2. The main result

We follow the analytic conventions for groupoids; in particular, we identify objects and identity arrows. An ample groupoid $\mathscr {G}$ is a topological groupoid whose unit space $\mathscr {G}^{(0)}$ is locally compact, Hausdorff and totally disconnected, and whose range map $\mathop {\boldsymbol r}\colon \mathscr {G}\to \mathscr {G}^{(0)}$ is a local homeomorphism. In this paper, all ample groupoids will be assumed Hausdorff. In this case, the unit space $\mathscr {G}^{(0)}$ is a clopen subspace of $\mathscr {G}$ . An open subset $U\subseteq \mathscr {G}$ is a (local) bisection if $\mathop {\boldsymbol d}|_U,\mathop {\boldsymbol r}|_U$ are injective. The compact bisections form a basis for the topology on $\mathscr {G}$ .

If R is a commutative ring with unit, the algebra $R\mathscr {G}$ of $\mathscr {G}$ [Reference Steinberg6] consists of the compactly supported locally constant functions $f\colon \mathscr {G}\to R$ under convolution,

$$ \begin{align*}f\ast g(\gamma)=\sum_{\mathop{\boldsymbol r}(\alpha)=\mathop{\boldsymbol r}(\gamma)}f(\alpha)g(\alpha^{-1} \gamma).\end{align*} $$

The complex algebra $\mathbb C\mathscr {G}$ is a $\ast $ -algebra with $f^\ast (\gamma ) = \overline {f(\gamma ^{-1})}$ . From now on, R will always be a subring of $\mathbb C$ closed under complex conjugation. The algebra $R\mathscr {G}$ is then a $\ast $ -subalgebra of $\mathbb C\mathscr {G}$ , and so we can talk about things like projections and unitaries in $R\mathscr {G}$ . For instance, if U is a compact open subset of $\mathscr {G}^{(0)}$ , then the indicator function $1_U$ is a projection. By a $\ast $ -algebra homomorphism $\varphi \colon R\mathscr {G}\to R\mathscr H$ , we mean an R-algebra homomorphism such that $\varphi (f^\ast )=\varphi (f)^\ast $ for all f.

Let us denote by $D(R\mathscr {G})$ the subalgebra of $R\mathscr {G}$ consisting of functions supported on $\mathscr {G}^{(0)}$ . Then $D(R\mathscr {G})$ is a commutative $\ast $ -subalgebra of $R\mathscr {G}$ , often referred to as the diagonal subalgebra. Note that $D(R\mathscr {G})$ is spanned over R by the projections $1_U$ with $U\subseteq \mathscr {G}^{(0)}$ compact open.

A homomorphism $\varphi \colon R\mathscr {G}\to R\mathscr H$ of groupoid algebras is diagonal preserving if $\varphi (D(R\mathscr {G}))\subseteq D(R\mathscr H)$ . We say that an isomorphism $\varphi $ is a diagonal-preserving isomorphism if $\varphi $ and $\varphi ^{-1}$ are diagonal preserving or, equivalently, $\varphi $ is an isomorphism taking $D(R\mathscr {G})$ onto $D(R\mathscr H)$ .

An element $n\in R\mathscr {G}$ is a normaliser of $D(R\mathscr {G})$ if there is an element $n'$ with ${nn'n=n}$ , $n'nn'=n'$ and $nD(R\mathscr {G})n'\cup n'D(R\mathscr {G})n\subseteq D(R\mathscr {G})$ . It is easy to see that any $f\in R\mathscr {G}$ whose support is a bisection is a normaliser. In [Reference Steinberg7], $\mathscr {G}$ was defined to satisfy the local bisection hypothesis with respect to R if the only normalisers are those which support a bisection. For example, a group G satisfies the local bisection hypothesis with respect to R if and only if the group ring $RG$ has only trivial units. It was shown in [Reference Steinberg7] that if there is a dense set of units $x\in \mathscr {G}^{(0)}$ such that the group ring over R of the isotropy group at x of the interior of the isotropy bundle of ${\mathscr {G}}$ has only trivial units, then $\mathscr {G}$ satisfies the local bisection hypothesis. This includes all boundary path groupoids of graphs and higher rank graphs. The main theorem of [Reference Steinberg7] admits the following theorem as a special case.

The papers [Reference Carlsen2, Reference Johansen and Sørensen5] investigated the case when every $\ast $ -algebra homomorphism must automatically be diagonal preserving in the setting of Leavitt path algebras. We consider here the general case.

For $n\geq 1$ , let $\mathcal R_n$ be the discrete groupoid associated to the universal equivalence relation on $\{1,\ldots , n\}$ . So $\mathcal R_n$ has these n objects and a unique isomorphism $(i,j)$ from j to i. Multiplication follows the rule

$$ \begin{align*}(i,j)(k,\ell) = \begin{cases} (i,\ell) & \text{if}\ j=k,\\ \text{undefined} & \text{else},\end{cases}\end{align*} $$

and the inversion is given by $(i,j)^{-1} = (j,i)$ . It is not difficult to see that $R\mathcal R_n$ is $\ast $ -isomorphic to $M_n(R)$ via $f\mapsto [f((i,j))]$ with inverse $A\mapsto f_A$ where $f_A((i,j))=A_{ij}$ . The diagonal subalgebra $D(R\mathcal R_n)$ consists of those functions supported on the diagonal and is sent via the above isomorphism onto the subalgebra $D_n(R)$ of diagonal matrices. This explains the nomenclature.

The following theorem greatly generalises and expands on [Reference Carlsen2, Reference Johansen and Sørensen5], where only Leavitt path algebras were considered.

Proof. The first implication is in [Reference Carlsen2], but we repeat the proof for the reader’s convenience. If $r_1=\sum _{i=1}^n |r_i|^2$ , then $r_1\geq 0$ . Therefore,

$$ \begin{align*}|1-r_1|^2+\sum_{i=2}^n|r_i|^2+\sum_{i=1}^n |r_i|^2 = 1-2r_1+|r_1|^2+\sum_{i=2}^2|r_i|^2+\sum_{i=1}^n|r_i|^2 = 1.\end{align*} $$

If any $r_i\neq 0$ , then $r_1>0$ , and so we must have $r_2=\cdots =r_n=0$ by item (1). Assume now item (2) and let $f\in R\mathscr {G}$ be a projection. Then $f=ff^*$ . Let $x\in \mathscr {G}^{(0)}$ . Then

$$ \begin{align*}f(x) = \sum_{\mathop{\boldsymbol r}(\gamma)=x}f(\gamma)f^*(\gamma^{-1})=\sum_{\mathop{\boldsymbol r}(\gamma)=x}|f(\gamma)|^2.\end{align*} $$

Thus, by item (2), if $\mathop {\boldsymbol r}(\gamma )=x$ and $\gamma \neq x$ , then $f(\gamma )=0$ . Therefore, f is supported on $\mathscr {G}^{(0)}$ and hence $f\in D(R\mathscr {G})$ as $\mathscr {G}$ is Hausdorff. It is immediate that item (3) implies item (4) as $D(R\mathscr {G})$ is spanned over R by projections and each projection in $R\mathscr H$ is diagonal by item (3). Trivially, item (4) implies item (5). Recalling that $R\mathcal R_n\cong M_n(R)$ via a $\ast $ -isomorphism taking $D(R\mathcal R_n)$ onto $D_n(R)$ , for item (5) implies item (6), it suffices to observe that conjugation by a unitary matrix is a $\ast $ -automorphism, and the normaliser in $GL_n(R)$ of $D_n(R)$ is the group of monomial matrices. Finally, suppose that item (6) holds and that $v\in R^n$ is a unit vector (which we view as a column vector). Then $vv^*$ is a projection, and so $U=I-2vv^*$ is a self-adjoint unitary matrix. Suppose that $v_i\neq 0\neq v_j$ with $i\neq j$ . Then $U_{ij} = -2v_i\overline {v_j}\neq 0$ , ${U_{ii} = 1-2|v_i|^2}$ and $U_{jj}= 1-2|v_j|^2$ . Thus, if U is monomial, then we must have $|v_i|^2=|v_j|^2=1/2$ . However, then

$$ \begin{align*} \begin{bmatrix} v_1 & -\overline v_2\\ v_2 & \overline v_1 \end{bmatrix} \end{align*} $$

is unitary and not monomial, contradicting item (6). Thus, v has a single nonzero entry. This completes the proof.

We remark that it was claimed without proof in [Reference Carlsen2] that there are rings satisfying item (2) but not item (1). In fact, it is easy to see directly that item (2) implies item (1) since if $1=\sum _{i=1}^n|r_i|^2$ , where without loss of generality $r_1\neq 0$ , then $|r_1|^2 = \sum _{i=1}^n|r_1r_i|^2$ , and so $r_i=0$ for $i\geq 2$ by item (2).

Those rings satisfying the equivalent conditions of Theorem 2.2 were called kind in [Reference Carlsen2] and were said to have a unique partition of the unit in [Reference Johansen and Sørensen5]. We prefer ‘kind’. Rings with the property that $c_1^2+\cdots +c_n^2=1$ implies $c_i=\pm 1$ for a unique i, and $c_j=0$ otherwise, were studied in [Reference Ishibashi4] under the name L-rings. Many of the observations in that paper about L-rings apply to kind rings. Note that a subring of $\mathbb R$ is kind if and only if it is an L-ring. Any L-ring which is a subring of the complex numbers closed under complex conjugation must be kind. There are, however, kind rings that are not L-rings. For instance, $\mathbb Z[i]$ is kind (see Proposition 2.3), but $2^2+i^2+i^2+i^2=1$ , so it is not an L-ring.

Proof. If $1/n\in R$ , then $(1/n,1/n,\ldots ,1/n)\in R^{n^2}$ is a unit vector, and so R is not kind. This proves item (1). The second item is clear since if $1=|r_1|^2+\cdots +|r_n|^2$ , then $|r_i|^2\geq 1$ for at most one value of i. Item (3) is clear since a unit vector has finitely many entries. For item (4), first note that $R[B]$ is closed under complex conjugation since $B\subseteq \mathbb R$ . We may assume by item (3) that B is finite and then, by induction, it suffices to handle the case $R[a]$ where $a\in \mathbb R$ is transcendental over F. Suppose that $(f_1(a),\ldots , f_n(a))\in R[a]^n$ is a unit vector with $f_i\in R[x]$ . Then since $a\in \mathbb R$ , we have $|f_i(a)|^2 = (f_i\overline f_i)(a)$ , and $f_i\overline f_i$ is a polynomial over R with real coefficients and with leading coefficient the square of the absolute value of the leading coefficient of $f_i$ . Therefore, if some $f_i$ is a nonconstant polynomial, then $g(x)=f_1\overline f_1+\cdots +f_n\overline f_n-1$ is a nonzero polynomial of degree twice the maximum degree of the $f_i$ with $g(a)=0$ . This is a contradiction since a is transcendental over F. Thus, $f_1,\ldots , f_n$ are constant polynomials, and so $(f_1(a),\ldots , f_n(a))\in R^n$ and hence has exactly one nonzero entry since K is kind. Finally, if $\sqrt {n}\notin F$ and $(a_1+b_1\sqrt {n},\ldots , a_m+b_m\sqrt {n})\in R[\sqrt {n}]^m$ is a unit vector, then

$$ \begin{align*}1=\sum_{i=1}^m (|a_i|^2+|b_i|^2|n|)+\sqrt{n}\sum_{i=1}^m (\pm a_i\overline b_i+\overline a_ib_i).\end{align*} $$

Since $\sqrt {n}\notin F$ , we must have $\sum _{i=1}^m (\pm a_i\overline b_i+\overline a_ib_i)=0$ . Since $|n|$ is a positive integer, we deduce using the fact that R is kind that there is a unique i with either $a_i\neq 0$ or $b_i\neq 0$ . We conclude that $R[\sqrt {n}]$ is kind.