,T_1,T_1^*)$ and $f:(H_0,T_0,T_0^*)\to (H_2,T_2,T_2^*)$ is an embedding. It suffices to show that f can be extended to a substructure of $(H_1,T_1,T_1^*)$ whose domain properly contains $H_0$ . For this purpose, we may assume, without loss of generality, that $H_{T_1,\mathrm {fin}}\subseteq H_0$ . It suffices then to show that, given $x\in H_1\ominus H_0$ , we may find an extension of f to an embedding from a substructure of $(H_1,T_1,T_1^*)$ containing x in its domain to $(H_2,T_2,T_2^*)$ . Since $H_0$ is reducing for $T_0^*$ , we have that $p(T)(x)\in H_1\ominus H_0$ for all $*$ -polynomials p. Consequently, it suffices to find $y\in H_2\ominus f(H_0)$ such that $\langle p(T_1)(x),x\rangle =\langle p(T_2)(y),y\rangle $ for all $*$ -polynomials p. Fix finitely such polynomials $p_1,\ldots ,p_m$ and $\epsilon>0$ ; setting $\alpha _i:=\langle p_i(T_1)(x),x\rangle $ , by saturation, it suffices to find $y\in H_2\ominus f(H_0)$ such that $|\langle p_i(T_2)(y),y\rangle -\alpha _i|<\epsilon $ for all $i=1,\ldots ,m$ .

Set $\mu $ to be the scalar measure on the spectrum corresponding to x, that is, $\mu (I)=\langle E_1(I)x,x\rangle $ for every Borel set $I\subseteq K$ ; here $E_1$ is the spectral measure associated with $T_1$ . Choose a sufficiently fine grid partition of the plane, by which we mean a partition of the plane into small squares, where, say, the left and bottom edge of each square is included in the set, while the top and right edge are not included. Let $I_1,\ldots ,I_t$ be the elements of the partition that intersect K. For each $k=1,\ldots ,t$ , we define an element $y_k\in E_2(I_k)$ (where $E_2$ is the spectral projection associated with $T_2$ ) as follows. If $E_2(I_k\cap K)$ has finite rank, then set $y_k:=0$ . If $E_2(I_k\cap K)$ has infinite rank, then, by saturation, its dimension is greater than the dimension of $f(H_0)$ , whence we may find a unit vector $y_k\in E_2(I_k)\cap (H_2\ominus f(H_0))$ . Set $y:=\sum _{k=1}^t \mu (I_k)y_k$ , which is an element of $H_2\ominus f(H_0)$ . If $E_2(I_k)$ has finite rank, then $E_1(I_k)\subseteq H_{T_1,\mathrm {fin}}$ ; since $x\in H_1\ominus H_{T_1,\mathrm {fin}}$ , we have $\langle E_2(I_k)y,y\rangle =0=\langle E_1(I_k)x,x\rangle $ . If $E_2(I_k)$ has infinite rank, we have

$$ \begin{align*}\langle E_2(I_k)y,y\rangle=\langle \mu(I_k)y_k,y_k\rangle=\mu(I_k)=\langle E_1(I_k)x,x\rangle.\end{align*} $$

For each $k=1,\ldots ,t$ , let $\lambda _k$ be an element of $I_k\cap K$ . We then have

$$ \begin{align*}\langle \sum_{k=1}^t p_i(\lambda_k)E_2(I_k)y,y\rangle=\langle \sum_{k=1}^t p_i(\lambda_k)E_1(I_k)x,x\rangle.\end{align*} $$

On the other hand, if the grid is sufficiently fine so that, for all $i=1,\ldots ,m$ , $|p_i(a)-p_i(b)|<\epsilon $ whenever a and b belong to the same $I_k$ , then by Fact 2.4, we have $\|p_i(T_j)-\sum _{k=1}^t p_i(\lambda _k)E_j(I_k)\|<\epsilon /2$ for $j=1,2$ . Consequently, this y is as desired.

Proof As already mentioned above, an inclusion $(H_1,T_1)\subseteq (H_2,T_2)$ between models of $\Sigma _K$ must be such that $H_1$ is $T_2$ -reducing and $T_1^*=T_2^*\upharpoonright \! H_1$ . Consequently, we have that $(H_1,T_1,T_1^*)\subseteq (H_2,T_2,T_2^*)$ as $L^*$ -structures. Since both of these structures are models of $\Sigma _K^*$ , Theorem 4.1 implies that this inclusion is elementary as $L^*$ -structures, whence the original inclusion $(H_1,T_1)\subseteq (H_2,T_2)$ is elementary.

Question 4.3. Do the theories $\Sigma _K$ have quantifier-elimination?

Proof We use the notation from the proof of Theorem 4.1. The only place in the proof of Theorem 4.1 where the symbol for the adjoint was used was the conclusion that the subspace $H_0$ was reducing for $T_1$ . If $T_1$ is self-adjoint, then $H_0$ is obviously reducing for $T_1$ . If $T_1$ is unitary, then it suffices to show that one may extend the embedding $f:(H_0,T_0)\to (H_2,T_2)$ to the smallest closed subspace of $H_1$ containing $H_0$ and closed under $T_1$ and $T_1^*$ , for then one may assume that $H_0$ is actually reducing for $T_1$ and the remainder of the proof of Theorem 4.1 remains the same. To verify this claim, first note that every element in the smallest subspace H of $H_1$ containing $H_0$ and closed under $T_1$ and $T_1^*$ is of the form $x:=\sum _{i=0}^n T_1^{-i}b_i$ , with $b_0,\ldots ,b_n\in H_0$ . We claim that we may extend f to a function $f':H\to H_2$ by defining $f'(x):=\sum _{i=0}^n T_2^{-i}f(b_i)$ for x as in the previous sentence. To see that this is well-defined, suppose that $x=0$ . Then $\sum _{i=0}^n T_1^{n-i}b_i=0$ , whence $\sum _{i=0}^n T_2^{n-i}f(b_i)=0$ since f intertwines $T_1$ and $T_2$ . Applying $T_2^{-n}$ to both sides yields $\sum _{i=0}^n T_2^{-i}f(b_i)=0$ , showing that the extended function $f'$ is well-defined. Reversing these steps and using that f is injective shows that $f'$ is also injective. Finally, note that the extended function $f'$ is linear and preserves the inner product, whence $f'$ extends to an embedding of $\bar H$ into $H_2$ ; moreover, it is clear that this extension of $f'$ to the closed space $\bar H$ still intertwines $T_1$ and $T_2$ , establishing the claim and completing the proof.

Proof It suffices to note that every model of $\Sigma _{\mathrm {normal}}$ embeds into a model of $\Sigma _{\mathbb D}$ , which follows from the fact that, for any two models $(H_1,T_1),(H_2,T_2)$ of $\Sigma _{\mathrm {normal}}$ , one has $\sigma (T_1\oplus T_2)=\sigma (T_1)\cup \sigma (T_2)$ .

Proof Once again, it suffices to see that any model of $\Sigma _K$ embeds in a model of $\Sigma _{K,m_\infty }$ . This follows by simply adding a model of $\Sigma _{K,m_\infty }$ as a direct summand.

Proof Since $\Sigma ^*$ is a definitional expansion of $\Sigma $ , we may as well assume that we work in the language $L^*$ where the theory of $(H,T,T^*)$ has quantifier elimination. If $\vec a$ and $\vec a'$ have the same type in $(H,T)$ over B, then they also have the same type over $\mathrm {dcl}(B)$ ; since $\langle B\rangle _0\subseteq \mathrm {dcl}(B)$ , they have the same type over $\langle B\rangle _0$ , whence (1) follows. In an elementary extension of $(H,T)$ , $\vec a$ and $\vec a'$ are conjugate by an automorphism fixing B and thus $\langle B\rangle _0$ ; this automorphism must then map $\vec c$ to $\vec c'$ by (1), whence (2) follows.

Conversely, assume that (1) and (2) holds. By quantifier elimination, it suffices to check that $\langle p(T,T^*)a_i,a_j\rangle =\langle p(T,T^*)a_i',a_j'\rangle $ for all $*$ -polynomials p and all $i,j=1,\ldots ,n$ . The desired equality follows from (1) and (2) by noting that

$$ \begin{align*}\langle p(T,T^*)a_i,a_j\rangle=\langle p(T,T^*)P_B(a_i),P_B(a_j)\rangle+\langle p(T,T^*)c_i,c_j\rangle\end{align*} $$

and the analogous equation obtained by replacing $a_i$ and $a_j$ with $a_i'$ and $a_j'$ .

Proof First assume that $x\in \langle B \rangle $ . If $x\in H_{\mathrm {fin},T}$ , then $x\in \mathrm {acl}(\emptyset )\subseteq \mathrm {acl}(B)$ . If $x=p(T,T^*)b$ , then $x\in \mathrm {dcl}(B)$ since the graph of $T^*$ is definable. Since $\mathrm {acl}(B)$ is always a closed subspace, we see that $\langle B\rangle \subseteq \mathrm {acl}(B)$ .

Assume now that $x\not \in \langle B \rangle $ . Write $x=x_B+x_1$ , where $x_B=P_{\langle B\rangle }(x)$ ; note that $x_1\neq 0$ . It suffices to show that, in an elementary extension of $(H,T)$ , there are infinitely many realizations of $\mathrm {tp}(x_1/B)$ separated by a uniform distance $\delta>0$ . Write $H=\langle B\rangle \oplus H_1$ , where $H_1=\langle B\rangle ^\perp $ . Since $\langle B\rangle $ is reducing for T, so is $H_1$ . Let $T_1=T\upharpoonright _{H_1}$ . Set $(H_\omega ,T_\omega ):=\bigoplus _{i<\omega }(H_1,T_1)$ and set $(H',T')=(\langle B\rangle \oplus H_\omega ,T\upharpoonright \!\langle B\rangle \oplus T_\omega )$ . Since $\sigma (T_1)\subseteq \sigma (T)$ , we have $\sigma (T')\subseteq \sigma (T)$ ; since we can identify $(H,T)$ with a substructure of $(H',T')$ , we also have $\sigma (T)\subseteq \sigma (T')$ and thus $\sigma (T)=\sigma (T')$ . Furthermore, the eigenspaces associated with the isolated points in $\sigma (T)$ belong to $\langle B\rangle $ , whence they have the same dimension in both spaces. It follows that $(H,T)\equiv (H',T')$ . Moreover, we get using quantifier elimination that $(H,T)\preceq (H',T')$ . Let $x_i$ be the copy of $x_1$ in the i-th component of the sum $\bigoplus _{i<\omega }(H_1,T_1)$ . Then $\mathrm {tp}(x_B+x_i/B)=\mathrm {tp}(x_B+x_1/B)$ and $d(x_b+x_i,x_B+x_j)=\sqrt {2}\|x_1\|$ whenever $i\neq j$ . Thus the family $\{x_B+x_i:i<\omega \}$ is uniformly separated and the proof is complete.

Proof It is enough to prove that $\text {dcl}(B)\subseteq \langle B\rangle _0$ . Suppose $v\not \in \langle B\rangle _0$ ; we show $v\notin \mathrm {dcl}(B)$ . Write $v=P_{B}(v)+w$ (recall that $P_{B}(v)$ stands for the projection onto the closed space $\langle B\rangle _0$ ) and note that $w\not =0$ . It suffices to show that $\mathrm {tp}(P_B(c)+w/B)=\mathrm {tp}(P_B(c)-w/B)$ , as from this it follows that $v\not \in \text {dcl}(B)$ . However, this follows immediately from Proposition 5.1 by noting that $\langle p(T,T^*)w,w\rangle =\langle p(T,T^*)(-w),-w\rangle $ for all polynomials $p(X,Y)\in \mathbb C[X,Y]$ .

Proof Fix a net $(p_i)$ from $S_n(\Sigma )$ ; we must show that $p_i\to p$ in the logic topology if and only if $\Phi (p_i)\to \Phi (p)$ in $M(K)^{n^2}$ . To see this, first suppose that $p_i\to p$ in the logic topology and fix realizations $\vec c^i$ and $\vec c$ of $p_i$ and p respectively; we must show that $\int fd\mu _{c_j^i,c_k^i}\to \int fd\mu _{c_j,c_j}$ for all $j,k=1,\ldots ,n$ and all continuous functions $f:K\to \mathbb {C}$ . However, since $*$ -polynomials are dense in the space of all continuous functions on K and $\int p(z,\bar z)d\mu _{c_j^i,c_k^i}=\langle p(T,T^*)c_j^i,c_k^i\rangle $ and $\int p(z,\bar z)d\mu _{c_j,c_k}=\langle p(T,T^*)c_j,c_k\rangle $ , the result follows by our assumption that $p_i \to p$ in the logic topology. The backwards direction follows from quantifier-elimination by reversing the reasoning in the first part of the proof.

Fact 5.6. Suppose that $\mathcal {E}$ is a collection of Borel subsets of $\mathbb {C}$ with the finite intersection property and such that any open subset of $\mathbb {C}$ can be written as a countable union of elements from $\mathcal {E}$ . Then for any $\mu _n,\mu \in M(\mathbb {C})$ , we have that $\mu _n\to \mu $ in the weak*-topology if and only if $\mu _n(B)\to \mu (B)$ for each $B\in \mathcal {E}$ .

Proof Fix $\mu \in M(K)_+$ and $n\in \mathbb {N}$ . Fix also $(H,T)\models \Sigma $ . Fix $n\geq 1$ and let $B_1,\ldots ,B_{t(n)}$ denote those $B_{k,n}$ ’s that intersect K. For each $i=1,\ldots ,t(n)$ , let $x^n_i\in E(I_i)(H)$ be such that $\|x_i^n\|=\sqrt {\mu (I_i\cap K)}$ . Setting $x_n:=\sum _{i=1}^{t(n)} x_i^n$ and $p_n:=\mathrm {tp}(x_n)$ , we have that $\mu _{p_m}(I_i)=\mu _p(I)$ for all $i=1,\ldots ,t(n)$ and $m\geq n$ . Fact 5.6 implies that $\mu _{p_n}\to \mu $ in the weak*-topology. Supposing, without loss of generality, that $\mu $ is a probability measure, we have that $\|x_n\|=1$ for all n. Using Proposition 5.5 and the fact that the set of types of unit vectors in $S_1(\Sigma )$ is compact, we have that the image under $\Phi $ of such types is weak*-closed, whence $\mu $ is in the image of $\Sigma $ , as desired.

Question 5.8. For $n>1$ , can one characterize the image of $\Phi _n$ ?

Proof Let r denote the right hand side of the above display. We first show that $d(p,q)^2\geq r$ . Fix arbitrary realizations $a\models p$ and $b\models q$ in some common model $(H,T)$ of $\Sigma _{K,m}$ such that $d(p,q)=\|a-b\|$ . For any measurable partition $(A_n)$ of K, we have

$$ \begin{align*}d(p,q)^2 & = \|a-b\|^2 \\ &=\sum_n \|E(A_n)a-E(A_n)b\|^2\geq \sum_n \left| \|E(A_n)a\|-\|E(A_n)b\|\right|{}^2.\end{align*} $$

By taking the supremum over all partitions, we see that $d(p,q)^2\geq r$ .

We now show that $d(p,q)^2\leq r$ . Fix a measurable partition $(A_n)$ of K. For each n, let $x_n,y_n\in E(A_n)(H)$ be parallel vectors satisfying $\|x_n\|=\|E(A_n)c\|$ and $\|y_n\|=\|E(A_n)d\|$ . Then $\|x_n-y_n\|=\left | \|E(A_n)c\|-\|E(A_n)d\|\right |$ . Setting $x:=\sum _n x_n$ , $y=\sum _n y_n$ , and $p_n:=\mathrm {tp}(x_n)$ and $q_n=\mathrm {tp}(y_n)$ , we have

$$ \begin{align*}d(p_n,q_n)^2\leq \|x-y\|^2 & =\sum_n \|x_n-y_n\|^2 \\ &=\sum_n \left| \|E(A_n)(c)\|-\|E(A_n)(d)\|\right|{}^2\leq r.\end{align*} $$

By taking finer square partitions as in the proof of Proposition 5.7 (and the discussion preceding it), we have that $p_n\to p$ and $q_n\to q$ in the logic topology. Since the metric is lower semicontinuous in the logic topology, we conclude that $d(p,q)^2\leq r$ , as desired.

Question 5.12. Does the norm topology on $M(K)_+$ coincide with the d-topology.

Proof Since $\|x\|=1$ , $\mu _p$ defines a probability measure on the Borel subsets of K. Assume first that $\mu _p(\mathrm {Isol}(K))=1$ ; it suffices to show that p is realized in all models of $\Sigma _{K,m}$ . Take $(H,T)\models \Sigma _{K,m}$ ; for each $\lambda \in K$ , write $H_{\lambda }$ for the eigenspace associated with $\lambda $ in the structure $(H,T)$ . For each $\lambda \in \mathrm {Isol}(K)$ , $\dim (H_{\lambda })=m(\lambda )$ , so all eigenspaces associated with isolated eigenvalues have positive dimension. Since $\mu _p(\mathrm {Isol}(K))=1$ and $\mathrm {Isol}(K)$ is countable, there is a sequence $(\lambda _n)$ from $\mathrm {Isol}(K)$ such that $\mu _p=\sum \alpha _n \delta _{\lambda _n}$ , where $\delta _{\lambda _n}$ is the probability measure concentrated on the singleton $\{\lambda _n\}$ , $\alpha _n\geq 0$ , and $\sum _n \alpha _n=1$ . For each n, choose $v_n\in H_{\lambda _n}$ with $\|v_n\|=1$ and set $w=\sum _{n} \alpha _n v_n$ . Then $w\in H$ is a realization of p since $\langle E(\{\lambda _n\})w,w\rangle =\langle \alpha _nv_n,v_n\rangle =\alpha _n=\mu _p(\{\lambda _n\})$ .

For the other direction, suppose $\mu _p(\mathrm {Isol}(K))<1$ , whence $\mu _p(K')>0$ , where $K':=K\setminus \mathrm {Isol}(K)$ is the set of accumulation points of K; it suffices to find a model of $\Sigma _{K,m}$ which omits p. Let Q be a dense Borel subset of $K'$ for which $\mu _p(K'\setminus Q)>0$ . For each $\lambda \in K$ , let $(H_{\lambda },T_{\lambda })$ be the structure satisfying $\dim (H_\lambda )=m(\lambda )$ and $T_\lambda (v)=\lambda v$ for every $v\in H_\lambda $ (recalling that $m(\lambda )=\infty $ when $\lambda \in K'$ ). Let $(H,T):=\bigoplus _{\lambda \in \mathrm {Isol}(K)\cup Q}(H_\lambda ,T_\lambda )$ . Then $(H,T)$ is a model of $\Sigma _{K,m}$ which omits p.

Proof Assume that $\mu _v(A)=0$ , so $\langle E(A)v,v\rangle =0$ . Since $E(A)$ is a projection, it satisfies $E(A)^2=E(A)$ and $E(A)^*=E(A)$ . Then we have $0\hspace{-1.1pt}=\hspace{-1.1pt}\langle E(A)v,v\rangle\hspace{-1.1pt} =\hspace{-1.1pt}\langle E(A)(E(A)v),v\rangle\hspace{-1.1pt} =\hspace{-1.1pt}\langle E(A)v,E^*(A)v\rangle\hspace{-1.1pt} =\hspace{-1.1pt}\langle E(A)v,E(A)v\rangle $ , whence $\|E(A)v\|=0$ . It follows that $\langle E(A)v,w\rangle =0$ .

Proof Clearly, if $p(\vec x)$ is principal, so is $p_i(x_i)$ . Conversely, assume that for each $i=1,\ldots ,n$ , we have $\mu _{p_i}(\mathrm {Isol}(K))=r_i^2$ , where $\|x_i\|=r_i$ belongs to $p_i$ ; we show that p is realized in all models of $\Sigma _{K,m}$ . Note that for any Borel set $A\subseteq K$ with $A\cap \mathrm {Isol}(K)=\emptyset $ , we have $\mu _{p_i}(A)=0$ ; by Lemma 5.16 the measure $\mu _{p_i,p_j}$ is absolutely continuous with respect to $\mu _{p_i}$ , so we also have $\mu _{p_i,p_j}(A)=0$ . Take $(H,T)\models \Sigma _{K,m}$ and a saturated elementary extension $(\hat H,\hat T)\succeq (H,T)$ . Take a realization $(v_1,\dots ,v_n)\in \hat H $ of $p(\vec x)$ . Since the measures $\mu _{p_i,p_j}$ are supported in $\mathrm {Isol}(K)$ , $p(\vec x)$ is determined by $\{\mu _{p_i,p_j}(\{\lambda \})=\langle P_{\lambda }(v_i),P_{\lambda }(v_j))\rangle : \lambda \in \mathrm {Isol}(K)\}$ together with the conditions $\|x_i\|=r_i$ . Since for each $\lambda \in \mathrm {Isol}(K)$ we have $m(\hat H_{\lambda })=m( H_{\lambda })$ , we can find $(w_1,\dots ,w_n)\in H$ with $\langle P_{\lambda }(w_i),P_{\lambda }(w_j))\rangle =\langle P_{\lambda }(v_i),P_{\lambda }(v_j))\rangle $ and $\|w_i\|=r_i$ for all $i,j=1,\ldots ,n$ and $\lambda \in \mathrm {Isol}(K)$ . It follows that $(w_1,\dots ,w_n)\in H$ is a realization of $p(\vec x)$ .

Proof The theory $\Sigma _{K,m}$ is $\omega $ -categorical if and only if every type $p(\vec x)\in S_n(\Sigma _{K,m})$ is principal; by Corollary 5.17, the latter condition is equivalent to the statement that K only has isolated points, which, since K is compact, is equivalent to the condition that K is finite.

Alternative proof of Corollary 5.18

Assume first that K is finite. The separable model of $\Sigma _{K,m}$ is necessarily then isomorphic to $\bigoplus _{\lambda \in K}(H_\lambda ,T_\lambda )$ , where $\dim (H_\lambda )=m(\lambda )$ and $T_\lambda x=\lambda x$ for all $x\in H_\lambda $ .

Assume now that K is infinite and take a separable model $(H,T)$ of $\Sigma _{K,m}$ . Since K is compact and infinite, it has an accumulation point $\lambda _0$ . Consider $H_{\lambda _0}$ , the corresponding eigenspace in $(H,T)$ . If $\dim (H_{\lambda _0})>0$ , then the structure $(H',T')=(H,T)\ominus (H_0,T\upharpoonright _{H_0})$ is not isomorphic to $(H,T)$ (the eigenspaces corresponding to $\lambda _0$ have different dimension) but $(H',T')$ is still a model of $\Sigma _{K,m}$ . If $\dim (H_{\lambda _0})=0$ , then by Upward Löwenheim–Skolem, we can choose $(H,T)\preceq (H',T')$ separable with $\dim (H^{\prime }_{\lambda _0})>0$ ; once again, $(H,T)$ and $(H',T')$ are not isomorphic since the eigenspaces corresponding to $\lambda _0$ have different dimension.

Proof We verify the usual list of axioms that guarantee a theory is stable and that the abstract independence relation characterizes nonforking independence. Clearly, the notion of independence satisfies transitivity and finite character.

Stationarity: Consider tuples $\vec a=(a_1,\dots ,a_n),\vec a'=(a_1',\dots ,a_n')\in H^n$ and sets $B\subseteq C\subseteq H$ with B algebraically closed. Furthermore suppose that $\mathrm {tp}(\vec a/B)=\mathrm {tp}(\vec a'/B)$ and that , . For each $i=1,\ldots , n$ , write $a_i=P_{B}(a_i)+c_i$ and $a_i'=P_{B}(a_i)+c_i'$ with $c_i,c_i'\perp B$ . Note that $\mathrm {tp}(c_1,\dots ,c_n/\emptyset )=\mathrm {tp}(c_1',\dots ,c_n'/\emptyset )$ . Since and , we have $c_i,c_i'\perp \langle B\cup C\rangle $ . It follows that $\mathrm {tp}(\vec a/B\cup C)=\mathrm {tp}(\vec a'/B\cup C)$ .

Symmetry: Consider $\vec a=(a_1,\dots ,a_n)\in H^n$ and $\vec c=(c_1,\dots ,c_k)\in H^k$ and assume that . For any $j=1,\ldots , n$ , any polynomials $p_1(T,T^*),\dots , p_k(T,T^*)$ , and any $b\in \langle B\rangle $ , we have

$$ \begin{align*}a_j-P_{\langle B\rangle}(a_j)\perp p_1(T,T^*)c_1+\dots+p_k(T,T^*)c_{k}+b.\end{align*} $$

Since the family of polynomials is closed under applying T and $T^*$ and B is also invariant under T and $T^*$ , we also have, for any $s\geq 1$ and any $b\in B$ , that

$$ \begin{align*}T^s(a_j)-P_{\langle B\rangle}(T^{s}a_j)\perp p_1(T,T^*)c_1+\dots+p_k(T,T^*)c_{k}+b,\end{align*} $$

$$ \begin{align*}(T^*)^s(a_j)-P_{\langle B\rangle}((T^*{s})a_j)\perp p_1(T,T^*)c_1+\dots+p_k(T,T^*)c_{k}+b.\end{align*} $$

Thus for any polynomial $t_j(T,T^*)$ and any $b\in B$ , we also have

$$ \begin{align*}t_j(T,T^*)a_j-P_{\langle B\rangle}(t_j(T,T^*)a_j)\perp p_1(T,T^*)c_1+\dots+p_\ell(T,T^*)c_{k}+b\end{align*} $$

and thus,

$$ \begin{align*}\sum_{j=1}^n(t_j(\! T,T^*)a_j-P_{\langle B\rangle}(t_j(T,T^*)a_j\! ))\perp p_1(T,T^*)c_1+\dots+p_\ell(T,T^*)c_{\ell}+b.\end{align*} $$

In particular, for any $i=1,\ldots ,k$ and any $b\in B$ we get

$$ \begin{align*}c_i,P_{\langle B\rangle}(c_i),b \perp \sum _j (t_j(T,T^*)a_j-P_{\langle B\rangle}(t_j(T,T^*)a_j))\end{align*} $$

and thus $c_i-P_{\langle B\rangle }(c_i) \perp \sum _j t_j(T,T^*)a_j+b$ . This shows that for any $i=1,\ldots ,k$ we have $P_{\langle B \cup \{a_j \ : \ j=1,\ldots , n\}\rangle }(c_i)=P_{\langle B \rangle }(c_i)$ and thus .

Existence: Let $B\subseteq C \subseteq H$ and let $\vec a\in H^n$ . We will show that $\mathrm {tp}(\vec a/B)$ has a free extension to C in an elementary extension of $(H,T)$ (and thus in $(H,T)$ by saturation). Without loss of generality, we may assume that B is algebraically closed. Set $D:=B^\perp $ , so D is a closed subspace of H invariant under T, $T^*$ and it is orthogonal to $H_{\mathrm {fin},T}$ . Let $(D',T')$ be a copy of $(D,T\upharpoonright _{D})$ . Then $(H,T)\oplus (D',T')$ is an elementary extension of $(H,T)$ . Write $a_i=b_i+d_i$ , where $b_i=P_{B}(a_i)$ and $d_i=P_{D}(a_i)$ . Let $d_i'$ be the copy of $d_i$ in the space $D'$ and let $a_i'=b_i+d_i'$ . Then $\mathrm {tp}(a_1',\dots ,a_n'/B)=\mathrm {tp}(a_1,\dots ,a_n/B)$ and for each $i=1,\ldots , n$ , we have $P_{H}(a_i')=P_{B}(a_i')$ . In particular, $P_{\langle C\rangle }(a_i')=P_{B}(a_i')$ for each $i=1,\ldots ,n$ , and so , as desired.

Local character: Fix $B\subseteq H$ and $\vec a=(a_1,\ldots ,a_n)\in H^n$ . If we set $D=\{P_{\langle B\rangle }(a_1),\dots ,P_{\langle B\rangle }(a_n)\}$ , then by the definition of independence, we have . Since $D\subseteq \langle B\rangle $ and D is finite, there is $B_0\subseteq B$ countable with $D\subseteq \langle B_0\rangle $ . It follows that .

Proof Fix $B_0\subseteq H$ and set $B=\langle B_0 \rangle $ . By the definition of independence, it suffices to prove that for any single element $a\in H$ and any $\epsilon>0$ , there are a finite $B_\epsilon \subseteq B_0$ and $a'\in H$ such that $d(a,a')<\epsilon $ and . Write $a=P_B(a)+a_1$ , where $a_1\in B^\perp $ . By Theorem 6.2, we have .

Set $h=P_{H_{\mathrm {fin},T}}(a)$ . Since $P_{B}(x)\in B$ , given $\epsilon>0$ , we can find $n\geq 0$ and a sum $p_1(T,T^*)(b_1)+\dots +p_n(T,T^*)(b_n)$ , where $b_1,\dots ,b_n\in B_0$ , such that

$$ \begin{align*}\|P_B(x)-(h+p_1(T,T^*)(b_1)+\dots+p_n(T,T^*)(b_n))\|<\epsilon.\end{align*} $$

Set $B_\epsilon =\{a_1,\dots ,a_n\}$ and $a'=(h+p_1(T,T^*)(b_1)+\dots +p_n(T,T^*)(b_n))+a_1$ . Then $d(a,a')=d(P_{B}(a),P_{B}(a'))<\epsilon $ and .