# Simplicity, bounded normal generation, and automatic continuity of groups of
unitaries

Abhinav Chand  and  Leonel Robert

###### Abstract.

We show that the commutator subgroup of the group of unitaries connected to
the identity in a simple unital C*-algebra is simple modulo its center. We
then go on to investigate the role of “regularity properties” in the structure
of the special unitary group of a C*-algebra. Under mild assumptions, we show
that this group has the invariant automatic continuity property and a unique
polish group topology. Strengthening our assumptions in the case of simple
C*-algebras, we show that the special unitary group modulo its center has
bounded normal generation. These results apply to all simple purely infinite
C*-algebras and too all simple nuclear C*-algebras in the “classifiable
class”. We show with counterexamples how our conclusions may in general fail
if no regularity conditions are imposed on the C*-algebra.

##  1\. Introduction

Let $A$ be a unital C*-algebra. Let $U_{0}(A)$ denote the connected component
of the identity of the unitary group of $A$. Let $DU_{0}(A)$ denote the
commutator (or derived) subgroup of $U_{0}(A)$. If $A$ is the C*-algebra of
$n\times n$ matrices $M_{n}(\mathbb{C})$, then $DU_{0}(A)$—i.e., the special
unitary group $SU(n)$—is perfect and simple modulo its center. We are thus led
to ask whether these properties hold in more generality. If $A$ has no
1-dimensional representations, then indeed $DU_{0}(A)$ is perfect
([RobertNormal, Theorem 6.2]). On the other hand, since non-commutative closed
two-sided ideals of $A$ give rise to non-central normal subgroups of
$DU_{0}(A)$, for the simplicity question it is natural to assume that $A$
itself be a simple C*-algebra. This question, and its counterpart with
invertible elements instead of unitaries, has been investigated by several
authors, going back to Kadison in the 50s, working in the setting of von
Neumann algebra factors, followed by de la Harpe, Skandalis, Thomsen, Ng,
Ruiz, among others (see [kad1, delaHarpe, delaHarpe-Skandalis3, thomsen, ng-
ruiz]). In[RobertNormal] the second author settled the question of simplicity
modulo the center in the setting of invertible elements: the commutator
subgroup of the connected component of the identity of the group of invertible
elements of a simple unital C*-algebra is simple modulo its center. In the
case of unitaries, the methods used in [RobertNormal] also indicate a path to
a proof, but key technical questions on Lie ideals were left unanswered in
[RobertNormal].

The starting point of this paper is to address the technical points left
unsolved in [RobertNormal]. First, working in the additive setting, we prove
that for a simple unital C*-algebra $A$, a noncentral abelian subgroup $V$ of
$iA_{\mathrm{sa}}$ (the skewadjoint elements of $A$) that is invariant under
conjugation by $DU_{0}(A)$ must contain $[A,A]\cap iA_{\mathrm{sa}}$ (the
skewadjoint elements in the linear span of the commutators). This partially
answers a question raised by Marcoux in [marcoux09, Question 4]. Then, using
techniques from [RobertNormal], we pass from the “additive” to the
“multiplicative” setting to prove the following theorem.

######  Theorem A.

If $A$ is simple and unital then $DU_{0}(A)$ is simple modulo its center.

While proving these results we take care to establish much finer versions that
apply to non-simple C*-algebras, and that have a quantitative nature (phrased
in terms of generating sets of normal subgroups). Our aim is to apply these
finer results to questions on automatic continuity and bounded normal
generation. In doing this, we follow in the footsteps of parallel results
obtained by Dowerk and Thom for von Neumann algebra factors in [DowerkThom,
DowerkThom2]. These applications still revolve around $DU_{0}(A)$, but in
situations where it naturally carries a Banach-Lie group structure, as it
agrees with the group $SU_{0}(A)$.

We denote by $SU_{0}(A)$ the subgroup of $U_{0}(A)$ generated by
$\\{e^{ih}:h\in\overline{[A,A]}\cap A_{\mathrm{sa}}\\}$. It is not difficult
to show that $SU_{0}(A)$ is a path connected normal subgroup of $U_{0}(A)$
containing $DU_{0}(A)$ and contained in the kernel of the de la Harpe-
Skandalis determinant. We regard $SU_{0}(A)$ as the Banach-Lie subgroup of
$U_{0}(A)$ associated to the Lie algebra of skewadjoint commutators
$\overline{[A,A]}\cap iA_{\mathrm{sa}}$. As such, $SU_{0}(A)$ comes endowed
with the topology having for basis of neighborhoods at the identity the sets

$\\{e^{ih}:h\in\overline{[A,A]}\cap
A_{\mathrm{sa}},\,\|h\|<\epsilon\\}\quad(\epsilon>0).$

The exponential length on $SU_{0}(A)$ is defined as

$\mathrm{el}_{SU_{0}(A)}(u)=\Big{\\{}\inf\sum_{i=1}^{n}\|h_{i}\|:u=\prod_{j=1}^{n}e^{h_{j}}\hbox{
and }h_{j}\in\overline{[A,A]}\cap A_{\mathrm{sa}}\hbox{ for all }j\Big{\\}}.$

This gives rise to a metric $(u,v)\mapsto\mathrm{el}(u^{*}v)$ that is left and
right invariant and induces the topology on $SU_{0}(A)$ ([ando, Theorem A]).
Note: the topology on $SU_{0}(A)$ is in general not the same as the topology
induced by its inclusion in $U_{0}(A)$ (where the topology on the latter is
that induced by the norm). In fact, $SU_{0}(A)$ can be dense in $U_{0}(A)$ in
the norm topology, e.g., if $A$ is simple, infinite dimensional, and of real
rank zero.

Under the assumption that $[A,A]$ is a closed set, which is present for many
C*-algebras, we have that $DU_{0}(A)=SU_{0}(A)$, and so $DU_{0}(A)$ is a
complete metric space.

Let us say that a C*-algebra $A$ has _bounded commutators generation_ if
$[A,A]$ is closed, and furthermore there exist $m\in\mathbb{N}$ and $C>0$ such
that if $h\in\overline{[A,A]}$ then

$h=\sum_{i=1}^{m}[x_{i},y_{i}]$

for some $x_{j},y_{j}\in A$ such that $\|x_{j}\|\cdot\|y_{j}\|\leq C\|h\|$ for
all $j$.

Bounded commutators generation is not an uncommon property of C*-algebras. We
shall return to this point later. But first, let us state our automatic
continuity result.

Following Dowerk and Thom ([DowerkThom, Definition 8.8]), let us say that a
topological group $G$ has the invariant automatic continuity property if any
group homomorphism $\phi\colon G\to H$, where $H$ is a topological separable
SIN group, is continuous. (A SIN group is one with a basis of neighborhoods of
the identity invariant under conjugation.)

######  Theorem B.

Let $A$ be a unital C*-algebra with bounded commutators generation and a full
square zero element (i.e., $x\in A$ such that $x^{2}=0$ and the ideal
generated by $x$ is $A$). The following are true:

  1. (i)

$SU_{0}(A)$ has the invariant automatic continuity property.

  2. (ii)

If $A$ is separable, then $SU_{0}(A)$ admits a unique polish group topology.

Following Dowerk and Thom ([DowerkThom, Definition 2.1]), let us say that a
simple group $G$ has bounded normal generation (BNG) if for any $g\in
G\backslash\\{1\\}$ there exists $n\in\mathbb{N}$ such that $G=(\\{hg^{\pm
1}h^{-1}:h\in G\\})^{n}$.

We introduce here a weaker version of the BNG property for topological groups.
Let us say that a topological group $G$ has local bounded normal generation if
for any $g\in G\backslash\\{1\\}$ there exists $n\in\mathbb{N}$ such that
$(\\{hg^{\pm 1}h^{-1}:h\in G\\})^{n}$ is a neighborhood of the identity.

######  Theorem C.

Let $A$ be a simple unital C*-algebra with bounded commutators generation. The
following are true:

  1. (i)

$SU_{0}(A)/Z(SU_{0}(A))$ is a simple group with local BNG.

  2. (ii)

If $SU_{0}(A)$ is bounded as a metric space, then $SU_{0}(A)/Z(SU_{0}(A))$ has
BNG.

If $A$ is a traceless unital C*-algebra, then by a theorem of Pop $A=[A,A]$
and $A$ has bounded commutators generation. It follows that in this case
$DU_{0}(A)=SU_{0}(A)=U_{0}(A)$, and Theorems B and C apply directly to
$U_{0}(A)$ (endowed with the norm topology). Let us single out the well
studied case of purely infinite simple C*-algebras: If $A$ is a unital,
separable, simple, purely infinite C*-algebra, then $U_{0}(A)$ has the
automatic invariant continuity property and a unique polish group topology.
Moreover, since the exponential length of simple purely infinite C*-algebras
is finite ([phillips]), $U_{0}(A)/\mathbb{T}$ is a simple group with bounded
normal generation.

Combining the invariant automatic continuity property with results on the
structure of Lie algebra isomorphisms by Ara-Mathieu and Brešar, we show that
the group structure of $U_{0}(A)$ suffices as a classification invariant for
prime traceless C*-algebras. More specifically, if $A$ and $B$ are separable
prime traceless unital C*-algebras containing full square zero elements, and
$U_{0}(A)\cong U_{0}(B)$ as groups, then $A$ is either isomorphic or anti-
isomorphic to $B$ (Corollary 6.5). The question whether C*-algebras can be
classified up to (anti-)isomorphism by their unitary groups has been studied
in [giordano], but it remains in general not well understood.

In contrast to the traceless case, if $A$ has at least one tracial state, then
$U_{0}(A)$ has discontinuous automorphisms. This is well-known in the case of
the classical unitary group (i.e. $A=M_{n}(\mathbb{C})$), and the argument can
be adapted to an arbitrary C*-algebra with a tracial state (Theorem 6.6). So
we must turn to $SU_{0}(A)$ in this case. In order to apply Theorems B and C,
we must ensure that the C*-algebra has the property of bounded commutators
generation. For unital C*-algebras with tracial states, this property is known
to hold assuming some form of regularity on the C*-algebra. Here are some
classes of C*-algebras with bounded commutators generation (see Theorem 6.1):

  * •

C*-algebras of finite decomposition rank (in the sense of Kirchberg and
Winter) and no finite dimensional representations.

  * •

C*-algebras with strict comparison of positive elements by traces and almost
divisible Cuntz semigroup.

  * •

Exact C*-algebras tensorially absorbing the Jiang-Su algebra.

Theorems B and C (i) can be readily applied to the C*-algebras in these
classes (the existence of full square zero elements holds automatically in
each case).

By Theorem C (ii), in order to guarantee BNG we must verify the boundedness of
$SU_{0}(A)$ (which is also necessary to have BNG). We prove below that if $A$
is a simple C*-algebra with strict comparison of positive elements by traces,
stable rank one, and finite exponential rank, then $SU_{0}(A)$ is bounded. All
necessary hypotheses are present for the simple nuclear C*-algebras recently
classified in the Elliott program: If $A$ is a separable simple unital
C*-algebra with finite nuclear dimension and in the UCT class, then
$SU_{0}(A)$ has the automatic invariant continuity property and a unique
polish group topology, and $SU_{0}(A)$ modulo its center is a simple group
with BNG (Corollary 7.6).

In the final section of the paper we turn to the construction of
counterexamples. These show that bounded commutators generation cannot be
altogether dropped in Theorem C. In [RobertCommutators] the second author
constructed examples of simple unital C*-algebras without bounded commutators
generation. Essentially the same construction allows us to prove the following
theorem.

######  Theorem D.

There exists a simple unital AH C*-algebra $A$ with a unique tracial state and
the following properties:

  1. (i)

$DU_{0}(A)/Z(DU_{0}(A))$ is a simple group without BNG.

  2. (ii)

The inclusion of $DU_{0}(A)$ in $SU_{0}(A)$ is proper.

  3. (iii)

There exist exponentials $e^{ih_{1}},e^{ih_{2}},\ldots\in SU_{0}(A)$ such that
$\mathrm{el}_{U_{0}(A)}(e^{ih_{n}})\to\infty$, where
$\mathrm{el}_{U_{0}(A)}(\cdot)$ denotes the exponential length in $U_{0}(A)$.
In particular, $SU_{0}(A)$ is unbounded as a metric space.

To prove this theorem we first find topological obstructions to “short
exponential length on commutators” in the unitary groups of homogeneous
C*-algebras. A simple C*-algebra is then obtained putting the result for
homogeneous C*-algebras through the Villadsen machine. We note that, although
a general recipe for obtaining C*-algebras of infinite exponential length and
having more than one tracial state was provided in [phillips], examples with a
unique tracial state are harder to come by and did not previously exist in the
literature. Theorem D shows that [phillips, Question 4.4] has a negative
answer.

##  2\. Preliminaries

Let us introduce some of the notation that will be used throughout the paper.

Let $A$ be a unital C*-algebra. Let $A_{\mathrm{sa}}$ and $A_{+}$ denote the
sets of selfadjoint and positive elements of $A$, respectively. Let $GL(A)$
and $U(A)$ denote the groups of invertible and unitary elements of $A$. Let
$GL_{0}(A)$ and $U_{0}(A)$ denote the connected component of the identity in
$GL(A)$ and $U(A)$. The commutator, or derived, subgroups of $GL_{0}(A)$ and
$U_{0}(A)$ are denoted by $DGL_{0}(A)$ and $DU_{0}(A)$.

Given $x,y\in A$, let us denote their commutator $xy-yx$ by $[x,y]$. The
notation $[X,Y]$, when applied to subsets $X,Y$ of $A$, denotes the linear
span of all $[x,y]$ with $x\in X$ and $y\in Y$. Thus, $[A,A]$ denotes the
linear span of all the commutator elements of $A$.

In the remainder of this section we prove a number of technical results that
will be needed in the next section.

Let us recursively define polynomials in noncommuting variables (nc-
polynomials) $\pi_{n}$, for $n\geq 0$, by $\pi_{0}(x)=x$ and

(2.1)
$\pi_{n}(x_{1},\ldots,x_{2^{n}})=[\pi_{n-1}(x_{1},\ldots,x_{2^{n-1}}),\pi_{n-1}(x_{2^{n-1}+1},\ldots,x_{2^{n}})],$

for $n\geq 1$.

######  Lemma 2.1.

In an associative algebra, the nc-polynomial
$[a\pi_{3}(x_{1},\ldots,x_{8})b,c]$ (in the variables
$a,b,c,x_{1},\ldots,x_{8}$) is a sum of terms of the forms

  1. (a)

$[x_{i},[p_{1},p_{2}]]$, where $p_{1},p_{2}$ are nc-polynomials on
$a,b,c,x_{1},\ldots,x_{8}$,

  2. (c)

$[[x_{i},[q_{1},q_{2}]],[r_{1},r_{2}]]$, where $q_{1},q_{2},r_{1},r_{2}$ are
nc-polynomials on $a,b,c,x_{1},\ldots,x_{8}$.

###### Proof.

This is [RobertNormal, Lemma 4.1]. ∎

We call a set $X\subseteq A$ full if the closed two-sided ideal that it
generates is $A$. We call $X$ fully noncentral if it is not contained in the
center of $A$, nor it is mapped onto the center of any non-trivial quotient of
$A$. Put differently, $X$ is fully noncentral if $[X,A]$ is a full set.

######  Lemma 2.2.

Let $X$ be a fully noncentral subset of $A$ invariant under conjugation by
$DU_{0}(A)$. Then $\pi_{3}(X^{8})$ is a full subset of $A$.

###### Proof.

Let $W$ denote the closed linear span of $X$. Let us show that $W$ is a Lie
ideal of $A$. Observe that $W$ is a fully noncentral closed subspace invariant
under conjugation by $DU_{0}(A)$. Since

$\lim_{t\to 0}\frac{1}{it}(e^{it[x^{*},x]}he^{-it[x^{*},x]}-h)=[h,[x^{*},x]],$

and $e^{it[x^{*},x]}\in DU_{0}(A)$ ([RobertNormal, Theorem 6.2]), we have that
$[h,[x^{*},x]]\subseteq W$ for all $h\in W$ and $x\in A$. Further, since the
selfcommutators $[x^{*},x]$ span $[A,A]$, $[W,[A,A]]\subseteq W$, i.e., $W$ is
a Lie ideal of $[A,A]$. By [RobertLie, Theorem 1.15], this implies that $W$ is
a Lie ideal of $A$.

Let $I$ be the closed two-sided ideal generated by $\pi_{3}(W^{8})$. Let
$\widetilde{W}\subseteq A/I$ be the image of $W$ under the quotient map, so
that $\pi_{3}(\widetilde{W}^{8})=0$. This implies that
$[A/I,\widetilde{W}]=0$, by (repeated applications of) [herstein, Lemma 2].
Since $W$ is fully noncentral, we must have that $I=A$. That is,
$\pi_{3}(W^{8})$ is a full set. Since $\pi_{3}$ is multilinear, the closed
linear spans of $\pi_{3}(W^{8})$ and $\pi_{3}(X^{8})$ agree. Hence,
$\pi_{3}(X^{8})$ is a full set. ∎

The following theorem may be extracted from the results and methods in
[RobertLie], although it is not stated there explicitly.

######  Theorem 2.3.

Let $A$ be a unital C*-algebra without 1-dimensional representations. Let
$V\subseteq A$ be a fully noncentral subset invariant under conjugation by
elements in $DGL_{0}(A)$ and under multiplication by $-1$. Then there exist
$n\in\mathbb{N}$ and $C>0$ such that for all $c,d\in A$, with
$\|c\|\cdot\|d\|\leq 1$, $[c,d]\in\sum_{j=1}^{n}v_{j}$, for some $v_{j}\in V$
such that $\|v_{j}\|\leq C$ for all $j$.

###### Proof.

Since $V$ is fully noncentral, $\pi_{3}(V^{8})$ is a full set by Lemma 2.2.
Hence, we can write

$1=\sum_{j=1}^{m}x_{j}\pi_{3}(v_{j,1},\ldots,v_{j,8})y_{j},$

where $v_{j,k}\in V$ and $x_{j},y_{j}\in A$ for all $j,k$. Enlarging the
number of terms in the sum if necessary, we may assume that
$\|x_{j}\|,\|y_{j}\|\leq 1$ for all $j$. Let $L=\max_{j,k}\|v_{j,k}\|$. Let us
multiply by a contraction $c\in A$ on the left and take commutator with a
contraction $d\in A$, to obtain

$[c,d]=\sum_{j=1}^{m}[cx_{j}\pi_{3}(v_{j,1},\ldots,v_{j,8})y_{j},d].$

Applying Lemma 2.1 on the right-hand side, we expand each term
$[cx_{j}\pi_{3}(v_{j,1},\ldots,v_{j,8})y_{j},d]$ into a sum of terms of the
forms $[v_{j,k},[p_{1},p_{2}]]$ or $[[v_{j,k},[q_{1},q_{2}],[r_{1},r_{2}]]$.
Here the norms of the elements $p_{i},q_{i},r_{i}$ are bounded by a universal
polynomial on the constant $L$ (since they are nc-polynomials on $cx_{j}$,
$v_{j,k}$, $y_{j}$, and $d$). By [RobertNormal, Theorem 4.3], there exist
$m_{1}\in\mathbb{N}$ and $C_{1}>0$ such that each commutator $[s_{1},s_{2}]$
is a sum of a $m_{1}$ square zero elements whose norms are bounded by
$C_{1}\|s_{1}\|\cdot\|s_{2}\|$. Applying this to the commutators in the terms
$[v_{j,k},[p_{1},p_{2}]]$ and $[[v_{j,k},[q_{1},q_{2}],[r_{1},r_{2}]]$ we
obtain

(2.2)
$[c,d]=\sum_{j=1}^{m_{2}}[v_{j},z_{j}]+\sum_{j=1}^{m_{3}}[[v_{j}^{\prime},z_{j}^{\prime}],z_{j}^{\prime\prime}],$

where $v_{j},v_{j}^{\prime}\in V$ for all $j$ satisfy that
$\|v_{j}\|,\|v_{j}^{\prime}\|\leq L$ for all $j$, where
$z_{j},z_{j}^{\prime},z_{j}^{\prime\prime}$ are square zero elements whose
norms are bounded by a function of $L$ for all $j$, and where $m_{2},m_{3}$
are both bounded by a universal constant times $mm_{1}$. Observe that

$[v,z]=(1+z)v(1-z)+(1-z)v(1+z)-2v.$

If $z$ is a square zero element then the first two terms on right-hand side
are similarity conjugates of $v$, since $(1+z)^{-1}=1-z$. Further, $1+z\in
DGL_{0}(A)$ ([RobertNormal, Lemma 2.5]), so the conjugation is by an element
in $DGL_{0}(A)$. Using this on the right-hand side of (2.2), the terms
$[v_{j},z_{j}]$ and $[[v_{j}^{\prime},z_{j}^{\prime}],z_{j}^{\prime\prime}]$
can be expressed as sums of similarity conjugates of $\pm v_{j},\pm
v_{j}^{\prime}$ by elements in $DGL_{0}(A)$. This is a sum of elements in $V$
whose number of terms is independent of $c$ and $d$. Further, since the norms
of the square zero elements $z_{j},z_{j}^{\prime},z_{j}^{\prime\prime}$ are
bounded by a constant (a function of $L$), the norms of the resulting elements
in $V$ (conjugates of the $\pm v_{j}$s and $\pm v_{j}^{\prime}$s) are bounded
by a constant independent of $c$ and $d$. This proves the theorem. ∎

Let

$\mathcal{N}_{2}=\\{x\in A:x^{2}=0\\}.$

That is, $\mathcal{N}_{2}$ denotes the set of square zero elements of $A$.

######  Lemma 2.4.

Let $h\in A$ and $x\in\mathcal{N}_{2}$ be such that $\|x\|\leq 2$. Then $xhx$
is a sum of 8 terms in the set

(2.3) $H:=\\{uhu^{*},\,\pm iuhu^{*}:u\in DU_{0}(A)\\}.$

###### Proof.

Assume that $x\in\mathcal{N}_{2}$ and $\|x\|\leq 1$. We will show that $4xhx$
is a sum of 8 terms in the set (2.3), which is essentially what is claimed in
the lemma.

For each $\alpha\in\mathbb{C}$ of absolute value $1$, define

$u(\alpha)=\sqrt{1-(x^{*}x+xx^{*})}+i(\alpha
x+\overline{\alpha}x^{*})=a+ib(\alpha).$

Let us show that these are unitaries in $DU_{0}(A)$. It suffices to work in
the universal C*-algebra generated by a square zero contraction. This is the
C*-algebra $M_{2}(C_{0}(0,1])$, with square zero contraction
$x=\begin{pmatrix}0&t\\\ 0&0\end{pmatrix}$. In this case

$u(\alpha)=\begin{pmatrix}\sqrt{1-t^{2}}&i\alpha t\\\
i\overline{\alpha}t&\sqrt{1-t^{2}}\end{pmatrix}=\begin{pmatrix}\alpha&0\\\
0&\overline{\alpha}\end{pmatrix}\begin{pmatrix}\sqrt{1-t^{2}}&t\\\
-t&\sqrt{1-t^{2}}\end{pmatrix}\begin{pmatrix}\overline{\alpha}&0\\\
0&\alpha\end{pmatrix}.$

It suffices to write as a commutator the middle matrix on the right-hand side.
The change of variables $t=\sin(\theta)$ gives rise to an isomorphism from
$M_{2}(C_{0}(0,1])$ to $M_{2}(C_{0}(0,\pi/2])$. In the latter algebra, the
middle matrix on the right-hand side becomes the rotation matrix

$\begin{pmatrix}\cos\theta&\sin\theta\\\ -\sin\theta&\cos\theta\end{pmatrix},$

which is easily expressed as a commutator:

$\begin{pmatrix}\cos\theta&\sin\theta\\\
-\sin\theta&\cos\theta\end{pmatrix}=\left(\begin{pmatrix}\cos\theta/2&\sin\theta/2\\\
-\sin\theta/2&\cos\theta/2\end{pmatrix},\begin{pmatrix}\cos(\theta)&\sin(\theta)\\\
\sin(\theta)&-\cos(\theta)\end{pmatrix}\right),$

for all $0\leq\theta\leq\pi/2$

Back in the C*-algebra $A$, we have that

$u(\alpha)^{*}hu(\alpha)+u(\alpha)hu(\alpha)^{*}=aha+b(\alpha)hb(\alpha).$

The left-hand side is a sum of two element in the set $H$ (defined in (2.3)),
and $a$ and $b$ are as in the definition of $u(\alpha)$ above. Subtracting two
expressions of the form $aha+b(\alpha)hb(\alpha)$ for two different values of
$\alpha$ we get that

$(\alpha_{1}x+\overline{\alpha_{1}}x^{*})h(\alpha_{1}x+\overline{\alpha_{1}}x^{*})-(\alpha_{2}x+\overline{\alpha_{2}}x^{*})h(\alpha_{2}x+\overline{\alpha_{2}}x^{*})$

is a sum of 4 elements of the set $H$. Using that $\alpha_{1}$ and
$\alpha_{2}$ are in the unit circle, this simplifies to

$(\alpha_{1}^{2}-\alpha_{2}^{2})xhx+(\overline{\alpha_{1}}^{2}-\overline{\alpha_{2}}^{2})x^{*}hx^{*}.$

Hence, $\alpha xhx+\overline{\alpha}x^{*}hx^{*}$ is a sum of 4 elements in $H$
for any $|\alpha|\leq 2$. With $\alpha=2$ and $\alpha=2i$ we get
$2(xhx+x^{*}hx^{*})$ and $2i(xhx-x^{*}hx^{*})$, respectively. Hence, $4xhx$ as
a sum of 8 elements in $H$. ∎

######  Lemma 2.5.

Let $B$ be a C*-algebra (possibly nonunital). Let $b\in B$ be of norm at most
1 such that $eb=be=b$ for some $e\in B_{+}$ of norm 1. Let

$h=\begin{pmatrix}0&b\\\ 0&0\end{pmatrix}\in M_{2}(B).$

Let $x,y\in B$ be of norm at most 1. Then

$\begin{pmatrix}0&xb^{2}y\\\ 0&0\end{pmatrix}$

is a sum of $33540$ terms in the set

(2.4) $H=\\{\pm uhu^{*},\,\pm iuhu^{*}:u\in DU_{0}(M_{2}(B))\\}.$

###### Proof.

Regard $B$ embedded in $M_{2}(B)$ in the top left corner. Set
$W=\begin{pmatrix}0&1\\\ 0&0\end{pmatrix}$, which we regard as an element of
the multiplier algebra of $M_{2}(B)$. Observe that for $x\in B$ we have that

$xW=\begin{pmatrix}0&x\\\ 0&0\end{pmatrix}.$

Let $x\in B$ be such that $\|x\|\leq 4$. We have that

$\begin{pmatrix}0&xbx\\\ 0&0\end{pmatrix}=\begin{pmatrix}0&x/2\\\
0&0\end{pmatrix}\begin{pmatrix}0&0\\\ 2e&0\end{pmatrix}\begin{pmatrix}0&b\\\
0&0\end{pmatrix}\begin{pmatrix}0&0\\\ 2e&0\end{pmatrix}\begin{pmatrix}0&x/2\\\
0&0\end{pmatrix}.$

Applying Lemma 2.4 twice we conclude that $xbxW$ (element on the left-hand
side) is a sum of $64$ terms in the set $H$ (defined in (2.4)). Observe that
this applies in particular to $b^{2}W=b^{\frac{1}{2}}(b)b^{\frac{1}{2}}W$.

Applying the conclusion from the previous paragraph with $x+e$ in place of
$x$, where $\|x\|\leq 3$, we get that

$(x+e)b(x+e)W=(xbx+b+(xb+bx))W$

is a sum of $64$ terms in the set $H$. Hence, $(xb+bx)W$ is a sum of $2\cdot
64+1=129$ terms in the set $H$ for all $\|x\|\leq 3$.

For $\|x\|\leq 1$ we have $\|xb-bx\|\leq 2$. Applying the conclusion from the
previous paragraph to

$(xb-bx)b+b(xb-bx)=xb^{2}-b^{2}x,$

we get that $(xb^{2}-b^{2}x)W$ is a sum of $129$ terms in $H$ for $\|x\|\leq
1$. Also, applying the same arguments to $b^{2}$, the element
$(xb^{2}+b^{2}x)W$ is a sum of $129$ unitary conjugates of $\pm b^{2}W,\pm
ib^{2}W$ by unitaries in $DU_{0}(M_{2}(B))$. Further, as observed above,
$b^{2}$ is a sum of $64$ elements in $H$, i.e., conjugates of $\pm bW$ and
$\pm ibW$ by unitaries in $DU_{0}(M_{2}(B))$. Hence, $(xb^{2}+b^{2}x)W$ is a
sum of $64\cdot 129$ elements in $H$. Now adding $xb^{2}+b^{2}x$ and
$xb^{2}-b^{2}x$ we get that $xb^{2}W$ is a sum of $65\cdot 129=8385$ elements
in $H$ for $\|x\|\leq 1$.

Let $u\in U(B)$. (Here we use the convention that if $B$ is non-unital the
unitaries in $U(B)$ are chosen in $B^{\sim}$ and of the form $1+v$, with $v\in
B$.) We have that

$\begin{pmatrix}0&ucu\\\ 0&0\end{pmatrix}=\begin{pmatrix}u&0\\\
0&u^{*}\end{pmatrix}\begin{pmatrix}0&c\\\
0&0\end{pmatrix}\begin{pmatrix}u^{*}&0\\\ 0&u\end{pmatrix}.$

The unitary $\begin{pmatrix}u&0\\\ 0&u^{*}\end{pmatrix}$ belongs to
$DU_{0}(M_{2}(B))$. (Proof: It suffices to assume that $\|u-1\|<1$, as any
unitary is a product of such $u$. In this case the diagonal unitary is a
single commutator by [delaHarpeSkandalis2, Lemma 5.17].) Thus, $(ucu)W$ is a
unitary conjugate of $cW$ by a unitary in $DU_{0}(M_{2}(B))$. Hence,
$uxb^{2}uW$ is a sum of $8385$ elements in $H$ (the same number as $xb^{2}W$)
for any $u\in U(B^{\sim})$. Since $x$ is arbitrary such that $\|x\|\leq 1$,
replacing $x$ by $u^{*}x$ we get that $(xb^{2}u)W$ is a sum of $8385$ elements
in $H$ for any $\|x\|\leq 1$ and $u\in U(B)$. But every element in $B$ of norm
$\leq 1$ is a sum of four unitaries in $U(B)$. Hence, $(xb^{2}y)W$ is a sum of
$4\cdot 8385=33540$ elements in $H$ for $\|x\|,\|y\|\leq 1$ , equivalently,
for $\|x\|\cdot\|y\|\leq 1$. ∎

The following estimates around the exponential function are easy to derive. We
include their proof for the convenience of the reader.

######  Lemma 2.6.

Let $0<|t|<1$, $x,y\in A$, and $v\in A$.

  1. (i)

We have that

$[v,x]=\frac{1}{t}(e^{tx}ve^{-tx}-v)+t\Delta_{1}.$

where $\|\Delta_{1}\|\leq\|v\|\cdot e^{2\|x\|}$.

  2. (ii)

We have that

$[[v,x],y]=\frac{1}{t^{2}}(e^{ty}\Big{(}e^{tx}ve^{-tx}-v\Big{)}e^{-ty})-\frac{1}{t}(e^{tx}ve^{-tx}-v)+t\Delta_{2}$

where

$\|\Delta_{2}\|\leq\|v\|\cdot(e^{2\|x\|+2\|y\|}+2\|y\|e^{2\|x\|})$

###### Proof.

(i) Recall that

$e^{x}ve^{-x}=v+[v,x]+\frac{1}{2!}[[v,x],x]+\frac{1}{3!}[[[v,x],x],x]]+\cdots$

(Put differently, $\mathrm{Ad}_{e^{x}}=e^{\mathrm{ad}_{x}}$; see [hofmann-
morris, Proposition 5.16]). Replacing $x$ by $tx$ we get

$e^{tx}ve^{-tx}=v+t[v,x]+t^{2}\Delta_{1},$

where

$\|\Delta_{1}\|\leq\|v\|\frac{e^{2\|x\||t|}-1-2|t|\|x\|}{t^{2}}\leq\|v\|e^{2\|x\|}.$

Isolating $[v,x]$ the formula of the lemma readily follows.

(ii) Using (i), we have

$\displaystyle[[v,x],y]$
$\displaystyle=[\frac{1}{t}(e^{tx}ve^{-tx}-v)+t\Delta_{1},y]$
$\displaystyle=[\frac{1}{t}(e^{tx}ve^{-tx}-v),y]+t[\Delta_{1},y]$

Using (i) again on the first term on the right side we get

$[[v,x],y]=\frac{1}{t}\left(e^{ty}\Big{(}\frac{1}{t}(e^{tx}ve^{-tx}-v)\Big{)}e^{-ty}-\frac{1}{t}(e^{tx}ve^{-tx}-v)\right)+t\widetilde{\Delta}_{1}+t[\Delta_{1},y].$

where

$\|\widetilde{\Delta}_{1}\|\leq\Big{\|}\frac{1}{it}(e^{tx}ve^{-tx}-v)\Big{\|}e^{2\|y\|}\leq\|v\|e^{2\|x\|}e^{2\|y\|}.$

Combining this estimate with (i), the estimate for
$\Delta_{2}=\widetilde{\Delta}_{1}+[\Delta_{1},y]$ readily follows. ∎

##  3\. Additive setting

Let $A$ be a unital C*-algebra. Let

(3.1) $\mathcal{D}=\\{[a,b]:a,b\in A_{\mathrm{sa}},\,\|a\|,\|b\|\leq
1\\}\subseteq iA_{\mathrm{sa}}.$

Given $n\in\mathbb{N}$ and a set $X\subseteq A$ we denote by $\sum^{n}X$ the
set of sums of $n$ elements of $X$.

In this section we prove the following theorem.

######  Theorem 3.1.

Let $A$ be a unital C*-algebra containing a full square zero element. Let
$V\subseteq iA_{\mathrm{sa}}$ be a fully noncentral set invariant under
conjugation by $DU_{0}(A)$ and multiplication by $-1$. Then there exists
$n\in\mathbb{N}$ such that $\mathcal{D}\subseteq\sum^{n}V$.

For our applications to automatic continuity we will need a finer version of
this theorem which says that the function $V\mapsto n$, assigning to a set $V$
as in the theorem the least $n$ such that $\mathcal{D}\subseteq\sum^{n}V$, is
locally bounded in a suitable sense. We state this theorem next. Given sets
$X,Y\subseteq A$, we write $X\subseteq_{\epsilon}Y$ to indicate that
$X\subseteq\\{x\in A:d(x,Y)\leq\epsilon\\}$.

######  Theorem 3.2.

Let $A$ and $V\subseteq iA_{\mathrm{sa}}$ be as in Theorem 3.1. Then there
exist $n\in\mathbb{N}$, a finite set $F\subseteq V$, and $\epsilon>0$, such
that if $V^{\prime}\subseteq iA_{\mathrm{sa}}$ is a fully noncentral set
invariant under conjugation by $DU_{0}(A)$ and multiplication by $-1$, and
$F\subseteq_{\epsilon}V^{\prime}$, then
$\mathcal{D}\subseteq\sum^{n}V^{\prime}$.

The proofs of the preceding theorems will follow after a series of lemmas.

######  Lemma 3.3.

Let $A$ and $V\subseteq iA_{\mathrm{sa}}$ be as in Theorem 3.1. Then there
exists $M\in\mathbb{N}$ such that

(3.2) $1=\sum_{j=1}^{M}x_{j}v_{j}y_{j},$

where $x_{j},y_{j}\in A$, $\|x_{j}\|,\|y_{j}\|<1$, and
$v_{j}\in\pi_{3}(V^{8})$ for all $j=1,\ldots,M$. (Here $\pi_{3}$ is the nc-
polynomial defined in (2.1).)

###### Proof.

Since $V$ is fully noncentral, $\pi_{3}(V^{8})$ is a full set, by Lemma 2.2.
Therefore, the unit $1\in A$ is expressible as a finite sum of terms of the
form $xvy$, with $x,y\in A$ and $v\in\pi_{3}(V^{8})$. Enlarging the number of
terms if necessary we may assume that $\|x\|,\|y\|<1$ for each of these terms.
The lemma is thus proved. ∎

For the remainder of this section we assume that $A$ is a unital C*-algebra
containing a full square zero element and that $V$ is a fully noncentral
subset of $iA_{\mathrm{sa}}$ invariant under conjugation by $DU_{0}(A)$ and
multiplication by $-1$. We fix $M\in\mathbb{N}$ associated to $V$ through
equation (3.2).

In the next three lemmas we further assume that

(3.3) $V\subseteq\\{x\in A:\|x\|<1\\}.$

We drop this assumption in the proof of Theorem 3.2.

######  Lemma 3.4.

Any $z\in\mathcal{D}$ is expressible as a sum of terms of the forms

  1. (a)

$[v,x]$, with $v\in V$ and $x\in\mathcal{D}$,

  2. (b)

$[[v,y],z]$, with $v\in V$ and $y,z\in\mathcal{D}$,

Moreover, the number of terms of the form (a) is $L_{1}M$, and the number of
terms of the form (b) is $L_{2}M$, where $L_{1},L_{2}\in\mathbb{N}$ are
universal constants.

###### Proof.

In equation (3.2) multiply by $c\in A_{\mathrm{sa}}$ of norm $\leq 1$ on the
left, and take commutator with $d\in A_{\mathrm{sa}}$ also of norm $\leq 1$,
to obtain

$[c,d]=\sum_{j=1}^{M}[cx_{j}v_{j}y_{j},d].$

Fix $1\leq j\leq M$. Say $v_{j}=\pi_{3}(v_{j,1},\ldots,v_{j,8})$, with
$v_{j,k}\in V$ for $k=1,\ldots,8$. By Lemma 2.1,
$[(cx_{j})\pi_{3}(v_{j,1},\ldots,v_{j,8})y_{j},d]$ expands into a sum of terms
of the forms $[v_{j,k},[r_{1},r_{2}]]$ and
$[[v_{j,k},[s_{1},s_{2}]],[t_{1},t_{2}]]$, where
$r_{1},r_{2},s_{1},s_{2},t_{1},t_{2}$ are polynomials on
$cx_{j},v_{j,1},\ldots,v_{j,8},y_{j}$. Since $cx_{j}$, $v_{j,k}$, and $y_{j}$
all belong to the open unit ball, the norms of
$r_{1},r_{2},s_{1},s_{2},t_{1},t_{2}$ are bounded by a universal constant. Let
$k_{1}\in\mathbb{N}$ be the number of terms of the form
$[v_{j,k},[r_{1},r_{2}]]$ in the expansion of
$[(cx_{j})\pi_{3}(v_{j,1},\ldots,v_{j,8})y_{j},d]$ from Lemma 2.1, and let
$k_{2}\in\mathbb{N}$ be the number of terms of the form
$[[v_{j,k},[s_{1},s_{2}]],[t_{1},t_{2}]]$.

Consider one term of the form $[v_{j,k},[r_{1},r_{2}]]$. Let us decompose both
$r_{1}$ and $r_{2}$ into the sum of a selfadjoint and a skewadjoint element.
The skewadjoint part of $[v_{j,k},[r_{1},r_{2}]]$ is thus expressed as the sum
of 2 terms of the form $[v_{j,k},[r_{1}^{\prime},r_{2}^{\prime}]]$, with
$r_{1}^{\prime},r_{2}^{\prime}\in A_{\mathrm{sa}}$. Since the norms of
$r_{1}^{\prime}$ and $r_{2}^{\prime}$ are bounded by a universal constant, we
can write $[v_{j,k},[r_{1}^{\prime},r_{2}^{\prime}]]=N_{1}[v_{j,k},x]$, where
$x\in\mathcal{D}$ and $N_{1}\in\mathbb{N}$ a universal natural number.

We handle similarly $[[v_{j,k},[s_{1},s_{2}]],[t_{1},t_{2}]]$: First express
its skewadjoint part as a sum of 8 terms of the form
$[[v_{j,k},[s_{1}^{\prime},s_{2}^{\prime}]],[t_{1}^{\prime},t_{2}^{\prime}]]$,
where $s_{1}^{\prime},s_{2}^{\prime},t_{1}^{\prime},t_{2}^{\prime}\in
A_{\mathrm{sa}}$, then choose a universal $N_{2}\in\mathbb{N}$ such that
$[[v_{j,k},[s_{1}^{\prime},s_{2}^{\prime}]],[t_{1}^{\prime},t_{2}^{\prime}]]=N_{2}[[v_{j,k},y],z]$
for $y,z\in\mathcal{D}$. Applying this decomposition across all terms
$[cx_{j}v_{j}y_{j},d]$, we express $[c,d]$ as a sum of obtain $2k_{1}N_{1}M$
terms of the form (a) and $8k_{2}N_{2}M$ terms of the form (b). ∎

######  Lemma 3.5.

Let $N\in\mathbb{N}$. There exists $n\in\mathbb{N}$ such that for each
$z\in\mathcal{D}$ we have that

$\Big{\|}z-\sum_{j=1}^{n}v_{j}\Big{\|}<\frac{1}{N}$

for some $v_{1},\ldots,v_{n}\in V$. Moreover, $n\leq CN^{2}M^{3}$, where $C$
is universal constant.

###### Proof.

We start by writing $z$ as a sum of $L_{1}M+L_{2}M$ terms of the forms (a) and
(b) of Lemma 3.4.

Let us consider a term of the form $[v,x]$, with $v\in V$ and
$x\in\mathcal{D}$. By Lemma 2.6 (i),

$[v,x]=\frac{1}{t}(e^{tx}ve^{-tx}-v)+t\Delta_{1},$

where $\|\Delta_{1}\|\leq\|v\|e^{2\|x\|}$. Observe that
$\frac{1}{t}(e^{tx}ve^{-tx}-v)$ is the sum of two elements of $\frac{1}{t}V$,
since $e^{tx}\in DU_{0}(A)$ (by [RobertNormal, Theorem 6.2]) and $V$ is
invariant under conjugation by $DU_{0}(A)$ and multiplication by $-1$. Since
$\|v\|<1$ (as we have assume that $V$ is contained in the open unit ball) and
$\|x\|\leq 2$, $\|\Delta_{1}\|$ is bounded by $C_{1}=e^{4}$.

Let us consider a term of the form $[[v,y],z]$, with $v\in V$ and
$y,z\in\mathcal{D}$. By Lemma 2.6 (ii),

$[[v,y],z]=\frac{1}{t^{2}}e^{tz}\Big{(}e^{ty}ve^{-ty}-v\Big{)}e^{-tz}-\frac{1}{t}(e^{ty}ve^{-ty}-v)+t\Delta_{2}$

where

$\|\Delta_{2}\|\leq\|v\|\cdot(e^{2\|y\|+2\|z\|}+2\|z\|e^{2\|y\|}).$

Again, since $e^{ty}$ and $e^{tz}$ belong to $DU_{0}(A)$, we have expressed
$[[v,y],z]$ as a sum of two elements in $\frac{1}{t^{2}}V$, two elements in
$\frac{1}{t}V$, plus the error term $t\Delta_{2}$. Since $\|v\|<1$ and
$\|y\|,\|z\|\leq 2$, we can choose a universal constant $C_{2}$ bounding
$e^{2\|y\|+2\|z\|}+2\|z\|e^{2\|y\|}$. Then $\|\Delta_{2}\|\leq C_{2}$.

Adding all the equations for the terms of the forms $[v,x]$ and $[[v,y],z]$
derived above, we get

(3.4)
$z=\frac{1}{t}\sum_{j=1}^{2(L_{1}+L_{2})M}v_{j}+\frac{1}{t^{2}}\sum_{j=1}^{2L_{2}M}v_{j}^{\prime}+t\Delta,$

where $v_{j},v_{j}^{\prime}\in V$ for all $j$ and $\|\Delta\|\leq C_{3}M$ for
$C_{3}=\max(C_{1},C_{2})$. Let $N^{\prime}\in\mathbb{N}$ be such that

$C_{3}MN<N^{\prime}\leq C_{3}MN+1.$

Set $t=1/N^{\prime}$ in (3.4). Then on the right-hand side of (3.4) we have a
sum of

$2(L_{1}+L_{2})MN^{\prime}+2L_{2}M(N^{\prime})^{2}$

elements in $V$ plus the error term $\|t\Delta\|<1/N$. Since $N^{\prime}\leq
C_{3}MN+1$, the number of terms in $V$ is bounded by $CM^{3}N^{2}$, for a
suitable universal constant $C$. The lemma is thus proved. ∎

Next, we get all “compactly supported” square zero elements in
$\sum^{n}(V+iV)$ for large enough $n$. Let $x\in A$ be a square zero element.
Write $x=|x^{*}|W$, where $W$ is a partial isometry in the bidual $A^{**}$.
(That is, start with $x^{*}=W^{*}|x^{*}|$, the polar decomposition of $x^{*}$
in $A^{**}$, and take adjoints.) Let $f\in C_{c}(0,\|x\|]$, where
$C_{c}(0,\|x\|]$ denotes the functions of compact support in $C_{0}(0,\|x\|]$.
Define $x_{f}=f(|x^{*}|)W$, which is an element of $xAx$ and so a square zero
element in $A$. Now define

$\mathcal{N}_{2}^{1,c}=\\{x_{f}:x\in\mathcal{N}_{2}\hbox{ and }f\in
C_{c}(0,1]\hbox{ such that $\|x\|\leq 1$ and $0\leq f\leq 1$}\\}.$

######  Lemma 3.6.

There exists $n\in\mathbb{N}$ such that
$\mathcal{N}_{2}^{1,c}\subseteq\sum^{n}(V+iV)$. Moreover, $n\leq
C^{\prime}M^{3}$, where $C^{\prime}$ is a universal constant.

###### Proof.

Let $x\in\mathcal{N}_{2}$, with $\|x\|\leq 1$. Write $x=bW$, with $b=|x^{*}|$
and $W$ a partial isometry in $A^{**}$. Let $f\in C_{c}(0,1]$ be such that
$0\leq f\leq 1$, and consider the element $x_{f}=f(b)W$ in
$\mathcal{N}_{2}^{1,c}$. We show below that $x_{f}\in\sum^{n}(V+iV)$, with $n$
independent of $x$ and as in the statement of the lemma.

Choose $g\in C_{c}(0,1]$ such that $0\leq g\leq 1$ and
$f=(g^{2}-\frac{1}{2})_{+}$. (To define $g$, let $\delta>0$ be such that
$f|_{(0,\delta]}=0$. Now set $g(t)=\sqrt{f(t)+1/2}$ for $t\geq\delta$,
$g(t)=0$ for $t<\delta/2$, and let $g$ be linear on the interval
$[\delta/2,\delta]$.)

Any $y\in\mathcal{N}_{2}$ can be expressed as a commutator by writing
$y=[w|y|^{\frac{1}{2}},|y|^{\frac{1}{2}}]$, where $y=w|y|$ is the polar
decomposition of $y$ in the bidual $A^{**}$. If $\|y\|\leq 1$, then
$y_{1}=w|y|^{\frac{1}{2}}$ and $y_{2}=|y|^{\frac{1}{2}}$ also have norm $\leq
1$. Decomposing $y_{1}$ into its selfadjoint and skewadjoint parts, we deduce
that $y\in\mathcal{D}+i\mathcal{D}$ for all $y\in\mathcal{N}_{2}$ of norm
$\leq 1$.

Consider $\frac{1}{2}g(b)W$, which is a square zero element in $A$. As argued
in the previous paragraph, there exist $z_{1},z_{2}\in\mathcal{D}$ such that
$\frac{1}{2}g(b)W=z_{1}+iz_{2}$. Applying Lemma 3.5 with $N=20$ to $z_{1}$ and
$z_{2}$ we obtain $n\in\mathbb{N}$ (depending only on $V$) and
$v\in\sum^{n}(V+iV)$ such that

$\|\frac{1}{2}g(b)W-v\|<\frac{1}{10}.$

Moreover, $n\leq 400CM^{3}$, where $C$ is the universal constant from Lemma
3.5.

Choose positive contractions $e,e^{\prime}\in C^{*}(b)_{+}$ such that
$eg(b)=g(b)$ and $e^{\prime}e=e$. (They are guaranteed to exist since $g$
vanishes on a neighborhood of 0.) Set $h=W^{*}eW$, which is an element of
$\mathrm{her}(x^{*}x)$. (The mapping $a\mapsto W^{*}aW$ is a C*-algebra
isomorphism from $\mathrm{her}(xx^{*})$ to $\mathrm{her}(x^{*}x)$.)
Multiplying $\frac{1}{2}g(b)W-v$ on the left by $e$ and on the right by $h$ we
get that

$\|\frac{1}{2}g(b)W-evh\|<\frac{1}{10}.$

Set $v^{\prime}=evh$. Then $v^{\prime}=yy^{*}vy^{*}y$, where
$y=e^{\frac{1}{2}}W\in A$ is a square zero element of norm at most $1$. It
then follows from Lemma 2.4, applied twice, that $v^{\prime}$ is a sum of $64$
conjugates of $\pm v$ and $\pm iv$ by unitaries in $DU_{0}(A)$. Hence,
$v^{\prime}\in\sum^{64n}(V+iV)$.

Observe that $v^{\prime}=cW$, where $c=ev(W^{*}e)\in A$ is such that
$e^{\prime}c=ce^{\prime}=c$, and in particular, $c\in\mathrm{her}(xx^{*})$.
Let us show that $c^{*}W$ belongs to $\sum^{8n}(V+iV)$. Since $W^{*}e$ is a
square zero element in $A$ of norm $\leq 1$, $(W^{*}e)v(W^{*}e)$ is a sum of 8
conjugates of $\pm v$ and $\pm iv$ by unitaries in $DU_{0}(A)$, by Lemma 2.4.
Hence, $(W^{*}e)v(W^{*}e)$ belongs to $\sum^{8n}(V+iV)$. But

$c^{*}W=(W^{*}e)^{*}v^{*}eW=((W^{*}e)v(W^{*}e))^{*}.$

Since $V+iV$ is a selfadjoint set, $c^{*}W$ is an element of
$\sum^{8n}(V+iV)$.

We have that

$\|\frac{1}{2}g(b)-c\|=\|\frac{1}{2}g(b)W-cW\|=\|\frac{1}{2}g(b)W-v^{\prime}\|<\frac{1}{10}.$

Set $c^{\prime}=c+c^{*}$. Then $\|g(b)-c^{\prime}\|<1/5$. From
$\|g(b)-c^{\prime}\|<1/5$ and $\|g(b)\|\leq 1$ we easily deduce that
$\|(g(b)^{2}-(c^{\prime})^{2}\|<\frac{1}{2}$. Hence, by the Kirchberg-Rørdam
Lemma ([KirchbergRordam2, Lemma 2.2]), there exists a contraction $d\in B$
such that

$f(b)=(g(b))^{2}-\frac{1}{2})_{+}=d(c^{\prime})^{2}d^{*}.$

We now apply Lemma 2.5 in the C*-algebra $D:=\mathrm{her}(x^{*}x+xx^{*})\cong
M_{2}(\mathrm{her}(xx^{*}))$ to get the square zero element $x_{f}=f(b)W$
expressed as a sum of 33540 unitary conjugates of $c^{\prime}W$, where the
unitaries are in $DU_{0}(D+\mathbb{C}1_{A})\subseteq DU_{0}(A)$. On the other
hand, since $cW\in\sum^{64n}(V+iV)$ and $c^{*}W\in\sum^{8n}(V+iV)$, we have
that

$c^{\prime}W=(c+c^{*})W\in\sum^{72n}(V+iV).$

It follows that $x_{f}$ is a sum of $33540\cdot 72n$ elements in $V$. Since
$n\leq 400CM^{3}$, the lemma readily follows. ∎

Recall that we assume throughout this section that $A$ is a unital C*-algebra
containing a full square zero element. These hypotheses come into play in the
proof of the following lemma.

######  Lemma 3.7.

There exists $m\in\mathbb{N}$ such that
$\mathcal{D}\subseteq\sum^{m}\mathcal{N}_{2}^{1,c}$.

###### Proof.

By assumption, there exists a full element $x_{0}\in\mathcal{N}_{2}$. Hence,
the positive element $x_{0}x_{0}^{*}$ is also full. We can express this as a
relation in the Cuntz semigroup of $A$: there exists $k\in\mathbb{N}$ such
that $[1]_{\mathrm{Cu}}\leq k[x_{0}x_{0}^{*}]_{\mathrm{Cu}}$. Here
$[a]_{\mathrm{Cu}}$ denotes the Cuntz class of a positive element $a\in A$ in
the Cuntz semigroup $\mathrm{Cu}(A)$.

Consider the set

$\mathcal{Y}=\\{x\in\mathcal{N}_{2}:[1]\leq k[xx^{*}]_{\mathrm{Cu}}\\}.$

Observe that $\mathcal{Y}$ is non-empty as $x_{0}\in\mathcal{Y}$.

_Claim_ : There exists $n_{0}\in\mathbb{N}$ such that each $z\in\mathcal{D}$
is a sum of $n_{0}$ elements of $\mathcal{Y}$ of norm at most 1. Proof: Let us
show first that $\mathcal{Y}$ is invariant under multiplication by nonzero
scalars, fully noncentral, and invariant under similarity. Clearly,
$\mathcal{Y}$ is closed under multiplication by non-zero scalars, as the Cuntz
class of a positive element does not change under multiplication by a positive
scalar. Moreover, $\mathcal{Y}$ is fully noncentral, for if $x_{0}$ maps into
the center in some quotient $A/I$, then it maps to 0, as the only central
square zero element is zero. But then $I=A$, by the fullness of $x_{0}$. Let
us prove invariance under similarity. Let $x\in\mathcal{N}_{2}$ and let $s\in
A$ be invertible. Set $y=sxs^{-1}$, which is a square zero element. Let us
show that $yy^{*}$ and $xx^{*}$ are Cuntz equivalent positive elements. We
have

$yy^{*}=s^{-1}xss^{*}x^{*}(s^{-1})*\precsim_{\mathrm{Cu}}xss^{*}x^{*}\leq\|s\|^{2}xx^{*}.$

This shows that $yy^{*}\precsim_{\mathrm{Cu}}xx^{*}$. Since $x$ is also
similar to $y$, $xx^{*}\precsim_{\mathrm{Cu}}yy^{*}$. Thus
$[xx^{*}]_{\mathrm{Cu}}=[yy^{*}]_{\mathrm{Cu}}$, and in particular if
$x\in\mathcal{Y}$ then $y\in\mathcal{Y}$ as well. We can now apply Theorem
2.3. It follows from this theorem that there exist $n_{0}\in\mathbb{N}$ and
$C>0$ such that every $z\in\mathcal{D}$ is a sum of $n_{0}$ elements in
$\mathcal{Y}$ of norm $\leq C$. Since $\mathcal{Y}$ is closed under
multiplication by nonzero scalars, we can assume that $C=1$, enlarging $n_{0}$
if necessary. This proves the claim.

In view of the previous claim, to prove the lemma it suffices to show that
there exists $n_{1}\in\mathbb{N}$ such that each element in $\mathcal{Y}$ or
norm $\leq 1$ is a sum of $n_{1}$ elements of $\mathcal{N}_{2}^{1,c}$.

Let $x\in\mathcal{Y}$ be such that $\|x\|\leq 1$. Let $B=\mathrm{her}(xx^{*})$
and write $x=bW$, with $b=|x^{*}|\in B$ and $W\in A^{**}$ a partial isometry.

Since $[1]_{\mathrm{Cu}}\leq k[b]_{\mathrm{Cu}}$, there exists $\epsilon>0$
such that $[1]_{\mathrm{Cu}}\leq k[(b-\epsilon)_{+}]_{\mathrm{Cu}}$. Then

$1=\sum_{j=1}^{k}x_{j}(b-\epsilon)_{+}x_{j}^{*},$

for some $x_{1},\ldots,x_{k}\in A$. Multiplying by $b^{\frac{1}{2}}$ on the
left and by $b^{\frac{1}{2}}W$ on the right we obtain that

(3.5)
$x=bW=\sum_{j=1}^{k}b^{\frac{1}{2}}x_{j}(b-\epsilon)_{+}x_{j}^{*}b^{\frac{1}{2}}W.$

Let $f\in C_{c}(0,1]$ be such that $0\leq f\leq 1$ and $f(t)=1$ for
$t\geq\epsilon$. Set $b^{\prime}=f(b)$, and observe that
$b^{\prime}(b-\epsilon)_{+}=(b-\epsilon)_{+}$. Then, for the terms on the
right-hand side of (3.5) we have that

$\displaystyle b^{\frac{1}{2}}x_{i}(b-\epsilon)_{+}x_{j}^{*}b^{\frac{1}{2}}W$
$\displaystyle=(b^{\frac{1}{2}}x_{j}((b-\epsilon)_{+})^{\frac{1}{2}})(b^{\prime})^{2}(((b-\epsilon)_{+})^{\frac{1}{2}}x_{j}^{*}b^{\frac{1}{2}})W$
$\displaystyle=r(b^{\prime})^{2}r^{*}W,$

where $r=b^{\frac{1}{2}}x_{j}((b-\epsilon)_{+})^{\frac{1}{2}}\in B$. Observe
that $\|r\|\leq 1$, since $\|b\|\leq 1$ and
$x_{j}(b-\epsilon)_{+}x_{j}^{*}\leq 1$. We now apply Lemma 2.5 in the
C*-algebra $D=\mathrm{her}(x^{*}x+xx^{*})\cong M_{2}(B)$. By this lemma we can
express $r(b^{\prime})^{2}r^{*}W$ as a sum of 33540 conjugates of $\pm
b^{\prime}W$ and $\pm ib^{\prime}W$ by unitaries in
$DU_{0}(D+\mathbb{C}1_{A})\subseteq DU_{0}(A)$. Since
$b^{\prime}W\in\mathcal{N}_{2}^{1,c}$, every term on the right-hand side of
(3.5) is a sum of $33540$ elements in $\mathcal{N}_{2}^{1,c}$. Thus, $x$ is a
sum of $33540k$ such terms. The lemma is thus proved. ∎

###### Proofs of Theorems 3.1 and 3.2.

Clearly it suffices to prove Theorem 3.2. Let $V$ be a set as in this theorem.
Suppose additionally that $V$ is contained in the open unit ball. Combining
Lemmas 3.6 and 3.7 we conclude that there exists $n\in\mathbb{N}$ such that
$\mathcal{D}\subseteq\sum^{n}(V+iV)$, and comparing skewadjoint parts we
further deduce that $\mathcal{D}\subseteq\sum^{n}V$. Here $n\leq
C^{\prime}mM^{3}$, where $m$ is as in Lemma 3.7, $C^{\prime}$ is the universal
constant from Lemma 3.6, and $M$ is associated to $V$ as in Lemma 3.3.

Consider the equation

$1=\sum_{j=1}^{M}x_{j}\pi_{3}(v_{j,1},\ldots,v_{j,8})y_{j},$

with $\|x_{j}\|,\|y_{j}\|<1$ for all $j$ and $v_{j,k}\in V$ for all $j,k$.
Clearly then we can choose $\epsilon>0$ such that if
$\|v_{j,k}-v_{j,k}^{\prime}\|<\epsilon$ for some elements $v_{j,k}^{\prime}\in
A$ with $j=1,\ldots,M$ and $k=1,\ldots,8$, then

$w=\sum_{j=1}^{M}x_{j}\pi_{3}(v_{j,1}^{\prime},\ldots,v_{j,8}^{\prime})y_{j}$

is invertible. Moreover, we can decrease $\epsilon>0$ if necessary so that
$\|w^{-1}x_{j}\|<1$ and $\|v_{j,k}^{\prime}\|<1$ for all $j,k$. Let
$F=\\{v_{j,k}:j=1,\ldots,M,\,k=1,\ldots,8\\}$. Now let $V^{\prime}\subseteq
iA_{\mathrm{sa}}$ be a fully noncentral set invariant under conjugation by
unitaries in $DU_{0}(A)$ and multiplication by $-1$. Suppose further that
$F\subseteq_{\epsilon}V^{\prime}$. Choose $v_{j,k}^{\prime}\in V^{\prime}$
such that $\|v_{j,k}-v_{j,k}^{\prime}\|<\epsilon$ for all $j,k$. Let

$W^{\prime}=\\{\pm uv_{j,k}^{\prime}u^{*}:j=1,\ldots,M,\,k=1,\ldots,8,\,u\in
DU_{0}(A)\\}\subseteq V^{\prime}.$

Then $W^{\prime}$ is a fully noncentral subset of $iA_{\mathrm{sa}}$ invariant
under conjugation by unitaries in $DU_{0}(A)$ and multiplication by $-1$, and
$W^{\prime}$ is contained in the open unit ball. The equation

$1=\sum_{i=1}^{M}(w^{-1}x_{j})\pi_{3}(v_{j,1}^{\prime},\ldots,v_{j,8}^{\prime})y_{j}$

shows that it shares the constant $M$ from Lemma 3.3 with $V$. Hence,

$\mathcal{D}\subseteq\sum^{n}W^{\prime}\subseteq\sum^{n}V^{\prime}.$

This proves the theorem in the case that $V$ is contained in the open unit
ball.

Finally, suppose that $V$ is as in Theorem 3.2, but it is not necessarily
contained in the unit ball. By Lemma 3.3,

$1=\sum_{i=1}^{M}x_{j}\pi_{3}(v_{j,1},\ldots,v_{j,8})y_{i},$

for some $M\in\mathbb{N}$, $v_{j,k}\in V$, and $\|x_{j}\|,\|y_{j}\|<1$. Let
$L=1+\max_{j,k}\|v_{j,k}\|$. Define the set

$W=\\{\pm uv_{j,k}u^{*}:j=1,\ldots,M,\,k=1,\ldots,8,\,u\in
DU_{0}(A)\\}\subseteq V.$

Then $\frac{1}{L}W$ is a fully noncentral subset of $iA_{\mathrm{sa}}$
invariant under conjugation by unitaries in $DU_{0}(A)$ and multiplication by
$-1$, and additionally $W$ is contained in the open unit ball. Hence,
$\mathcal{D}\subseteq\frac{1}{L}\sum^{n}W$ for some $n\in\mathbb{N}$. Since
$\frac{1}{L}\mathcal{D}\subseteq\mathcal{D}$, we also have that
$\mathcal{D}\subseteq\sum^{n}W$. Similarly, if a finite set
$F\subseteq\frac{1}{L}W$ and $\epsilon>0$ satisfy the requisite property
relative to $\frac{1}{L}W$, then $LF\subseteq W$ and $L\epsilon$ have the same
property relative to $W$. Thus, the validity of the theorem for $\frac{1}{L}W$
implies that the theorem is valid for $W$, whence also for $V$. ∎

######  Remark 3.8.

If $A$ is a simple unital C*-algebra other than $\mathbb{C}$, then it contains
a non-zero (hence full) square zero element, by Glimm’s halving lemma. If $A$
has no 1-dimensional representations and stable rank one, then again the
existence of a full square zero element is guaranteed by [APRT, Theorem 9.1].
So in these cases we can drop the existence of a full square zero element from
the hypotheses of the previous theorem (the fullness of the set $[H,A]$
already precludes the existence of 1-dimensional representations). If $A$ has
no finite dimensional representations and real rank zero, then again it
contains a full square zero element by [elliott-rordam, Corollary 2.4]. It is
an open problem, known as the “Global Glimm halving problem", whether every
unital C*-algebra without finite dimensional representations must contain a
full square zero element.

##  4\. The multiplicative setting

Throughout this section $A$ denotes a unital C*-algebra.

######  Lemma 4.1.

For all $n\in\mathbb{N}$ there exists $\epsilon>0$ such that if
$h_{1},\ldots,h_{n}\in[A,A]\cap A_{\mathrm{sa}}$ are such that
$\|h_{k}\|<\epsilon$ for $k=1,\ldots,n$, then

$e^{i(\sum_{k=1}^{n}h_{k})}=\prod_{k=1}^{n}e^{ih_{k}^{\prime}},$

where $h_{k}^{\prime}$ is a conjugate of $h_{k}$ by a unitary in $DU_{0}(A)$
for all $k$.

###### Proof.

The case $n=2$ is proven in [RobertNormal, Theorem 5.2]. More specifically,
there exists $\epsilon>0$ such that if $\|h_{1}\|,\|h_{2}\|<\epsilon$ then

$e^{i(h_{1}+h_{2})}=e^{ir}e^{ih_{1}}e^{-ir}e^{is}e^{ih_{2}}e^{-is},$

where $r,s\in A_{\mathrm{sa}}$ are such that $r\in h_{1}+[A,A]$ and $s\in
h_{2}+[A,A]$. Since $h_{1},h_{2}\in[A,A]$, it follows that $r,s\in[A,A]$, and
so $e^{ir},e^{is}\in DU_{0}(A)$ ([RobertNormal, Theorem 6.2]).

The general case follows by induction and the $n=2$ case. ∎

Given $X\subseteq A$ and $n\in\mathbb{N}$, we denote by $X^{n}$ the set of
products of $n$ elements of $X$.

Given $a,b\in GL(A)$, we denote their multiplicative commutator
$aba^{-1}b^{-1}$ by $(a,b)$.

For a unitary $u\in U(A)$, $a\in A_{\mathrm{sa}}$, and
$0<|t|<\frac{\ln(2)}{2\|a\|}$ define

$W_{t}(u,a)=\frac{1}{it}\log((u,e^{ita})).$

Extend $W_{t}(u,a)$ to $t=0$ by continuity by setting $W_{0}(u,a)=uau^{*}-a$.
We have that $W_{t}(u,a)\in[A,A]\cap A_{\mathrm{sa}}$ for all $t$ by
[RobertNormal, Theorem 5.1] (essentially by the Campbell-Baker-Hausdorff-
Dynkin formula).

Recall that we call a set $H\subseteq A$ fully noncentral if $[H,A]$ is a full
subset of $A$. Recall also that we denote by $\mathcal{D}$ the set
$\\{[a,b]:a,b\in A_{\mathrm{sa}},\,\|a\|,\|b\|\leq 1\\}$. We denote this set
by $\mathcal{D}_{A}$ if reference to the underlying C*-algebra is necessary. A
set $H\subseteq GL(A)$ is called symmetric if $H=H^{-1}$.

######  Theorem 4.2.

Let $A$ be a unital C*-algebra containing a full square zero element. Let
$H\subseteq U_{0}(A)$ be a fully noncentral symmetric set invariant under
conjugation by $DU_{0}(A)$. Then there exists $n\in\mathbb{N}$ such that
$\\{e^{z}:z\in\mathcal{D}\\}\subseteq H^{n}$.

###### Proof.

Let $B=C([0,1],A)$. Define $V\subseteq iB_{\mathrm{sa}}$ as the smallest set
containing the functions

$\Big{\\{}[0,1]\ni t\mapsto iW_{t}(u,a):u\in H,\,a\in[A,A]\cap
A_{\mathrm{sa}},\,\|a\|\leq\frac{\ln(2)}{4}\Big{\\}},$

and invariant under conjugation by unitaries in $DU_{0}(B)$ and multiplication
by $-1$. Let us show that $V$ is fully noncentral. Suppose for the sake of
contradiction that $V$ becomes central on a nonzero quotient of $B$, and
assume without loss of generality that this quotient is simple. Then the
quotient map factors through a point evaluation homomorphism. Thus, there
exists $t_{0}\in[0,1]$ such that the image of $V$ under the point evaluation
at $t=t_{0}$ is not fully noncentral. Hence, for a nonzero quotient $A/I$, the
set

$\Big{\\{}iW_{t_{0}}(u,a):u\in H,\,a\in[A,A]\cap
A_{\mathrm{sa}},\,\|a\|\leq\frac{\ln(2)}{4}\Big{\\}}$

is mapped to the center of $A/I$ by the quotient map. Since
$\frac{1}{t}W_{t_{0}}(u,ta)\to uau^{*}-a$ as $t\to 0$, $uau^{*}-a$ is mapped
to the center of $A/I$ for all $u\in H$ and $a\in[A,A]\cap A_{\mathrm{sa}}$.
Let $\bar{u},\bar{a}$ denote the images of $u\in H$ and $a\in[A,A]\cap
A_{\mathrm{sa}}$ in $A/I$. Since $\bar{u}$ commutes with
$\bar{u}\bar{a}\bar{u}^{*}-\bar{a}$, $\bar{u}\bar{a}\bar{u}^{*}-\bar{a}$ is
quasinilpotent by the Kleinecke-Shirokov theorem ([kleinecke]). Since it is
also a selfadjoint element, $\bar{u}\bar{a}\bar{u}^{*}-\bar{a}=0$. This
readily implies that $\bar{u}$ commutes with $[A/I,A/I]$; in particular,
$[\bar{u},[\bar{u},A/I]]=0$ for all $u\in H$. By Herstein’s theorem [herstein,
Theorem 1], $H$ is mapped to the center in $A/I$, which contradicts that $H$
is fully noncentral.

The C*-algebra $B$ contains full square zero elements, namely, any constant
function $t\mapsto x$, where $x\in A$ is a full square zero element. Since $B$
and $V$ satisfy the requisite hypotheses from Theorem 3.1, there exists
$n_{1}\in\mathbb{N}$ such that for each $z\in\mathcal{D}$ we have that

$z=\sum_{j=1}^{n_{1}}(-1)^{k_{j}}iW_{t}(u_{j}(t),a_{j}(t))\quad(0\leq t\leq
1)$

where the left-hand side is regarded as a constant function in $B$, and where
$u_{j}(t)\in H$ and $a_{j}(t)\in[A,A]\cap A_{\mathrm{sa}}$ is of norm
$\leq\frac{\ln(2)}{4}$ for all $j$ and all $t$. Multiplying by $t$ on both
sides and exponentiating we get

(4.1)
$e^{tz}=e^{\sum_{j=1}^{n_{1}}(-1)^{k_{j}}\log((u_{j}(t),e^{ita_{j}(t)}))}.$

Recall that $\log((u_{j}(t),e^{ita_{j}(t)}))\in[A,A]\cap iA_{\mathrm{sa}}$
(e.g., by [RobertNormal, Theorem 5.1]). Let $\delta>0$ be such that Lemma 4.1
is valid for $n_{1}$ elements of norm $<\delta$. Let $\epsilon>0$ be such that

$\|\log((u,e^{ita}))\|<\delta$

for all $\|a\|\leq\frac{\ln(2)}{4}$, $0\leq t<\epsilon$, and all unitaries
$u$. Then, applying Lemma 4.1 on the right-hand side of (4.1) we get

(4.2)
$e^{tz}=\prod_{j=1}^{n_{1}}(u_{j}^{\prime}(t),e^{ita_{j}^{\prime}(t)})^{\pm
1}.$

for all $0\leq t<\epsilon$, where $u_{j}^{\prime}(t)$ and $a_{j}^{\prime}(t)$
are conjugates of $u_{j}(t)$ and $a_{j}(t)$ by unitaries in $DU_{0}(A)$ for
all $j$. Since $H$ is invariant under conjugation by $DU_{0}(A)$,
$u_{j}^{\prime}(t)\in H$ for all $j$. Furthermore, since $a_{j}(t)\in[A,A]\cap
A_{\mathrm{sa}}$, $e^{ita_{j}^{\prime}(t)}\in DU_{0}(A)$ (by [RobertNormal,
Theorem 6.2]). So the right-hand side of (4.2) belongs to $H^{2n_{1}}$. Choose
$n_{2}\in\mathbb{N}$ such that $1/n_{2}<\epsilon$. Then
$e^{\frac{1}{n_{2}}z}\in H^{2n_{1}}$ for all $z\in\mathcal{D}$, and so
$e^{z}\in H^{2n_{2}n_{1}}$ for all $z\in\mathcal{D}$. ∎

###### Proof of Theorem A.

If $A=\mathbb{C}$, the theorem is trivial. Assume that $A\neq\mathbb{C}$.
Then, by Glimm’s lemma, $A$ contains a nonzero (and necessarily full) square
zero element.

Let $G$ be a normal subgroup of $DU_{0}(A)/Z(DU_{0}(A))$ not consisting solely
of the identity element. Let $H$ denotes the preimage of $G$ in $DU_{0}(A)$.
Then $H$ is a fully noncentral normal subgroup of $DU_{0}(A)$. By the previous
theorem, $H$ contains $e^{z}$, with $z\in\mathcal{D}$. Since these elements
generate $DU_{0}(A)$, by [RobertNormal, Theorem 6.2], $H=DU_{0}(A)$. It
follows that $G=DU_{0}(A)/Z(DU_{0}(A))$. ∎

######  Theorem 4.3.

Let $A$ be a unital C*-algebra containing a full square zero element. Let
$h\in A_{\mathrm{sa}}$ be fully noncentral. For each $\alpha\in\mathbb{R}$
define

$W_{\alpha}=\\{ue^{\pm i\alpha h}u^{*}:u\in DU_{0}(A)\\}.$

Then there exist $n\in\mathbb{N}$ and $\delta>0$ such that for each
$0<|\alpha|<\delta$ the set $W_{\alpha}^{n}$ contains a set of the form
$\\{e^{\delta^{\prime}z}:z\in\mathcal{D}\\}$ for some $\delta^{\prime}>0$.

###### Proof.

Assume without loss of generality that $\|h\|=1$.

The set

$V=\\{[uhu^{*},a]:u\in DU_{0}(A),\,a\in[A,A]\cap
A_{\mathrm{sa}},\,\|a\|\leq\frac{\ln 2}{4}\\}\subseteq iA_{\mathrm{sa}}$

is fully noncentral, invariant under conjugation by $DU_{0}(A)$ and under
multiplication by $-1$. (If $[h,[A,A]]$ is mapped to the center in a non-zero
quotient, then $h$ is also mapped to the center, by Herstein’s [herstein,
Theorem 1], violating that $h$ is fully noncentral.) Set $B=C([0,1],A)$ and
define $\widetilde{V}\subseteq iB_{\mathrm{sa}}$ by

$\widetilde{V}=\\{\bar{u}v\bar{u}^{*}:\bar{u}\in DU_{0}(B),\,v\in V\\},$

where $V$ is regarded as a set of constant functions in $B$. The set
$\widetilde{V}$ is again fully noncentral, as it is mapped onto $V$ by any
point evaluation map, and it is invariant under conjugation by $DU_{0}(B)$ and
under multiplication by $-1$. Let $n_{1}\in\mathbb{N}$,
$F\subseteq\widetilde{V}$, and $\epsilon_{1}>0$, be associated to
$\widetilde{V}$ as in Theorem 3.2, i.e., such that if
$F\subseteq_{\epsilon_{1}}\widetilde{V}^{\prime}$ for a fully noncentral set
$\widetilde{V}^{\prime}\subseteq iB_{\mathrm{sa}}$ invariant under conjugation
by $DU_{0}(B)$ and multiplication by $-1$, then
$\mathcal{D}_{B}\subseteq\sum^{n_{1}}\widetilde{V}^{\prime}$.

For $\alpha\neq 0$, define $\widetilde{V}_{\alpha}\subseteq iB_{\mathrm{sa}}$
as the smallest set containing the functions

$\Big{\\{}t\mapsto\frac{i}{\alpha}W_{t}(e^{i\alpha h},a):a\in[A,A]\cap
A_{\mathrm{sa}},\|a\|\leq\frac{\ln 2}{4}\Big{\\}}$

and invariant under conjugation by unitaries in $DU_{0}(B)$ and multiplication
by $-1$. Then $\widetilde{V}_{\alpha}$ is symmetric, invariant under
conjugation by $DU_{0}(B)$, and fully noncentral. Applying the Campbell-Baker-
Hausdorff-Dynkin formula three times, we have that for small enough $\alpha$

$\displaystyle\frac{i}{\alpha}W_{t}(e^{i\alpha h},a)$
$\displaystyle=\frac{1}{\alpha t}\log(e^{i\alpha h}e^{iat}e^{-i\alpha
h}e^{-iat})$ $\displaystyle=\frac{1}{\alpha t}\log(e^{i\alpha
h+iat+\frac{1}{2}[i\alpha h,iat]+\cdots}\cdot e^{-i\alpha
h-iat+\frac{1}{2}[i\alpha h,iat]+\cdots})$ $\displaystyle=[a,h]+\Delta,$

where $\|\Delta\|\leq C|\alpha t|$, for a constant $C$. (Recall that we have
assumed that $\|h\|=1$.) Thus, for small enough $\alpha$,
$\widetilde{V}_{\alpha}$ almost contains any constant function of the form
$t\mapsto[a,h]$, with $\|a\|\leq\frac{\ln(2)}{4}$, as well as any conjugation
of one such function by a unitary in $DU_{0}(B)$. Hence, with $\epsilon>0$ and
$F\subseteq\widetilde{V}$ as in the previous paragraph, we have that
$F\subseteq_{\epsilon}\widetilde{V}_{\alpha}$ for small enough $\alpha$. Let
$\delta>0$ be such that this holds for $0<|\alpha|<\delta$, and fix any such
$\alpha$. Then $z\in\sum^{n_{1}}\widetilde{V}_{\alpha}$ for all
$z\in\mathcal{D}_{A}$, where $z$ is understood as a constant function in $B$.
Thus,

$z=\sum_{j=1}^{n_{1}}(-1)^{k_{j}}\frac{i}{\alpha}W_{t}(u_{j}(t)e^{i\alpha
h}u_{j}(t)^{*},u_{j}(t)a_{j}u_{j}(t)^{*})\quad(0\leq t\leq 1),$

where $u_{j}\in DU_{0}(B)$ and $a_{j}\in[A,A]\cap A_{\mathrm{sa}}$ is such
that $\|a_{j}\|\leq\frac{\ln 2}{4}$ for all $j$. Multiplying by $\alpha t$ on
both sides and exponentiating we get

$e^{\alpha tz}=e^{\sum_{j=1}^{n_{1}}(-1)^{k_{j}}\log((u_{j}(t)e^{i\alpha
h}u_{j}(t)^{*},u_{j}(t)e^{ita_{j}}u_{j}(t)^{*}))}.$

As in the proof of Theorem 4.2, we now find $\delta^{\prime}>0$ such that
Lemma 4.1 can be applied to $n_{1}$ selfadjoint elements of norm
$<\delta^{\prime}$, and then find $\epsilon>0$ such that

$\|\log((u,e^{ita})\|<\delta^{\prime},$

for all $0<t\leq\epsilon$, unitary $u$, and $\|a\|\leq\frac{\ln(2)}{4}$. Then

$e^{\alpha
tz}=\prod_{j=1}^{n_{1}}(u_{j}^{\prime}(t),e^{ita_{j}^{\prime}(t)})^{\pm 1}.$

for $0<t\leq\epsilon$, where $u_{j}^{\prime}(t)$ is a conjugate of $e^{i\alpha
h}$ by a unitary in $DU_{0}(A)$ for all $j$ and all $0<t\leq\epsilon$. From
$a_{j}^{\prime}(t)\in[A,A]\cap A_{\mathrm{sa}}$ we get that
$e^{ita_{j}^{\prime}(t)}\in DU_{0}(A)$ for all $j$. It follows that
$W_{\alpha}^{2n_{1}}$ contains $\\{e^{\alpha\epsilon z}:z\in\mathcal{D}\\}$. ∎

Under the hypotheses of the previous theorem, fully noncentral selfadjoint
elements are guaranteed to exist.

######  Lemma 4.4.

If $x\in A$ is a full square zero element then $[x^{*},x]$ is a fully
noncentral selfadjojnt element.

###### Proof.

Let $h=[x^{*},x]$, where $x$ is a full square zero element of $A$. If $h+I$ is
central in a quotient $A/I$, then in particular $x+I$ commutes with
$[(x+I)^{*},x+I]$. By the Kleinecke-Shirokov theorem [kleinecke],
$[(x+I)^{*},x+I]$ is quasinilpotent, which in turn entails that it is zero,
since it is also selfadjoint. Hence, $[x^{*},x]\in I$. By functional calculus,
$x^{*}x=([x^{*},x])_{+}\in I$, and so $I=A$ by the fullness of $x$. ∎

##  5\. The special unitary group

Let us briefly recall the definition of the de la Harpe-Skandalis determinant.
We refer the reader to [delaHarpeSkandalis] for further details.

Let $A$ be a unital C*-algebra. Let $M_{\infty}(A)$ denote the *-algebra of
infinite matrices with entries in $A$ whose entries are all but finitely many
equal to zero. Regard $M_{n}(A)$ embedded in $M_{\infty}(A)$ as the $n\times
n$ top left corner matrices, so that
$M_{\infty}(A)=\bigcup_{n=1}^{\infty}M_{n}(A)$. Endow $M_{\infty}(A)$ with the
inductive limit locally convex topology and its unitization
$M_{\infty}(A)+\mathbb{C}1_{\infty}$ with the product of the topologies of
$M_{\infty}(A)$ and $\mathbb{C}$. Let $GL_{0}^{\infty}(A)\subseteq
M_{\infty}(A)+\mathbb{C}1$ denote the connected component of the identity in
the group of invertible elements. Regard $GL_{0}^{n}(A)$ (i.e.,
$GL_{0}(M_{n}(A))$) embedded in $GL_{0}^{\infty}(A)$ by the map $a\mapsto
a+(1_{\infty}-1_{n})$, where $1_{n}$ is the identity in $M_{n}(A)$. We endow
$GL_{0}^{\infty}(A)$ with the topology that it inherits from its inclusion in
$M_{\infty}(A)+\mathbb{C}1_{\infty}$. In this topology, a compact set is
always contained in $M_{n}(A)+\mathbb{C}1_{\infty}$ for some $n$. In
particular, a path $\alpha\colon[t_{1},t_{2}]\to GL_{0}^{\infty}(A)$ is always
contained in $GL_{0}^{n}(A)$ for some $n$. Let $U_{0}^{\infty}(A)$ denote the
unitary elements in $GL_{0}^{\infty}(A)$.

Let $T\colon A\to A/\overline{[A,A]}$ denote the quotient map, where the
codomain is regarded as a Banach space under the quotient norm. We call $T$
the universal trace map. Define $Tr\colon M_{\infty}(A)\to A$ as addition
along the main diagonal: $Tr((a_{ij})_{i,j})=\sum_{i}a_{ii}$. We extend $T$ to
$M_{\infty}(A)$ by setting $T((a_{i,j})_{i,j})=T(Tr((a_{i,j})_{i,j}))$ for all
$(a_{i,j})_{i,j}\in M_{\infty}(A)$.

Let $\alpha\colon[t_{1},t_{2}]\to GL_{0}^{\infty}(A)$ be a smooth path. Its de
la Harpe-Skandalis determinant is defined as

$\widetilde{\Delta}_{T}(\alpha)=\frac{1}{2\pi
i}\int_{t_{1}}^{t_{2}}T(\alpha^{\prime}(t)\alpha(t)^{-1})dt\in
A/\overline{[A,A]}.$

By the results of [delaHarpeSkandalis], if two paths $\alpha_{1}$ and
$\alpha_{2}$ both connect $1$ to an invertible element $a\in
GL_{0}^{\infty}(A)$, then

$\widetilde{\Delta}_{T}(\alpha_{1})-\widetilde{\Delta}_{T}(\alpha_{2})\in 2\pi
i\\{T(p)-T(q):p,q\in M_{\infty}(A)\hbox{ projections}\\}.$

Given an invertible element $a\in GL_{0}^{\infty}(A)$, its de la Harpe-
Skandalis determinant $\Delta_{T}(a)$ is defined as the image of
$\widetilde{\Delta}_{T}(\alpha)$ in the quotient

$(A/\overline{[A,A]})\,/\,(2\pi i\\{T(p)-T(q):p,q\in M_{\infty}(A)\hbox{
projections}\\}),$

where $\alpha$ is any path in $GL_{0}^{\infty}(A)$ connecting $1$ to $a$.

We are interested in $\ker\Delta_{T}\cap U_{0}(A)$, which is a normal subgroup
of $U_{0}(A)$. It can be described as follows: Let $u\in U_{0}(A)$, and let
$u=\prod_{j=1}^{n}e^{ih_{j}}$, with $h_{j}\in A_{\mathrm{sa}}$ for all $j$, be
any representation of $u$ as a product of exponentials. Then
$u\in\ker\Delta_{T}$ if and only if

(5.1) $\sum_{j=1}h_{j}\in\overline{[A,A]}+2\pi\\{Tr(p)-Tr(q):p,q\in
M_{\infty}(A)\hbox{ projections}\\}.$

Let $\tau\colon A\to\mathbb{C}$ be a tracial state. The de la Harpe-Skandalis
determinant $\Delta_{\tau}$ associated to $\tau$ is defined similarly to
$\Delta_{T}$, with $\tau$ in place of $T$: Given a smooth path
$\alpha\colon[t_{1},t_{2}]\to GL_{0}^{\infty}(A)$,

$\widetilde{\Delta}_{\tau}(\alpha)=\frac{1}{2\pi
i}\int_{t_{1}}^{t_{2}}\tau(\alpha^{\prime}(t)\alpha^{-1}(t))\in\mathbb{C}.$

Given $a\in GL_{0}^{\infty}(A)$ we choose any path
$\alpha\colon[t_{1},t_{2}]\to GL_{0}^{\infty}(A)$ connecting $1$ to $a$ and
define $\Delta_{\tau}(a)$ as the equivalence class of
$\widetilde{\Delta}_{\tau}(\alpha)$ in the quotient

$\mathbb{C}/2\pi i\\{\tau(p)-\tau(q):p,q\in M_{\infty}(A)\hbox{
projections}\\}.$

(Described in terms of the pairing of the group $K_{0}(A)$ with the tracial
states of $A$, we are taking the quotient of $\mathbb{C}$ by the group $2\pi
i\tau(K_{0}(A))$.) The group $\ker\Delta_{\tau}\cap U_{0}(A)$ consists of the
unitaries in $U_{0}(A)$ such that in any one (and all) representations as a
product of exponentials $\prod_{j=1}^{n}e^{ih_{j}}$, with $h_{j}\in
A_{\mathrm{sa}}$ for all $j$, one has that

(5.2) $\tau\Big{(}\sum_{j=1}^{n}h_{j}\Big{)}\in\\{\tau(p)-\tau(q):p,q\in
M_{\infty}(A)\hbox{ projections}\\}.$

######  Lemma 5.1.

Let $u\in U_{0}(A)$. The following are equivalent:

  1. (i)

There is a path $\eta\colon[t_{1},t_{2}]\to U_{0}(A)$ starting at 1 and ending
at $u$ such that $\widetilde{\Delta}_{T}(\eta)=0$.

  2. (ii)

$u$ belongs to the subgroup of $U_{0}(A)$ generated by
$\\{e^{ih}:h\in\overline{[A,A]}\cap A_{\mathrm{sa}}\\}$.

  3. (iii)

$u=\prod_{k=1}^{n}e^{ih_{k}}$, for some $h_{1},\ldots,h_{n}$ such that
$\sum_{k=1}^{n}h_{k}\in\overline{[A,A]}\cap A_{\mathrm{sa}}$.

  4. (iv)

There is a path $\eta\colon[0,1]\to U_{0}(A)$ starting at 1 and ending at $u$
such that $\widetilde{\Delta}_{T}(\eta|_{[0,t]})=0$ for all $t\in[0,1]$.

###### Proof.

(i) $\Rightarrow$ (ii): Choose $t_{1}=x_{0}<x_{1}<\cdots<x_{N}=t_{2}$, a
sufficiently fine partition of the interval $[t_{1},t_{2}]$ such that
$\|u(x_{k})-u(x_{k+1})\|<1$ for all $j$. Then $u=\prod_{k=1}^{N}e^{ih_{k}}$,
where $e^{ih_{k}}=u_{x_{k-1}}^{-1}u_{x_{k}}$ for all $k$. By the proof of
[delaHarpeSkandalis, Lemma 3], if the partition
$t_{1}=x_{0}<x_{1}<\cdots<x_{N}=t_{2}$ is sufficiently fine, then
$\sum_{k=1}^{N}h_{k}\in\overline{[A,A]}$

(ii) $\Rightarrow$ (iii) This is obvious.

(iii) $\Rightarrow$ (iv) The path $\eta(t)=\prod e^{ith_{k}}$, with
$t\in[0,1]$, is such that

$\widetilde{\Delta}_{T}(\eta|_{[0,t]})=T(t(\sum_{k=1}^{n}ih_{k}))=0$

for all $t\in[0,1]$.

(iv) $\Rightarrow$ (i) This is obvious. ∎

Let $SU_{0}(A)$ denote the subgroup of $U_{0}(A)$ of unitaries satisfying any
of the equivalent conditions of Lemma 5.1. Clearly, $SU_{0}(A)$ is a path
connected normal subgroup of $U_{0}(A)$ contained in $\ker\Delta_{T}\cap
U_{0}(A)$. By Lemma 5.1 (ii), $SU_{0}(A)$ is the Banach-Lie subgroup of $U(A)$
associated to the Lie algebra of skewadjoint commutators $\overline{[A,A]}\cap
iA_{\mathrm{sa}}$. We refer the reader to [hofmann-morris, Chapter 5] for
background on Banach-Lie groups.

We regard $SU_{0}(A)$ endowed with the topology with basis of neighborhoods at
the identity

$\\{e^{ih}:h\in\overline{[A,A]}\cap
A_{\mathrm{sa}},\,\|h\|<\epsilon\\}\quad(\epsilon>0).$

Then $\overline{[A,A]}\ni h\mapsto e^{ih}\in SU_{0}(A)$ is continuous
([hofmann-morris, Theorem 5.52]). The exponential length on $SU_{0}(A)$ is
defined as

(5.3)
$\mathrm{el}_{SU_{0}(A)}(u)=\Big{\\{}\inf\sum_{j=1}^{n}\|h_{j}\|:u=\prod_{j=1}^{n}e^{ih_{j}}\hbox{
and }h_{j}\in\overline{[A,A]}\cap A_{\mathrm{sa}}\hbox{ for all }j\Big{\\}}.$

This gives rise to a metric $(u,v)\mapsto\mathrm{el}(u^{*}v)$ that is left and
right invariant and induces the topology on $SU_{0}(A)$ (see [ando,
Proposition 3.2]). It is easily established that $SU_{0}(A)$ is complete under
this metric. Moreover, by [ando, Proposition 3.9], the metric induced by
$\mathrm{el}_{SU_{0}(A)}$ agrees with the “Finsler metric” defined by

$d_{SU_{0}(A)}(u,1)=\inf\Big{\\{}\int_{0}^{1}\|u^{\prime}(t)\|dt:t\mapsto
u(t)\hbox{ smooth path connecting 1 to $u$ in }SU_{0}(A).\Big{\\}}$

######  Remark 5.2.

The topology on $SU_{0}(A)$ need not be that induced by the topology on
$U_{0}(A)$. Take for example $A$ to be an infinite dimensional simple
C*-algebra of real rank zero with at least one tracial state (e.g. the CAR
C*-algebra $M_{2^{\infty}}$). Then $SU_{0}(A)$ is a norm dense proper subgroup
of $U_{0}(A)$ ([elliott-rordam0]). In particular, $SU_{0}(A)$ is not complete
under the uniform structure induced by the norm topology.

We summarize various properties of $SU_{0}(A)$ in the following theorem:

######  Theorem 5.3.

Let $A$ be a unital C*-algebra. The following are true:

  1. (i)

$DU_{0}(A)$ is a dense subgroup of $SU_{0}(A)$. If $\overline{[A,A]}=[A,A]$,
then $DU_{0}(A)=SU_{0}(A)$.

  2. (ii)

If $A$ is traceless, then $DU_{0}(A)=SU_{0}(A)=U_{0}(A)$, and the topology on
$SU_{0}(A)$ is that induced by the norm on $A$.

  3. (iii)

If $A$ is separable, then $SU_{0}(A)$ agrees with the path connected component
of 1 in $\ker\Delta_{T}\cap U_{0}(A)$.

  4. (iv)

If $A$ has stable rank one, then $SU_{0}(A)=\ker\Delta_{T}\cap U_{0}(A)$.

  5. (v)

If $A$ has real rank zero, then $SU_{0}(A)=\ker\Delta_{T}\cap U_{0}(A)$.

  6. (vi)

If $A$ is simple, then $SU_{0}(A)$ is topologically simple modulo its center.

###### Proof.

(i) This follows at once from the continuity of $\overline{[A,A]}\ni h\mapsto
e^{ih}\in SU_{0}(A)$ and the fact that $DU_{0}(A)$ is generated by the set
$\\{e^{ih}:h\in[A,A]\cap A_{\mathrm{sa}}\\}$ ([RobertNormal, Theorem 6.2]).

(ii) By Pop’s theorem [Pop, Theorem 1], if $A$ is a traceless unital
C*-algebra then $A=[A,A]$. Since the set $\\{e^{ih}:h\in[A,A]\cap
A_{\mathrm{sa}}\\}$ is a generating set of $DU_{0}(A)$, it follows that
$U_{0}(A)=DU_{0}(A)$, and a fortiori $SU_{0}(A)=U_{0}(A)$.

(iii) _Claim_ : If $[t_{1},t_{2}]\ni t\mapsto u(t)$ is a continuous path in
$\ker\Delta_{T}\cap U_{0}(A)$ such that $\|u(t)-1\|<1$ for all $t$ and
$u(t_{1})=1$, then $\log(u(t))\in\overline{[A,A]}$ for all
$t\in[t_{1},t_{2}]$. Proof: Let $h(t)=\log(u(t))$ for all $t\in[t_{1},t_{2}]$,
which is defined since $\|u(t)-1\|<1$, and such that $h(t_{1})=0$. Let $\tau$
be a tracial state. Then $t\mapsto\tau(h(t))$ varies continuously, and by the
description of $\ker\Delta_{T}\cap U_{0}(A)$ (see (5.1)), ranges in the set

$2\pi i\\{\tau(p)-\tau(q):p,q\in M_{\infty}(A)\hbox{ projections}\\}.$

This set is countable, since $A$ is separable and the value of a trace on a
projection is invariant under homotopy. Hence, $t\mapsto\tau(h(t))$ is
constant, and so it is 0 as $h(t_{1})=0$. This shows $\tau(h(t))=0$ for all
tracial states $\tau$, which in turn implies that $h(t)\in\overline{[A,A]}$
for all $t$. This proves the claim.

In virtue of the claim just established, if $u\in\ker\Delta_{T}\cap U_{0}(A)$
is connected to 1 through a path $t\mapsto u(t)$ inside $\ker\Delta_{T}\cap
U_{0}(A)$ and such that $\|u(t)-1\|<1$ for all $t$, then the path lies
entirely in $SU_{0}(A)$. We can drop the assumption that $\|u(t)-1\|<1$ by
partitioning the path into small enough pieces.

(iv) Let $u\in\ker\Delta_{T}\cap U_{0}(A)$, and write
$u=\prod_{k=1}^{n}e^{ih_{k}}$, where $h_{k}\in A_{\mathrm{sa}}$ for all $k$.
Then

$\sum_{k=1}^{n}h_{k}=h+2\pi iTr(p)-2\pi iTr(q),$

for some $h\in\overline{[A,A]}\cap A_{\mathrm{sa}}$ and projections $p,q\in
M_{\infty}(A)$. By [RobertNormal, Lemma 6.1], $u$ is equal modulo commutators
(in $DU_{0}(A)$) to $e^{ih}e^{2\pi iTr(p)}e^{-2\pi iTr(q)}$. Since
$DU_{0}(A)\subseteq SU_{0}(A)$ and $e^{ih}\in SU_{0}(A)$, it remains to prove
that any unitary of the form $v=e^{2\pi iTr(p)}$, with $p\in M_{n}(A)$ a
projection, belongs to $SU_{0}(A)$. Consider the projection loop $[0,1]\ni
t\mapsto e^{2\pi itp}$ in $M_{n}(A)$. By Rieffel’s theorems [rieffel1,
Corollary 8.6] and [rieffel2, Proposition 2.6], the loop $[0,1]\ni t\mapsto
e^{2\pi itp}$ is homotopic to a loop $\beta\colon[0,1]\to U_{0}(A)$ (here we
have used the stable rank one hypothesis). Now $\eta(t)=e^{2\pi
itTr(p)}\beta^{-1}(t)$ is a path in $U_{0}(A)$ connecting $1$ to $e^{2\pi
iTr(p)}$ and

$\widetilde{\Delta}_{T}(\eta)=\Delta(t\mapsto e^{2\pi
itTr(p)})-\Delta(\beta)=2\pi iTr(p)-2\pi iTr(p)=0.$

It follows by Lemma 5.1 that $u\in SU_{0}(A)$, as desired.

(v) Let $u\in\ker\Delta_{T}\cap U_{0}(A)$. By Lin’s [LinRR0, Theorem 5],
$u=e^{ih_{1}}e^{ih_{2}}$, where $\|h_{1}\|\leq\pi$ and the norm of $h_{2}$ can
be made arbitrarily small. Choosing $h_{2}$ of a sufficiently small norm we
have that $e^{ih_{1}}e^{ih_{2}}=e^{i(h_{1}+h_{2})}e^{ic}$, where
$c\in\overline{[A,A]}$ ([ker-det, Lemma 2.2]). Since $e^{ic}\in SU_{0}(A)$, it
remains to show that if $e^{ih}\in\ker\Delta_{T}\cap U_{0}(A)$, then
$e^{ih}\in SU_{0}(A)$. We have that $h=h^{\prime}+2\pi(Tr(p)-Tr(q))$, with
$h^{\prime}\in\overline{[A,A]}\cap A_{\mathrm{sa}}$ and $p,q\in M_{\infty}(A)$
projections. As in the proof of (iv), $e^{ih}$ is equivalent modulo
commutators to $e^{h^{\prime}}e^{2\pi iTr(p)}e^{-2\pi Tr(q)}$, so it remains
to show that $e^{2\pi iTr(p)}$ belongs to $SU_{0}(A)$ for any projection $p\in
M_{n}(A)$. By Zhang’s [zhang, Theorem 1.1], the Murray-von Neumann monoid of
projections of a C*-algebra of real rank zero has the Riesz decomposition
property. Hence, there exists $v\in M_{\infty}(A)$ such that $p=v^{*}v$ and
$q=vv^{*}$ is a diagonal matrix with projections $q_{1},\ldots,q_{n}\in A$
along the main diagonal. Observe that $Tr(v^{*}v)-Tr(vv^{*})\in[A,A]$. So
$e^{2\pi iTr(q)}$ is equivalent modulo $DU_{0}(A)$ to $e^{2\pi
i(\sum_{k=1}^{n}q_{k})}$, which in turn is equivalent modulo $DU_{0}(A)$ to
$\prod_{k=1}^{n}e^{2\pi iq_{k}}=1$ ([RobertNormal, Lemma 6.1]), thus
completing the proof.

(vi) Let $H\subseteq SU_{0}(A)$ be a closed normal subgroup not contained in
the center of $SU_{0}(A)$ (thus, not contained in the center of $A$ either).
Then $DU_{0}(A)\subseteq H$, by Corollary A. Since $DU_{0}(A)$ is dense in
$SU_{0}(A)$ by part (i), and $H$ is closed, $H=SU_{0}(A)$. ∎

######  Remark 5.4.

Consider the inclusions

$DU_{0}(A)\subseteq SU_{0}(A)\subseteq\ker\Delta_{T}\cap U_{0}(A).$

Both inclusions may be proper, though it is often the case that all three sets
agree (e.g., by a combination of the results in the preceding theorem). If
$A=M_{n}(C(X))$, for $X$ a compact Hausdorff space, then $DU_{0}(A)=SU_{0}(A)$
by [thomsen, Proposition 1.3], and it is not difficult to show that
$SU_{0}(A)=\ker\Delta_{T}\cap U_{0}(A)$. The examples that we construct in
Section 8 below show failure of the first inclusion for some simple AH
C*-algebras.

The question whether $SU_{0}(A)=\ker\Delta_{T}\cap U_{0}(A)$—equivalently, in
the separable case, whether $\ker\Delta_{T}\cap U_{0}(A)$ is a path connected
subset of $U_{0}(A)$—is closely related to whether the C*-algebra $A$ has
determinant unitary rank equal to 1, in the sense of Gong, Lin, and Xue [DUR].
In [DUR, Theorem 5.13], an example is given of a homogeneous C*-algebra with
determinant unitary rank $>1$. This is also an example where
$SU_{0}(A)\neq\ker\Delta_{T}\cap U_{0}(A)$. Briefly described,
$A=pM_{4}(C(S^{4}))p$, where $S^{4}$ denotes the 4-dimensional sphere and
$p\in M_{4}(C(S^{4}))$ is a suitably chosen rank 2 projection. By the results
in [DUR] there exists a projection $q\in M_{2}(A)$ equivalent to the trivial
rank one projection such that for no loop $\alpha\colon[0,1]\to U_{0}(A)$ can
one have $\widetilde{\Delta}_{T}(\alpha)=2\pi iTr(q)+\overline{[A,A]}$. It
follows that $u=e^{2\pi iTr(q)}\in U_{0}(A)$ does not belong to $SU_{0}(A)$,
since a path $\eta$ of zero determinant connecting $1$ to $u$ concatenated
with the path $[0,1]\ni t\mapsto e^{2\pi i(1-t)Tr(q)}$ would provide one such
loop. On the other hand, $u$ is clearly in the kernel of the determinant.

##  6\. Automatic continuity

Let us recall the definition of the property of bounded commutators generation
given in the introduction. Let $A$ be a unital C*-algebra. We say that $A$ has
bounded commutators generation if there exist $n\in\mathbb{N}$ and $C>0$ such
that for all $h\in\overline{[A,A]}$ we have

$h=\sum_{j=1}^{n}[x_{j},y_{j}]$

for some $x_{j},y_{j}\in A$ such that $\|x_{j}\|\cdot\|y_{j}\|\leq C\|h\|$ for
all $j$.

In the following theorem we gather some classes of C*-algebras known to have
bounded commutators generation:

######  Theorem 6.1.

The following classes of unital C*-algebras have bounded commutators
generation:

  1. (i)

traceless C*-algebras,

  2. (ii)

C*-algebras having finite decomposition rank and without finite dimensional
representations,

  3. (iii)

pure C*-algebras (i.e., with almost unperforated and almost divisible Cuntz
semigroup) where every bounded 2-quasitrace is a trace,

  4. (iv)

exact C*-algebras tensorially absorbing the Jiang-Su C*-algebra.

###### Proof.

(i) This is Pop’s [Pop, Theorem 1].

(ii) This follows from [RobertCommutators, Theorem 1.2].

(iii) This follows from [NgRobertpure, Theorem 4.10].

(iv) This follows from (iii), as a Jiang-Su stable C*-algebra is pure.
Moreover, exactness implies that bounded 2-quasitraces are traces, by
Haagerup’s theorem ([haagerup]). Thus, the conditions in (iii) are met. ∎

Recall that we denote by $\mathcal{D}$ the subset of $[A,A]\cap
iA_{\mathrm{sa}}$ defined in (3.1).

######  Lemma 6.2.

Let $A$ be a unital C*-algebra containing a full square zero element and
having bounded commutators generation. Then there exists $N\in\mathbb{N}$ such
that

$\\{e^{\epsilon z}:z\in\mathcal{D}\\}^{N}$

is a neighborhood of the identity in $SU_{0}(A)$ for all $\epsilon>0$.

###### Proof.

By the property of bounded commutators generation, there exists
$N\in\mathbb{N}$ such that if $h\in\overline{[A,A]}\cap A_{\mathrm{sa}}$ is of
norm $\leq 1$, then

$ih=\sum_{j=1}^{N}z_{j}$

for some $z_{j}\in\mathcal{D}$. Let $\epsilon_{0}>0$ be as in Lemma 4.1
applied to $n=N$ elements. Let $0<\epsilon<\epsilon_{0}$. Then

$e^{\epsilon ih}=\prod_{j=1}^{N}e^{\epsilon z_{j}^{\prime}},$

where $z_{j}^{\prime}\in\mathcal{D}$ for all $j$. Hence, the set
$\\{e^{\epsilon z}:z\in\mathcal{D}\\}^{N}$ contains all $e^{ih}$, with
$h\in\overline{[A,A]}\cap A_{\mathrm{sa}}$ of norm $\leq\epsilon$, and it is
thus a neighborhood of the identity in $SU_{0}(A)$. If on the other hand
$\epsilon\geq\epsilon_{0}$, then $\\{e^{\epsilon z}:z\in\mathcal{D}\\}$
contains $\\{e^{\frac{\epsilon_{0}}{2}z}:z\in\mathcal{D}\\}$, and so again
$\\{e^{\epsilon z}:z\in\mathcal{D}\\}^{N}$ is a neighborhood of the identity.
∎

######  Theorem 6.3.

Let $A$ be a unital C*-algebra containing a full square zero element and
having bounded commutators generation. Let $h\in\overline{[A,A]}\cap
A_{\mathrm{sa}}$ be fully noncentral. For each $\alpha\in\mathbb{R}$, let

$H_{\alpha}=\\{ue^{\pm i\alpha h}u^{*}:u\in SU_{0}(A)\\}.$

Then there exist $n\in\mathbb{N}$ and $\delta>0$ such that the sets
$H_{\alpha}^{n}$, with $0<\alpha\leq\delta$, form a basis of neighborhoods of
the identity in $SU_{0}(A)$.

###### Proof.

By Theorem 4.3, there exist $n_{1}\in\mathbb{N}$ and $\delta>0$ such that for
each $0<\alpha\leq\delta$ the set $H^{n_{1}}_{\alpha}$ contains
$\\{e^{\epsilon z}:z\in\mathcal{D}\\}$ for some $\epsilon>0$. Then, with
$N\in\mathbb{N}$ as in Lemma 6.2, $H_{\alpha}^{n_{1}N}$ contains
$\\{e^{\epsilon z}:z\in\mathcal{D}\\}^{N}$, and it is thus a neighborhood of
the identity in $SU_{0}(A)$.

Set $n=n_{1}N$. Let $V\subseteq SU_{0}(A)$ be an arbitrary neighborhood of the
identity in $SU_{0}(A)$. Choose $W$, symmetric neighborhood of the identity
invariant under conjugation and such that $W^{n}\subseteq V$. Then,
$e^{i\alpha h}\in W$ for a small enough $0<\alpha\leq\delta$. It follows that
$H_{\alpha}^{n}\subseteq V$. Thus, the sets $H_{\alpha}^{n}$ form a basis of
neighborhoods of the identity in $SU_{0}(A)$, as desired. ∎

Let us recall the invariant Steinhaus property, introduced by Dowerk and Thom
([DowerkThom]). Let $G$ be a topological group. A subset $W$ of $G$ is called
(left) countably syndetic if countably many left translates of $W$ cover $G$;
i.e., $G=\bigcup_{k=1}^{\infty}g_{k}W$ for some $g_{1},g_{2},\ldots\in G$. Let
$n\in\mathbb{N}$. The group $G$ is said to have the invariant Steinhaus
property with exponent $n\in\mathbb{N}$ if for any $W\subseteq G$ that is
countably syndetic, symmetric ($W=W^{-1}$), and invariant under conjugation,
$W^{n}$ is a neighborhood of the identity.

######  Theorem 6.4.

Let $A$ be a unital C*-algebra containing a full square zero elements and
having bounded commutators generation. Then $SU_{0}(A)$ has the invariant
Steinhaus property.

###### Proof.

Let us choose $h\in\overline{[A,A]}\cap A_{\mathrm{sa}}$ fully noncentral
(i.e., $h=[x^{*},x]$, with $x$ a full square zero element; see Lemma 4.4). Let
$H_{\alpha}$, $n\in\mathbb{N}$, and $\delta>0$ be as in Theorem 6.3. Let us
show that $SU_{0}(A)$ has the invariant Steinhaus property with exponent $2n$.

Let $W$ be a subset of $SU_{0}(A)$ that is symmetric, invariant under
conjugation, and countably syndetic. Say
$SU_{0}(A)=\bigcup_{k=1}^{\infty}g_{k}W$ for some $g_{1},g_{2},\ldots$ in
$SU_{0}(A)$. Then for some $k$ the set $g_{k}W\cap\\{e^{i\alpha
h}:\alpha\in\mathbb{R}\\}$ is uncountable. Using that $W$ is symmetric, it
follows that $W^{2}$ contains $e^{i\alpha_{0}h}$ for some
$0<\alpha_{0}\leq\delta$. Hence, $H_{\alpha_{0}}^{n}\subseteq W^{2n}$. Thus,
$W^{2n}$ is a neighborhood of the identity, by Theorem 6.3. ∎

###### Proof of Theorem B.

(i) By the previous theorem, $SU_{0}(A)$ has the invariant Steinhaus property,
which by [DowerkThom, Proposition 8.10] implies that $SU_{0}(A)$ has the
invariant automatic continuity property.

(ii) Let $\phi\colon U_{0}(A)\to U_{0}(A)$ be a group automorphism. Let us
show that $\phi$ is continuous, from which the desired result immediately
follows. By [kallman, Corollary 3], to show that a group isomorphism of polish
groups $\psi\colon G_{1}\to G_{2}$ is continuous, it suffices to show that for
all $U\subseteq G_{2}$ ranging in a basis of neighborhoods of the identity,
$\psi^{-1}(U)$ is an analytic set in $G_{2}$.

Let $h\in\overline{[A,A]}\cap A_{\mathrm{sa}}$ be fully noncentral (guaranteed
to exist by Lemma 4.4). Let $H_{\alpha}$, $n\in\mathbb{N}$, and $\delta>0$ be
as in Theorem 6.3, so that $(H_{\alpha}^{n})_{0<\alpha\leq\delta}$ is a basis
neighborhoods of the identity in $SU_{0}(A)$. Let
$v_{\alpha}=\phi^{-1}(e^{i\alpha h})$. Then

$\phi^{-1}(H_{\alpha}^{n})=\\{uv_{\alpha}^{\pm 1}u^{-1}:u\in
SU_{0}(A)\\}^{n}.$

This set is clearly analytic. Hence, $\phi$ is continuous. ∎

######  Corollary 6.5.

Let $\phi\colon U_{0}(A)\to U_{0}(B)$ be a group isomorphism, where $A$ and
$B$ are prime, unital, traceless C*-algebras containing full square zero
elements. Then $\phi$ is the restriction to $U_{0}(A)$ of either an
isomorphism or an anti-isomorphism between $A$ and $B$.

###### Proof.

Since $A$ and $B$ are traceless, $SU_{0}(A)=U_{0}(A)$ and
$SU_{0}(B)=U_{0}(B)$. So Theorem B applies directly to $U_{0}(A)$ and
$U_{0}(B)$. By the invariant automatic continuity property of these groups,
$\phi$ is a homeomorphism. Recall that $U_{0}(A)$ and $U_{0}(B)$ are the
Banach-Lie groups of the Lie algebras of skewadjoint elements
$iA_{\mathrm{sa}}$ and $iB_{\mathrm{sa}}$. Thus, by the functoriality of the
Lie algebra of a Banach-Lie group ([hofmann-morris, Theorem 5.42]), there
exists a Lie algebras homomorphism $\psi\colon iA_{\mathrm{sa}}\to
iB_{\mathrm{sa}}$ such that $\phi(e^{ih})=e^{\psi(ih)}$ for all $h\in
A_{\mathrm{sa}}$. Moreover, since $\phi$ is an isomorphism, so is $\psi$. Let
us extend $\psi$ to $A$ by setting

$\psi(a+ib)=-i\psi(ia)+\psi(ib),$

for $a,b\in A_{\mathrm{sa}}$. We readily check that $\psi$ is again a Lie
algebras isomorphism between $A$ and $B$. By [ara-mathieu, Theorem 6.5.24]
(alternatively, by [bresar]), $\phi$ is either an isomorphism or an anti-
isomorphism of C*-algebras. In either case we get that

$\phi(e^{ih})=e^{\psi(ih)}=\psi(e^{ih})$

for all $h\in A_{\mathrm{sa}}$. It follows that $\phi$ is the restriction of
$\psi$ to $U_{0}(A)$. ∎

If $A$ is a unital traceless C*-algebra containing a full square zero element,
then the equality $SU_{0}(A)=U_{0}(A)$ and Theorem B imply that $U_{0}(A)$
does not have discontinuous automorphisms. This is in contrast with the
tracial case.

######  Theorem 6.6.

Let $A$ be a separable unital C*-algebra with at least one tracial state. Then
$U_{0}(A)$ admits discontinuous automorphisms.

###### Proof.

We adapt the proof of the same result for $U_{0}(M_{n}(\mathbb{C}))$ from
[kallman].

Let $\tau\colon A\to\mathbb{C}$ be a tracial state. Let
$\Gamma\subseteq\mathbb{T}$ denote the group of scalar unitaries in the kernel
of $\Delta_{\tau}$. From the description of the kernel of $\Delta_{\tau}$
(5.2) we see that

$\Gamma=\\{e^{2\pi i\theta}:\theta\in\tau(K_{0}(A))\\}.$

Since $A$ is separable, this group is countable (as $K_{0}(A)$ is countable).
Let us choose a discontinuous automorphism
$\alpha\colon\mathbb{T}\to\mathbb{T}$ that fixes $\Gamma$. (To get one, let
$V$ be the vector subspace of $\mathbb{R}$ spanned by $\tau(K_{0}(A))$, where
the scalar field is $\mathbb{Q}$. Choose a basis of $V$, a fortiori a
countable set, and extend it to a Hamel basis of $\mathbb{R}$. Using this
basis, choose a $\mathbb{Q}$-linear transformation
$f\colon\mathbb{R}\to\mathbb{R}$ that is the identity on $\tau(K_{0}(A))$, but
that is otherwise discontinuous. Now define $\alpha(e^{2\pi i\theta})=e^{2\pi
if(\theta)}$, which is well defined since $f$ is the identity on
$\mathbb{Z}\subseteq\tau(K_{0}(A))$.)

Any unitary $v\in U_{0}(A)$ is expressible in the form $zu$, where
$z\in\mathbb{T}$ and $u\in\ker\Delta_{\tau}$. To see this, choose any path
$\eta\colon[t_{1},t_{2}]\to U_{0}(A)$ connecting $1$ to $v$, and set
$z=e^{2\pi i\theta}$, with $\theta=\tilde{\Delta}_{\tau}(\eta)$. Then $v=zu$
with $u\in\ker\Delta_{\tau}$.

Let us define $\tilde{\alpha}\colon U_{0}(A)\to U_{0}(A)$ by
$\tilde{\alpha}(zu)=\alpha(z)u$, where $z\in\mathbb{T}$ and
$u\in\ker\Delta_{\tau}$. We readily verify that this is a well defined map: If
$z_{1}u_{1}=z_{2}u_{2}$, with $z_{1},z_{2}\in\mathbb{T}$ and
$u_{1},u_{2}\in\ker\Delta_{\tau}$, then

$z_{1}z_{2}^{-1}=u_{2}u_{1}^{-1}\in\Gamma.$

Since $\alpha$ is the identity on $\Gamma$,
$\alpha(z_{1}z_{2}^{-1})=u_{2}u_{1}^{-1}$, which in turn implies that
$\alpha(z_{1})u_{1}=\alpha(z_{2})u_{2}$. Hence $\tilde{\alpha}$ is well
defined. It is then easily shown that $\tilde{\alpha}$ is a bijective group
homomorphism. Since the restriction of $\tilde{\alpha}$ to the scalar
unitaries agrees with $\alpha$, $\tilde{\alpha}$ is necessarily discontinuous.
∎

##  7\. Bounded normal generation

######  Theorem 7.1.

Let $A$ be a unital C*-algebra containing a full square zero element and
having bounded commutators generation. Let $H\subseteq SU_{0}(A)$ be a
symmetric, fully noncentral set invariant under conjugation. Then there exists
$n\in\mathbb{N}$ such that $H^{n}$ is a neighborhood of the identity in
$SU_{0}(A)$.

###### Proof.

By Theorem 4.2, there exists $n_{1}\in\mathbb{N}$ such that $H^{n_{1}}$
contains $\\{e^{z}:z\in\mathcal{D}\\}$, where $\mathcal{D}\subseteq
iA_{\mathrm{sa}}$ as defined in (3.1). On the other hand, by Lemma 6.2 there
exists $N\in\mathbb{N}$ such that $\\{e^{z}:z\in\mathcal{D}\\}^{N}$ is a
neighborhood of the identity in $SU_{0}(A)$. It follows that $H^{n_{1}N}$ is
also a neighborhood of the identity, thus proving the theorem. ∎

Specializing the theorem above to simple C*-algebras, we obtain Theorem C from
the introduction.

###### Proof of Theorem C.

If $A=\mathbb{C}$, the theorem is trivial. Assume thus that this is not the
case. Observe then that $A$ contains nonzero square zero elements, by Glimm’s
halving lemma, and these are automatically full since $A$ is simple.

(i) Let $g$ be an element in $SU_{0}(A)/Z(SU_{0}(A))$ distinct from the
identity. Let $w\in SU_{0}(A)$ be any lift of $g$, necessarily noncentral (and
fully noncentral, as $A$ is simple). Let $H=\\{uw^{\pm 1}u^{*}:u\in
SU_{0}(A)\\}$. By Theorem 7.1, there exists $n\in\mathbb{N}$ such that $H^{n}$
contains a neighborhood of the identity in $SU_{0}(A)$. Then $H^{n}$ is mapped
onto a neighborhood of the identity by the quotient map. This establishes
local bounded normal generation for $SU_{0}(A)/Z(SU_{0}(A))$.

(ii) Suppose that $SU_{0}(A)$ is bounded as a metric space. Claim: For any
neighborhood of the identity $V\subseteq SU_{0}(A)$ there exists
$m\in\mathbb{N}$ such that $SU_{0}(A)=V^{m}$. Proof: By [ando, Theorem A], the
boundedness of $SU_{0}(A)$ under the exponential length metric implies that
$SU_{0}(A)$ is coarsely bounded. This, by [rosendalOB, Theorem 1.4], implies
that there exist a finite set $F\subseteq SU_{0}(A)$ and $k\in\mathbb{N}$ such
that $SU_{0}(A)=(FV)^{k}$. Since $SU_{0}(A)$ is also connected, the finite set
$F$ is contained in $V^{k^{\prime}}$ for a large enough $k^{\prime}$. Hence
$SU_{0}(A)=V^{(k^{\prime}+1)k}$, proving the claim.

Let $g$ be an element in $SU_{0}(A)/Z(SU_{0}(A))$ distinct from the identity.
Let us apply the claim just established to $H^{n}$, with $H$ and $n$ as in the
proof of (i). Then $U_{0}(A)=H^{nm}$. Passing to the quotient, we obtain
bounded normal generation for $SU_{0}(A)/Z(SU_{0}(A))$. ∎

A class of C*-algebras to which all results of this section and the previous
section apply is the purely infinite simple C*-algebras.

######  Corollary 7.2.

Let $A$ be a simple, unital, purely infinite C*-algebra. Then
$U_{0}(A)/\mathbb{T}$ has bounded normal generation, $U_{0}(A)$ has the
automatic invariant continuity property, and if $A$ is separable then
$U_{0}(A)$ has a unique polish group topology.

###### Proof.

Since $A$ is traceless, $A=[A,A]$ and $A$ has bounded commutators generation,
by Pop’s theorem. Hence, $SU_{0}(A)=U_{0}(A)$. Moreover, by [phillips,
Proposition 9], the exponential length of $A$ is at most $\pi$. Hence,
$U_{0}(A)$ is bounded. The previous theorem then implies that
$U_{0}(A)/\mathbb{T}$ has bounded normal generation. The automatic invariant
continuity property and, in the separable case, the uniqueness of the polish
group topology, follow from Theorem B. ∎

In Theorem 6.1 we have already recalled various classes of C*-algebras that
have bounded commutators generation. We now turn to the question of
boundedeness of $SU_{0}(A)$, in order to produce more examples of C*-algebras
where the group $SU_{0}(A)/Z(SU_{0}(A))$ has bounded normal generation.

######  Lemma 7.3.

Let $A$ be a unital C*-algebra. Then
$\mathrm{el}_{SU_{0}(A)}(e^{i[x^{*},x]})\leq\pi$ for all
$x\in\mathcal{N}_{2}$.

###### Proof.

If $x=0$ the lemma is trivially true, so assume that this is not the case.
Write $x=Cy$, with $y\in\mathcal{N}_{2}$ of norm 1 and $C=\|x\|$. It suffices
to prove the lemma in the case that $y$ is the canonical generator in the
universal C*-algebra generated by a square zero element of norm $\leq 1$. This
C*-algebra is $M_{2}(C_{0}(0,1])$, with $y=\begin{pmatrix}0&t\\\
0&0\end{pmatrix}$. Let us thus assume that we are in this set-up. Then

$e^{i[x^{*},x]}=e^{iC^{2}[y^{*},y]}=\begin{pmatrix}e^{iC^{2}t^{2}}&0\\\
0&e^{-iC^{2}t^{2}}\end{pmatrix}.$

This is a unitary in the unitization of $M_{2}(C_{0}(0,1])$.

Let $0<t_{1}<t_{2}<\cdots<t_{n}\leq 1$ be the points where $Ct^{2}$ is an
integer multiple of $\pi$. Choose $\delta>0$ such that the intervals
$(t_{k}-2\delta,t_{k}+2\delta)$, with $k=1,\ldots,n$, are pairwise disjoints.
Define $f\in C_{0}((0,1])_{+}$ as follows:

$f(t)=\begin{cases}Ct_{k}^{2}&\hbox{if $t\in(t_{k}-\delta,t_{k}+\delta)$ for
some $k$,}\\\ Ct^{2}&\hbox{if
$t\notin\bigcup_{k}(t_{k}-2\delta,t_{k}+2\delta)$,}\\\ \hbox{linear}&\hbox{for
$t\in(t_{k}-2\delta,t_{k}-\delta)$ and $t\in(t_{k}+\delta,t_{k}+2\delta)$ and
all $k$.}\end{cases}$

Let $\epsilon>0$. Choose $\delta$ smaller if necessary, so that the function
$f$ defined above satisfies that $|f(t)-Ct^{2}|<\epsilon$ for all $t\in[0,1]$.
Now let $g(t)=f(t)-Ct^{2}$ for $t\in[0,1]$, so that

$Ct^{2}=f(t)+g_{+}(t)-g_{-}(t).$

Set

$x^{\prime}=\begin{pmatrix}0&f^{\frac{1}{2}}\\\ 0&0\end{pmatrix},\quad
y_{1}=\begin{pmatrix}0&g_{+}^{\frac{1}{2}}\\\ 0&0\end{pmatrix},\quad
y_{2}=y_{1}=\begin{pmatrix}0&g_{-}^{\frac{1}{2}}\\\ 0&0\end{pmatrix}.$

It is straightforward to verify that

$[x^{*},x]=[(x^{\prime})^{*},x^{\prime}]+[y_{1}^{*},y_{1}]+[y_{2}^{*},y_{2}].$

Since the summands on the right-hand side commute, we have that

$e^{i[x^{*},x]}=e^{i[(x^{\prime})^{*},x^{\prime}]}e^{i[y_{1}^{*},y_{1}]}e^{i[y_{2}^{*},y_{2}]}.$

Observe that $\|y_{1}\|=\|g_{+}^{\frac{1}{2}}\|<\epsilon^{\frac{1}{2}}$, so
that $\mathrm{el}_{SU_{0}(A)}(e^{i[y_{1}^{*},y_{1}]})<\epsilon$. Similarly, we
have that $\mathrm{el}_{SU_{0}(A)}(e^{i[y_{2}^{*},y_{2}]})<\epsilon$.

To complete the proof, let us show that $e^{i[(x^{\prime})^{*},x^{\prime}]}$
is unitarily equivalent to $e^{i[(x^{\prime\prime})^{*},x^{\prime\prime}]}$
for some $x^{\prime\prime}$ such that $\|x^{\prime\prime}\|\leq\sqrt{\pi}$.
Define $f_{1}\in C(0,1]$ such that $0\leq f_{1}(t)\leq\pi$ for all
$t\in[0,1]$, and either $f(t)-f_{1}(t)$ or $f(t)+f_{1}(t)$ is an integer
multiple of $\pi$ for all $t$. Let

$x^{\prime\prime}=\begin{pmatrix}0&f_{1}^{\frac{1}{2}}\\\ 0&0\end{pmatrix}.$

Then $\|x^{\prime\prime}\|\leq\sqrt{\pi}$ and

$e^{i[(x^{\prime})^{*},x^{\prime}]}=\begin{pmatrix}e^{if}&0\\\
0&e^{-if}\end{pmatrix},\quad
e^{i[(x^{\prime\prime})^{*},x^{\prime\prime}]}=\begin{pmatrix}e^{if_{1}}&0\\\
0&e^{-if_{1}}\end{pmatrix}.$

These two unitaries are unitarily equivalent. This stems from the fact that
the unordered pairs $\\{e^{if(t)},e^{-if(t)}\\}$ and
$\\{e^{if_{1}(t)},e^{-if_{1}(t)}\\}$ agree for all $t$, and further, the
functions $e^{if(t)}$ agree $e^{-if(t)}$ on an open neighborhood of the values
of $t$ at which they cross. A unitary $t\mapsto w(t)$ conjugating
$e^{i[(x^{\prime})^{*},x^{\prime}]}$ and
$e^{i[(x^{\prime\prime})^{*},x^{\prime\prime}]}$ is gotten by setting $w(t)$
equal to either the identity matrix or the matrix $\begin{pmatrix}0&1\\\
1&0\end{pmatrix}$ for $t$ outside the intervals $(t_{k}-\delta,t_{k}+\delta)$,
and inside these intervals $t\mapsto w(t)$ is defined as any path connecting
these two matrices. ∎

######  Lemma 7.4.

Let $A$ be a unital C*-algebra of nuclear dimension at most 1. Then
$\mathrm{el}_{SU_{0}(A)}(e^{ih})\leq 4\pi$ for all $h\in\overline{[A,A]}\cap
A_{\mathrm{sa}}$.

###### Proof.

Let $h\in\overline{[A,A]}\cap A_{\mathrm{sa}}$. By [ker-det, Theorem 3.2],
there exist sequences $h_{n}^{(1)},h_{n}^{(2)},h_{n}^{(3)},h_{n}^{(4)}$ in
$\overline{[A,A]}\cap A_{\mathrm{sa}}$ such that

  1. (1)

$h_{n}^{(1)}+h_{n}^{(2)}+h_{n}^{(3)}+h_{n}^{(4)}\to h$,

  2. (2)

$[h_{n}^{(1)}+h_{n}^{(2)},h_{n}^{(3)}+h_{n}^{(4)}]\to 0$,
$[h_{n}^{(1)},h_{n}^{(2)}]\to 0$, $[h_{n}^{(3)},h_{n}^{(4)}]\to 0$,

  3. (3)

$h_{n}^{(k)}=[(x_{n}^{(k)})^{*},x_{n}^{(k)}]$ with
$x_{n}^{k}\in\mathcal{N}_{2}$ for $k=1,2,3,4$ and all $n$.

We now use [ker-det, Lemma 2.2], which states that if $a_{n},b_{n}$ are
bounded sequences of selfadjoint elements such that $[a_{n},b_{n}]\to 0$ then
$e^{i(a_{n}+b_{n})}e^{-ia_{n}}e^{-ib_{n}}\to 0$ in $SU_{0}(A)$. Using this
lemma (repeatedly) we deduce that

$e^{ih_{n}^{(1)}}e^{ih_{n}^{(2)}}e^{ih_{n}^{(3)}}e^{ih_{n}^{(4)}}\to e^{ih}$

in $SU_{0}(A)$. On the other hand, by Lemma 7.3, the exponential length in
$SU_{0}(A)$ of $e^{ih_{n}^{(k)}}$ is bounded by $\pi$. The lemma thus follows.
∎

Let $A$ be a unital C*-algebra. The exponential rank of $A$ is defined as the
least $m\in\mathbb{N}$ such that any unitary $u\in U_{0}(A)$ is a product of
at most $m$ exponential unitaries (i.e., of the form $e^{ih}$ with $h\in
A_{\mathrm{sa}}$). If no such $m$ exists then the exponential rank of $A$ is
set equal to $\infty$. If $A$ has exponential rank $m+1$, and the products of
$m$ exponentials form a dense subset of $U_{0}(A)$, then $A$ is also said to
have exponential rank $m+\epsilon$.

Next, let us recall the property of strict comparison by traces in a simple
unital C*-algebra. (There is also a version of this property for arbitrary
C*-algebras; e.g., see [NgRobertpure].) Let $A$ be a simple unital C*-algebra.
Then $A$ said to have strict comparison of positive elements by traces if for
all $a,b\in(A\otimes\mathcal{K})_{+}$ such that $d_{\tau}(a)<d_{\tau}(b)$ for
all tracial states $\tau$ on $A$ we have that $a\precsim_{\mathrm{Cu}}b$.
(Here $d_{\tau}(c):=\lim_{n}\tau(c^{\frac{1}{n}})$ and
$\precsim_{\mathrm{Cu}}$ denotes the Cuntz comparison relation, i.e.,
$d_{n}^{*}bd_{n}\to a$ for some sequence $d_{n}\in A\otimes\mathcal{K}$.) This
property, introduced by Blackadar in [blackadar], is a C*-algebraic analogue
of the “comparison of projections by traces" properties of factors.

######  Theorem 7.5.

Let $A$ be a simple unital C*-algebra. Suppose that $A$ has stable rank one,
strict comparison by traces, and finite exponential rank. Then $SU_{0}(A)$ is
bounded. Consequently, $SU_{0}(A)/Z(SU_{0}(A))$ has BNG.

###### Proof.

If $A$ is a matrix algebra $M_{n}(\mathbb{C})$, the theorem is well known. Let
us thus assume that $A$ is non-elementary. Then, under the hypotheses of the
theorem, $A$ is a pure C*-algebra, i.e., its Cuntz semigroup is almost
unperforated and almost divisible. Almost unperforation of the Cuntz semigroup
is implied by strict comparison of positive elements by traces ([ERS]), which
$A$ is assumed to have. For simple non-elementary unital C*-algebras of stable
rank one, almost divisibility in the Cuntz semigroup follows automatically
from almost unperforation, by Thiel’s [Thiel, Theorem 8.11]. Since $A$ is pure
and has strict comparison of positive elements by traces, it has bounded
commutators generation (Theorem 6.1). Thus, the BNG property for
$SU_{0}(A)/Z(SU_{0}(A))$ will indeed follow once we have shown that
$SU_{0}(A)$ is bounded.

To show that $SU_{0}(A)$ is bounded we shall apply the “special subalgebra
technique" also employed in [ker-det], [ART], [jacelon]. More specifically, by
[ker-det, Theorem 4.1], there exists a C*-subalgebra $B\subseteq A$ such that

  1. (1)

$[B,B]=[A,A]\cap B$,

  2. (2)

$B$ has nuclear dimension 1 (in fact, $B$ is of the form $C\otimes W$, where
$C$ is an AF C*-algebra and $W$ is the Jacelon-Razak algebra),

  3. (3)

every non-invertible selfadjoint element $h\in A$ with connected spectrum is
approximately unitarily equivalent to some $h^{\prime}\in B$.

The proof now proceeds along a series of claims, each time expanding the set
of elements in $SU_{0}(A)$ on which the exponential length
$\mathrm{el}_{SU_{0}(A)}$ is bounded.

_Claim 1_ : $\mathrm{el}_{SU_{0}(A)}$ is bounded on the set of elements of the
form $e^{ih}$, with $h\in[B,B]\cap B_{\mathrm{sa}}$. Proof: The C*-algebra
$B^{\sim}$ has nuclear dimension 1. The claim now follows from Lemma 7.4.

_Claim 2_ : $\mathrm{el}_{SU_{0}(A)}$ is bounded on the set of elements of the
form $e^{ih}$, with $h\in[A,A]\cap A_{\mathrm{sa}}$ non-invertible and with
connected spectrum. Proof: Let $h\in[A,A]\cap A_{\mathrm{sa}}$ be non-
invertible and with connected spectrum. Then, by the properties of $B$ listed
above, $h$ is approximately unitarily equivalent to some $h^{\prime}\in B$.
Say $u_{n}h^{\prime}u_{n}^{*}\to h$. Then

$\mathrm{el}_{SU_{0}(A)}(e^{iu_{n}h^{\prime}u_{n}^{*}})\to\mathrm{el}_{SU_{0}(A)}(e^{ih}),$

while on the other hand

$\mathrm{el}_{SU_{0}(A)}(e^{iu_{n}h^{\prime}u_{n}^{*}})=\mathrm{el}_{SU_{0}(A)}(u_{n}e^{ih^{\prime}}u_{n}^{*})=\mathrm{el}_{SU_{0}(A)}(e^{ih^{\prime}}),$

since $\mathrm{el}_{SU_{0}(A)}$ is invariant under automorphisms of $A$.
Hence,
$\mathrm{el}_{SU_{0}(A)}(e^{ih})=\mathrm{el}_{SU_{0}(A)}(e^{ih^{\prime}})$.
The claim now follows from the previous claim.

_Claim 3_ : $\mathrm{el}_{SU_{0}(A)}$ is bounded on the set of elements of the
form $e^{ih}$, with $h\in[A,A]\cap A_{\mathrm{sa}}$. Proof: Consider a unitary
$e^{ih}$, with $h\in[A,A]\cap A_{\mathrm{sa}}$. By [ker-det, Lemma], there
exist $h^{\prime}\in[A,A]\cap A_{\mathrm{sa}}$ and $x\in\mathcal{N}_{2}$ such
that $h^{\prime}$ is a selfadjoint element with connected spectrum,
$h^{\prime}$ and $[x^{*},x]$ belong to $C^{*}(h)$, and

$\|h-(h^{\prime}+[x^{*},x])\|<1.$

Since $h$, $h^{\prime}$, and $[x^{*},x]$ commute with each other,
$e^{ih}=e^{ih^{\prime}}e^{[x^{*},x]}e^{ic}$, where $c\in[A,A]\cap
A_{\mathrm{sa}}$ and $\|c\|\leq 1$. Observe that $\mathrm{el}_{SU_{0}(A)}$ is
uniformly bounded on the second and third factors (by Lemma 7.3). It remains
to show that $\mathrm{el}_{SU_{0}(A)}$ is bounded on the set of exponentials
$e^{ih}$ where $h\in[A,A]\cap A_{\mathrm{sa}}$ has connected spectrum. If
$\|h\|\leq 2\pi$, we are done. If $\|h\|>2\pi$, then there is a translate of
$h$ by an integer multiple of $2\pi\cdot 1$ that is non-invertible, and again
we are done, by the previous claim. This proves the claim.

_Claim 4_ : $\mathrm{el}_{SU_{0}(A)}$ is bounded on commutators of the form
$(e^{ih},w)$, with $h\in A_{\mathrm{sa}}$ and $w\in U_{0}(A)$: Proof: Consider
an element of the form $(e^{ih},w)=e^{ih}e^{-iwhw^{*}}$. By [KirchbergRordam,
Lemma 6.4] there exists an approximately central sequence of selfadjoint
elements $(h_{n})_{n}$ such that $h-h_{n}\in[A,A]$ and $\|h_{n}\|\leq\|h\|$
for all $n$. Then, using [ker-det, Lemma 2.2], for large enough $n$ we have
that

$\displaystyle e^{ih}=e^{ic_{1}}e^{i(h-h_{n})}e^{ih_{n}},\quad
e^{ih_{n}}e^{-iwh_{n}w^{*}}=e^{ic_{2}}$

for some $c_{1},c_{2}\in[A,A]\cap A_{\mathrm{sa}}$ of norm $\leq 1$. Hence,

$e^{ih}e^{-iwhw^{*}}=e^{ic_{1}}e^{i(h-h_{n})}e^{ic_{2}}e^{-iw(h-h_{n})w^{*}}e^{-iwc_{1}w^{*}}.$

By the previous claim, $\mathrm{el}_{SU_{0}(A)}$ is uniformly bounded on the
terms $e^{i(h-h_{n})}$ and $e^{-iw(h-h_{n})w^{*}}$. The claim thus follows.

The following claim completes the proof of the theorem.

_Claim 5_ : $\mathrm{el}_{SU_{0}(A)}$ is bounded on all $SU_{0}(A)$. Let $R$
denote the exponential rank of $A$. The C*-algebra $A$ fulfills the
assumptions of [ker-det, Theorem 1.1]. Thus, by this theorem, any element of
$SU_{0}(A)$ can be expressed as a product of $7R+29$ commutators
$(v_{k},w_{k})$, with $v_{k},w_{k}\in U_{0}(A)$. It thus suffices to show that
$\mathrm{el}_{SU_{0}(A)}$ is bounded on the set of commutators $(v,w)$, with
$v,w\in U_{0}(A)$. Consider a pair $v,w\in U_{0}(A)$. Since $A$ has
exponential rank $R$, we can write $v=\prod_{k=1}^{R}e^{ih_{k}}$, where
$h_{k}\in A_{\mathrm{sa}}$ for all $k$. Repeatedly using the identity
$(xy,w)=(x,y)(y,xw)(x,w)$, we can express the commutator
$(\prod_{k=1}^{R}e^{ih_{k}},w)$ as a product of commutators of the form
$(e^{ih_{k}},w^{\prime})$. It thus suffices to prove that
$\mathrm{el}_{SU_{0}(A)}$ is bounded on all commutators of the form
$(e^{ih},w)$, with $h\in A_{\mathrm{sa}}$ and $w\in U_{0}(A)$. This is what we
have proven in the previous claim. ∎

Let $\mathcal{Z}$ denote the Jiang-Su C*-algebra. A C*-algebra $A$ is called
$\mathcal{Z}$-stable if $A\otimes\mathcal{Z}\cong A$.

######  Corollary 7.6.

Let $A$ be a unital separable simple nuclear $\mathcal{Z}$-stable C*-algebra
satisfying the UCT. Then $SU_{0}(A)/Z(SU_{0}(A))$ has BNG.

###### Proof.

By Rørdam’s dichotomy [RordamZ], a simple unital $\mathcal{Z}$-stable
C*-algebra is either purely infinite or of stable rank one. The purely
infinite case was dealt with in Corollary 7.2. Let us thus assume that $A$ has
stable rank one. Then $\mathcal{Z}$-stability implies strict comparison by
2-quasitraces, and nuclearity implies that 2-quasitraces are traces by
Haagerup’s theorem ([haagerup]). So $A$ has strict comparison of positive
elements by traces. By the previous theorem, it remains to check that $A$ has
finite exponential rank. We prove this next. (Note: There is abundant evidence
that classifiable C*-algebras have exponential rank $1+\epsilon$, e.g.
[LinTR1], but we will content ourselves here with a finite bound.)

By the classification theorem of Gong, Lin, and Niu in [GLN1], together with
[EGLN2], [TWW], and the work on the Toms-Winter in [CETWW], the C*-algebra $A$
is an inductive limit of subhomogeneous C*-algebras $(A_{n})_{n=1}^{\infty}$
with spectra of dimension at most 2. Since the exponential rank is easily seen
not to increase by more than “$\epsilon$” when passing to an inductive limit,
we will be done once we have shown that the C*-algebras
$(A_{n})_{n=1}^{\infty}$ have uniformly bounded exponential rank.

The C*-algebra $A_{n}$ is a pullback of the following form (see [GLN1, Section
13] and [XinLi, Section 2]):

$\textstyle{A_{n}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\psi}$$\scriptstyle{\phi}$$\textstyle{C([0,1],E)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\mathrm{ev_{0}}\oplus\mathrm{ev_{1}}}$$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\beta}$$\textstyle{E\oplus
E}$

In this diagram

  * •

$B=pM_{N}(C(X))p$ where $p$ is a projection and $X$ is a compact metric space
of dimension at most $2$,

  * •

$E$ is finite dimensional,

  * •

$\mathrm{ev}_{0}$ and $\mathrm{ev}_{1}$ are the point evaluations at the
endpoints,

  * •

$\beta$ is a unital homomorphism.

Let us show that these C*-algebras have uniformly bounded exponential rank.
Let $u\in U_{0}(A_{n})$, and let $[0,1]\ni t\mapsto u_{t}$ be a path
connecting $1$ to $u$. Consider the path $t\mapsto\phi(u_{t})$ in $U_{0}(B)$.
By [phillipsHowMany, Theorem 4.7], there exists a constant $C(3)$ bounding the
exponential ranks of the C*-algebras that are a corner of a matrix algebra
over a compact matrix space of dimension $\leq 3$. Applied to the C*-algebra
$C([0,1],B)$, we get that there exist selfadjoint elements $h_{k}\in
C([0,1],B)$, for $k=1,\ldots,C(3)$, such that

$\phi(u_{t})=\prod_{k=1}^{C(3)}e^{ih_{k}(t)}$

for all $t\in[0,1]$. The map $\phi$ is surjective (since
$\mathrm{ev}_{0}\oplus\mathrm{ev}_{1}$ is), and so the induced map
$C([0,1],A_{n})\to C([0,1],B)$ is also surjective. Let $\tilde{h}_{k}\in
C([0,1],A_{n})$ be a selfadjoint lift of $h_{k}$ for all $k$. Set

$v=u\prod_{k=1}^{C(3)}e^{-i\tilde{h}_{k}(1)}.$

Observe then $v$ is connected to $1$ in $U_{0}(A_{n})$ by a path that is
mapped to $1\in B$ by $\phi$. To complete the proof we show that $v$ is a
product of two exponentials. Observe that $\psi(v)\in C([0,1],E)$ is a unitary
satisfying the endpoint conditions $\psi(v)(0)=\psi(v)(1)=1_{E}$. It thus
belongs to the subalgebra $I=\\{f\in
C([0,1,E]):f(0)=f(1)\in\mathbb{C}1_{E}\\}$. Moreover, $\psi(v)$ is connected
to $1$ by a path of unitaries in $I$. Since $I$ has exponential rank
$1+\epsilon$ by [phillipsFU, Proposition 2.8], there exist selfadjoint
elements $g_{1},g_{2}\in I$ such that $\psi(v)=e^{ig_{1}}e^{ig_{2}}$. Let
$\tilde{g}_{1},\tilde{g}_{2}\in A_{n}$ be their lifts that are mapped to $1\in
B$ by $\phi$. Then $v=e^{i\tilde{h}_{1}}e^{i\tilde{g}_{2}}$, as desired. ∎

######  Remark 7.7.

In [dowerkthesis] Dowerk introduced the property of “topological bounded
normal generation”. A topological group $G$ has this property if for any $g\in
G$ distinct from the unit there exists $n\in\mathbb{N}$ such that $\\{hg^{\pm
1}h^{-1}:h\in G\\}^{n}$ is dense in $G$. In [DowerkLeMaitre] examples are
given of nonsimple groups with topological bounded normal generation. The
C*-algebras covered in Corollary 7.6 provide us with new examples of such
groups. Take $A$ an infinite dimensional simple unital AF C*-algebra. Since
$A$ has at least one tracial state, there are noncentral unitaries in
$U_{0}(A)$ with nonzero determinant. Hence, $U_{0}(A)/\mathbb{T}$ is not
simple, as $DU_{0}(A)/\mathbb{T}$ is a proper normal subgroup of
$U_{0}(A)/\mathbb{T}$. In particular, $U_{0}(A)/\mathbb{T}$ does not have BNG.
Let us show that $U_{0}(A)/\mathbb{T}$ has topological bounded normal
generation. Let $u\in U_{0}(A)$ be noncentral. Then $u^{\prime}=(u,e^{ih})$ is
noncentral for some $h\in A_{\mathrm{sa}}$. By the bounded normal generation
of $DU_{0}(A)$, there exists $n$ such that

$DU_{0}(A)=(\\{v(u^{\prime})^{\pm 1}v^{*}:v\in
DU_{0}(A)\\})^{n}\subseteq(\\{vu^{\pm 1}v^{*}:v\in DU_{0}(A)\\})^{2n}.$

Since $A$ has real rank zero, $U_{0}(A)=\overline{DU_{0}(A)}^{\|\cdot\|}$
([elliott-rordam0]). Thus, $(\\{vu^{\pm 1}v^{*}:v\in DU_{0}(A)\\})^{2n}$ is
dense in $U_{0}(A)$. This shows that $U_{0}(A)/\mathbb{T}$ has topological
bounded normal generation.

##  8\. Counterexamples

Let us first introduce notation and recall a known lemma on the exponential
length in $U_{0}(A)$. Let $A$ be a unital C*-algebra. Given $u\in U_{0}(A)$,
we define its exponential length as

$\mathrm{el}_{U_{0}(A)}(u)=\inf\Big{\\{}\sum_{j=1}^{n}\|h_{j}\|:u=\prod_{j=1}^{n}e^{ih_{j}},\,h_{j}\in
A_{\mathrm{sa}}\hbox{ for all }j\Big{\\}}.$

Clearly, for $u\in SU_{0}(A)$ we have that
$\mathrm{el}_{U_{0}(A)}(u)\leq\mathrm{el}_{SU_{0}(A)}(u)$ (the latter defined
in (5.3)). These two quantities are often different. We call
$(u,v)\mapsto\mathrm{el}_{U_{0}(A)}(u^{*}v)$ the exponential distance in
$U_{0}(A)$.

######  Lemma 8.1.

Let $m\in\mathbb{N}$ and $0<\theta<\pi$. Let $A$ be a unital C*-algebra and
let $u\in U_{0}(A)$ be such that $\mathrm{el}_{U_{0}(A)}(u)<m\theta$. Then
there exist $h_{1},\ldots,h_{m}\in A_{\mathrm{sa}}$ such that
$u=\prod_{j=1}^{m}e^{ih_{j}}$ and $\|h_{j}\|<\theta$ for all $j$.

###### Proof.

This is essentially [Ringrose, Theorem 2.6]. By assumption,

$u=\prod_{j=1}^{n}e^{if_{j}},$

were $f_{j}\in A_{\mathrm{sa}}$ for all $j$ and
$\sum_{j=1}^{n}\|f_{j}\|<m\theta$. Increasing $n$ if necessary, we can assume
that $\|f_{j}\|<\delta$ for all $j$, where we choose $\delta>0$ such that
$0<\frac{m\theta}{\theta-\delta}<m+1$. Let us now break-up the list
$e^{if_{1}},\ldots,e^{if_{n}}$ into groups of consecutive exponentials such
that the norms of the exponents in each group add up to less than $\theta$,
but $\geq\theta-\delta$. It is then easily derived that the number of groups
obtained in this way must be at most $m$. The product of the exponentials in
each of these groups can be written as $e^{ih}$, with $\|h\|<\theta$, by
[Ringrose, Corollary 2.2]. We thus have
$u=\prod_{j=1}^{m^{\prime}}e^{ih_{j}}$, with $\|h_{j}\|<\theta$ for all $j$
and $m^{\prime}\leq m$. Setting $h_{j}=0$ for $m^{\prime}<j\leq m$ the lemma
readily follows. ∎

###  8.1. Homogeneous C*-algebras

Let $X$ be a compact Hausdorff space (below we will fix $X=(S^{2})^{n}$). Let
us introduce notation and recall some standard facts around the projections in
$M_{\infty}(C(X))$.

Let $p,q\in M_{\infty}(C(X))$ be projections. Then $p$ and $q$ are said to be
Murray-von Neumann equivalent if $p=v^{*}v$ and $q=vv^{*}$ for some $v\in
M_{\infty}(C(X))$. We denote the Murray-von Neumann class of a projection
$p\in M_{\infty}(C(X))$ by $[p]$.

Suppose that $p\in M_{m}(C(X))$ and $q\in M_{n}(C(X))$. Let us denote by
$p\oplus q$ the projection

$\begin{pmatrix}p&0\\\ 0&q\end{pmatrix}\in M_{n+m}(C(X)).$

One has that $[p\oplus q]$ depends only on $[p]$ and $[q]$. Setting
$[p]+[q]:=[p\oplus q]$ defines the addition operation in the Murray-von Neuman
monoid of projections.

Let us denote by $\mathrm{her}(p)$ the C*-algebra $pM_{\infty}(C(X))p$, i.e.,
the hereditary C*-subalgebra generated by $p$. We adopt a 2x2 matrix notation
for the elements of $\mathrm{her}(p\oplus q)$. More specifically, we represent
$a\in\mathrm{her}(p\oplus q)$ as the matrix

$\begin{pmatrix}pap&paq\\\ qap&qaq\end{pmatrix}.$

We will make reference to the complex vector bundle associated to a
projection. The complex vector bundle on $X$ associated to $p$ is $(X,E,\pi)$,
where

$E=\\{(x,v)\in X\times\mathbb{C}^{m}:p(x)v=v\\},$

and $\pi\colon E\to X$ is the projection onto the first coordinate. We denote
this vector bundle by $\eta_{p}$. The isomorphism class of $\eta_{p}$ depends
only on $[p]$.

Let $p\in M_{\infty}(C(X))$ be a projection. Let us denote by $e(p)\in
H^{*}(X)$ the Euler class of the vector bundle $\eta_{p}$. If $p$ has rank
$k$, then $\eta_{p}$ has dimension $k$ over $\mathbb{C}$, and $2k$ over
$\mathbb{R}$, so $e(p)\in H^{2k}(X)$.

Let $n\in\mathbb{N}$. From this point on we fix $X=(S^{2})^{n}$, where $S^{2}$
denotes the 2-dimensional sphere.

Let us recall a few standard facts on the homology and cohomology groups (with
integer coefficients) of $X$. By Kunneth’s theorem,

$H^{*}(X)\cong
Z[\alpha_{1},\alpha_{2},\ldots,\alpha_{n}]\,/\,(\alpha_{j}^{2}:j=1,\ldots,n),$

where $\alpha_{1},\ldots,\alpha_{n}\in H^{2}(X)$ are induced by the
projections $\pi_{j}\colon X\to S^{2}$ onto each sphere factor and the choice
of a generator in $H^{2}(S^{2})$.

For each choice of indices $i_{1},i_{2},\ldots,i_{k}$, let
$S_{i_{1},i_{2},\ldots,i_{k}}\subseteq X$ denote the cartesian product with
$S^{2}$ at the indices $i_{1},i_{2},\ldots,i_{k}$ and a fixed point (choose
any) at the remaining indices. The inclusion
$S_{i_{1},i_{2},\ldots,i_{k}}\subseteq X$ gives rise to homology classes
$[S_{i_{1},i_{2},\ldots,i_{k}}]\in H_{*}(X)$, image of the fundamental
homology class of $S_{i_{1},i_{2},\ldots,i_{k}}$ in $H^{*}(X)$. By Poincare
duality, we have that $H_{n-k}(X)\cong H^{k}(X)$ for $k=0,\ldots,n$, where the
isomorphism is given by $\alpha\mapsto\alpha\cap[X]$ (cap product with the
fundamental homology class of $X$). Moreover, we can compute that

$(\alpha_{i_{1}}\alpha_{i_{2}}\cdots\alpha_{i_{k}})\cap[X]=[S_{i_{1}^{\prime},i_{2}^{\prime},\ldots,i_{n-k}^{\prime}}],$

where the indices on the right side run through the complement of the indices
on the left side.

Let $\kappa\colon X\to[-1,1]$ be defined as

(8.1) $\kappa((x_{1},y_{1},z_{1}),\dots,(x_{n},y_{n},z_{n}))=x_{1}.$

######  Theorem 8.2.

Let $k,m\in\mathbb{N}$. Let $p,q\in M_{\infty}(C(X))$ be projections such that
$[p]\leq k[1]$, the Euler class $e(q)$ belongs to the subring of $H^{*}(X)$
generated by $\alpha_{2},\ldots,\alpha_{n}$, and further $e(q)^{5km}\neq 0$.
Consider a selfadjoint element $h\in\mathrm{her}(p\oplus q)$ of the form

$h=\begin{pmatrix}\kappa p&0\\\ 0&w\end{pmatrix}.$

Then the exponential distance in $U_{0}(\mathrm{her}(p\oplus q))$ from
$e^{ith}$ to the set

$(\\{(v,w):v,w\in U_{0}(\mathrm{her}(p\oplus q))\\})^{m}$

is at least $\min(\frac{t}{k},(m+1)\pi)$ for all $t\geq 0$.

###### Proof.

Since we may replace $q$ by a Murray-von Neumann equivalent projection, let us
assume that $q$ is a smooth function on $X$.

Since $[p]\leq k[1]$, $p$ is Murray-von Neumann subequivalent to $1_{k}$ (the
unit in $M_{k}(C(X))$). Say $p=v^{*}v$ and $p^{\prime}:=vv^{*}\leq 1_{k}$ for
some $v\in M_{\infty}(C(X))$. Then $a\mapsto(v\oplus q)a(v^{*}\oplus q)$ is a
C*-algebra isomorphism from $\mathrm{her}(p\oplus q)$ to
$\mathrm{her}(p^{\prime}\oplus q)$. Since the selfadjoint $h^{\prime}=vhv^{*}$
has the same form as $h$ (with $p$ replaced by $p^{\prime}$), we may prove the
theorem with $p^{\prime}$ in place of $p$. Let us thus assume that $p\leq
1_{k}$, i.e., $p\in M_{k}(C(X))$. Then $\mathrm{her}(p\oplus q)$ is a
hereditary subalgebra of $\mathrm{her}(1_{k}\oplus q)$, and so
$h\in\mathrm{her}(1_{k}\oplus q)$. It will suffice to prove the estimated
exponential distance for $e^{ith}$ in the C*-algebra $\mathrm{her}(1_{k}\oplus
q)$, since an approximation of $e^{ith}$ by products of $m$ commutators in
$U_{0}(\mathrm{her}(p\oplus q))$ yields an approximation for $e^{ith}$ in
$U_{0}(\mathrm{her}(1_{k}\oplus q))$ within at most the same distance. Note
that

$e^{ih}=\begin{pmatrix}e^{i\kappa p}+1_{k}-p&0\\\ 0&e^{iw}\end{pmatrix}$

in $\mathrm{her}(1_{k}\oplus q)$.

We shall show that for all $0<t_{0}<k(m+1)\pi$ and $t\geq t_{0}$ the
exponential distance from $e^{ith}$ to $(\\{(v,w):v,w\in
U_{0}(\mathrm{her}(1_{k}\oplus q))\\})^{m}$ is $\geq t_{0}/k$. That the
distance from $e^{it}$ to that same set is at least
$\min(\frac{t}{k},(m+1)\pi)$ for all $t\geq 0$ is easily derived from this.

Fix $0<t_{0}<k(m+1)\pi$ and $t\geq t_{0}$. Suppose for the sake of
contradiction that the exponential distance from $e^{ith}$ to the set
$(\\{(v,w):v,w\in U_{0}(\mathrm{her}(1_{k}\oplus q))\\})^{m}$ is $<t_{0}/k$.
Then there exist $v_{j},w_{j}\in U_{0}(\mathrm{her}(1_{k}\oplus q))$ for
$j=1,\ldots,m$ such that the exponential length of

$e^{-ith}(v_{1},w_{1})\cdots(v_{m},w_{m})$

is $<t_{0}/k$. By Lemma 8.1 applied with $\theta=\frac{t_{0}}{k(m+1)}<\pi$,
the displayed unitary is expressible as a product
$\prod_{j=1}^{m+1}e^{ih_{j}}$, where $h_{j}\in\mathrm{her}(1_{k}\oplus
q)_{\mathrm{sa}}$ and $\|h_{j}\|<\frac{t_{0}}{k(m+1)}$ for all $j$. Omitting
$e^{ih_{m+1}}$ we deduce that

(8.2) $\|e^{ith}-(v_{1},w_{1})\cdots(v_{m},w_{m})\cdot e^{ih_{1}}\cdots
e^{ih_{m}}\|<|e^{i\frac{t_{0}}{k(m+1)}}-1|.$

Recall that we represent elements of $\mathrm{her}(1_{k}\oplus q)$ by a 2x2
matrix

$\begin{pmatrix}a&b\\\ c&d\end{pmatrix},$

where $a\in M_{k}(C(X))$, $b^{t},c\in qM_{\infty}(C(X))1_{k}$, and
$d\in\mathrm{her}(q)$. Observe that we can regard the first $k$ columns of
$b^{t}$ and $c$ as sections of the vector bundle $\eta_{q}$ associated to $q$.
Now, say,

$v_{j}=\begin{pmatrix}a_{j}^{(1)}&b_{j}^{(1)}\\\
c_{j}^{(1)}&d_{j}^{(1)}\end{pmatrix},\quad
w_{j}=\begin{pmatrix}a_{j}^{(2)}&b_{j}^{(2)}\\\
c_{j}^{(2)}&d_{j}^{(2)}\end{pmatrix}\quad
h_{j}=\begin{pmatrix}a_{j}^{(3)}&(b_{j}^{(3)})^{*}\\\
b_{j}^{(3)}&c_{j}^{(3)}\end{pmatrix}\quad(j=1,\ldots,m).$

Taken all together, the columns of the off-diagonal elements

(8.3)
$(b_{j}^{(1)})^{t},\,c_{j}^{(1)},\,(b_{j}^{(2)})^{t},\,c_{j}^{(2)},b_{j}^{(3)}$

give rise to $5km$ sections of $\eta_{q}$, equivalently, a single section of
$\eta_{q}^{\oplus 5km}$.

The set of smooth sections of $\eta_{q}^{\oplus 5km}$ that are transversal to
the zero section is uniformly dense in the set of all sections, by Thom’s
transversality theorem ([Thom, Theorem I.5]). By an approximation, let us
perturb the off-diagonal elements (8.3) so that the resulting section of
$\eta_{q}^{\oplus 5km}$ is transversal to the zero section, while (8.2) is
still valid. Let $Z\subseteq X$ denote the zero set of the off-diagonal
elements (8.3). The transversality to the zero section implies that $Z$ is an
orientable submanifold of $X$ whose homology class $[Z]$ in $H_{*}(X)$ is
equal to the Poincare dual of the Euler class of $\eta_{q}^{\oplus 5km}$
([BottTu, Proposition 12.8]). This Euler class is $e(q^{\oplus
5km)})=e(q)^{5km}$. By assumption, $e(q)^{5km}$ is a nonempty sum of monomials
$\alpha_{i_{1}}\alpha_{i_{2}}\cdots\alpha_{i_{l}}$, where $i_{j}\neq 1$ for
all $j$. Using that

$\alpha_{i_{1}}\alpha_{i_{2}}\cdots\alpha_{i_{l}}\cap[X]=[S_{i_{1}^{\prime},i_{2}^{\prime},\ldots,i_{n-l}^{\prime}}],$

where the indices on the right side run through the complement of the indices
on the left side, we obtain that $[Z]$ is a sum of homology classes coming
from subproducts of spheres of the form
$[S_{1,i_{2}^{\prime},\ldots,i_{n-l}^{\prime}}]$ (i.e., all the terms include
the index $1$). Let us decompose $Z$ into connected components—which are also
oriented submanifolds—and pick a connected component $Z^{\prime}$. Then
$[Z^{\prime}]$ is again a sum of homology classes coming from subproducts of
spheres, where all the terms include the first sphere, i.e., have the form
$[S_{1,i_{2},\ldots,i_{l}}]$.

_Claim_ : The projection map $\pi_{1}\colon X\to S^{2}$ onto the first sphere
maps $Z^{\prime}\subseteq X$ onto $S^{2}$. Proof: Suppose it does not. Then
$Z^{\prime}$ is contained in $X^{\prime}=(S^{2}\backslash\\{x\\})\times
S_{2}\times\cdots\times S_{2}$ for some $x\in S^{2}$. The inclusion of
$X^{\prime}$ in $X$ induces a map $H_{*}(X^{\prime})\to H_{*}(X)$ whose range
is contained in the subgroup generated by the classes
$[S_{j_{1},j_{2},\cdots,j_{l}}]$ with $j_{1}\neq 1$. This is impossible, since
the homology class $[Z^{\prime}]$ belongs to the range of this map, as the
inclusion of $Z^{\prime}$ in $X$ factors through the inclusion of $X^{\prime}$
in $X$.

Having established our claim, let us restrict (8.2) to the set $Z^{\prime}$.
Since the elements $v_{j},w_{j},h_{j}$ are simultaneously of diagonal form on
$Z^{\prime}$, comparing the top left corners we get that

$\|e^{it\kappa}p+(1_{k}-p)-(a_{1}^{(1)},a_{1}^{(2)})\cdots(a_{m}^{(1)},a_{m}^{(2)})\cdot
e^{ia_{1}^{(3)}}\cdots e^{ia_{m}^{(3)}}|\|<|e^{i\frac{t_{0}}{k(m+1)}}-1|,$

where this relation is now taken in $M_{k}(C(Z^{\prime}))$. Here $a_{j}^{(1)}$
and $a_{j}^{(2)}$ are unitaries in $M_{k}(C(Z^{\prime}))$, and $a_{j}^{(3)}$
is selfadjoint such that $\|a_{j}^{(3)}\|<\frac{t_{0}}{k(m+1)}$ for all $j$.
This in turn implies that

$e^{it\kappa}p+(1_{k}-p)=(a_{1}^{(1)},a_{1}^{(2)})\cdots(a_{m}^{(1)},a_{m}^{(2)})\cdot
e^{ia_{1}^{(3)}}\cdots e^{ia_{m}^{(3)}}\cdot e^{ib},$

where $b\in M_{k}(C(Z^{\prime}))$ is selfadjoint and
$\|b\|<\frac{t_{0}}{k(m+1)}$ ([Ringrose, Proposition 2.4]). Evaluating at an
arbitrary $x\in Z^{\prime}$ and comparing the determinants of both sides we
get

$\mathrm{rank}(p)t\kappa(x)-\mathrm{Tr}(b(x))-\sum_{j=1}^{m}\mathrm{Tr}(a_{j}^{(3)}(x))\in
2\pi\mathbb{Z}$

for all $x\in Z^{\prime}$. Since $Z^{\prime}$ is connected,

$\mathrm{rank}(p)t\kappa(x)-\mathrm{Tr}(b(x))-\sum_{j=1}^{m}\mathrm{Tr}(a_{j}^{(3)}(x))$

is constant for $x\in Z^{\prime}$. This is impossible for $t\geq t_{0}$.
Indeed, on one hand the variation of $\mathrm{rank}(p)t\kappa(x)$ on the set
$Z^{\prime}$ is $2\cdot\mathrm{rank}(p)t\geq 2t_{0}$, as $\kappa$ maps
$Z^{\prime}$ onto $[-1,1]$. On the other hand, the variation of
$\mathrm{Tr}(b(x))+\sum_{j=1}^{m}\mathrm{Tr}(a_{j}^{(3)}(x))$ on all $X$ is
less than $2t_{0}$, as $a_{j}^{(3)}(x)$ and $b(x)$ are $k\times k$ matrices of
norm $<\frac{t_{0}}{k(m+1)}$. ∎

###  8.2. Simple inductive limit

Let us now describe simple inductive limits. We follow the example from
[RobertCommutators, Section 6], which in turn is based on Villadsen’s “second
type” construction in [villadsen].

######  Theorem 8.3.

There exists a simple unital AH C*-algebra with a unique tracial state and the
following property: For each $m\in\mathbb{N}$ there exists
$h\in\overline{[A,A]}\cap A_{\mathrm{sa}}$ of norm $\leq 1$ and such that

  1. (i)

$e^{ih}$ is not contained in the norm closure of the set $(\\{(v,w):v,w\in
U_{0}(A)\\})^{m}$,

  2. (ii)

the exponential distance from $e^{iCh}$ to the set $(\\{(v,w):v,w\in
U_{0}(A)\\})^{m}$ is $\geq(m+1)\pi$ for large enough $C>0$.

###### Proof.

Let $(k_{n})_{n=1}^{\infty}$ be an increasing sequence of natural numbers. We
will use this sequence to construct the C*-algebra $A$ as an inductive limit.
Then, by letting $(k_{n})_{n=1}^{\infty}$ grow sufficiently fast, we will show
that $A$ has the desired properties.

For each $n=1,\ldots$, let $X_{n}=(S^{2})^{k_{n}}$. Let $P_{n}\in
M_{2^{n}}(C(X_{n}))$ be the rank one projection $P_{n}=P^{\otimes k_{n}}$,
where $P\in M_{2}(C(S^{2}))$ is a projection whose first Chern class is a
generator of $H^{2}(S^{2})$. Let us now form the product
$Y_{n}=\prod_{j=1}^{n}X_{j}$. Observe that $Y_{n}$ is the cartesian product of
$(\sum_{j=1}^{n}k_{j})$ 2-dimensional spheres.

Consider the projection $p_{n}\in M_{\infty}(C(Y_{n}))$ defined by

$p_{n}(y)=P_{1}(x_{1})^{\oplus l_{1}}\oplus P_{2}(x_{2})^{\oplus
l_{2}}\oplus\cdots\oplus P_{n}(x_{n})^{\oplus l_{n}},$

where $y=(x_{1},x_{2},\dots,x_{n})\in Y_{n}$. Let us set the numbers $l_{n}$
recursively such that $l_{1}=1$ and $l_{n+1}=\mathrm{rank}(p_{n})$ for $n\geq
1$. (In fact, this yields $l_{n}=2^{n-1}$ for $n\geq 1$.) Choose for each $n$
a point $c_{n}\in Y_{n}$.

Let $\phi_{n}\colon\mathrm{her}(p_{n})\to\mathrm{her}(p_{n+1})$ be a
homomorphism defined as follows:

$\phi_{n}(f)(y,x_{n+1})=\begin{pmatrix}f(y)&0\\\ 0&f(c_{n})\otimes
P_{n+1}(x_{n+1})\end{pmatrix},$

for all $(y,x_{n+1})\in Y_{n}\times X_{n+1}$. On the right-hand side of this
formula we have used the 2x2 matrix notation for
$\mathrm{her}(p_{n+1})=\mathrm{her}(p_{n}\oplus P_{n+1}^{\oplus l_{n+1}})$.
Also, $f(c_{n})\otimes P_{n+1}$ is regarded as an element in
$\mathrm{her}(P_{n+1}^{\oplus l_{n+1}})$ via the identifications
$\mathrm{her}(P_{n+1}^{\oplus l_{n+1}})\cong
M_{l_{n+1}}(\mathbb{C})\otimes\mathrm{her}(P_{n+1})$ and
$\mathrm{her}(p_{n}(c_{n}))\cong M_{l_{n+1}}(\mathbb{C})$ (recall that $p_{n}$
has rank $l_{n+1}$).

It is known that choosing the points $c_{n}\in Y_{n}$ suitably, one can
arrange for the inductive limit C*-algebra
$A:=\varinjlim(\mathrm{her}(p_{n}),\phi_{n})$ to be simple and have a unique
tracial state (see [villadsen]). The uniqueness of tracial state comes from
the fact that if $\sigma$ is a tracial state on $\mathrm{her}(p_{m+n})$ then
$\sigma\circ\phi_{m,m+n}$ is an average of $2^{n}$ traces out of which
$2^{n}-1$ do not depend on $\sigma$ (coming from point evaluations).

Let us now describe how to recursively define the sequence
$(k_{n})_{n=1}^{\infty}$ so that the resulting inductive limit C*-algebra has
the desired properties. Define $k_{1}=1$. Assume that $k_{1},\ldots,k_{n}$
have been chosen. Choose $M_{n}\in\mathbb{N}$ such that $[p_{n}]\leq
M_{n}[1_{Y_{n}}]$. Now choose $k_{n+1}\geq 5nM_{n}l_{n+1},k_{n}$, and continue
this process ad infinitum.

To prove the theorem, it suffices to find for each $m\in\mathbb{N}$ a
selfadjoint element $a_{m}$ in
$\overline{[\mathrm{her}(p_{m+1}),\mathrm{her}(p_{m+1})]}$ of norm $1$ such
that

  1. (a)

$e^{ia_{m}}$ is within a distance of at least $\epsilon_{m}>0$ from any
product of $m$ commutators in $U_{0}(\mathrm{her}(p_{m+1}))$,

  2. (b)

$e^{iC_{m}a_{m}}$ is within distance $\geq(m+1)\pi$ from any product of $m$
commutators in $U_{0}(\mathrm{her}(p_{m+1}))$ for a large enough $C_{m}$.

Moreover, these properties are not destroyed by moving $a_{m}$ along the
inductive limit.

Fix $m\in\mathbb{N}$. Recall that $Y_{m}$ is a product of 2-dimensional
spheres. Let $\pi_{1}\colon Y_{m}\to S^{2}$ be the projection onto the first
of the sphere factors in $Y_{m}$. Let $\kappa\colon Y_{m}\to[-1,1]$ be defined
as in (8.1). Let $a_{m}\in\mathrm{her}(p_{m+1})$ be given by

$a_{m}(y,x_{m+1})=\begin{pmatrix}\kappa(y)p_{m}(y)&0\\\
0&-\kappa(y)P_{m+1}^{\oplus l_{m+1}}(x_{m+1})\end{pmatrix},$

for all $(y,x_{m+1})\in Y_{m+1}$, Since $a_{m}$ has pointwise zero trace, it
is in the kernel of every bounded trace. Hence,
$a_{m}\in\overline{[\mathrm{her}(p_{m+1}),\mathrm{her}(p_{m+1})]}$. Observe
also that $\|a_{m}\|=1$.

Let us verify that the conditions are met to apply Theorem 8.2. Recall that
$M_{m}\in\mathbb{N}$ is such that $[p_{m}]\leq M_{m}[1_{Y_{m}}]$. Let
$\widetilde{P}_{m+1}\in M_{\infty}(C(Y_{m+1}))$ be defined as the pull back of
$P_{m+1}$ along the projection onto $X_{m+1}$, i.e.,

(8.4) $\widetilde{P}_{m+1}(y,x_{m+1})=P_{m+1}(x_{m+1}),\quad\hbox{for
$(y,x_{m+1})\in Y_{m}\times X_{m+1}$.}$

A routine Euler class computation shows that

$e(\widetilde{P}_{m+1})=\sum_{j=1}^{k_{m+1}}\alpha_{j,m+1},$

where we have denoted by $\alpha_{j,m+1}\in H^{2}(Y_{m+1})$ the cohomology
classes coming from projecting onto the $k_{m+1}$ 2-dimensional spheres in
$X_{m+1}=(S^{2})^{k_{m+1}}$. Observe that $e(\widetilde{P}_{m+1})$ belongs to
the subring generated by elements in $H^{2}(Y_{m+1})$ arising from the sphere
factors in $X_{m+1}$, and thus omits all the generators coming from the sphere
factors in $Y_{m}$. Moreover, from our choice of the sequence $(k_{n})_{n}$,
$k_{m+1}\geq 5mM_{m}l_{m+1}$, and so

$e(\widetilde{P}_{m+1}^{\oplus
l_{m+1}})^{5mM_{m}}=e(\widetilde{P}_{m+1})^{5mM_{m}l_{m+1}}\neq 0.$

We can therefore apply Theorem 8.2 with $k=M_{m}$ to conclude that the
exponential distance from $e^{ia_{m}}$ to the set of products of $m$
commutators in $U_{0}(\mathrm{her}(p_{m+1}))$ is at least
$\epsilon_{m}=\frac{1}{M_{m}}$. Moreover, also by Theorem 8.2, the exponential
distance from $e^{iC_{m}a_{m}}$, with $C_{m}=M_{m}(m+1)\pi$, to the set of
products of $m$ commutators in $U_{0}(\mathrm{her}(p_{m+1}))$ is at least
$(m+1)\pi$.

Next we show that moving $a_{m}$ along the inductive limit does not change its
properties. Let $n\in\mathbb{N}$. Consider the image of
$a_{m}\in\mathrm{her}(p_{m+1})$ in $\mathrm{her}(p_{m+n})$ by the connecting
homomorphism
$\phi_{m+1,m+n}\colon\mathrm{her}(p_{m+1})\to\mathrm{her}(p_{m+n})$. We regard
$\mathrm{her}(p_{m+n})$ as $\mathrm{her}(p_{m}\oplus q_{m,n})$, with

(8.5) $q_{m,n}(x)=P_{m+1}(x_{m+1})^{\oplus l_{m+1}}\oplus\cdots\oplus
P_{m+n}(x_{m+n})^{\oplus l_{m+n}},$

for $x=(x_{m+1},\ldots,x_{m+n})\in Y_{n+m}$. Then

$\phi_{m+1,m+n}(a_{m})(y,x)=\begin{pmatrix}\kappa(y)p_{m}(y)&\\\
&*\end{pmatrix},$

for $(y,x)\in Y_{m}\times(X_{m+1}\times\cdots\times X_{m+n})$. A routine Euler
class computation shows that $e(q_{m,n})$ belongs to a subring of
$H^{*}(Y_{m+n})$ generated by cohomology classes in $H^{2}(Y_{m+n})$ coming
from the sphere factors in the product $X_{m+1}\times\cdots\times X_{m+n}$.
Moreover, from our choice of $(k_{n})_{n}$, $e(q_{m,n})^{5mM_{m}}\neq 0$.
Hence, applications of Theorem 8.2 show that the exponential distance from
$e^{i\phi_{m+n}(a_{m})}$ to the products of $m$ commutators in
$U_{0}(\mathrm{her}(p_{m+n})$ is still $\geq\frac{1}{M_{m}}$, and the distance