SIMPLICITY OF LEAVITT PATH ALGEBRAS VIA GRADED RING THEORY

Abstract

Suppose that R is an associative unital ring and that $E=(E^0,E^1,r,s)$ is a directed graph. Using results from graded ring theory, we show that the associated Leavitt path algebra $L_R(E)$ is simple if and only if R is simple, $E^0$ has no nontrivial hereditary and saturated subset, and every cycle in E has an exit. We also give a complete description of the centre of a simple Leavitt path algebra.

1 Introduction

The Leavitt path algebra of a row-finite graph, over a field, was introduced in [2, 5] and has since then been successively generalised (see, for example, [3, 20]). The Leavitt path algebra of an arbitrary directed graph, over a unital ring, was introduced in [12]. For an account of the development of the field of Leavitt path algebras, we refer the reader to [1]. Here is our first main result.

Theorem 1.1. Suppose that R is an associative unital ring and that $E=(E^0,E^1,r,s)$ is a directed graph. The Leavitt path algebra $L_R(E)$ is simple if and only if R is simple, $E^0$ has no nontrivial hereditary and saturated subset, and every cycle in E has an exit.

Characterisations of simple Leavitt path algebras over fields have previously been established in [19, Theorem 6.18], [3, Theorem 3.1] and [11, Theorem 3.5]. Theorem 1.1 generalises all of those results, and also partially generalises [20, Theorem 7.20]. Our second main result, stated below, completely describes the centre of a simple Leavitt path algebra. It generalises [6, Theorem 4.2] from the case where R is a field and E is a row-finite graph.

Theorem 1.2. Suppose that R is an associative unital ring and that $E=(E^0,E^1,r,s)$ is a directed graph. Furthermore, suppose that $L_R(E)$ is a simple Leavitt path algebra. The following assertions hold.

1. (a) If $L_R(E)$ is not unital, then $Z(L_R(E)) = \{0\}$ .

2. (b) If $L_R(E)$ is unital, then $Z(L_R(E)) = Z(R) \cdot 1_{L_R(E)}$ .

Whereas earlier proofs of Theorems 1.1 and 1.2 (when R is a field) use ad hoc arguments, specifically designed for graph algebras, we use the general theory of graded rings to obtain our results. This makes our proofs shorter and, we believe, clearer. Indeed, we show that $L_R(E)$ is graded simple if and only if R is simple and $E^0$ has no nontrivial hereditary and saturated subset (see Proposition 3.6). We also show that every cycle in E has an exit if and only if the centre of each corner subring of $L_R(E)$ at a vertex has degree zero (see Proposition 3.14).

We point out that there are various generalisations of Leavitt path algebras in the literature (see, for example, [1, Section 5] and [9, 18]). A simplicity result for Steinberg algebras was obtained in [8], and when translated to Leavitt path algebras, one recovers Theorem 1.1 in the special case where R is a commutative unital ring. Note that [6, Theorem 4.2] was generalised to Kumjian–Pask algebras in [7], and in [10], Steinberg algebra techniques were used to give a complete description of the centre of a general (not necessarily simple) Leavitt path algebra $L_R(E)$ , where R is a commutative unital ring.

2 Simple $\mathbb {Z}$ -graded rings

Let $\mathbb {Z}$ denote the rational integers and write $\mathbb {N} := \{ 1,2,3,\ldots \}$ . Suppose that S is a ring. By this, we mean that S is associative but not necessarily unital. If S is unital, then we let $1_S$ denote the multiplicative identity of S. Furthermore, we let $Z(S)$ denote the centre of S, that is, the set of all $s \in S$ satisfying $st = ts$ for every $t \in S$ . Recall that S is said to be $\mathbb {Z}$ -graded if, for each $n \in \mathbb {Z}$ , there is an additive subgroup $S_n$ of S such that $S = \oplus _{n\in \mathbb {Z}} S_n$ and $S_n S_m \subseteq S_{n+m}$ , for all $n,m\in \mathbb {Z}$ . In that case, each element $s\in S$ may be written as $s=\sum _{n\in \mathbb {Z}} s_n$ , where $s_n \in S_n$ is zero for all but finitely many $n\in \mathbb {Z}$ . The support of s is defined as the finite set $\operatorname {\mathrm {Supp}}(s):=\{n\in \mathbb {Z} \mid s_n \neq 0\}$ . An ideal I of a $\mathbb {Z}$ -graded ring S is said to be graded if $I=\oplus _{n\in \mathbb {Z}} (I \cap S_n)$ . If $\{0\}$ and S are the only graded ideals of S, then S is said to be graded simple.

We recall some properties of graded rings.

Lemma 2.1. Suppose that S is a unital $\mathbb {Z}$ -graded ring.

1. (a) The ring $Z(S)$ is $\mathbb {Z}$ -graded with respect to the grading $Z(S)_n$ which is defined by $Z(S)_n := Z(S) \cap S_n$ for $n \in \mathbb {Z}$ .

2. (b) If S is a field, then $S = S_0$ .

Proof. Item (a) is [15, page 15, Exercise 8] and item (b) is [15, Remark 1.3.10].

Next, we state a special case of [17, Theorem 1.2] and [13, Theorem 5]. For the convenience of the reader, we include a shortened version of the proof from these sources adapted to the situation at hand.

Proposition 2.2. Suppose that S is a unital $\mathbb {Z}$ -graded ring. Then, the following assertions are equivalent:

1. (i) S is simple;

2. (ii) S is graded simple and $Z(S)$ is a field;

3. (iii) S is graded simple and $Z(S) \subseteq S_0$ .

Proof. (i) $\Rightarrow $ (ii) is clear and (ii) $\Rightarrow $ (iii) follows from Lemma 2.1. Now we show that (iii) $\Rightarrow $ (i). Suppose that S is graded simple and that $Z(S) \subseteq S_0$ . Let I be a nonzero ideal of S. We wish to show that $1_S \in I$ . Amongst all nonzero elements of I, choose s such that $|\!\operatorname {\mathrm {Supp}}(s)|$ is minimal. Take $m \in \operatorname {\mathrm {Supp}}(s)$ . Since S is graded simple, there are $n \in \mathbb {N}$ and homogeneous elements $p_1,\ldots ,p_n,q_1,\ldots ,q_n \in S$ , such that $\sum _{i=1}^n p_i s_m q_i = 1_S$ and $p_i s_m q_i \in S_0 \setminus \{0\}$ for every $i\in \{1,\ldots ,n\}$ . Write $t := \sum _{i=1}^n p_i s q_i$ . Note that $t \in I$ , $t_0 = 1_S$ and $|\!\operatorname {\mathrm {Supp}}(t)| \leq |\!\operatorname {\mathrm {Supp}}(s)|$ . Take $z \in \mathbb {Z}$ and $x \in S_z$ . Then, $tx-xt \in I$ and, since $t_0 = 1_S$ , it follows that $|\!\operatorname {\mathrm {Supp}}(tx - xt)| < |\!\operatorname {\mathrm {Supp}}(t)|$ . By the assumptions on s, we get $|\!\operatorname {\mathrm {Supp}}(tx - xt)| = 0$ and hence $xt = tx$ . Thus, $t \in Z(S) \subseteq S_0$ . We conclude that $1_S = t_0 = t \in I$ .

Let S be a ring. Recall from [4] (see also [16]) that a set U of idempotents in S is called a set of local units for S if for every $n \in \mathbb {N}$ and all $s_1,\ldots ,s_n \in S$ , there is some $e \in U$ such that $e s_i = s_i e = s_i$ for every $i \in \{1,\ldots ,n\}$ .

Remark 2.3. Suppose that S is a $\mathbb {Z}$ -graded ring. If $e\in S_0$ is an idempotent, then the corner subring $eSe$ inherits a natural $\mathbb {Z}$ -grading defined by $(eSe)_n:= e S_n e$ for $n\in \mathbb {Z}$ .

For future reference, we recall the following two results.

Proposition 2.4. Suppose that S is a $\mathbb {Z}$ -graded ring equipped with a set of local units $U \subseteq S_0$ . Then, S is (graded) simple if and only if, for every $f \in U$ , the ring $f S f$ is (graded) simple.

Proof. First we show the ‘only if’ statement. Suppose that S is (graded) simple and that $f \in U$ . Let J be a nonzero (graded) ideal of $f S f$ . By (graded) simplicity of S, it follows that $SJS = S$ . Thus, $fSf = fSJSf = (fSf)J(fSf) \subseteq J$ and hence $J = fSf$ . Next, we show the ‘if’ statement. Suppose that $fSf$ is (graded) simple for every $f \in U$ . Let I be a nonzero (graded) ideal of S. Take a nonzero (homogeneous) $x \in S$ . Take a nonzero (homogeneous) $y \in I$ and $f \in U$ with $fx = xf = x$ and $fy = yf = y$ . By (graded) simplicity of $fSf$ , it follows that $I \supseteq f SyS f = fSf \ni x$ . Thus, $I = S$ .

Proposition 2.5. Suppose that S is a $\mathbb {Z}$ -graded ring equipped with a set of local units and that $f \in S_0$ is a nonzero idempotent. If S is graded simple and $f S f$ is simple, then S is simple.

Proof. Suppose that S is graded simple and that $f S f$ is simple. Let I be a nonzero ideal of S. Take a nonzero $s \in I$ and write $s = \sum _{n \in \operatorname {\mathrm {Supp}}(s)} s_n$ . Fix $m \in \operatorname {\mathrm {Supp}}(s)$ and define $J := S s_m S$ . Then, J is a nonzero graded ideal of S. By graded simplicity of S, it follows that $J = S$ and, in particular, that $f \in J$ . Note that $f \in fJf$ . Since $f \neq 0$ , it follows that there exist nonzero homogeneous $y,z \in S$ such that $fy s_m zf$ is nonzero and $\deg (y)+\deg (z) = -m$ . Now, define $s' := fy s zf$ . By the construction of $s'$ , it follows that $s' \in I \cap fSf$ and that $s'$ is nonzero. In particular, $I \cap fSf \neq \{ 0 \}$ . Hence, by simplicity of $fSf$ , we see that $I \cap fSf = fSf$ . Thus, $f \in I$ . Note that $SfS$ is a nonzero graded ideal of S. Hence, by graded simplicity of S, we have $I \supseteq SfS = S$ . This shows that $I = S$ .

3 Simple Leavitt path algebras

Let R be an associative unital ring and let $E = (E^0,E^1,r,s)$ be a directed graph. Recall that r (range) and s (source) are maps $E^1 \to E^0$ . The elements of $E^0$ are called vertices and the elements of $E^1$ are called edges. The elements of $E^1$ are called real edges, while for $f\in E^1$ , we call $f^*$ a ghost edge. The set $\{f^* \mid f \in E^1\}$ will be denoted by $(E^1)^*$ . A path $\mu $ in E is a sequence of edges $\mu = \mu _1 \ldots \mu _n$ such that $r(\mu _i)=s(\mu _{i+1})$ for $i\in \{1,\ldots ,n-1\}$ . In that case, $s(\mu ):=s(\mu _1)$ is the source of $\mu $ , $r(\mu ):=r(\mu _n)$ is the range of $\mu $ and $|\mu |:=n$ is the length of $\mu $ . If $\mu =\mu _1\ldots \mu _n$ is a (real) path in E, then we let $\mu ^*:=\mu _n^*\ldots \mu _1^*$ denote the corresponding ghost path. For any vertex $v \in E^0$ , we put $s(v):=v$ and $r(v):=v$ . We let $r(f^*)$ denote $s(f)$ and we let $s(f^*)$ denote $r(f)$ . For $n \geq 2$ , we define $E^n$ to be the set of paths of length n and $E^* := \bigcup _{n\geq 0} E^n$ the set of all finite paths.

Following Hazrat [12], we make the following definition.

Definition 3.1. The Leavitt path algebra of E with coefficients in R, denoted by $L_R(E)$ , is the algebra generated by the sets $\{v \mid v\in E^0\}$ , $\{f \mid f\in E^1\}$ and $\{f^* \mid f\in E^1\}$ with the coefficients in R, subject to the relations:

1. (1) $uv = \delta _{u,v} v$ for all $u,v \in E^0$ ;

2. (2) $s(f)f=fr(f)=f$ and $r(f)f^*=f^*s(f)=f^*$ for all $f\in E^1$ ;

3. (3) $f^*f'=\delta _{f,f'} r(f)$ for all $f,f'\in E^1$ ;

4. (4) $\sum _{f\in E^1, s(f)=v} ff^* = v$ for every $v\in E^0$ for which $s^{-1}(v)$ is nonempty and finite.

Here, elements of the ring R commute with the generators.

Remark 3.2. (a) The Leavitt path algebra $L_R(E)$ carries a natural $\mathbb {Z}$ -grading. Indeed, put $\deg (v):=0$ for each $v\in E^0$ . For each $f\in E^1$ , we put $\deg (f):=1$ and $\deg (f^*):=-1$ . By assigning degrees to the generators in this way, we obtain a $\mathbb {Z}$ -grading on the free algebra $F_R(E) = R \langle v, f, f^* \mid v \in E^0, f \in E^1 \rangle $ . Moreover, the ideal coming from relations (1)–(4) in Definition 3.1 is graded. Using this, it is easy to see that the natural $\mathbb {Z}$ -grading on $F_R(E)$ carries over to a $\mathbb {Z}$ -grading on the quotient algebra $L_R(E)$ .

(b) The set $\{\sum _{v\in F} v \mid F \text { is a finite subset of } E^0\}$ is a set of local units for $L_R(E)$ . If $E^0$ is finite, then $L_R(E)$ is unital and $1_{L_R(E)}=\sum _{v\in E^0} v$ .

(c) Motivated by Definition 3.1(2), for $u\in E^0$ , we write $u^*:=u$ .

Definition 3.3. Let $E=(E^0,E^1,r,s)$ be a directed graph. A subset $H \subseteq E^0$ is said to be hereditary if $s(f) \in H$ implies $r(f) \in H$ for any $f\in E^1$ . A hereditary subset $H \subseteq E^0$ is called saturated if $\{r(f) \in H \mid f \in E^1 \text { and } s(f)=v \} \subseteq H$ implies $v\in H$ whenever $v\in E^0$ satisfies $0 < |s^{-1}(v)| < \infty $ .

Remark 3.4. Note that $\varnothing $ and $E^0$ are always hereditary and saturated subsets of $E^0$ . They are referred to as trivial.

Lemma 3.5. Every element in $E^0 \cup E^1 \cup (E^1)^*$ is nonzero in $L_R(E)$ , and the set of real (respectively ghost) paths is linearly independent in the left R-module $L_R(E)$ and in the right R-module $L_R(E)$ .

Proof. The proof of [20, Proposition 4.9] immediately carries over to the case where R is a noncommutative unital ring. The same holds for the proof of [20, Proposition 3.4] in case $E^0$ and $E^1$ are countable sets. Otherwise, the proof may be adapted by taking $\aleph $ to be an infinite cardinal at least as large as $\text {card}(E^0 \cup E^1)$ and defining $Z:=\oplus _{\aleph } R$ (with the notation of [20, Proposition 3.4]).

Proposition 3.6. The Leavitt path algebra $L_R(E)$ is graded simple if and only if R is simple and $E^0$ has no nontrivial hereditary and saturated subset.

Proof. First we show the ‘if’ statement. Suppose that R is simple and that $E^0$ has no nontrivial hereditary and saturated subset. Let I be a nonzero graded ideal of $L_R(E)$ . Consider the set $H_I := \{ v \in E^0 \mid k v \in I \text { for some nonzero } k \in R \}$ . By the same argument as in [20, Lemma 5.1], $H_I$ is nonempty. Furthermore, since R is simple, it follows that $H_I = \{ v \in E^0 \mid v \in I \}$ . We wish to show that $H_I$ is hereditary and saturated. To this end, take $v\in H_I$ . Suppose that $e \in E^1$ with $s(e) = v$ . Then, $r(e) = e^* e = e^* v e \in ~I$ . Thus, $H_I$ is hereditary. Now, take $v\in E^0$ such that $0 < |s^{-1}(v)| < \infty $ , and suppose that $r(s^{-1}(v)) \subseteq H_I$ . For each $e \in s^{-1}(v)$ , we have $r(e) \in H_I$ and hence $e e^* = e r(e) e^* \in I$ . Thus, $v = \sum _{e \in s^{-1}(v)} ee^* \in I$ and $v \in H_I$ . Therefore, $H_I$ is saturated. By our assumption, $H_I = E^0$ . This shows that I must contain all the local units of $L_R(E)$ and thus $I = L_R(E)$ . Hence, $L_R(E)$ is graded simple.

Now, we show the ‘only if’ statement. Suppose that $L_R(E)$ is graded simple. Let J be a proper ideal of R. We wish to show that $J = \{ 0 \}$ . To this end, let $L_J(E)$ denote the graded ideal of $L_R(E)$ consisting of all elements of $L_R(E)$ with coefficients coming from J. Consider the natural ring homomorphism $\varphi : L_R(E) \to L_{R/J}(E)$ . Clearly, $\varphi $ is well defined. Note that $L_J(E) \subseteq \ker (\varphi )$ . Choose some $u \in E^0$ . By Lemma 3.5, applied to $L_{R/J}(E)$ , it follows that $u \notin \mathrm{ker}(\varphi )$ and hence $u \notin L_J(E)$ . Thus, $L_J(E)$ is a proper graded ideal of $L_R(E)$ . By graded simplicity of $L_R(E)$ , it follows that $L_J(E) = \{ 0 \}$ . Thus, in particular, $Ju = \{ 0 \}$ . By Lemma 3.5 applied to $L_R(E)$ , we see that $J = \{ 0 \}$ .

Let H be a proper hereditary and saturated subset of $E^0$ . Following [2, 3], we let $F:=(F^0,F^1,r,s)$ be the graph consisting of all vertices not in H and all edges whose range is not in H. For $v \in E^0$ , define $\Psi (v) := v$ if $v \in F^0$ , and $\Psi (v) := 0$ otherwise. For $e \in E^1$ , define $\Psi (e) := e$ if $e \in F^1$ , and $\Psi (e) := 0$ otherwise. Furthermore, define $\Psi (e^*) := e^*$ if $e^* \in (F^1)^*$ , and $\Psi (e^*) := 0$ otherwise. The argument in [2, 3] shows that this yields a well-defined ring homomorphism $\Psi : L_R(E) \to L_R(F)$ . Clearly, $\Psi $ is graded. Thus, the ideal $I := \ker (\Psi )$ of $L_R(E)$ is graded. Note that $F^0$ is nonempty, because H is proper, and hence $I \neq L_R(E)$ . By our assumption, $I=\{0\}$ . By the construction of $\Psi $ , it follows that $H \subseteq I$ . Thus, $H=\varnothing $ .

Definition 3.7. Define an additive map $\mathcal {L} : L_R(E) \to L_R(E)$ by requiring that $\mathcal {L}(\lambda \alpha \beta ^*) = \lambda \beta \alpha ^*$ for all $\lambda \in R$ and $\alpha ,\beta \in E^*$ .

Remark 3.8. The map $\mathcal {L}$ is an isomorphism of additive groups such that $\mathcal {L}((L_R(E))_N) = (L_R(E))_{-N}$ for every $N \in \mathbb {Z}$ .

Lemma 3.9. Suppose that $u \in E^0$ . The map $\mathcal {L}$ restricts to an isomorphism of additive groups $\mathcal {L}\lvert _{Z(u L_R(E) u)} : Z(u L_R(E) u) \to Z(u L_R(E) u)$ . In particular, the equality $\mathcal {L}((Z(u L_R(E) u))_N) = (Z(u L_R(E) u))_{-N}$ holds for every $N \in \mathbb {Z}$ .

Proof. Let $x = \sum _{j=1}^m \lambda _j \alpha _j \beta _j^* \in Z(u L_R(E) u)$ , where $\lambda _j \in R$ , $\alpha _j,\beta _j \in E^*$ and $s(\alpha _j)=s(\beta _j)=u$ for $j \in \{1,\ldots ,m\}$ . Take $r \in R$ . Then, $0 = xru-ru x = \sum _{j=1}^m (\lambda _j r - r \lambda _j) \alpha _j \beta _j^*$ . Therefore, we have $0 = \mathcal {L}(0) = \sum _{j=1}^m (\lambda _j r - r \lambda _j) \mathcal {L}(\alpha _j \beta _j^*) = \sum _{j=1}^m ( \lambda _j r - r \lambda _j ) \beta _j \alpha _j^* = \mathcal {L}(x) ru - ru \mathcal {L}(x)$ . Thus, $\mathcal {L}(x) ru = ru \mathcal {L}(x)$ . Take $\gamma ,\delta \in E^*$ with $s(\gamma ) = s(\delta ) = u$ . Then, $0 = x \gamma \delta ^* - \gamma \delta ^* x = \sum _{j=1}^m \lambda _j (\alpha _j \beta _j^* \gamma \delta ^* - \gamma \delta ^* \alpha _j \beta _j^*)$ . Therefore, we have $0 = \mathcal {L}(0) = \sum _{j=1}^m \lambda _j \mathcal {L}(\alpha _j \beta _j^* \gamma \delta ^* \kern1.4pt{-}\kern1.4pt \gamma \delta ^* \alpha _j \beta _j^*) \kern1.4pt{=}\kern1.4pt \sum _{j=1}^m \lambda _j (\delta \gamma ^* \beta _j\alpha _j^* \kern1.4pt{-}\kern1.4pt \beta _j\alpha _j^* \delta \gamma ^*) \kern1.4pt{=}\kern1.4pt \delta \gamma ^* \mathcal {L}(x) \kern1.4pt{-}\kern1.4pt \mathcal {L}(x) \delta \gamma ^*$ . Thus, $\mathcal {L}(x)\delta \gamma ^* = \delta \gamma ^* \mathcal {L}(x)$ . Finally, $\mathcal {L}(x) r \delta \gamma ^* = \mathcal {L}(x) ru \delta \gamma ^* = ru \mathcal {L}(x) \delta \gamma ^* = ru \delta \gamma ^* \mathcal {L}(x) = r \delta \gamma ^* \mathcal {L}(x)$ . This shows that $\mathcal {L}(x) \in Z(u L_R(E) u)$ .

Lemma 3.10. Suppose that $u,v \in E^0$ and that $\alpha \in E^*$ is such that $s(\alpha ) = u$ and $r(\alpha ) =~v$ . If $x \in Z(u L_R(E) u)$ , then $\alpha ^* x \alpha \in Z(v L_R(E) v)$ .

Proof. Let $x \in Z(u L_R(E) u)$ . Take $y \in v L_R(E) v$ . Since $\alpha y \alpha ^* \in u L_R(E) u$ , it follows that $y \alpha ^* x \alpha = v y \alpha ^* x \alpha = \alpha ^* \alpha y \alpha ^* x \alpha = \alpha ^* x \alpha y \alpha ^* \alpha = \alpha ^* x \alpha y v = \alpha ^* x \alpha y$ . Thus, $\alpha ^* x \alpha \in Z(v L_R(E) v)$ .

Definition 3.11 [20]

Let $E=(E^0,E^1,r,s)$ be a directed graph. A cycle in E is a path $\mu \in E^*\setminus E^0$ such that $s(\mu )=r(\mu )$ . An edge $f \in E^1$ is said to be an exit for the cycle $\mu =\mu _1 \ldots \mu _n$ if $s(f)=s(\mu _i)$ but $f \neq \mu _i$ for some $i \in \{1, 2, \ldots , n\}$ .

Remark 3.12. The definition of a cycle in a directed graph varies in the literature on Leavitt path algebras. In contrast to the most common definition of a cycle (see [2, page 320], following [20], we allow a cycle to ‘intersect’ itself. In Theorem 1.1, the condition that ‘every cycle in E has an exit’ appears. That condition is commonly known as Condition (L). It is easy to see that Condition (L) is satisfied with the first definition of a cycle [2] if and only if it is satisfied with the second definition of a cycle [20].

Remark 3.13. Let x be a nonzero element of $L_R(E)$ . It is clear from the definition of $L_R(E)$ that x can be represented as a finite sum $x = \sum _{i=1}^n r_i \alpha _i \beta _i^*$ , where $r_i \in R \setminus \{ 0 \}$ and $\alpha _i,\beta _i \in E^*$ . Following [20, Definition 4.8], we define the real degree (respectively ghost degree) of this representation as $\max \{ \deg (\alpha _i) \mid 1 \leq i \leq n \}$ (respectively $\max \{ \deg (\beta _i) \mid 1 \leq i \leq n \}$ ). Note that, in general, the real degree and ghost degree of x depend on the particular choice of representation. If, however, x has a representation in only real (respectively ghost) edges, that is, if $x = \sum _{i=1}^n r_i \alpha _i$ (respectively $x = \sum _{i=1}^n r_i \beta _i^*$ ), then, by Lemma 3.5, the real (respectively ghost) degree is independent of the choice of representation of x in real (respectively ghost) edges.

Proposition 3.14. Every cycle in E has an exit if and only if for every $u \in E^0$ , the inclusion $Z(u L_R(E) u) \subseteq (u L_R(E) u)_0$ holds.

Proof. First, we show the ‘if’ statement by contrapositivity. Suppose that there is a cycle $p \in E^* \setminus E^0$ without any exit. Set $u:=s(p)$ and write $p^0:=u$ . Take $r \in R$ and $\alpha ,\beta \in E^*$ with $s(\alpha ) = s(\beta ) = u$ and $r(\alpha )=r(\beta )$ . Since p has no exit, there are $m,n \in \mathbb {N} \cup \{0\}$ and $\gamma \in E^*$ such that $\alpha = p^m \gamma $ and $\beta = p^n \gamma $ . Note that $\gamma \gamma ^*=u=pp^*$ . This yields $p r\alpha \beta ^* = p r p^m \gamma \gamma ^* (p^*)^n = r p^{m+1} (p^*)^n$ and $r\alpha \beta ^* p = r p^m \gamma \gamma ^* (p^*)^n p = r p^m (p^*)^n p$ . If $n=0$ , then $p^{m+1}(p^*)^n=p^{m+1}= p^m (p^*)^n p$ , and if $n>0$ , then $p^{m+1}(p^*)^n = p^m pp^* (p^*)^{n-1} = p^m (p^*)^{n-1} p^*p = p^m (p^*)^{n} p$ . In either case, we get $p r\alpha \beta ^* = r\alpha \beta ^* p$ . Thus, $p \in Z(u L_R(E) u) \setminus (u L_R(E) u)_0$ .

Now we show the ‘only if’ statement. Suppose that every cycle in E has an exit. Take $u \in E^0$ . We wish to show that $Z(u L_R(E) u) \subseteq (u L_R(E) u)_0$ . By Lemma 2.1(a) and Lemma 3.9, it is enough to show that $(Z(u L_R(E) u))_N = \{ 0 \}$ for every negative integer N.

We now adapt parts of the proof of [3, Theorem 3.1] to our situation. Take $N < 0$ . Seeking a contradiction, suppose that the set

$$ \begin{align*} M:= \{ (u,x) \mid u \in E^0 \text{ and } \ x \in (Z(uL_R(E)u))_N \setminus \{ 0 \} \} \end{align*} $$

is nonempty. If $(u,x),(v,y) \in M$ , then we write $(u,x) \leq (v,y)$ if x has a representation in $L_R(E)$ of real degree less than or equal to all real degrees of representations of y in $L_R(E)$ . We write $(u,x) = (v,y)$ whenever $(u,x) \leq (v,y)$ and $(v,y) \leq (u,x)$ . Clearly, $\leq $ is a total order on M which therefore has a minimal element $(u,x)$ . Choose a minimising representation $x = \sum _{i=1}^n e_i a_i + b$ , where $e_1,\ldots ,e_n \in E^1$ are all distinct, each $a_i \in L_R(E)$ is either zero or nonzero and representable as an element of smaller real degree than that of x, and b is a polynomial (possibly zero) in only ghost paths whose source and range equals u. Take $i\in \{1,\ldots ,n\}$ . Write $v_i := r(e_i)$ . By Lemma 3.10, $e_i^* x e_i \in (Z(v_i L_R(E) v_i))_N$ . Since $e_i^* x e_i$ is of smaller real degree than x, it follows that $e_i^* x e_i = 0$ . Further, since $x \in (Z(u L_R(E) u))_N$ , it follows that $e_i^* x = e_i^* e_i e_i^* x = e_i^* x e_i e_i^* = 0$ . Thus, $0 = e_i^* x = a_i + e_i^* b$ and hence $a_i = -e_i^* b$ .

Now, $0 \neq x = (u - \sum _{i=1}^n e_i e_i^*)b$ . Thus, $u \neq \sum _{i=1}^n e_i e_i^*$ and $b\neq 0$ . This implies that there is some $f \in E^1 \setminus \{e_1,\ldots ,e_n\}$ with $s(f) = u$ . Furthermore, $f^* x = f^* b$ , and, by Lemma 3.5, $f^* b \neq 0$ since it is a sum of distinct ghost paths. Write $v:=r(f)$ . By Lemma 3.10, it follows that $f^* x f \in (Z(v L_R(E) v))_N$ . Observing that $0 \neq f^* x = f^* f f^* x = f^* x f f^*$ , we get $f^* x f \neq 0$ . Note that the real degree of $f^* x f$ is less than or equal to the real degree of x. Hence, by the assumption made on $(u,x)$ , and possibly after replacing $(u,x)$ by $(v,f^*xf)$ , we may assume that $a_i=0$ for every $i\in \{1,\ldots ,n\}$ . Therefore, suppose that $x = \sum _{j=1}^m r_j \beta _j^*$ for some nonzero $r_j \in R$ and some distinct paths $\beta _j \in E^{-N}$ with $s(\beta _j)=r(\beta _j) = u$ . Take $k \in \{ 1,\ldots ,m \}$ . By Lemma 3.10, it follows that $r_k \beta _k^* = \beta _k^* x \beta _k \in Z(u L_R(E) u)$ . By assumption, the cycle $\beta _k$ has an exit at some $w \in E^0$ . Thus, there are $\gamma ,\delta \in E^*$ and $\epsilon \in E^1$ such that $\beta _k = \gamma \delta $ , $r(\gamma ) = s(\epsilon )=w$ and $\epsilon ^* \delta = 0$ . By Lemma 3.10, it follows that $r_k (\delta \gamma )^* = r_k \gamma ^* \delta ^* \gamma ^*\gamma = \gamma ^* r_k \beta _k^* \gamma \in Z(w L_R(E) w)$ . We now reach a contradiction, because $0 \neq \epsilon \epsilon ^* r_k (\delta \gamma )^* = r_k (\delta \gamma )^* \epsilon \epsilon ^* = 0$ .

Now, we prove our main result.

Proof of Theorem 1.1

First, we show the ‘only if’ statement. Suppose that $L_R(E)$ is simple. Then $L_R(E)$ is graded simple and hence, by Proposition 3.6, it follows that R is simple and that $E^0$ has no nontrivial hereditary and saturated subset. Furthermore, Proposition 2.4 implies that $uL_R(E)u$ is simple for every $u \in E^0$ , and hence, by Proposition 2.2, $Z(uL_R(E)u) \subseteq (uL_R(E)u)_0$ for every $u \in E^0$ . Thus, by Proposition 3.14, every cycle in E has an exit.

Now we show the ‘if’ statement. Suppose that R is simple, $E^0$ has no nontrivial hereditary and saturated subset, and every cycle in E has an exit. By Proposition 3.6, $L_R(E)$ is graded simple. Take $u \in E^0$ . It follows from Proposition 3.14 that $Z(uL_R(E)u) \subseteq (uL_R(E)u)_0$ . Furthermore, by Proposition 2.4, $uL_R(E)u$ is graded simple. Thus, by Proposition 2.2, $uL_R(E)u$ is simple. Hence, by Proposition 2.5, $L_R(E)$ is simple.

4 The centre of a simple Leavitt path algebra

In this section, we prove Theorem 1.2 using results from the previous sections together with some auxiliary observations.

Remark 4.1. Let $E=(E^0,E^1,r,s)$ be a directed graph.

(a) Take $v \in E^0$ . We write $w \leq v$ , for $w \in E^0$ , if there is $\mu \in E^*$ with $s(\mu )=v$ and $r(\mu )=w$ . The set $T(v) := \{ w \in E^0 \mid w \leq v \}$ is the smallest hereditary subset of $E^0$ containing v.

(b) Suppose that $X \subseteq E^0$ . Put $T(X) := \bigcup _{x \in X} T(x)$ . The hereditary saturated closure $\overline {X}$ of X is defined as the smallest hereditary and saturated subset of $E^0$ containing X. One can show (see [6, page 626] and the references therein) that $\overline {X} = \bigcup _{n=0}^{\infty } X_n$ , where $X_0 := T(X)$ and $X_n := \{ y \in E^0 \mid 0 < |s^{-1}(y)| < \infty \ \mbox {and} \ r( s^{-1}(y) ) \subseteq X_{n-1} \} \cup X_{n-1}$ for $n \geq 1$ .

The following result can be proved by induction (see [14, Proposition 14.11] and [20, Lemma 5.2]).

Proposition 4.2. Suppose that R is an associative unital ring and that $E=(E^0,E^1,r,s)$ is a directed graph. If $a \in (L_R(E))_0$ is nonzero, then there exist $\alpha , \beta \in E^*$ , $v \in E^0$ and a nonzero $k \in R$ such that $\alpha ^* a \beta = kv$ .

Now, we prove our second main result.

Proof of Theorem 1.2

Write $S := L_R(E)$ . If S is not unital, then it follows immediately from [21, Ch. 1, Section 3.3] that $Z(S)=\{0\}$ . This proves item (a). Now, we show item (b). Suppose that S is unital, that is, $E^0$ is finite. Take a nonzero $x \in Z(S)$ . By Proposition 2.2, it follows that $x \in S_0$ . Therefore, by Proposition 4.2, there are $\alpha ,\beta \in E^*$ , $v \in E^0$ and a nonzero $k \in R$ such that $\alpha ^* x \beta = kv$ . From this equality, the grading and the fact that $x \in Z(S)$ , it follows that $\alpha =\beta $ and $r(\alpha )=v$ . Hence, $vx = \alpha ^*\alpha x = \alpha ^* x\alpha = \alpha ^* x \beta = kv$ . The equality $vx = kv$ implies that $k \in Z(R)$ . Put $X := \{ v \}$ . Then, $\overline {X}$ is a nonempty hereditary and saturated subset of $E^0$ . By Theorem 1.1, $\overline {X} = E^0$ . We claim that this implies that $wx = kw$ for every $w \in E^0$ . Let us assume, for a moment, that this claim holds. Then, $x = 1_S \cdot x = \sum _{w \in E^0} wx = \sum _{w \in E^0} kw = k \cdot \sum _{w \in E^0} w = k \cdot 1_S \in Z(R) \cdot 1_S$ . Thus, $Z(S) \subseteq Z(R) \cdot 1_S$ . Clearly, $Z(R) \cdot 1_S \subseteq Z(S)$ holds.

Now we show the claim. We will use induction to prove that for every $n \geq 0$ , the implication $w \in X_n \Rightarrow wx = kw$ holds. From this, the claim follows. Base case: $n=0$ . Suppose that $w\in X_0$ , that is, $w \leq v$ . Then, there is a path $\delta $ from v to w. This gives $wx = \delta ^* \delta x = \delta ^* v \delta x = \delta ^* vx \delta = \delta ^* kv \delta = k \delta ^* v \delta = k \delta ^* \delta = k w$ . Induction step: Suppose that $wx = kw$ for every $w \in X_{n-1}$ . Take $y \in X_n \setminus X_{n-1}$ and note that $0 < |s^{-1}(y)| < \infty $ and $r( s^{-1}(y) ) \subseteq X_{n-1}$ . Then, $yx = \sum _{e \in s^{-1}(y)} e e^* x = \sum _{e \in s^{-1}(y)} e r(e) x e^* = \sum _{e \in s^{-1}(y)} e k r(e) e^* = k \sum _{e \in s^{-1}(y)} e e^* = ky$ .