Proof. Since $\vec {u}$ and $\vec {v}$ are not parallel, we have $ \left \lvert { \left \langle {\vec {u}, \vec {v}} \right \rangle } \right \rvert < 1 = \cos (0)$ by the Cauchy–Schwarz inequality. We induct on the least $n \in \mathbb {N}$ such that $ \left \lvert { \left \langle {\vec {u}, \vec {v}} \right \rangle } \right \rvert \le \cos (\pi /2^{n+1})$ (i.e., when $\mathbb {F} = \mathbb {R}$ , the least n such that the acute angle between the lines spanned by $\vec {u}$ and $\vec {v}$ is $\ge 90^\circ /2^n$ ).

First, suppose $n = 0$ , i.e., $\vec {u} \perp \vec {v}$ . By conjugating $\mathbb {F}^3$ by a suitable linear isometry, we may assume that $\vec {u} = \vec {e}_1$ and $\vec {v} = \vec {e}_2$ are the standard basis vectors. Let ${\vec {w} = (a,b,c) \in \mathbb {F}^3}$ . Then there is an isometry $\mathbb {F}^3 \to \mathbb {F}^3$ fixing $\vec {v}$ taking $\vec {u} \mapsto (a,0,d)$ for some $d \in \mathbb {F}$ (e.g., $d = \sqrt {1- \left \lvert {a} \right \rvert ^2}$ ), followed by an isometry fixing $\vec {u}$ taking $(a,0,d) \mapsto (a,b,c) = \vec {w}$ .

Now suppose n is such that the claim is known for any $\vec {u}, \vec {v}$ with $ \left \lvert { \left \langle {\vec {u}, \vec {v}} \right \rangle } \right \rvert \le \cos (\pi /2^{n+1})$ ; we must show the same when n is replaced with $n+1$ . By conjugating as above, we may assume $\vec {v} = \vec {e}_1$ and $\vec {u} = ( \left \langle {\vec {u},\vec {v}} \right \rangle ,a,0),$ where $a := \sqrt {1- \left \lvert { \left \langle {\vec {u},\vec {v}} \right \rangle } \right \rvert ^2}$ . Observe that for any $\theta \in \mathbb {R}$ , there is an isometry $f_\theta : \mathbb {F}^3 \to \mathbb {F}^3$ fixing $\vec {v}$ and taking $\vec {u} \mapsto ( \left \langle {\vec {u},\vec {v}} \right \rangle ,a\cos (\theta ),a\sin (\theta )) =: \vec {u}_\theta $ . When $\theta = 0$ , we have $ \left \langle {\vec {u},\vec {u}_\theta } \right \rangle = \left \langle {\vec {u},\vec {u}} \right \rangle = 1$ ; when $\theta = \pi $ , we have

$$ \begin{align*} \left\lvert { \left\langle {\vec{u},\vec{u}_\theta} \right\rangle } \right\rvert &= \left\lvert { \left\langle {\vec{u},\vec{v}} \right\rangle } \right\rvert ^2 - a^2 = 2 \left\lvert { \left\langle {\vec{u},\vec{v}} \right\rangle } \right\rvert ^2 - 1 \\ &\le 2\cos(\pi/2^{n+2})^2 - 1 = \cos(\pi/2^{n+1}). \end{align*} $$

Thus there is some $0 \le \theta \le \pi $ such that $ \left \lvert { \left \langle {\vec {u},\vec {u}_\theta } \right \rangle } \right \rvert = \cos (\pi /2^{n+1})$ . By the induction hypothesis, there is a finite composition of isometries fixing either $\vec {u}$ or $\vec {u}_\theta $ and taking $\vec {u} \mapsto \vec {w}$ ; the isometries that fix $\vec {u}_\theta $ are a $f_\theta $ -conjugate of an isometry that fixes $\vec {u}$ , so we are done.

Proof of Lemma 3.4

We may assume that $A_0 \cap A_1$ is codimension $1$ in both $A_0$ and $A_1$ . Indeed, assuming this special case has been shown, we may deduce the general case as follows. If $A_0 \subseteq A_1$ or $A_1 \subseteq A_0$ , then the claim is trivial. If $A_0 \cap A_1$ is codimension $1$ in $A_0$ and codimension $n> 0$ in $A_1$ , then letting $\vec {v}_0, \dotsc , \vec {v}_{n-1}$ be a basis for $A_1 \cap (A_0 \cap A_1)^\perp $ , we have that x is supported on the linear span of $A_0 \cup \{\vec {v}_0, \dotsc , \vec {v}_{n-2}\}$ , and on $A_1$ , which have intersection of codimension $1$ in each, whence by the special case, x is supported on said intersection, the linear span of $(A_0 \cap A_1) \cup \{\vec {v}_0, \dotsc , \vec {v}_{n-2}\}$ ; by induction on n, we eventually get that x is supported on $A_0 \cap A_1$ . Finally, in general if $A_0 \cap A_1$ is codimension $m> 0$ in $A_0$ , we have that x is supported on $A_0$ and on the span of $A_1$ together with $m-1$ basis vectors in $A_0 \cap (A_0 \cap A_1)^\perp $ ; by the previous case, we get that x is supported on the span of $A_0 \cap A_1$ together with said $m-1$ vectors; now induct on m.

So suppose $A_0$ is spanned by $(A_0 \cap A_1) \cup \{\vec {u}\}$ , and $A_1$ by $(A_0 \cap A_1) \cup \{\vec {v}\}$ , where $\vec {u} \in A_0 \cap (A_0 \cap A_1)^\perp $ and $\vec {v} \in A_1 \cap (A_0 \cap A_1)^\perp $ are unit vectors. Let $f, g : A \rightrightarrows B \in {\mathsf {Hilb_r}}$ agree on $A_0 \cap A_1$ ; we must show that $Uf(x) = Ug(x)$ .

First, suppose g is surjective. Put $\vec {w} := g^{-1}(f(\vec {u}))$ ; then $\vec {w}$ is a unit vector, such that

$$ \begin{align*} \vec{w} &\perp A_0 \cap A_1, \end{align*} $$

since

$$ \begin{align*} g(\vec{w}) = f(\vec{u}) &\perp f(A_0 \cap A_1) = g(A_0 \cap A_1). \end{align*} $$

By the preceding lemma applied to the subspace spanned by $\vec {u}, \vec {v}$ , and $\vec {w}$ , which is orthogonal to $A_0 \cap A_1$ , find linear isometric automorphisms $h_0, \dotsc , h_{k-1} : A \cong A$ each of which fixes either $(A_0 \cap A_1) \cup \vec {u}$ or $(A_0 \cap A_1) \cup \vec {v}$ , hence either $A_0$ or $A_1$ , such that $(h_0 \circ \dotsb \circ h_{k-1})(\vec {u}) = \vec {w}$ . Then in the sequence

$$ \begin{align*} g,\; g \circ h_0,\; g \circ h_0 \circ h_1,\; g \circ h_0 \circ h_1 \circ h_2,\; \dotsc,\; g \circ h_0 \circ \dotsb \circ h_{k-1},\; f : A \to B, \end{align*} $$

each consecutive pair of maps agrees on either $A_0$ or $A_1$ , so $Ug(x) = Uf(x)$ since x is supported on $A_0$ and on $A_1$ . (This argument is a more involved version of [Reference Lieberman, Rosický and Vasey7, Lemma 14].)

If g is not surjective, note that $f(A) + g(A) \subseteq B$ has the same dimension as A. Let $h : A \cong f(A) + g(A)$ be an arbitrary isometric isomorphism, let ${i : f(A) + g(A) \hookrightarrow B}$ be the subspace inclusion, and let $f', g' : A \rightrightarrows f(A) + g(A)$ be the codomain restrictions of $f, g$ . Then $Uf'(x) = Uh(x) = Ug'(x)$ by the surjective case, whence $Uf(x) = Ui(Uf'(x)) = Ui(Ug'(x)) = Ug(x)$ .

Proof. Since U preserves directed colimits and A is the directed colimit of its finite-dimensional subspaces $A_0 \subseteq A$ , there is some such $A_0$ , with inclusion ${i : A_0 \to A}$ , as well as $x_0 \in UA_0$ such that $x = Ui(x_0)$ . In particular, it follows by Remark 3.3 that x is supported on $A_0$ (in $A_0$ , hence in A). Let $\operatorname {\mathrm {supp}}_A(x) \subseteq A$ be a finite-dimensional subspace of minimal dimension on which x is supported; by Lemma 3.4, it is in fact the unique minimal support of x in A.

Proof. Following [Reference Lieberman, Rosický and Vasey7, Theorem 18], pick $\lambda> 2^{\aleph _0}$ of countable cofinality such that U maps a Hilbert space of dimension $\lambda $ to a set of cardinality $\lambda $ . (For the reader’s convenience, we sketch an elementary argument for the existence of such $\lambda $ that does not use any accessible category theory. Let $\lambda _0$ be the supremum of $2^{\aleph _0}$ and the cardinalities of $U(\mathbb {F}^n)$ for each $n \in \mathbb {N}$ . Then for any Hilbert space A of dimension ${\lambda \ge \lambda _0}$ , without loss of generality, $A = \ell ^2(\lambda )$ , A is the directed colimit of the subspaces $\ell ^2(F)$ for all finite subsets $F \subseteq \lambda $ ; for each such F, we have $ \left \lvert {U(\ell ^2(F))} \right \rvert \le \lambda _0 \le \lambda $ ; and there are $\lambda $ -many such F, hence $ \left \lvert {U(\ell ^2(\lambda ))} \right \rvert = \left \lvert{\operatorname {\mathrm {colim}}_F U(\ell ^2(F))}\right \rvert \le \lambda \cdot \lambda = \lambda. $ It suffices to take any $\lambda \ge \lambda _0$ of countable cofinality, e.g., $\lambda = \sup \{\lambda _0, \lambda _0^+, \lambda _0^{++}, \dotsc \}$ .)

Now let A be a Hilbert space of dimension $\lambda $ . Suppose for contradiction that there were $x \in UA$ with $\operatorname {\mathrm {supp}}_A(x) \subseteq A$ a nontrivial finite-dimensional subspace, say of dimension $1 \le n < \aleph _0$ . Observe that $ \left \lvert {A} \right \rvert = \left \lvert {\ell ^2(\lambda )} \right \rvert = \lambda ^{\aleph _0}$ , and A is the union of all subspaces $B \subseteq A$ of dimension n, each of which has cardinality $2^{\aleph _0} < \lambda < \lambda ^{\aleph _0}$ ; thus there are $\lambda ^{\aleph _0}$ distinct subspaces $B \subseteq A$ of dimension n. However, for any such $B \subseteq A$ of dimension n, there is a linear isometric automorphism $f : A \to A$ taking $\operatorname {\mathrm {supp}}_A(x)$ to B, and so there must be $y = Uf(x) \in UA$ such that $\operatorname {\mathrm {supp}}_A(y) = B$ (by the preceding remark). This is a contradiction as $ \left \lvert {UA} \right \rvert = \lambda < \lambda ^{\aleph _0}$ by Kőnig’s theorem. Thus every $x \in UA$ has trivial support, which yields the desired claim by Remark 3.2.

Proof. f and g are the directed colimits of their respective restrictions $f_0, g_0$ to all subspaces $A_0 \subseteq A$ of dimension exactly $\lambda $ , and $Uf_0 = Ug_0$ for all such $A_0$ by the preceding Proposition; thus since U preserves directed colimits, $Uf = Ug$ .

Proof. First, suppose $\dim (B) = \dim (A)$ . Let $x, y \in UA$ , and suppose $Uf(x) = Uf(y) \in UB$ . Since A is infinite-dimensional, it is the directed colimit of all finite-dimensional $A_0 \subseteq A$ , and so there are $x_0, y_0 \in UA_0$ for some such $A_0$ , with inclusion $i : A_0 \to A$ , such that $x = Ui(x_0)$ and $y = Ui(y_0)$ . Since A and B have the same infinite dimension, and $A_0 \subseteq A$ is finite-dimensional, there is an isomorphism $A \cong B$ that takes the inclusion $i : A_0 \hookrightarrow A$ to the embedding $fi : A_0 \hookrightarrow A \hookrightarrow B$ :

By assumption, $x_0$ and $y_0$ map to the same thing in $UB$ along the top-right composite, hence also along the left-bottom, hence also along the left, i.e., $x = Ui(x_0) = Ui(y_0) = y$ .

This proves the result for $\dim (B) = \dim (A)$ . If $\dim (B)> \dim (A)$ , write ${f : A \to B}$ as a directed colimit of $f : A \to B_0$ , where $B_0 \subseteq B$ contains $\operatorname {\mathrm {im}}(f)$ and has the same dimension as A.

Proof. From Corollary 3.9, we know U maps parallel pairs of morphisms between sufficiently high-dimensional Hilbert spaces $\ell ^2(\lambda )$ to the same morphism in ${\mathsf {Set}}$ . For two arbitrary maps $f, g : A \rightrightarrows B \in {\mathsf {Hilb_r}}$ between infinite-dimensional spaces, consider the square

Then $U(f \oplus {\mathrm {id}}) = U(g \oplus {\mathrm {id}})$ , and the inclusion $B \hookrightarrow B \oplus \ell ^2(\lambda )$ is mapped to an injection by Lemma 3.10, hence $Uf = Ug$ . So U is constant on all parallel pairs of morphisms in ${\mathsf {Hilb_r}}^{\ge \aleph _0}$ .

This implies that U maps every $f : A \to B \in {\mathsf {Hilb_r}}$ with $\dim (A) \ge \aleph _0$ to a bijection in ${\mathsf {Set}}$ . To see this: if $\dim (B) = \dim (A)$ , then there exists a linear isometric isomorphism $g : A \cong B$ ; thus $Uf = Ug$ is a bijection. If $\dim (B)> \dim (A)$ , as in the preceding proof, write f as the directed colimit of its codomain restrictions to $B_0 \subseteq B$ of the same dimension as A.

We now define a natural isomorphism $\phi : U(\ell ^2) \cong U|_{{\mathsf {Hilb_r}}^{\ge \aleph _0}}$ , from the constant functor with value $U(\ell ^2)$ to the restriction of U to all infinite-dimensional Hilbert spaces A. Let $\phi _A : U(\ell ^2) \to UA$ be $Uf$ for any embedding $f : \ell ^2 \hookrightarrow A$ . This is a natural transformation, since for any morphism $g : A \to B \in {\mathsf {Hilb_r}}$ , the square

commutes, as both composites are U applied to a morphism $\ell ^2 \to B$ , hence equal by the first part of this proof; and each $\phi _A$ is invertible by the second part of this proof.

Proof. For any $A \in {\mathsf {Hilb}}$ , let

$$\begin{align*} i &:&\kern-20pt A &\to A \oplus \ell^2 &&\kern-20pt = \text{first coordinate injection}, \\j &:&\kern-20pt A \oplus \ell^2 &\to A \oplus A \oplus \ell^2 &&\kern-20pt = \text{first and third coordinate injection}, \\k &:&\kern-20pt A \oplus \ell^2 &\to A \oplus A \oplus \ell^2 &&\kern-20pt = \text{second and third coordinate injection}, \\p &:&\kern-20pt A \oplus A \oplus \ell^2 &\to A &&\kern-20pt = \text{second coordinate projection}; \end{align*}$$

then

$$ \begin{align*} \begin{aligned} U(0_A) = U(pji) &= Up \circ Uj \circ Ui \\ &= Up \circ Uk \circ Ui \quad \text{by Theorem }{3.11} \\ &= U({\mathrm{id}}_A) = {\mathrm{id}}_{UA}, \end{aligned} \end{align*} $$

i.e., U takes the zero map on every $A \in {\mathsf {Hilb}}$ to the identity. Now whisker the natural transformations

by U to get that U is naturally isomorphic to the constant functor with value $U(0)$ .

Proof. By [Reference Lieberman, Rosický and Vasey7, Lemma 14], there is a well-defined support $\operatorname {\mathrm {supp}}_A(x) \subseteq A$ for any infinite-dimensional $A \in {\mathsf {Hilb_m}}$ and $x \in UA$ , which is a finite-dimensional subspace of A. By the proofs of Proposition 3.8 and Corollary 3.9, it follows that U is constant on all parallel pairs $f, g : A \rightrightarrows B \in {\mathsf {Hilb_m}}$ , where A is of sufficiently high dimension $\ge \lambda $ for some infinite cardinal $\lambda $ . By Lemma 3.10, we still have that U maps every morphism in the subcategory ${\mathsf {Hilb_r}}^{\ge \aleph _0} \subseteq {\mathsf {Hilb_m}}^{\ge \aleph _0}$ to an injection. The proof of Theorem 3.11 (which only uses Lemma 3.10 for a linear isometric embedding $A \hookrightarrow A \oplus \ell ^2(\lambda )$ ) now applies.

Proof. Let $V : {\mathsf {Hilb_r}} \to {\mathsf {Set}}$ take the trivial space to $\varnothing $ and every other space A to $U(\ell ^2(\kappa ) \otimes A)$ , where $\otimes $ denotes the Hilbert space tensor product (see [Reference Pedersen10, Exercise E3.2.19]), with the obvious extension to morphisms. Then by Theorem 3.11, V is essentially constant when restricted to ${\mathsf {Hilb_r}}^{\ge \aleph _0}$ .

In particular, U takes every morphism $f : A \to B \in {\mathsf {Hilb_r}}^{\ge \kappa }$ which is isomorphic to a morphism of the form ${\mathrm {id}}_{\ell ^2(\kappa )} \otimes g$ for some $g : C \to D \in {\mathsf {Hilb_r}}^{\ge \aleph _0}$ to a bijection $V(g)$ ; such f are easily seen (by taking an orthonormal basis for A, then extending it to one for B, and using $\kappa \ge \aleph _0$ ) to consist of precisely all linear isometries which either are surjective or have $\ge \kappa $ -codimensional image. But every $f : A \to B \in {\mathsf {Hilb_r}}^{\ge \kappa }$ may be composed with some such morphism (e.g., the injection $B \hookrightarrow B \oplus \ell ^2(\kappa )$ ) to yield another such morphism; thus, U takes every morphism in ${\mathsf {Hilb_r}}^{\ge \kappa }$ to a bijection.

Moreover, U takes every endomorphism $f : A \to A$ which is conjugate to ${\mathrm {id}}_{\ell ^2(\kappa )} \otimes g$ for some $g : C \to C \in {\mathsf {Hilb_r}}^{\ge \aleph _0}$ to the identity. But every $f : A \to A \in {\mathsf {Hilb_r}}^{\ge \kappa }$ fits into a commutative square

where $i(\vec {v}) := \vec {e}_0 \otimes \vec {v}$ ; thus U takes every endomorphism in ${\mathsf {Hilb_r}}^{\ge \kappa }$ to the identity. As in the proof of Theorem 3.11, this easily implies that U is constant on any parallel pair $f, g : A \rightrightarrows B \in {\mathsf {Hilb_r}}^{\ge \kappa }$ (by assuming $\dim (A) = \dim (B)$ via codomain restriction and composing with some $h : B \cong A$ ), which together with the above yields a natural isomorphism $U(\ell ^2(\kappa )) \cong U$ .

Proof. Follows from Theorem 3.11 and the faithful forgetful functors ${\mathsf {Hilb_r}} \to {\mathsf {Ban_r}} \to {\mathsf {Met_r}}$ . For the last part, use Corollary 4.3.