Proof (1): Set $\mathcal {T} := \{ B \in \operatorname {\mathrm {Sub}}_{\mathrm {sep}}(A) : B\cap I \in \mathcal {S} \}$ . It is easy to see that $\mathcal {T}$ is $\sigma $ -complete. To show that it is cofinal, let $B_0\in \operatorname {\mathrm {Sub}}_{\mathrm {sep}}(A)$ . We will inductively find increasing sequences $(I_k)_k$ in $\mathcal {S}$ and $(B_k)_k$ in $\operatorname {\mathrm {Sub}}_{\mathrm {sep}}(A)$ such that

$$\begin{align*}B_0\cap I \subseteq I_0 \subseteq B_1\cap I \subseteq I_1 \subseteq \cdots. \end{align*}$$

Assume that we have obtained $B_k$ for some $k\in \mathbb {N}$ . Then, $B_k\cap I\in \operatorname {\mathrm {Sub}}_{\mathrm {sep}}(I)$ , and since $\mathcal {S}$ is cofinal in $\operatorname {\mathrm {Sub}}_{\mathrm {sep}}(I)$ , we obtain $I_k\in \mathcal {S}$ such that $B_k\cap I\subseteq I_k$ . Then, let $B_{k+1}$ be the sub- $C^*\!$ -algebra of A generated by $B_k$ and $I_k$ .

Set $B:=\overline {\bigcup _k B_k}$ , which belongs to $\operatorname {\mathrm {Sub}}_{\mathrm {sep}}(A)$ and contains $B_0$ . We have $B\cap I = \overline {\bigcup _k I_k}$ . Since $\mathcal {S}$ is $\sigma $ -complete, $B\cap I$ belongs to $\mathcal {S}$ . Thus, B belongs to $\mathcal {T}$ , as desired.

Statement (2) is shown similarly.

Proof $\sigma $ -completeness: Let $\mathcal {T}\subseteq \mathcal {S}$ be a countable directed family, and set $C:=\overline {\bigcup \{B:B\in \mathcal {T}\}}$ . To show that C is d-homogeneous, let $\varrho $ be an irreducible representation of C. Since C is d-subhomogeneous (as a subalgebra of A), the dimension of $\varrho $ is at most d. If $\dim (\varrho )<d$ , then the restriction of $\varrho $ to each $B\in \mathcal {T}$ is zero, whence $\varrho =0$ , a contradiction.

In [Reference Brown and PedersenBP09, Section 2.2], Brown and Pedersen introduce the topological dimension of type $\mathrm {I} C^*\!$ -algebras. Given a homogeneous $C^*\!$ -algebra D, the topological dimension $\operatorname {\mathrm {topdim}}(D)$ is equal to $\operatorname {\mathrm {locdim}}(\operatorname {\mathrm {Prim}}(D))$ . Hence, each $B\in \mathcal {T}$ satisfies $\operatorname {\mathrm {topdim}}(B)=\operatorname {\mathrm {locdim}}(\operatorname {\mathrm {Prim}}(B))\leq l$ . By [Reference ThielThi13, Lemma 3], a continuous trace $C^*\!$ -algebra (in particular, a homogeneous $C^*\!$ -algebra) has topological dimension at most l whenever it is approximated by sub- $C^*\!$ -algebras with topological dimension at most l. Hence,

$$\begin{align*}\operatorname{\mathrm{locdim}}(\operatorname{\mathrm{Prim}}(C)) = \operatorname{\mathrm{topdim}}(C) \leq l, \end{align*}$$

which verifies that C belongs to $\mathcal {S}$ , as desired.

Cofinality: Let $B_0\subseteq A$ be a separable sub- $C^*\!$ -algebra. We identify A with $C(X)\otimes M_d$ . Let $e_{jk}\in M_d$ , $j,k=1,\ldots ,d$ , be matrix units. Let $C(Y)\subseteq C(X)$ be a separable, unital sub- $C^*\!$ -algebra such that $f\in C(X)$ belongs to $C(Y)$ whenever $f\otimes e_{jk}\in B_0$ for some $j,k$ . Using that the real rank satisfies (D6), let $C(Z)\subseteq C(X)$ be a separable sub- $C^*\!$ -algebra containing $C(Y)$ such that $\operatorname {\mathrm {rr}}(C(Z))\leq \operatorname {\mathrm {rr}}(C(X))$ . Then,

$$\begin{align*}\dim(Z)=\operatorname{\mathrm{rr}}(C(Z))\leq\operatorname{\mathrm{rr}}(C(X))=\dim(X)\leq l, \end{align*}$$

and it follows that $C(Z)\otimes M_d\subseteq C(X)\otimes M_d$ has the desired properties.

Proof As in the proof of Lemma 3.4, we obtain that $\mathcal {S}$ is $\sigma $ -complete.

Cofinality: Let $B_0\subseteq A$ be a separable sub- $C^*\!$ -algebra. Let $I\subseteq A$ be the ideal generated by $B_0$ . Then, I is d-homogeneous and $X:=\operatorname {\mathrm {Prim}}(I)$ is $\sigma $ -compact. We view I as a $C_0(X)$ -algebra with all fibers isomorphic to $M_d$ . Since the $M_d$ -bundle associated with I is locally trivial, and since X is $\sigma $ -compact, we can choose a sequence of compact subsets $X_0,X_1,X_2,\ldots \subseteq X$ that cover X such that $I(X_j)\cong C(X_j)\otimes M_d$ for each $j\in \mathbb {N}$ .

Given j, let $\pi _j\colon I\to C(X_j)\otimes M_d$ be the corresponding quotient map, and set

$$\begin{align*}\mathcal{S}_j := \big\{ B\in\operatorname{\mathrm{Sub}}_{\mathrm{sep}}(I) : \pi_j(B)\ d\text{-homogeneous}, \operatorname{\mathrm{locdim}}(\operatorname{\mathrm{Prim}}(\pi_j(B)))\leq l \big\}. \end{align*}$$

Applying Lemmas 3.2(2) and 3.4, we obtain that $\mathcal {S}_j$ is $\sigma $ -complete and cofinal. It follows that $\mathcal {S}:=\bigcap _{j=0}^\infty \mathcal {S}_j$ is $\sigma $ -complete and cofinal as well. Choose $B\in \mathcal {S}$ satisfying $B_0\subseteq B$ .

To verify that B is d-homogeneous, let $\varrho $ be an irreducible representation of B. Since B is d-subhomogeneous, we have $\dim (\varrho )\leq d$ . Extend $\varrho $ to an irreducible representation $\varrho '$ of I (a priori on a possibly larger Hilbert space). Then, there exists $x\in X$ such that $\varrho '$ is isomorphic to the quotient map to the fiber at x. Let $j\in \mathbb {N}$ such that $x\in X_j$ . Since B belongs to $\mathcal {S}_j$ , it exhausts the fiber at x, and we deduce that $\dim (\varrho )\geq d$ .

To see that $\operatorname {\mathrm {locdim}}(\operatorname {\mathrm {Prim}}(B))\leq l$ , let $K\subseteq \operatorname {\mathrm {Prim}}(B)$ be a compact subset. For each j, let $F_j\subseteq \operatorname {\mathrm {Prim}}(B)$ be the closed subset corresponding to the quotient $\pi _j(B)$ of B. Since B belongs to $\mathcal {S}_j$ , we have $\operatorname {\mathrm {locdim}}(F_j)\leq l$ . Hence, $\dim (K\cap F_j)\leq l$ . We have $K=\bigcup _j (K\cap F_j)$ , and therefore

$$\begin{align*}\dim(K) = \sup_j \dim(K\cap F_j) \leq l \end{align*}$$

by the Countable Sum Theorem (see [Reference PhillipsPea75, Theorem 3.2.5, p. 125]; see also the introduction to Section 5).

Proof Let us first assume that $\mathbf {a}\in \operatorname {\mathrm {Gen}}_{n+1}(A)_{\mathrm {sa}}$ . For $x\in X$ , let $\pi _x\colon A\to A(x)$ be the quotient map onto the fiber at x. Since $\pi _x$ is a surjective ${}^*$ -homomorphism, it maps $\operatorname {\mathrm {Gen}}_{n+1}(A)_{\mathrm {sa}}$ to $\operatorname {\mathrm {Gen}}_{n+1}(A(x))_{\mathrm {sa}}$ , which verifies (a). Similarly, for distinct points $x,y\in X$ , the map $\pi _x\oplus \pi _y\colon A\to A(x)\oplus A(y)$ is a surjective ${}^*$ -homomorphism. It follows that $(\mathbf {a}(x),\mathbf {a}(y))=(\pi _x\oplus \pi _y)(\mathbf {a})\in \operatorname {\mathrm {Gen}}_{n+1}(A(x)\oplus A(y))_{\mathrm {sa}}$ . To verify (b), assume that $\alpha \colon A(x)\to A(y)$ is an isomorphism satisfying $\alpha (\mathbf {a}(x))=\mathbf {a}(y)$ . Then,

$$\begin{align*}C^*((\mathbf{a}(x),\mathbf{a}(y))) = \big\{ (d,\alpha(d))\in A(x)\oplus A(y) : d\in A(x) \big\} \neq A(x)\oplus A(y), \end{align*}$$

which contradicts that $(\mathbf {a}(x),\mathbf {a}(y))$ generates $A(x)\oplus A(y)$ . Thus, no such $\alpha $ exists.

Conversely, let us assume that (a) and (b) are satisfied. Set $B:=C^*(\mathbf {a})$ . We need to prove $B=A$ . This follows from [Reference Tikuisis, White and WinterTW14, Lemma 3.2] once we show that B exhausts the fiber $A(x)$ for each $x\in X$ , and that for distinct $x,y\in X$ , there exists $b\in B$ such that $b(x)$ is full in $A(x)$ and $b(y)=0$ . The exhaustion of fibers follows directly from (a).

Let $x,y\in X$ be distinct, and set $C:=(\pi _x\oplus \pi _y)(B)\subseteq A(x)\oplus A(y)$ . Note that C is the sub- $C^*\!$ -algebra of $A(x)\oplus A(y)$ generated by $(\mathbf {a}(x),\mathbf {a}(y))$ . If $C\neq A(x)\oplus A(y)$ , using that $A(x)$ and $A(y)$ are simple, it follows from [Reference Thiel and WinterThi21, Lemma 5.10] that there exists an isomorphism $\alpha \colon A(x)\to A(y)$ such that

$$\begin{align*}C = \big\{ (d,\alpha(d))\in A(x)\oplus A(y) : d\in A(x) \big\}, \end{align*}$$

which implies that $\alpha (\mathbf {a}(x))=\mathbf {a}(y)$ . Since this contradicts (b), we deduce that $C=A(x)\oplus A(y)$ . Hence, there exists $b\in B$ such that $b(x)$ is full in $A(x)$ and $b(y)=0$ .

Notation 4.2 For $d\geq 2$ and $n\in \mathbb {N}$ , we set

$$ \begin{align*} E_d^{n+1} :=(M_d)^{n+1}_{\mathrm{sa}}, \quad\text{ and }\quad G_d^{n+1} :=\operatorname{\mathrm{Gen}}_{n+1}(M_d)_{\mathrm{sa}} \subseteq E_d^{n+1}. \end{align*} $$

Note that $E_d^{n+1}$ is isomorphic to $\mathbb {R}^{(n+1)d^2}$ as topological vector spaces. In particular, $E_d^{n+1}$ is a (real) manifold with $\dim (E_d^{n+1})=(n+1)d^2$ .

We let ${\mathcal {U}}_d$ denote the unitary group of $M_d$ . It is a compact Lie group of dimension $d^2$ . Every automorphism of $M_d$ is inner, and the kernel of ${\mathcal {U}}_d\to \operatorname {\mathrm {Aut}}(M_d)$ is the group of central unitaries $\mathbb {T} 1\subseteq {\mathcal {U}}_d$ . Hence, $\operatorname {\mathrm {Aut}}(M_d)$ is naturally isomorphic to $\mathcal {PU}_d:={\mathcal {U}}_d/(\mathbb {T} 1)$ , the projective unitary group, which is a compact Lie group of dimension $d^2-1$ . Given $u\in {\mathcal {U}}_d$ , we use $[u]$ to denote its class in $\mathcal {PU}_d$ .

The action $\mathcal {PU}_d\curvearrowright M_d$ induces an action $\mathcal {PU}_d\curvearrowright E_d^{n+1}$ by setting

$$\begin{align*}[u].\mathbf{a} := (ua_0u^*,\ldots,ua_nu^*) \end{align*}$$

for $u\in {\mathcal {U}}_d$ and $\mathbf {a}=(a_0,\ldots ,a_n)\in E_d^{n+1}$ .

Proof Note that T contains exactly one element of depth zero, namely the conjugacy class of $\{1\}$ . Therefore,

$$\begin{align*}M\setminus M_{\mathrm{free}} = \bigcup_{t\in T, \operatorname{\mathrm{depth}}(t)\geq 1} M_t, \end{align*}$$

and it follows that

$$\begin{align*}\dim(M\setminus M_{\mathrm{free}}) = \max\big\{ \dim(M_t) : \operatorname{\mathrm{depth}}(t)\geq 1 \big\}. \end{align*}$$

To show that (1) implies (2), assume that $\dim (X)\geq \dim (M)-\dim (M\setminus M_{\mathrm {free}})$ . Choose $t\in T$ of depth $\geq 1$ such that $\dim (X)\geq \dim (M)-\dim (M_t)$ . As noted in [Reference Beggs and EvansBE91, Proposition 1.6], it follows that $C(X,M\setminus M_t)\subseteq C(X,M)$ is not dense, which implies that (1) fails.

Assuming (2), let us prove (1). Let $f\in C(X,M)$ and $\varepsilon>0$ . The proof is similar to that of Theorem 1.3 in [Reference Beggs and EvansBE91]. We inductively change f to avoid each $M_t$ , but instead of proceeding by the (co)dimension of the submanifolds, we use their depths.

It follows from the frontier condition that for each $t\in T$ , the set $\overline {M_t}\setminus M_t$ is contained in the union of submanifolds $M_s$ with $s\in T$ and $\operatorname {\mathrm {depth}}(s)>\operatorname {\mathrm {depth}}(t)$ . Let $t_1,\ldots ,t_K$ be an enumeration of the elements in T with depth $\geq 1$ , such that $\operatorname {\mathrm {depth}}(t_1)\geq \operatorname {\mathrm {depth}}(t_2)\geq \cdots \geq \operatorname {\mathrm {depth}}(t_K)$ . Note that $M_{t_1}$ is a closed submanifold (since $t_1$ has maximal depth and thus $\overline {M_{t_1}}\setminus M_{t_1}=\emptyset $ ), and for each $j\geq 2$ , the set $\overline {M_{t_j}}\setminus M_{t_j}$ is contained in $M_{t_1}\cup \ldots M_{t_{j-1}}$ . Furthermore, every $M_{t_j}$ is a submanifold of codimension $\geq \dim (X)+1$ .

By [Reference Beggs and EvansBE91, Lemma 1.4], if $Y\subseteq M$ is submanifold of codimension $\geq \dim (X)+1$ , if $\delta>0$ , and if $g\in C(X, M)$ satisfies $g(X)\cap (\overline {Y}\setminus Y)=\emptyset $ , then there exists $g'\in C(X,M)$ such that $d(g,g')\leq \delta $ and $g'(X)\cap \overline {Y}=\emptyset $ . Set $f_0:=f$ . We will inductively find $f_k\in C(X,M)$ such that, for each $k=1,\ldots ,K$ , we have

$$\begin{align*}d(f_{k-1},f_k) < \frac{\varepsilon}{2^k}, \quad\text{ and }\quad f_k(X)\cap \overline{M_{t_j}}=\emptyset \text{ for } j=1,\ldots,k. \end{align*}$$

First, using that the boundary of $M_{t_1}$ is empty, we can apply [Reference Beggs and EvansBE91, Lemma 1.4] to obtain $f_1\in C(X,M)$ such that

$$\begin{align*}d(f_0,f_1) < \frac{\varepsilon}{2}, \quad\text{ and }\quad f_1(X)\cap \overline{M_{t_1}}=\emptyset. \end{align*}$$

For $k\geq 2$ , assuming that we have chosen $f_{k-1}$ , let $\delta _k$ denote the (positive) distance between the compact set $f_{k-1}(X)$ and $\overline {M_{t_1}}\cup \cdots \cup \overline {M_{t_{k-1}}}$ . Applying [Reference Beggs and EvansBE91, Lemma 1.4], we obtain $f_k\in C(X,M)$ such that

$$\begin{align*}d(f_{k-1},f_k) < \min\left\{ \frac{\varepsilon}{2^k},\delta_k \right\}, \quad\text{ and }\quad f_k(X)\cap \overline{M_{t_k}}=\emptyset. \end{align*}$$

By choice of $\delta _k$ , it follows that $f_k(X)$ is disjoint from $\overline {M_{t_1}}\cup \cdots \cup \overline {M_{t_k}}$ .

Finally, the element $f_K$ belongs to $C(X,M_{\mathrm {free}})$ and satisfies $d(f,f_K)<\varepsilon $ .

Proof Given $u\in {\mathcal {U}}_d$ , we have $[u].\mathbf {a}=\mathbf {a}$ if and only if $uxu^*=x$ for every $x\in C^*(\mathbf {a})$ . This implies the formula for $\operatorname {\mathrm {Stab}}(\mathbf {a})$ .

If $\mathbf {a}\in G^{n+1}_d$ , then $C^*(\mathbf {a})'=\mathbb {C} 1$ , which implies that $\operatorname {\mathrm {Stab}}(\mathbf {a})$ is trivial. Conversely, assuming that $\mathbf {a}\in E_d^{n+1}\setminus G_d^{n+1}$ , let us verify that $\mathbf {a}$ has nontrivial stabilizer subgroup. Since $C^*(\mathbf {a})\neq M_d$ , we also have $C^*_1(\mathbf {a})\neq M_d$ . Using the bicommutant theorem, we deduce that $C^*_1(\mathbf {a})'$ is strictly larger than the center of $M_d$ . Using that $C^*(\mathbf {a})'=C^*_1(\mathbf {a})'$ , we obtain a noncentral unitary in $C^*(\mathbf {a})'$ .

Proof Let $B_1,B_2\in \operatorname {\mathrm {Sub}}_1(M_d)$ . If $u\in {\mathcal {U}}_d$ satisfies $uB_1u^*=B_2$ , then one checks $uB_1'u^*=B_2'$ . Using also that $B_1$ and $B_2$ agree with their bicommutants, we obtain

$$\begin{align*}B_1\sim B_2 \quad\Leftrightarrow\quad B_1'\sim B_2'. \end{align*}$$

Using that $C^*_1(\mathbf {a}) = C^*(\mathbf {a})"$ and $C^*_1(\mathbf {a})' = C^*(\mathbf {a})'$ , and similarly $C^*_1(\mathbf {b}) = C^*(\mathbf {b})"$ and $C^*_1(\mathbf {b})' = C^*(\mathbf {b})'$ , we need to show

$$ \begin{align*} \operatorname{\mathrm{Stab}}(\mathbf{a}) \sim \operatorname{\mathrm{Stab}}(\mathbf{b}) \quad\Leftrightarrow\quad C^*(\mathbf{a})' \sim C^*(\mathbf{b})'. \end{align*} $$

To prove the forward implication, we assume that $\operatorname {\mathrm {Stab}}(\mathbf {a}) \sim \operatorname {\mathrm {Stab}}(\mathbf {b})$ . Let $v\in {\mathcal {U}}_d$ such that $[v]\operatorname {\mathrm {Stab}}(\mathbf {a})[v]^{-1}= \operatorname {\mathrm {Stab}}(\mathbf {b})$ . Given $u\in {\mathcal {U}}(C^*(\mathbf {a})')$ , it follows from Lemma 4.7 that

$$\begin{align*}[vuv^*] \in \operatorname{\mathrm{Stab}}(\mathbf{b}) = \big\{ [w] : w\in{\mathcal{U}}(C^*(\mathbf{b})') \big\}. \end{align*}$$

Using that $\mathbb {T} 1\subseteq {\mathcal {U}}(C^*(\mathbf {b})')$ , we obtain $vuv^*\in {\mathcal {U}}(C^*(\mathbf {b})')$ . Since $C^*(\mathbf {a})'$ is spanned by its unitary elements, we get $vC^*(\mathbf {a})'v^*\subseteq C^*(\mathbf {b})'$ . The reverse inclusion is shown analogously, whence $vC^*(\mathbf {a})'v^*= C^*(\mathbf {b})'$ , that is, $C^*(\mathbf {a})'\sim C^*(\mathbf {b})'$ .

Conversely, if $C^*(\mathbf {a})' \sim C^*(\mathbf {b})'$ , let $v\in {\mathcal {U}}_d$ such that $uC^*(\mathbf {a})'v^*=C^*(\mathbf {b})'$ . Using Lemma 4.7, we get $[v]\operatorname {\mathrm {Stab}}(\mathbf {a})[v]^{-1}= \operatorname {\mathrm {Stab}}(\mathbf {b})$ , that is, $\operatorname {\mathrm {Stab}}(\mathbf {a}) \sim \operatorname {\mathrm {Stab}}(\mathbf {b})$ .

Proof Set $G:=\{\mathbf {a}\in B^{n+1}_{\mathrm {sa}} : C^*_1(\mathbf {a})=B\}$ .

Denseness: By [Reference Thiel and WinterThi21, Lemma 7.2], we have $\operatorname {\mathrm {gr}}(B)\leq 1\leq n$ . Since B is unital and separable, it follows from Theorem 2.2 that $\operatorname {\mathrm {Gen}}_{n+1}(B)_{\mathrm {sa}}\subseteq B^{n+1}_{\mathrm {sa}}$ is dense. Using that $\operatorname {\mathrm {Gen}}_{n+1}(B)_{\mathrm {sa}}\subseteq G$ , we get that G is also dense in $B^{n+1}_{\mathrm {sa}}$ .

Openness: Let $\mathcal {D}$ denote the family of sub- $C^*\!$ -algebras $D\subseteq B$ such that $D+\mathbb {C} 1_B$ is a proper sub- $C^*\!$ -algebra of B (that is, $C^*_1(D)\neq B$ ). Then,

$$\begin{align*}G = B^{n+1}_{\mathrm{sa}} \setminus \bigcup_{D\in\mathcal{D}} D^{n+1}_{\mathrm{sa}}. \end{align*}$$

Thus, we need to show that $\bigcup _{D\in \mathcal {D}} D^{n+1}_{\mathrm {sa}}$ is a closed subset of $B^{n+1}_{\mathrm {sa}}$ .

We let ${\mathcal {U}}(B)$ denote the unitary group of B. It naturally acts on $\mathcal {D}$ by setting $u.D:=uDu^*$ for $u\in {\mathcal {U}}(B)$ and $D\in \mathcal {D}$ . Since B is finite-dimensional, two sub- $C^*\!$ -algebras $D_1,D_2\subseteq B$ are unitarily equivalent if and only if $D_1\cong D_2$ and the inclusions induce the same maps in ordered $K_0$ -theory. It follows that the action ${\mathcal {U}}(B)\curvearrowright \mathcal {D}$ has only finitely many orbits, and we choose representatives $D_1,\ldots ,D_m\in \mathcal {D}$ . Then, $\mathcal {D}=\bigcup _{j=1}^m \bigcup _{u\in {\mathcal {U}}{B}} uD_ju^*$ .

For each j, since $D_j$ is a closed subset of B, it follows that $(D_j)^{n+1}_{\mathrm {sa}}$ is a closed subset of $B^{n+1}_{\mathrm {sa}}$ . Since B is finite-dimensional, ${\mathcal {U}}(B)$ is compact, and it follows that

$$\begin{align*}\bigcup_{D\in\mathcal{D}} D^{n+1}_{\mathrm{sa}} = \bigcup_{j=1}^m \bigcup_{u\in{\mathcal{U}}(B)} u(D_j)^{n+1}_{\mathrm{sa}} u^* \end{align*}$$

is closed, as desired.

Path-connectedness: We only sketch the argument for the case $B=M_d$ for some $d\,{\geq}\, 2$ . Let $\mathbf {a}\in \operatorname {\mathrm {Gen}}_{n+1}(M_d)_{\mathrm {sa}}$ . Using that the unitary group of $M_d$ is path-connected, and that $a_0$ is unitarily equivalent to a diagonal matrix, we find a path in $\operatorname {\mathrm {Gen}}_{n+1}(M_d)_{\mathrm {sa}}$ from $\mathbf {a}$ to some $\mathbf {b}$ such that $b_0$ is diagonal. By splitting multiple eigenvalues of $b_0$ and moving them away from zero, we find a path $(x_t)_{t\in [0,1]}$ inside the self-adjoint, diagonal matrices starting with $x_0=b_0$ and ending with some $x_1$ such that $x_1$ has k distinct, nonzero diagonal entries, and such that $b_0\in C^*(x_t)$ for each $t\in [0,1]$ . Then, $t\mapsto (x_t,b_1,\ldots ,b_n)$ defines a path inside $\operatorname {\mathrm {Gen}}_{n+1}(M_d)_{\mathrm {sa}}$ .

Let S denote the set of self-adjoint matrices in $M_d$ such that every off-diagonal entry is nonzero. Note that S is path-connected. Next, we let $(y_t)_{t\in [0,1]}$ be a path inside the self-adjoint matrices starting with $y_0=b_1$ , ending with some matrix $y_1$ that has the eigenvalues $1,2,\ldots ,d$ such that $y_t$ belongs to S for every $t\in (0,1]$ . Note that $x_1$ and $y_t$ generated $M_d$ for every $t\in (0,1]$ . It follows that $t\mapsto (x_1,y_t,b_2,\ldots ,b_n)$ defines a path inside $\operatorname {\mathrm {Gen}}_{n+1}(M_d)_{\mathrm {sa}}$ .

Conjugating by a suitable path of unitaries, we find a path in $\operatorname {\mathrm {Gen}}_{n+1}(M_d)_{\mathrm {sa}}$ from $(x_1,y_1,b_2,\ldots ,b_n)$ to some $\mathbf {c}=(c_0,c_1,\ldots ,c_n)$ such that $c_1=\operatorname {\mathrm {diag}}(1,2,\ldots ,d)$ . Arguing as above, we find a path in $\operatorname {\mathrm {Gen}}_{n+1}(M_d)_{\mathrm {sa}}$ that changes $c_0$ to the matrix $\tilde {c_0}$ with all entries $1$ . Then, $\tilde {c_0}$ and $c_1$ generate $M_d$ .

Then, $t\mapsto (\tilde {c_0},c_1,(1-t)c_2,\ldots ,(1-t)c_n)$ is a path in $\operatorname {\mathrm {Gen}}_{n+1}(M_d)_{\mathrm {sa}}$ connecting to $(\tilde {c_0},c_1,0\ldots ,0)$ . Thus, every $\mathbf {a}\in \operatorname {\mathrm {Gen}}_{n+1}(M_d)_{\mathrm {sa}}$ is path-connected to the same element.

Proof Set $B:=B(\mathbf {d},\mathbf {m})$ . Note that a tuple $\mathbf {a}\in E^{n+1}_d$ belongs to $E_{[\mathbf {d},\mathbf {m}]}$ if and only if $C^*_1(\mathbf {a})\sim B$ . Set

$$\begin{align*}F:= \big\{ \mathbf{a}\in E^{n+1}_d : C^*_1(\mathbf{a})=B \big\}. \end{align*}$$

By Lemma 4.9, F is connected. Since every orbit in $E_{[\mathbf {d},\mathbf {m}]}$ meets F, and since $\mathcal {PU}_d$ is connected, it follows that $E_{[\mathbf {d},\mathbf {m}]}$ is connected as well.

By [Reference BredonBre72, Theorem IV.3.8], if a compact Lie group L acts smoothly on a connected manifold M such that all orbits have the same type, then $\dim (M)=\dim (M/L)+\dim (L/K)$ , where K is the stabilizer subgroup of any element in M.

Let $K\subseteq \mathcal {PU}_d$ be the stabilizer subgroup of some element in F. By considering the restricted action $\mathcal {PU}_d\curvearrowright E_{[\mathbf {d},\mathbf {m}]}$ , we obtain that

$$\begin{align*}\dim( E_{[\mathbf{d},\mathbf{m}]} ) = \dim( E_{[\mathbf{d},\mathbf{m}]}/\mathcal{PU}_d ) + \dim( \mathcal{PU}_d/ K ). \end{align*}$$

Closed subgroups of Lie groups are again Lie groups. It follows that K is a Lie group as well. Since K is acting freely on the connected manifold $\mathcal {PU}_d$ with only one orbit type, we also get $\dim ( \mathcal {PU}_d/ K ) = \dim ( \mathcal {PU}_d ) - \dim (K)$ and thus

(4.1) $$ \begin{align} \dim( E_{[\mathbf{d},\mathbf{m}]} ) = \dim( E_{[\mathbf{d},\mathbf{m}]}/\mathcal{PU}_d ) + \dim( \mathcal{PU}_d ) - \dim(K). \end{align} $$

Set

$$\begin{align*}N:= \big\{ [u]\in\mathcal{PU}_d : uBu^*=B \big\}, \end{align*}$$

which is a closed subgroup of $\mathcal {PU}_d$ . Given $\mathbf {a}\in F$ and $[u]\in \mathcal {PU}_d$ , we have $[u].\mathbf {a}\in F$ if and only if $[u]\in N$ . It follows that N naturally acts on F. Furthermore, for each $\mathbf {a}\in F$ , the N-orbit $N.\mathbf {a}$ agrees with $\mathcal {PU}_d.\mathbf {a}\cap F$ . Since every $\mathcal {PU}_d$ -orbit in $E_{[\mathbf {d},\mathbf {m}]}$ meets F, we deduce that $E_{[\mathbf {d},\mathbf {m}]} / \mathcal {PU}_d \cong F / N$ . Note that $B^{n+1}_{\mathrm {sa}}$ is a linear space. By Lemma 4.9, F is an open subset of $B^{n+1}_{\mathrm {sa}}$ . It follows that F is a manifold satisfying

$$\begin{align*}\dim(F) = \dim(B^{n+1}_{\mathrm{sa}}) = (n+1)\sum_{j=1}^L d_j^2. \end{align*}$$

Analogous to (4.1), by considering the action of the compact Lie group N on F, we obtain

(4.2) $$ \begin{align} \dim( F ) = \dim( F/N ) + \dim( N ) - \dim(K). \end{align} $$

Note that N contains $\{[u]:u\in {\mathcal {U}}(B)\}$ , which implies that

$$\begin{align*}\dim(N)\geq \left(\sum_{j=1}^L d_j^2 \right) - 1. \end{align*}$$

Combining this estimate with (4.1) and (4.2), using that $E_{[\mathbf {d},\mathbf {m}]} / \mathcal {PU}_d \cong F / N$ , and that $[\mathbf {d},\mathbf {m}]\neq [(d),(1)]$ , we get

$$ \begin{align*} \dim( E_{[\mathbf{d},\mathbf{m}]} ) &= \dim( F ) + \dim( \mathcal{PU}_d ) - \dim( N ) \\ &\leq \left( (n+1)\sum_{j=1}^L d_j^2 \right) + \left( d^2 - 1 \right) - \left( \left(\sum_{j=1}^L d_j^2\right) -1 \right) \\ &= d^2 + n\sum_{j=1}^L d_j^2 \\ &\leq d^2 + n((d-1)^2+1) = (n+1)d^2 -2n(d-1). \end{align*} $$

For $[\mathbf {d},\mathbf {m}]=[(d-1,1),(1,1)]$ , we have $B(\mathbf {d},\mathbf {m})\cong M_{d-1}\oplus \mathbb {C}\subseteq M_d$ . In this case, we get $N=\{[u]\in \mathcal {PU}_d: u\in {\mathcal {U}}(M_{d-1}\oplus \mathbb {C})\}$ and thus $\dim (N)=(d-1)^2+1-1= (d-1)^2$ . It follows that

$$ \begin{align*} \dim( E_{[(d-1,1),(1,1)]} ) &= \dim( F ) + \dim( \mathcal{PU}_d ) - \dim( N ) \\ &= (n+1)((d-1)^2+1) + (d^2-1) - (d-1)^2 \\ &= (n+1)d^2 -2n(d-1). \\[-34pt] \end{align*} $$

Proof We use the notation from Paragraph 4.10. The orbit-type decomposition for the action $\mathcal {PU}_d\curvearrowright E^{n+1}_d$ is

$$\begin{align*}E^{n+1}_d = \bigcup_{[\mathbf{d},\mathbf{m}] \in T} E_{[\mathbf{d},\mathbf{m}]}, \quad E_{[\mathbf{d},\mathbf{m}]} := \big\{ \mathbf{a}\in E^{n+1}_d : C^*_1(\mathbf{a})\sim B(\mathbf{d},\mathbf{m}) \big\}. \end{align*}$$

Furthermore, $G^{n+1}_d = E_{[(d),(1)]}$ , which is the submanifold of orbits with trivial stabilizers. In the notation of Paragraph 4.4, with $M=E^{n+1}_d$ , we have $M_{\mathrm {free}}=G^{n+1}_d$ .

Applying Proposition 4.5, we obtain that $C(X,G^{n+1}_d)\subseteq C(X,E^{n+1}_d)$ is dense if and only if

$$\begin{align*}\dim(X) < \dim(E^{n+1}_d) - \dim(E^{n+1}_d\setminus G^{n+1}_d). \end{align*}$$

Since $E^{n+1}_d\setminus G^{n+1}_d$ is the finite union of $E_{[\mathbf {d},\mathbf {m}]}$ for $[\mathbf {d},\mathbf {m}]\neq [(d),(1)]$ , we obtain from Lemma 4.11

$$\begin{align*}\dim(E^{n+1}_d\setminus G^{n+1}_d) = \max_{[\mathbf{d},\mathbf{m}]\neq[(d),(1)]} \dim( E_{[\mathbf{d},\mathbf{m}]} ) = (n+1)d^2 -2n(d-1). \end{align*}$$

Now, the result follows using that $\dim (E^{n+1}_d)=(n+1)d^2$ .

Proof Set $X:=\operatorname {\mathrm {Prim}}(A)$ . Since X is compact, and since the $M_d$ -bundle associated with A is locally trivial, we can choose closed subsets $X_1,\ldots ,X_m\subseteq X$ that cover X such that $A(X_j)\cong C(X_j,M_d)$ for each j. Let $\pi _j\colon A\to C(X_j,M_d)$ be the corresponding quotient map, which induces a natural map $A^{n+1}_{\mathrm {sa}}\to C(X_j,M_d)^{n+1}_{\mathrm {sa}} \cong C(X_j,E^{n+1}_d)$ that we also denote by $\pi _j$ .

A tuple $\mathbf {a}\in A^{n+1}_{\mathrm {sa}}$ belongs to $\operatorname {\mathrm {Gen}}^{\mathrm {fiber}}_{n+1}(A)_{\mathrm {sa}}$ if and only if $\pi _j(\mathbf {a})$ belongs to $C(X_j,G^{n+1}_d)$ for each j. It follows from Lemma 4.9 that $G^{n+1}_d\subseteq E^{n+1}_d$ is open. Since X is compact, we obtain that $C(X_j,G^{n+1}_d)\subseteq C(X_j,E^{n+1}_d)$ is always open. Hence, $\operatorname {\mathrm {Gen}}^{\mathrm {fiber}}_{n+1}(A)_{\mathrm {sa}}\subseteq A^{n+1}_{\mathrm {sa}}$ is open.

Since the intersection of finitely many open dense sets is again dense, we see that (1) holds if and only if $C(X_j,G^{n+1}_d)\subseteq C(X_j,E^{n+1}_d)$ is dense for each j. By Lemma 4.12, this is in turn equivalent to $\dim (X_j)<2n(d-1)$ for each j. Using that $\dim (X)=\max _j\dim (X_j)$ , this is finally equivalent to (2).

Proof Using that the action $\mathcal {PU}_d\curvearrowright G^{n+1}_d$ is free, it follows that the quotient map $q\colon G^{n+1}_d\to G^{n+1}_d/\mathcal {PU}_d$ is the projection of a fiber bundle with base space $G^{n+1}_d/\mathcal {PU}_d$ and with fibers homeomorphic to $\mathcal {PU}_d$ . Using the homotopy lifting property for fiber bundles, we obtain $H\colon X\times [0,1]\to G^{n+1}_d$ such that

$$\begin{align*}q\circ H = F, \quad\text{ and }\quad H(x,0) = \tilde{F}(x,0),\quad \text{for } x\in X. \end{align*}$$

Next, we will correct H to agree with $\widetilde {F}$ on $Y\times [0,t]$ for some $t>0$ .

Given $(y,s)\in Y\times [0,1]$ , we have

$$\begin{align*}q(H(y,s)) = F(y,s) = q(\widetilde{F}(y,s)). \end{align*}$$

Let $c(y,s)\in \mathcal {PU}_d$ be the unique element such that $H(y,s)=c(y,s).\widetilde {F}(y,s)$ . This defines a map $c\colon Y\times [0,1]\to \mathcal {PU}_d$ . Using that the fiber bundle is locally trivial, we see that c is continuous. For every $y\in Y$ , we have $H(y,0)=\widetilde {F}(y,0)$ and therefore $c(y,0)=1$ . We extend c to a map $c\colon (Y\times [0,1])\cup (X\times \{0\})\to \mathcal {PU}_d$ by setting $c(x,0):=1$ for every $x\in X$ .

Every Lie group is a (metrizable) locally contractible, finite-dimensional space and therefore an absolute neighborhood extensor (see Theorems 1.2.7 and 4.2.33 in [Reference NagisavM01]). This allows us to extend c to a continuous map $\tilde {c}\colon U\to \mathcal {PU}_d$ defined on a neighborhood U of $(Y\times [0,1])\cup (X\times \{0\})\subseteq X\times [0,1]$ . Then, define $\widetilde {H}\colon U\to G^{n+1}_d$ by

$$\begin{align*}\widetilde{H}(x,s) := \tilde{c}(x,s).\widetilde{F}(x,s),\quad \text{for } (x,s)\in U\subseteq X\times[0,1]. \end{align*}$$

Choose $t>0$ such that $(Y\times [0,1])\cup (X\times [0,t])\subseteq U$ . Then, the restriction of $\widetilde {H}$ to $(Y\times [0,1])\cup (X\times [0,t])$ has the desired properties.

Proof Set $X:=\operatorname {\mathrm {Prim}}(A)$ . Since A is noncommutative, we have $\operatorname {\mathrm {gr}}(A)\geq 1$ by [Reference Thiel and WinterThi21, Proposition 5.7]. We also have $\left \lceil \frac {\dim (X)+1}{2d-2}\right \rceil \geq 1$ for every value of $\dim (X)$ . Thus, it is enough to show that, for every $n\geq 1$ , the following holds:

$$\begin{align*}\operatorname{\mathrm{gr}}(A)\leq n \quad\Leftrightarrow\quad \dim(X)<2n(d-1). \end{align*}$$

Recall that we use $E(X,G^{n+1}_d/\mathcal {PU}_d)$ to denote the set of injective continuous maps $X\to G^{n+1}_d/\mathcal {PU}_d$ . As explained in Paragraph 4.3, we have the following inclusions and maps:

Assume that $\operatorname {\mathrm {gr}}(A)\leq n$ . Since A is separable, it follows from Theorem 2.2 that $\operatorname {\mathrm {Gen}}_{n+1}(A)_{\mathrm {sa}}\subseteq A^{n+1}_{\mathrm {sa}}$ is dense. Since $\operatorname {\mathrm {Gen}}_{n+1}(A)_{\mathrm {sa}}\subseteq \operatorname {\mathrm {Gen}}^{\mathrm {fiber}}_{n+1}(A)_{\mathrm {sa}}$ , we deduce from Lemma 4.13 that $\dim (X)<2n(d-1)$ .

Conversely, assume that $\dim (X)<2n(d-1)$ . Applying Lemma 4.13, we see that $\operatorname {\mathrm {Gen}}^{\mathrm {fiber}}_{n+1}(A)_{\mathrm {sa}}\subseteq A^{n+1}_{\mathrm {sa}}$ is dense and open. Furthermore, by Proposition 4.1, a tuple $\mathbf {a}\in \operatorname {\mathrm {Gen}}^{\mathrm {fiber}}_{n+1}(A)_{\mathrm {sa}}$ belongs to $\operatorname {\mathrm {Gen}}_{n+1}(A)_{\mathrm {sa}}$ if and only if $\Psi (\mathbf {a})$ belongs to $E(X,G^{n+1}_d/\mathcal {PU}_d)$ . Thus, to verify $\operatorname {\mathrm {gr}}(A)\leq n$ , it suffices to show the following.

Let $\mathbf {a}\in \operatorname {\mathrm {Gen}}^{\mathrm {fiber}}_{n+1}(A)_{\mathrm {sa}}$ and $\varepsilon>0$ . Then, there exists $\mathbf {b}\in \operatorname {\mathrm {Gen}}^{\mathrm {fiber}}_{n+1}(A)_{\mathrm {sa}}$ such that

$$\begin{align*}\mathbf{b}=_\varepsilon\mathbf{a}, \quad and \quad \Psi(\mathbf{b})\in E(X,G^{n+1}_d/\mathcal{PU}_d). \end{align*}$$

By [Reference LuukkainenLuu81, Theorem 5.1], if M is a metrizable manifold with $2\dim (X)<\dim (M)$ , then $E(X,M)\subseteq C(X,M)$ is dense with respect to the metric $d(f,g)=\sup \{d_M(f(x),g(x)):x\in X\}$ , where $d_M$ is a metric inducing the topology on M.

By Lemma 4.9, $G^{n+1}_d$ is an open subset of $E^{n+1}_d$ and therefore is a manifold of dimension $(n+1)d^2$ . Furthermore, $\mathcal {PU}_d$ is a compact Lie group of dimension $d^2-1$ , acting freely on $G^{n+1}_d$ . Hence, as noted in the proof of Lemma 4.11, it follows from [Reference BredonBre72, Theorem IV.3.8] that $G^{n+1}_d/\mathcal {PU}_d$ is a manifold of dimension $(n+1)d^2- (d^2-1)=nd^2+1$ . By assumption, we have $\dim (X)<2n(d-1)$ , and thus

$$\begin{align*}2\dim(X) < 4n(d-1) \leq nd^2+1. \end{align*}$$

It follows that $E(X,G^{n+1}_d/\mathcal {PU}_d)$ is dense in $C(X,G^{n+1}_d/\mathcal {PU}_d)$ .

Set $f:=\Psi (\mathbf {a})$ . Then, $f\colon X\to G^{n+1}_d/\mathcal {PU}_d$ is a continuous map, which can be approximated arbitrarily closely by embeddings. To complete the proof, we need to show that one of these embeddings is realized as $\Psi (\mathbf {b})$ for some $\mathbf {b}\in A^{n+1}_{\mathrm {sa}}$ close to $\mathbf {a}$ . We will do this by successively applying our version of the homotopy extension lifting property proved in Lemma 4.14.

Every manifold is finite-dimensional and locally contractible and therefore an absolute neighborhood retract (ANR) (see [Reference NagisavM01, Theorem 4.2.33]). Given a homotopy $H\colon X\times [0,1]\to M$ and $t\in [0,1]$ , we let $H_t\colon X\to M$ be given by $H_t(x):=H(x,t)$ .

Step 1: We find a homotopy $F\colon X\times [0,1]\to G^{n+1}_d/\mathcal {PU}_d$ such that $F_0=f$ and such that $F_{1/k}$ belongs to $E(X,G^{n+1}_d/\mathcal {PU}_d)$ for every $k\geq 1$ .

Set $M:=G^{n+1}_d/\mathcal {PU}_d$ . We use that M is an ANR. Given $\delta>0$ , one says that $H\colon X\times [0,1]\to M$ is a $\delta $ -homotopy if $d(H_0,H_t)<\delta $ for all $t\in [0,1]$ . By [Reference NagisavM01, Theorem 4.1.1], for every $\delta>0$ , there exists $\gamma>0$ such that, for every $g\in C(X,M)$ satisfying $d(f,g)<\gamma $ , there exists a $\delta $ -homotopy $H\colon X\times [0,1]\to M$ with $H_0=f$ and $H_1=g$ . Given $n\in \mathbb {N}$ , we apply this for $\delta _n=\tfrac {1}{2^n}$ , to obtain $\gamma _n>0$ . Using that $E(X,M)\subseteq C(X,M)$ is dense, choose $g_n\in E(X,M)$ satisfying $d(f,g_n)<\gamma _n$ . By choice of $\gamma _n$ , we obtain a $\tfrac {1}{2^n}$ -homotopy $H^{(n)}\colon X\times [0,1]\to M$ satisfying $H^{(n)}_0=f$ and $H^{(n)}_1=g_n$ .

Next, we define $H\colon X\times [0,\infty )\to M$ by

$$\begin{align*}H(x,t) = \begin{cases} H^{(k)}(x,2k+1-t), &\text{ if } t\in[2k,2k+1], \\ H^{(k+1)}(x,t-2k-1), &\text{ if } t\in[2k+1,2k+2]. \end{cases} \end{align*}$$

Thus, H is the concatenation of the reverse of $H^{(1)}$ , followed by $H^{(2)}$ and its reverse, and so on, as shown in the following picture:

Note that $H(\_\,,2k)=g_k$ for each $k\in \mathbb {N}$ , and $\lim _{t\to \infty }H(x,t)=f(x)$ for every $x\in X$ . Let $\varrho \colon (0,1]\to [0,\infty )$ be a strictly decreasing, continuous map satisfying $\varrho (\tfrac {1}{k})=2k-2$ for $k\geq 1$ . Then, $F\colon X\times [0,1]$ defined by $F(x,0)=f(x)$ and $F(x,t)=H(x,\varrho (t))$ for $t\in (0,1]$ has the desired properties.

Step 2: Since X is compact, and since the $M_d$ -bundle associated with A is locally trivial, we can choose closed subsets $X_1,\ldots ,X_m\subseteq X$ that cover X such that $A(X_j)\cong C(X_j,M_d)$ for each j. Let $\pi _j\colon A\to C(X_j,M_d)$ be the corresponding quotient map. Abusing notation, we also use $\pi _j$ to denote the naturally induced map

$$\begin{align*}\pi_j\colon\ A^{n+1}_{\mathrm{sa}}\to C(X_j,M_d)^{n+1}_{\mathrm{sa}} \cong C(X_j,E^{n+1}_d). \end{align*}$$

Given $j,k\in \{1,\ldots ,m\}$ , both $\pi _j$ and $\pi _k$ induce an isomorphism between $A(X_j\cap X_k)$ and $C(X_j\cap X_k,M_d)$ . Let $c_{k,j}\colon X_j\cap X_k\to \mathcal {PU}_d=\operatorname {\mathrm {Aut}}(M_d)$ be the continuous map such that $c_{k,j}(x).\pi _j(e)(x)=\pi _k(e)(x)$ for every $e\in A$ and $x\in X_j\cap X_k$ . Then,

(4.3) $$ \begin{align} c_{k,j}(x).\pi_j(\mathbf{e})(x)=\pi_k(\mathbf{e})(x) \end{align} $$

for every $\mathbf {e}\in A^{n+1}_{\mathrm {sa}}$ and $x\in X_j\cap X_k$ .

Step 3: We will successively choose $t_1\geq t_2\geq \cdots \geq t_m>0$ and continuous maps

$$\begin{align*}H^{(k)}\colon X_k\times [0,t_k] \to G^{n+1}_d \end{align*}$$

such that

(4.4) $$ \begin{align} H^{(k)}(\_\,,0)=\pi_k(\mathbf{a}), \quad\text{ and }\quad q\circ H^{(k)} = F|_{X_k\times[0,t_k]}, \end{align} $$

and such that, for every $j\leq k$ and $(x,s)\in (X_j\cap X_k)\times [0,t_k]$ , we have

(4.5) $$ \begin{align} c_{k,j}(x).H^{(j)}(x,s)=H^{(k)}(x,s). \end{align} $$

We start by setting $t_1:=1$ . The map $\pi _1(\mathbf {a})\colon X_1\to G^{n+1}_d$ satisfies $q\circ \pi _1(\mathbf {a})=f|_{X_1}$ . Thus, $\pi _1(\mathbf {a})$ is a lift of $F|_{X_1\times \{0\}}$ . Using the homotopy lifting property for fiber bundles, we obtain $H^{(1)}\colon X_1\times [0,1]\to G^{n+1}_d$ such that

$$\begin{align*}H^{(1)}_0 = \pi_1(\mathbf{a}), \quad\text{ and }\quad q\circ H^{(1)} = F|_{X_1\times[0,1]}. \end{align*}$$

Next, assume that we have chosen $t_1\geq \cdots \geq t_{k-1}$ and $H^{(j)}$ for $j=1,\ldots ,k-1$ . Set $Y_k:= X_k\cap (X_1\cup \cdots \cup X_{k-1})$ , which is a closed subset of $X_k$ . We define $\widetilde {F}^{(k)}\colon (Y_{k}\times [0,t_{k-1}])\cup (X_{k}\times \{0\})\to G^{n+1}_d$ by

$$\begin{align*}\widetilde{F}^{(k)}(x,t) := \begin{cases} c_{k,j}(x).H^{(j)}(x,t), & \text{ if } x\in X_{k}\cap X_j, \text{for } j\leq k-1, \\ \pi_{k}(\mathbf{a})(x), & \text{ if } t=0. \end{cases} \end{align*}$$

It follows from (4.5) that $\widetilde {F}^{(k)}$ is well defined. Furthermore, using (4.4), we obtain that $q\circ \widetilde {F}^{(k)}$ and F agree on $(Y_{k}\times [0,t_{k-1}])\cup (X_k\times \{0\})$ . Applying Lemma 4.14, we obtain $t_k\in (0,t_{k-1}]$ and $H^{(k)}$ making the following diagram commute:

One checks that $H^{(k)}$ has the desired properties.

Step 4: Let $t\in [0,t_m]$ . For each $j\in \{1,\ldots ,m\}$ , the map $H^{(j)}_t\colon X_j\to G^{n+1}_d$ defines an element in $\mathbf {b}_t^{(j)}\in C(X_j,M_d)^{n+1}_{\mathrm {sa}}$ . Given $j\leq k$ in $\{1,\ldots ,m\}$ and $x\in X_j\cap X_k$ , it follows from (4.5) that

$$\begin{align*}c_{k,j}(x).\mathbf{b}_t^{(j)}(x) = \mathbf{b}_t^{(k)}(x). \end{align*}$$

Thus, $\mathbf {b}_t^{(1)},\ldots ,\mathbf {b}_t^{(m)}$ can be patched to give $\mathbf {b}_t\in A^{n+1}_{\mathrm {sa}}$ such that $\mathbf {b}_t^{(j)}=\pi _j(\mathbf {b}_t)$ for each j. One checks that each $\mathbf {b}_t^{(j)}$ depends continuously on t, which implies that the map $[0,t_m]\to A^{n+1}_{\mathrm {sa}}$ , $t\mapsto \mathbf {b}_t$ , is continuous. By construction, we have $\mathbf {b}_0=\mathbf {a}$ , and $\Psi (\mathbf {b}_t)=F_t$ for each $t\in [0,t_m]$ . Using that $\mathbf {a}$ belongs to $\operatorname {\mathrm {Gen}}^{\mathrm {fiber}}_{n+1}(A)_{\mathrm {sa}}$ , which is an open subset of $A^{n+1}_{\mathrm {sa}}$ , we can choose $k\geq 1$ such that

$$\begin{align*}\mathbf{a}=_\varepsilon\mathbf{b}_{1/k} \in \operatorname{\mathrm{Gen}}^{\mathrm{fiber}}_{n+1}(A)_{\mathrm{sa}}. \end{align*}$$

We have $\Psi (\mathbf {b}_{1/k})=F_{1/k}$ , which by construction of F (Step 1) belongs to $E(X,G^{n+1}_d/\mathcal {PU}_d)$ . It follows that $\mathbf {b}_{1/k}\in \operatorname {\mathrm {Gen}}_{n+1}(A)_{\mathrm {sa}}$ .

Proof Set $n:=\operatorname {\mathrm {gr}}(A)$ and $l:=\dim (\operatorname {\mathrm {Prim}}(A))$ , and then set

$$ \begin{align*} \mathcal{S}_1 &:= \big\{ B\in\operatorname{\mathrm{Sub}}_{\mathrm{sep}}(A) : 1\in B, \operatorname{\mathrm{gr}}(B)\leq n \big\}, \quad\text{ and }\quad \\ \mathcal{S}_2 &:= \big\{ B\in\operatorname{\mathrm{Sub}}_{\mathrm{sep}}(A) : B\ d\text{-homogeneous}, \operatorname{\mathrm{locdim}}(\operatorname{\mathrm{Prim}}(B))\leq l \big\}. \end{align*} $$

As noted in Paragraph 3.1, since $\operatorname {\mathrm {gr}}$ satisfies (D5) and (D6), it follows that $\mathcal {S}_1$ is $\sigma $ -complete and cofinal. By Proposition 3.5, $\mathcal {S}_2$ is $\sigma $ -complete and cofinal. Hence, $\mathcal {S}_1\cap \mathcal {S}_2$ is $\sigma $ -complete and cofinal as well.

Let $B\in \mathcal {S}_1\cap \mathcal {S}_2$ . Then, B is a unital, separable, d-homogeneous $C^*\!$ -algebra. Hence, $\dim (\operatorname {\mathrm {Prim}}(B))=\operatorname {\mathrm {locdim}}(\operatorname {\mathrm {Prim}}(B))$ , and by Lemma 4.15, we have

$$\begin{align*}\operatorname{\mathrm{gr}}(B) = \left\lceil\frac{\dim(\operatorname{\mathrm{Prim}}(B))+1}{2d-2}\right\rceil. \end{align*}$$

Thus, each $B\in \mathcal {S}_1\cap \mathcal {S}_2$ satisfies $\operatorname {\mathrm {gr}}(B)\leq \left \lceil \tfrac {l+1}{2d-2}\right \rceil $ . Since A is approximated by the family $\mathcal {S}_1\cap \mathcal {S}_2$ , we obtain $\operatorname {\mathrm {gr}}(A)\leq \left \lceil \tfrac {l+1}{2d-2}\right \rceil $ by Theorem 2.6.

To show the converse inequality, set

$$\begin{align*}m := \max \left\{ m_0\in\mathbb{N} : n \geq \left\lceil\frac{m_0+1}{2d-2}\right\rceil \right\}. \end{align*}$$

Then, each $B\in \mathcal {S}_1\cap \mathcal {S}_2$ satisfies $\operatorname {\mathrm {topdim}}(B)=\dim (\operatorname {\mathrm {Prim}}(B))\leq m$ . Arguing with the topological dimension as in the proof of Lemma 3.4, we deduce that $\dim (\operatorname {\mathrm {Prim}}(A))=\operatorname {\mathrm {topdim}}(A)\leq m$ , and thus $n\geq \left \lceil \frac {\dim (\operatorname {\mathrm {Prim}}(A))+1}{2d-2}\right \rceil $ , as desired.

Proof For $d=1$ , this follows from Theorem 2.7. So assume that $d\geq 2$ . By Proposition 2.4, we have $\operatorname {\mathrm {gr}}_0(A) \leq \operatorname {\mathrm {gr}}(A)$ . Let $K\subseteq X$ be a compact subset. The corresponding quotient $A(K)$ is a unital d-homogeneous $C^*\!$ -algebra with $\operatorname {\mathrm {Prim}}(A(K))\cong K$ . Using Proposition 4.16 at the first step, and using Theorem 2.5 at the last step, we get

$$\begin{align*}\left\lceil\frac{\dim(K)+1}{2d-2}\right\rceil = \operatorname{\mathrm{gr}}(A(K)) = \operatorname{\mathrm{gr}}_0(A(K)) \leq \operatorname{\mathrm{gr}}_0(A). \end{align*}$$

Since this holds for every compact subset of X, we deduce that

$$\begin{align*}\left\lceil\frac{\operatorname{\mathrm{locdim}}(X)+1}{2d-2}\right\rceil \leq \operatorname{\mathrm{gr}}_0(A) \leq \operatorname{\mathrm{gr}}(A). \end{align*}$$

To verify that $\operatorname {\mathrm {gr}}(A)\leq \left \lceil \tfrac {\operatorname {\mathrm {locdim}}(X)+1}{2d-2}\right \rceil $ , set $l:=\operatorname {\mathrm {locdim}}(X)$ , which we may assume to be finite. By Proposition 3.5, the collection

$$\begin{align*}\mathcal{S} := \big\{ B\in\operatorname{\mathrm{Sub}}_{\mathrm{sep}}(A) : B\ d\text{-homogeneous}, \operatorname{\mathrm{locdim}}(\operatorname{\mathrm{Prim}}(B))\leq l \big\} \end{align*}$$

is $\sigma $ -complete and cofinal. Let $B\in \mathcal {S}$ . We view B as a locally trivial $M_d$ -bundle over $Y:=\operatorname {\mathrm {Prim}}(B)$ . Since B is separable, Y is $\sigma $ -compact and thus $\dim (Y)=\operatorname {\mathrm {locdim}}(Y)\leq l<\infty $ . By [Reference ThielPhi07, Lemma 2.5], the $M_d$ -bundle has finite type. Applying [Reference ThielPhi07, Proposition 2.9], we obtain a locally trivial $M_d$ -bundle over the Stone–Čech-compactification $\beta Y$ extending the bundle associated with B. This means that there is a unital d-homogeneous $C^*\!$ -algebra D with $\operatorname {\mathrm {Prim}}(D)\cong \beta (Y)$ such that B is an ideal in D. Since Y is a normal space, we have $\dim (\beta Y)=\dim (Y)$ by [Reference PhillipsPea75, Proposition 6.4.3, p. 232]. Using Theorem 2.5 at the first step, and Proposition 4.16 at the second step, we get

$$\begin{align*}\operatorname{\mathrm{gr}}(B) \leq\operatorname{\mathrm{gr}}(D) = \left\lceil\frac{\dim(\beta Y)+1}{2d-2}\right\rceil \leq \left\lceil\frac{l+1}{2d-2}\right\rceil. \end{align*}$$

Since A is approximated by $\mathcal {S}$ , we obtain $\operatorname {\mathrm {gr}}(A)\leq \left \lceil \tfrac {l+1}{2d-2}\right \rceil $ by Theorem 2.6.

Proof For $d=1$ , this follows from [Reference Thiel and WinterThi21, Proposition 5.9]. So assume that $d\geq 2$ . Set $X:=\operatorname {\mathrm {Prim}}(A)$ , and $Y:=\operatorname {\mathrm {Prim}}(B)$ . Then, $A\oplus B$ is d-homogeneous with $\operatorname {\mathrm {Prim}}(A\oplus B)\cong X\sqcup Y$ , the disjoint union of X and Y. Applying Theorem 4.17 at the first and last steps, we obtain

$$ \begin{align*} \operatorname{\mathrm{gr}}(A\oplus B) &= \left\lceil\frac{\operatorname{\mathrm{locdim}}(X\sqcup Y)+1}{2d-2}\right\rceil = \left\lceil\frac{\max\{\operatorname{\mathrm{locdim}}(X),\operatorname{\mathrm{locdim}}(Y)\}+1}{2d-2}\right\rceil \\ &= \max\left\{ \left\lceil\frac{\operatorname{\mathrm{locdim}}(X)+1}{2d-2}\right\rceil, \left\lceil\frac{\operatorname{\mathrm{locdim}}(Y)+1}{2d-2}\right\rceil \right\} \\ &=\max\big\{ \operatorname{\mathrm{gr}}(A),\operatorname{\mathrm{gr}}(B) \big\}.\\[-32pt] \end{align*} $$

Proof For each $k\in \mathbb {N}$ , let $z_k$ denote the support projection of $I_k$ in $A^{**}$ , and let $\pi _k\colon A\to A/I_k$ denote the quotient map.

Claim: Let $\varphi \in A^*_+$ and $k\in \mathbb {N}$ . Then, $\|\varphi |_{I_k}\|=\varphi ^{**}(z_k)$ . To prove the claim, let $(h_\alpha )_\alpha $ denote an increasing, positive, contractive approximate unit of $I_k$ . Since $\varphi |_{I_k}$ is a positive functional on $I_k$ , we have $\|\varphi |_{I_k}\| = \lim _\alpha \varphi (h_\alpha )$ by [Reference BlackadarBla06, Proposition II.6.2.5]. Using also that $z_k$ is the weak*-limit of $(h_\alpha )_\alpha $ in $A^{**}$ , we get

$$\begin{align*}\|\varphi|_{I_k}\| = \lim_\alpha \varphi(h_\alpha) = \varphi^{**}(z_k), \end{align*}$$

which proves the claim.

Let $S(A)$ denote the set of states on A, which is a compact, convex subset of $A^*$ , and let $P(A)$ denote the pure states on A, which agrees with the set of extreme points in $S(A)$ . Given $a\in (A^{**})_{\mathrm {sa}}$ , we let $\widehat {a}\colon S(A)\to \mathbb {R}$ be given by

$$\begin{align*}\widehat{a}(\varphi) = \varphi^{**}(a), \end{align*}$$

for $\varphi \in S(A)$ . Then, $\widehat {a}$ is affine. If $a\in A_{\mathrm {sa}}$ , then $\widehat {a}$ is continuous. Given $k\in \mathbb {N}$ , let $(h_\alpha )_\alpha $ be an increasing approximate unit of $I_k$ . Then, $\widehat {z_k}$ is the pointwise supremum of the increasing net $(\widehat {h_\alpha })_\alpha $ of continuous functions, and therefore lower-semicontinuous.

To show that (1) implies (2), assume that $\bigcup _k \operatorname {\mathrm {hull}}(I_k)=\operatorname {\mathrm {Prim}}(A)$ . Let $\varphi \in P(A)$ . Since every pure state on A factors through an irreducible representation, there exists k such that $\varphi $ factors through $\pi _k$ . Let $\bar {\varphi }\in (A/I_k)^*$ such that $\varphi =\bar {\varphi }\circ \pi _k$ . We have $\pi _k^{**}(z_k)=0$ , and therefore

$$\begin{align*}\widehat{z_k}(\varphi) = \varphi^{**}(z_k) = \bar{\varphi}^{**}(\pi_k^{**}(z_k)) = 0. \end{align*}$$

Thus, $(\widehat {z_k})_k$ is a decreasing sequence of lower semicontinuous, affine functions with $\lim _{k\to \infty }\widehat {z_k}(\varphi )=0$ for each $\varphi \in P(A)$ . By [Reference AlfsenAlf71, Proposition 1.4.10, p. 36], we have $\lim _{k\to \infty }\widehat {z_k}(\varphi )=0$ for each $\varphi \in S(A)$ . Applying the claim, it follows that $\lim _{k\to \infty }\|\varphi |_{I_k}\|=0$ for every $\varphi \in S(A)$ . Now, (2) follows using that every functional in $A^*$ is a linear combination of four states, by [Reference BlackadarBla06, Theorem II.6.3.4, p. 106].

To show that (2) implies (1), assume that $\bigcup _k \operatorname {\mathrm {hull}}(I_k)\neq \operatorname {\mathrm {Prim}}(A)$ . We will show that (2) does not hold. Let $J\subseteq A$ be a primitive ideal with $J\notin \bigcup _k \operatorname {\mathrm {hull}}(I_k)$ , and let $\bar {\varphi }$ be a pure state on $A/J$ . Let $\pi \colon A\to A/J$ denote the quotient map. Set $\varphi :=\bar {\varphi }\circ \pi $ , which is a pure state on A. Let $k\in \mathbb {N}$ . In general, the restriction of a pure state to an ideal is either zero or again a pure state. Since $J\notin \operatorname {\mathrm {hull}}(I_k)$ , we have $\varphi |_{I_k}\neq 0$ , and thus $\|\varphi |_{I_k}\|=1$ . Thus, $\lim _{k\to \infty }\|\varphi |_{I_k}\|=1 \neq 0$ .

Proof We first reduce to the unital case. So assume that A is nonunital, let $\widetilde {A}$ denote its minimal unitization, and let $\bar {B}$ denote the sub- $C^*\!$ -algebra of $\widetilde {A}$ generated by B and the unit of $\widetilde {A}$ . For each $k\in \mathbb {N}$ , we consider $I_k$ as an ideal in $\widetilde {A}$ . Let $\pi _k\colon A\to A/I_k$ and $\pi _k^+\colon \widetilde {A}\to \widetilde {A}/I_k$ denote the quotient maps. Note that $\widetilde {A}/I_k$ is naturally isomorphic to $(A/I_k)^+$ , the forced unitization of $A/I_k$ . By assumption, $\pi _k(B)=\pi _k(A)$ . It follows that $\pi _k^+(\bar {B})=\pi _k^+(\widetilde {A})$ . Furthermore, $\operatorname {\mathrm {Prim}}(\widetilde {A})$ is the union of the hulls of the $I_k$ . Then, assuming that the results holds in the unital case, we obtain $\bar {B}=\widetilde {A}$ , which implies $B\,{=}\,A$ .

Thus, we may assume from now on that A is unital. To reach a contradiction, assume that $B\neq A$ . Using Hahn–Banach, we choose $\varphi \in A^*$ with $\varphi |_{B}\equiv 0$ and $\|\varphi \|=1$ . Apply Lemma 5.1 to obtain k such that $\|\varphi |_{I_k}\|<\tfrac {1}{8}$ . Since every functional is a linear combination of four states [Reference BlackadarBla06, Theorem II.6.3.4, p.106], we obtain $\psi _m\in (I_k)^*_+$ with $\varphi |_{I_k}=\sum _{m=0}^3 i^m \psi _m$ , and we may also ensure that $\|\psi _m\|\leq \|\varphi |_{I_k}\|<\tfrac {1}{8}$ . Using [Reference BlackadarBla06, Theorem II.6.4.16, p. 111], we can extend each $\psi _m$ to a positive functional $\tilde {\psi }_m\in A^*_+$ with $\|\tilde {\psi }_m\|=\|\psi _m\|$ . Set $\omega :=\varphi -\sum _{m=0}^3 i^m \tilde {\psi }_m$ . Then, $\omega \in A^*$ satisfies $\omega |_{I_k}\equiv 0$ and ${\|\varphi -\omega \|<\tfrac {1}{2}}$ .

Let $\bar {\omega }\in (A/I_k)^*$ satisfy $\omega =\bar {\omega }\circ \pi _k$ . Given $a\in A$ , use that $A/I_k=\pi _k(B)$ to choose $b\in B$ with $\pi _k(b)=\pi _k(a)$ and $\|b\|=\|\pi _k(a)\|$ (see [Reference BlackadarBla06, Proposition II.5.1.5]). Then, $\omega (a)=\omega (b)$ , and thus

$$\begin{align*}|\omega(a)| = |\omega(b)| \leq |\omega(b)-\varphi(b)|+|\varphi(b)| \leq \|\omega-\varphi\| \|b\| \leq \tfrac{1}{2} \|a\|. \end{align*}$$

Hence, $\|\omega \|\leq \tfrac {1}{2}$ , and so $1=\|\varphi \|\leq \|\varphi -\omega \|+\|\omega \|<1$ , which is a contradiction.

Proof Part 1: We verify the equality for $\operatorname {\mathrm {gr}}_0$ . For each k, set $B_k:=A/I_k$ and let $\pi _k\colon A\to B_k$ denote the quotient map. By Theorem 2.5, we have $\operatorname {\mathrm {gr}}_0(A)\geq \operatorname {\mathrm {gr}}_0(B_k)$ . It thus remains to prove $\operatorname {\mathrm {gr}}_0(A) \leq \sup _k \operatorname {\mathrm {gr}}_0(B_k)$ . Set $n:=\sup _k \operatorname {\mathrm {gr}}_0(B_k)$ , which we may assume to be finite. For each k, set