# ON THE K-THEORY OF ALGEBRAIC CUNTZ-PIMSNER RINGS

Thibaut Lescure

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Thibaut Lescure. ON THE K-THEORY OF ALGEBRAIC CUNTZ-PIMSNER RINGS. 2026. hal-05507689

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# ON THE K-THEORY OF ALGEBRAIC CUNTZ-PIMSNER RINGS

## THIBAUT LESCURE

**ABSTRACT.** We establish a long exact sequence for the homotopy K-theory groups of the algebraic Cuntz-Pimsner rings introduced by Carlsen and Ortega [CO11] by adapting Pimsner's original proof [Pim97] to Cuntz's formalism.

## 1. INTRODUCTION

In [Pim97], Pimsner introduced a class of $C^*$-algebras generalizing both crossed products by $\mathbb{Z}$ and graph $C^*$-algebras. The construction takes as input a base $C^*$-algebra $A$ and a right Hilbert module $\mathcal{H}$ together with a left action of $A$ on $\mathcal{H}$ by adjointable operators. By seeing the elements of the $\mathcal{H}$ as creation and annihilation operators on the Fock space one defines a $C^*$-algebra called the Toeplitz(-Pimsner) algebra $\mathcal{T}_{\mathcal{H}}$. If $\mathcal{I}$ is a $C^*$-ideal of $A$ that acts on the left on $\mathcal{H}$ by compact operators then quotienting by a suitable ideal yields the relative Cuntz-Pimsner algebra $\mathcal{O}_{\mathcal{H},\mathcal{I}}$. Using Kasparov's bivariant K-theory, Pimsner was able to build an inverse in K-theory to the inclusion map $A \to \mathcal{T}_{\mathcal{H}}$. This result together with computation of the maps gives a long exact sequence of K-theory groups generalizing both that of Pimsner-Voiculescu and that for graph $C^*$ algebras.

Carlsen and Ortega [CO11] introduced an analogue of Pimsner's constructions to the purely algebraic setting in which $A$ is replaced by a (non-unital, non-commutative) ring $R$ and $\mathcal{H}$ is replaced by an $R$-system $\mathcal{X} = (X, X', g)$ which is just a pair of two $R$-bimodules and an $R$-bimodule map $X' \otimes_R X \to R$. In [CT06], Cortiñas and Thom introduced an algebraic analogue of Kasparov's bivariant K-theory carrying the classical approach of Cuntz [CMR07] to a purely discrete setting. Their theory is related to Weibel's homotopy K-theory [Wei89] in the same way as operator K-theory is related to Kasparov's KK-theory. In this same article they were able to establish the exact sequence of Pimsner-Voiculescu to compute the homotopy K-theory of a crossed product of the form $R \times_{\mathbb{Z}} \mathcal{Z}$. In [ABC09] a long exact sequence of homotopy K-theory groups was established for Leavitt path algebras of a row-finite quiver over an arbitrary ring. These two classes of rings are particular cases of the so-called algebraic Cuntz-Pimsner rings.

We rewrite Pimsner's original proof using a suitable quasi-homomorphism as an inverse in $K$-theory to the inclusion $R \to \mathcal{T}_{\mathcal{X}}$. Instead of working directly with homotopy $K$-theory or its bivariant version, we work with a given homotopy invariant, split-exact and $M$-stable functor $E: \text{Rings} \to \text{Ab}$ (see [CMR07] or [CT06]). We need to ensure that the underlying module structures are sufficiently non-degenerate, in particular the $R$-system shall satisfy condition (FS) of [CO11] to ensure that the Toeplitz ring is universal. Moreover, for all $R$-system $\mathcal{X}$ and any $M$-stable functor $E$, if $\mathcal{K}_R(\mathcal{X})$ denotes the ring of finite-rank operators then there should be an induced map $E(\mathcal{K}_R(\mathcal{X})) \to E(R)$ and corner embeddings should be sent to isomorphisms by $E$.

For these reasons, we assume that $R$ has local units, that the $R$-system $\mathcal{X}$ satisfies condition (FS) and comes together with a functional homomorphism to a canonical $R$-system $R^{(I)}$ (see Definition 2.12). Under these assumptions we are able to solve these technical problems using the formalism introduced by Burgstaller [Bur25]. We then recall the construction of the Toeplitz and relative Cuntz-Pimsner rings and then proceed with the main computation. As a first result we get
**Theorem A.** Let $R$ be a ring with local units. Let $\mathcal{X}$ be an $R$-correspondence. Let $\mathcal{T}_{\mathcal{X}}$ be the corresponding Toeplitz ring. Every homotopy invariant, split-exact and $M$-stable functor $E$ sends the inclusion $R \to \mathcal{T}_{\mathcal{X}}$ to an isomorphism $E(R) \simeq E(\mathcal{T}_{\mathcal{X}})$.

Let $(E_n)_{n \in \mathbb{Z}}$: **Rings** $\to$ **Ab** be a sequence of homotopy invariant and $M$-stable functor which satisfy excision. This means that any extension of rings

$$I \longrightarrow R \longrightarrow R/I$$

induces a long exact sequence of the corresponding abelian groups

$$\dots \longrightarrow E_n(I) \longrightarrow E_n(R) \longrightarrow E_n(R/I) \longrightarrow E_{n-1}(I) \longrightarrow \dots$$

Using Theorem A and the fact that the Cuntz-Pimsner ring is a quotient of the Toeplitz ring we obtain

**Theorem B.** Let $R$ be a ring with local units. Let $\mathcal{X}$ be an $R$-correspondence. Let $\mathcal{I} \triangleleft R$ be a two-sided ideal that has local units and acts on $X$ on the left by compact operators. There is a long exact sequence $(n \in \mathbb{Z})$

$$\dots \longrightarrow E_n(\mathcal{I}) \xrightarrow{E_n(i)-E_n(\mathcal{X})} E_n(R) \xrightarrow{E_n(j)} E_n(\mathcal{O}_{\mathcal{X},\mathcal{I}}) \longrightarrow E_{n-1}(\mathcal{I}) \longrightarrow \dots$$

Here $j: R \to \mathcal{O}_{\mathcal{X},\mathcal{I}}$ and $i: \mathcal{I} \to R$ are the natural inclusions.

Theorem A and B are both true if we replace the sequence of functors $(E_n)_{n \in \mathbb{Z}}$ by either Weibel's homotopy K-theory [Wei89] or by periodic cyclic homology [CQ97].

## 2. PRELIMINARIES

### 2.1 Burgstaller's functional modules and $M$-stable functors

Recall that a ring $R$ is said to have local units if for any $r_1, \dots, r_n \in R$ there exists an idempotent $e \in R$ such that for all $1 \le i \le n$ we have $er_i = r_i e = r_i$. We fix a ring $R$ with local units.

**Definition 2.1.** A right functional $R$-module is a triple $(X, X', g)$ where $X$ is a right $R$-module, $X'$ is a left $R$-module and $g: X' \otimes_{\mathbb{Z}} X \to R$ is an $R$-bimodular map (often called the scalar product of the functional module). For all $f \in \operatorname{Hom}_R(X, X)$ we say that $f$ is adjointable with adjoint $f^* \in \operatorname{Hom}_R(X', X')$ if

$$\forall x \in X, \forall \phi \in X', g(\phi \otimes f(x)) = g(f^*(\phi) \otimes x)$$

We call $g$ non-degenerate if

$$(\forall \phi \in X', g(\phi \otimes x) = 0) \implies x = 0 \text{ and } (\forall x \in X, g(\phi \otimes x) = 0) \implies \phi = 0$$

We will write $g(\phi \otimes x) = \phi(x)$ for $x \in X$ and $\phi$ in $X'$. One easily defines the direct sum of two right functional $R$-modules $\mathcal{X} = (X, X', g)$ and $\mathcal{Y} = (Y, Y', h)$ as $\mathcal{X} \oplus \mathcal{Y} = (X \oplus Y, X' \oplus Y', g+h)$. The following proposition is easy to verify.

**Proposition 2.2.** For each $\mathcal{X} = (X, X', g)$ a right functional module over $R$, the set of adjointable endomorphisms of $\mathcal{X}$ forms a ring. If $g$ is non-degenerate then the adjoint is unique and then adjunction satisfies $(f_1 f_2)^* = f_2^* f_1^*$.

We denote by $\mathcal{L}_R(\mathcal{X})$ the ring of adjointable right $R$-modular endomorphisms on the functional module $\mathcal{X}$, with pointwise sum and composition as operations. We will always be working with functional modules with non-degenerate scalar product. Hence for two functional $R$-modules $\mathcal{X}$ and $\mathcal{Y}$ we will often write $f: \mathcal{X} \to \mathcal{Y}$ for an $R$-modular map $f: X \to Y$ admitting an adjoint.

**Definition 2.3.** Let $\mathcal{X} = (X, X', g)$ be a right functional module over $R$. We define the abelian group of compact operators of $\mathcal{X}$ to be $\mathcal{K}_R(\mathcal{X}) = X \otimes_R X'$. It is a ring with the product defined by $(x_1 \otimes \phi_1)(x_2 \otimes \phi_2) = x_1 \otimes (\phi_1(x_2) \cdot \phi_2)$.
$\mathcal{K}_R(\mathcal{X})$ acts on the left on $X$ by $(x \otimes \phi) \cdot y = x \cdot \phi(y)$ but also on $X'$ on the right by $\psi \cdot (x \otimes \phi) = \psi(x) \cdot \phi$. This gives a ring homomorphism from $\mathcal{K}_R(\mathcal{X})$ into the adjointable operators of $\mathcal{X}$.

**Definition 2.4.** [CO11] A right functional $R$-module $\mathcal{X} = (X, X', g)$ is said to satisfy condition (FS) if for all $x_1, \dots, x_n \in X$ and $\phi_1, \dots, \phi_n \in X'$ there exist $\Theta_1, \Theta_2 \in \mathcal{K}_R(\mathcal{X})$ such that $\Theta_1 \cdot x_i = x_i$ and $\phi_i \cdot \Theta_2 = \phi_i$ for all $1 \le i \le n$.

Recall that a right $R$-module $M$ is called non-degenerate if $M \cdot R = M$. If $R$ has local units then non-degeneracy is equivalent to asking that the right action map $M \otimes_R R \to M$ is bijective.

**Proposition 2.5.** Let $\mathcal{X}$ is a right functional module over $R$ satisfying condition (FS).

* The underlying scalar product of $\mathcal{X}$ is non-degenerate.

* The underlying $R$-module structures of $\mathcal{X}$ are non-degenerate.

* $\mathcal{K}_R(\mathcal{X})$ is embedded in $\mathcal{L}_R(\mathcal{X})$ as a two-sided ideal by the ring homomorphism $j : x \otimes \phi \mapsto \theta_{x,\phi}$ defined by $\theta_{x,\phi}(y) = x \cdot \phi(y)$.

*Proof.* Let $x \in X$ be such that for all $\phi \in X'$, $\phi(x) = 0$. There exists a $\Theta \in \mathcal{K}_R(\mathcal{X})$ such that $\Theta \cdot x = x$. We can write $\Theta = \sum_i x_i \otimes \phi_i$ thus $x = \sum_i x_i \cdot \phi_i(x) = 0$. Similarly if $\phi \in X'$ is such that $\forall x \in X, \phi(x) = 0$ then $\phi = 0$.

The formulas $f \circ \theta_{x,\phi} = \theta_{f(x),\phi}$ and $\theta_{x,\phi} \circ f = \theta_{x,f^*(\phi)}$ ensure that the image of $\mathcal{K}_R(\mathcal{X})$ is a two-sided ideal of $\mathcal{L}_R(\mathcal{X})$. Let $k \in \mathcal{K}_R(\mathcal{X})$ be such that $j(k) = 0$, i.e. $k \cdot x = 0$ for all $x \in X$. Write $k = \sum_i x_i \otimes \phi_i$. Let $e \in \mathcal{K}_R(\mathcal{X})$ be such that $\phi_i \cdot e = \phi_i$ for all $1 \le i \le n$. We have $ke = \sum_i x_i \otimes (\phi_i \cdot e) = k$. Write now $e = \sum_j y_j \otimes \psi_j$. Finally, $k = ke = \sum_j (k \cdot y_j) \otimes \psi_j = 0$. $\square$

In the rest of this article we will always be working with functional modules satisfying condition (FS). Hence we will identify the elements $x \otimes \phi \in X \otimes_R X'$ with $\theta_{x,\phi} \in \mathcal{L}_R(\mathcal{X})$.

**Example 2.6.** For any set $I$ we will denote by $R^{(I)}$ the $R$-bimodule of finitely supported sequences $(r_i)_{i \in I}$ of elements of $R$. We will also denote by $R^{(I)}$ the functional module $(R^{(I)}, R^{(I)}, \langle, \rangle_R)$ equipped with the scalar product given by $\langle(\alpha_i), (\beta_i)\rangle = \sum_i \alpha_i \beta_i$. We will make use of the identification $\mathcal{K}_R(R^{(I)}) \simeq M_I(R)$ with the ring of matrices with values in $R$ having only a finite number of non-zero values. When $R$ has local units this functional module satisfies condition (FS).

**Definition 2.7.** [Bur25] Let $\mathcal{X} = (X, X', g)$ and $\mathcal{Y} = (Y, Y', h)$ be two right functional $R$-modules. A functional homomorphism is the data of two $R$-module maps $U : X \to Y$ and $V : X' \to Y'$ such that

$$ \forall \phi \in X', x \in X, V(\phi)(U(x)) = \phi(x) $$

If the scalar product of $\mathcal{Y}$ is non-degenerate then the maps $U$ and $V$ are injective. (If $U(x) = 0$ then $\phi(x) = 0$ for all $\phi \in X'$).

**Example 2.8.** Assume that we have three right functional $R$-modules such that $\mathcal{Y} = \mathcal{X} \oplus \mathcal{P}$. In this case we just let $U : X \to Y$ and $V : X' \to Y'$ to be the inclusions associated to the splittings $Y = X \oplus P$ and $Y' = X' \oplus P'$. In other words if $\iota : \mathcal{X} \to \mathcal{Y}$ and $\pi : \mathcal{Y} \to \mathcal{X}$ are the obvious inclusion and projection, we let $U = \iota$ and $V = \pi^*$. We have $V(\phi)(U(x)) = (\pi^*(\phi))(\iota(x)) = \phi(\pi(\iota(x))) = \phi(x)$. Let $\phi_1, \dots, \phi_n \in X'$ and $y \in Y$. $V(\phi_i)(y) = (\pi^*(\phi_i))(y) = \phi_i(\pi(y))$. Hence $(\iota, \pi^*)$ is a functional homomorphism.

**Example 2.9.** Let $M$ be a finitely generated right $R$-module. $M$ is then projective if and only if it is a direct summand of a right $R$-module of the form $(eR)^n$ for an idempotent $e \in R$ (see [Abr83]). There is a right functional $R$-module $\mathcal{M} = (M, M^*, g)$, where $M^* = \text{Hom}_R(M, R)$ and $g$ is the evaluation map. We have $((eR)^n)^* \simeq (Re)^n$. By a slight abuse of notation we denote by $(eR)^n$ the right functional $R$-module $((eR)^n, (Re)^n, (\cdot, \cdot))$. There exists a right functional $R$-module $\mathcal{N}$ such that $\mathcal{M} \oplus \mathcal{N} \simeq (eR)^n$. Moreover there is an obvious functional homomorphism $(eR)^n \to R^n$ given by the inclusions. Hence there is a functional homomorphism $\mathcal{M} \to R^n$.
Let $(U, V) : \mathcal{X} = (X, X', g) \to \mathcal{Y} = (Y, Y', h)$ be a functional homomorphism between two right functional $R$-modules. The formula of Definition 2.7 makes the map $\iota_{\mathcal{X},\mathcal{Y}} = U \otimes V : \mathcal{K}_R(\mathcal{X}) \to \mathcal{K}_R(\mathcal{Y})$ a ring homomorphism. Moreover one can check that if $\mathcal{X}$ satisfies condition (FS) then $\iota_{\mathcal{X},\mathcal{Y}}$ is injective. Let **Rings** be the category of associative rings with ring homomorphisms.

**Definition 2.10.** Let $E : \mathbf{Rings} \to \mathbf{Ab}$ be a functor. $E$ is called $M$-stable if for any ring $S$, any set $I$ and any distinguished element $i \in I$ the diagonal embedding $\sigma_i : S \to M_I(S)$ at the coordinate $i$ is sent to an isomorphism.

According to [CMR07, Proposition 3.16] any such functor is invariant under inner isomorphisms, and thus $E(\sigma_i)$ does not depend on the choice of $i$.

Burgstaller [Bur25] recently extended this isomorphism to more general corner embeddings by using functional homomorphisms. For the reader's convenience, we recall his result together with its proof.

**Proposition 2.11.** Let $E : \mathbf{Rings} \to \mathbf{Ab}$ be an $M$-stable functor. Let $R$ be a ring with local units. Let $\mathcal{X}$ be a right functional $R$-module admitting a functional homomorphism to a functional module of the form $R^{(I)}$, $I$ being a set. The map induced by the upper left corner embedding $E(\iota_{R,R \oplus \mathcal{X}}) : E(R) \to E(\mathcal{K}_R(R \oplus \mathcal{X}))$ is an isomorphism.

*Proof.* Write $D_{\mathcal{X}} = \mathcal{K}_R(R \oplus \mathcal{X})$. The functional homomorphism $(U, V) : \mathcal{X} \to R^{(I)}$ gives another functional homomorphism $(id_R \oplus U, id_R \oplus V) : R \oplus \mathcal{X} \hookrightarrow R^{(I \cup a)}$ which induces a ring homomorphism $\rho = \iota_{\mathcal{X},R^{(I \cup a)}} : D_{\mathcal{X}} \to M_{I \cup \{a\}}(R)$. Denote by $e : R \hookrightarrow D_{\mathcal{X}}$ the upper left corner embedding induced by the split embedding of functional modules $R \hookrightarrow R \oplus \mathcal{X}$. Consider the following diagram

<figure><img src="image_2.png" /></figure>

Here $e = \iota_{R,R \oplus \mathcal{X}}$ denotes the corner embedding $R \to D_{\mathcal{X}}$. $i_a$ and $i'_a$ denote the embeddings into the corner of the coordinate labelled $a$ in the corresponding matrix ring. $I' \cup a'$ is just a copy of $I \cup a$ and $j_2$ embeds $M_{I \cup a}(D_{\mathcal{X}})$ via identification with the coordinates of $I' \cup a'$. $j_1$ is the lower right corner embedding. We have the following identifications $M_2(D_{\mathcal{X}}) \simeq \mathcal{K}_R((R \oplus \mathcal{X})^2)$ and $M_{I \cup a \cup I' \cup a'}(D_{\mathcal{X}}) \simeq \mathcal{K}_R((R \oplus \mathcal{X})^{I \cup a \cup I' \cup a'})$. $\phi$ is associated to the following split embedding of functional modules

$$ (R \oplus \mathcal{X})^2 \xrightarrow{id_{R \oplus \mathcal{X}} \oplus id_{R \oplus \mathcal{X}}} R \oplus \mathcal{X} \oplus R^{(I \cup a)} \xrightarrow{\alpha \oplus \beta} (R \oplus \mathcal{X})^{(I \cup a)} \oplus (R \oplus \mathcal{X})^{(I' \cup a')} $$

$\alpha : R \oplus \mathcal{X} \hookrightarrow (R \oplus \mathcal{X})^{(I \cup a)}$ is the embedding at the $a$-th coordinate. $\beta : R^{(I \cup a)} \hookrightarrow (R \oplus \mathcal{X})^{(I' \cup a)}$ is the embedding $R \hookrightarrow R \oplus \mathcal{X}$ at each coordinate.

A direct computation proves that this diagram commutes ($\beta$ induces the ring homomorphism $M_{I \cup a}(e)$). By $M$-stability the maps $E(i_a), E(i'_a), E(j), E(j_2)$ are isomorphisms of abelian groups. It remains to show that $E(e)$ is an isomorphism with inverse $E(i_a)^{-1}E(\rho)$. We have $E(i_a)^{-1}E(\rho)E(e) = \text{id}_{E(R)}$ because $i_a = \rho e$.

$$ \begin{aligned} E(e)E(i_a)^{-1}E(\rho) &= E(i'_a)^{-1}E(M_{I \cup a}(e))E(\rho) \\ &= E(i'_a)^{-1}E(j_2)^{-1}E(j_2)E(M_{I \cup a}(e))E(\rho) = E(i'_a)^{-1}E(j_2)^{-1}E(\phi)E(j_1) \end{aligned} $$

The trick is now to write $E(j_2) = E(j'_2)$ and $E(j_1) = E(j'_1)$ where $j'_1 : D_{\mathcal{X}} \hookrightarrow M_2(D_{\mathcal{X}})$ is the upper left corner embedding and $j'_2 : M_{I \cup a}(D_{\mathcal{X}}) \hookrightarrow M_{I \cup a \cup I' \cup a}(D_{\mathcal{X}})$ is the embedding obtained by identifying the coordinates $I \cup a$ in $I \cup a \cup I' \cup a'$. As $\phi j'_1 = j'_2 i'_a$ we finally get $E(i'_a)^{-1}E(j_2)^{-1}E(\phi)E(j_1) = E(i'_a)^{-1}E(j'_2)^{-1}E(\phi)E(j'_1) = \text{id}_{E(D_{\mathcal{X}})}$. $\square$
The notion of correspondence serves here as the minimal algebraic analogue of the classical notion of a $C^*$-correspondence.

**Definition 2.12.** Let $R$ and $S$ be two rings. An $R$-$S$ correspondence is the data of a quadruple $(\mathcal{X}, \Delta, \mathcal{U}, I)$ where $\mathcal{X}$ together with a ring homomorphism $R \to \mathcal{L}_S(\mathcal{X})$ (called the left action) such that

* $\mathcal{X}$ is a right functional $S$-module satisfying condition (FS)

* $\Delta : R \to \mathcal{L}_S(\mathcal{X})$ is a ring homomorphism (called the left action) such that the induced left $R$-module structure on $X$ and right $R$-module structure on $X'$ are non-degenerate : $\Delta(R) \cdot X = X$ and $X' \cdot \Delta(R) = X'$

* $I$ is a set and $\mathcal{U}: \mathcal{X} \to R^{(I)}$ is a functional homomorphism

An $R$-correspondence is an $R$-$R$ correspondence.

We will often simply write $\mathcal{X}$ for the correspondence $(\mathcal{X}, \Delta, \mathcal{U}, I)$. Choose an $R$-$S$ correspondence $\mathcal{X} = (X, X', g)$ with functional homomorphism $\mathcal{U}: \mathcal{X} \to S^{(I)}$ and an $S$-$T$ correspondence $\mathcal{Y} = (Y, Y', h)$ with functional homomorphism $\mathcal{U}': \mathcal{Y} \to T^{(J)}$. One defines the tensor product $\mathcal{X} \otimes_S \mathcal{Y}$ as the $R$-$T$ correspondence given by $\mathcal{X} \otimes_S \mathcal{Y} = (X \otimes_S Y, Y' \otimes_S X', k)$ with $k$ defined by the formula $(\psi \otimes \phi)(x \otimes y) = \psi(\phi(x) \cdot y)$ for all $x \in X, y \in Y, \phi \in X', \psi \in Y'$. The functional homomorphism underlying $\mathcal{X} \otimes_S \mathcal{Y}$ is given by $U \otimes U' : X \otimes_S Y \to T^{(I \times J)}$ and $V' \otimes V : Y' \otimes_S X' \to T^{(J \times I)}$.

Let $\mathcal{X}$ be an $R$-$S$ correspondence. Denote by $\Delta: R \to \mathcal{L}_S(\mathcal{X})$ the left action. Assume that $R$ is such that $\forall r \in R, \Delta(r) \in \mathcal{K}_S(\mathcal{X})$. Let $E: \text{Rings} \to \text{Ab}$ be an $M$-stable functor. As $E(\iota_{R,R \oplus \mathcal{X}})$ is an isomorphism we see that $\mathcal{X}$ induces a map $E(\mathcal{X}): E(R) \to E(S)$ by the following diagram

$$R \xrightarrow{\Delta} \mathcal{K}_S(\mathcal{X}) \xrightarrow{\iota_{\mathcal{X},S \oplus \mathcal{X}}} \mathcal{K}_S(S \oplus \mathcal{X}) \xleftarrow{\iota_{S,S \oplus \mathcal{X}}} S$$

For the applications, we will need another way of computing $E(\mathcal{X})$. The functional homomorphism of $\mathcal{X}$ yields the existence of a ring homomorphism $\iota_{\mathcal{X},S(I)}: \mathcal{K}_S(\mathcal{X}) \to M_I(S)$. As $E$ is an $M$-stable functor there is a map $E(\mathcal{K}_S(\mathcal{X})) \to E(S)$ obtained by composing with the inverse of the map induced by any corner embedding $\iota_{S,S(I)}: S \to M_I(S)$.

**Proposition 2.13.** $E(\iota_{S,S \oplus \mathcal{X}})^{-1} E(\iota_{\mathcal{X},S \oplus \mathcal{X}}) = E(\iota_{S,S(I)})^{-1} E(\iota_{\mathcal{X},S(I)})$.

*Proof.* The following diagram

$$\begin{tikzcd}
\mathcal{K}_S(\mathcal{X}) \arrow[r, "\iota_{\mathcal{X},S \oplus \mathcal{X}}"] \arrow[d, "h"'] & \mathcal{K}_S(S \oplus \mathcal{X}) \arrow[r, "\iota_{S,S \oplus \mathcal{X}}"] \arrow[d, "\rho"'] & S \arrow[l, "\iota_a"'] \\
M_{I \cup a}(S) & &
\end{tikzcd}$$

is commutative. $\rho = \iota_{S \oplus \mathcal{X}, S(I \cup a)}$ is induced by the splitting $S^{(I \cup a)} = (S \oplus \mathcal{X}) \oplus \mathcal{Y}$. $h = \iota_{\mathcal{X}, S(I \cup a)}$ is the composition of $\iota_{\mathcal{X}, S(I)}: \mathcal{K}_S(\mathcal{X}) \to M_I(S)$ with the embedding $\sigma = \iota_{S(I), S(I \cup a)}: M_I(S) \hookrightarrow M_{I \cup a}(S)$. $\iota_a = \iota_{S, S(I \cup a)}$ is the embedding at the coordinate $a$. Its commutativity yields $E(\iota_{S,S \oplus \mathcal{X}})^{-1} E(\iota_{\mathcal{X},S \oplus \mathcal{X}}) = E(i_a)^{-1} E(h)$. The diagram

$$\begin{tikzcd}
\mathcal{K}_S(\mathcal{X}) \arrow[r, "\iota_{\mathcal{X},S(I)}"] \arrow[d, "h"'] & M_I(S) \arrow[r, "\iota_{S,S(I)}"] \arrow[d, "\sigma"'] & S \arrow[l, "i_a"'] \\
M_{I \cup a}(S) & &
\end{tikzcd}$$

is also commutative. We get similarly $E(\iota_{S,S(I)})^{-1} E(\iota_{\mathcal{X},S(I)}) = E(i_a)^{-1} E(h)$. $\square$
## 2.2 Algebraic Cuntz-Pimsner rings

In this section $R$ is a ring with local units and $\mathcal{I}$ is a two-sided ideal of $R$ which also admits local units. Let $\mathcal{X} = (X, X', g)$ be an $R$-correspondence such that $\mathcal{I}$ acts on $X$ on the left by compact operators. This means that if $\Delta : R \to \mathcal{L}_R(X)$ is the left action then $\Delta(\mathcal{I}) \subset \mathcal{K}_R(X)$.

Let $T(X) = \bigoplus_{n \in \mathbb{N}} X^{\otimes n}$ be the $\mathbb{N}$-graded $R$-R bimodule known as the Fock space (where the tensor products are taken over $R$).

There is an $R$-correspondence

$$T(\mathcal{X}) = (T(X), T(X'), \tilde{g})$$

where $\tilde{g}(\phi_1 \otimes \cdots \otimes \phi_n, x_1 \otimes \cdots \otimes x_m) = 0$ if $n \neq m$, and

$$\tilde{g}(\phi_1 \otimes \cdots \otimes \phi_n, x_1 \otimes \cdots \otimes x_n) = \phi_1(\phi_2(\cdots (\phi_n(x_1)x_2) \cdots )x_n).$$

for all $\phi_k \in X'$ and $x_i \in X$. For $\phi \in X'$ and $x \in X$, define

$$T_x(p) := x \otimes p,$$

$$T_{\phi}(p) := \begin{cases} 0, & \text{if } \deg(p) = 0 \\ \phi(p_1) \cdot p_2 \otimes \cdots \otimes p_n & \text{if } p = p_1 \otimes \cdots \otimes p_n, n \ge 1. \end{cases}$$

Then $T_x, T_{\phi} \in \mathcal{L}_R(T(X))$ with adjoints given by

$$T_x^*(\psi) := \begin{cases} 0, & \text{if } \deg(\psi) = 0, \\ \psi_1 \otimes \cdots \otimes \psi_{n-1} \cdot \psi_n(x), & \text{if } \psi = \psi_1 \otimes \cdots \otimes \psi_n \end{cases}$$ $$T_{\phi}^*(\psi) := \psi \otimes \phi,$$

in $\mathcal{L}_R(T(X'))$.

**Definition 2.14.** The Toeplitz ring $\mathcal{T}_X$ of $\mathcal{X}$ is the subring generated by the $T_x, T_\phi, r \cdot \text{id}_{T(X)} \in \mathcal{L}_R(T(X))$, $\phi \in X'$, $x \in X$, $r \in R$.

Let $J_{\mathcal{X},\mathcal{I}}$ be the two-sided ideal of $\mathcal{T}_X$ generated by $\mathcal{I} \cdot P_0$ where $P_0 \in \mathcal{L}_R(T(X))$ is the map that sends $x$ to $0$ if $\deg(x) \ge 1$ and is the identity on $R$. The Cuntz-Pimsner ring of $\mathcal{X}$ with respect to the ideal $\mathcal{I}$ is the quotient ring $\mathcal{O}_{\mathcal{X},\mathcal{I}} = \mathcal{T}_{\mathcal{X}}/J_{\mathcal{X},\mathcal{I}}$.

$T(X)$ is an $\mathbb{N}$-graded ring and $\mathcal{T}_X$ and $\mathcal{O}_{\mathcal{X},\mathcal{I}}$ are $\mathbb{Z}$-graded rings. Observe that these rings have local units if the base ring $R$ has local units.

**Remark 2.15.** Be careful that $P_0$ is not in general an element of $\mathcal{T}_X$. The reason why $\mathcal{I} \cdot P_0$ is included in $\mathcal{T}_X$ is because $\mathcal{I}$ acts on $X$ by compact operators. Hence for all $i \in \mathcal{I}$,

$$\Delta(i) = \sum_i x_i \otimes \phi_i$$

and

$$i \cdot P_0 = i \cdot \text{id}_{T(X)} - \sum_i T_{x_i} T_{\phi_i} \in \mathcal{T}_X$$

**Proposition 2.16.** $J_{\mathcal{X},\mathcal{I}}$ is the ideal of compact operators of the right functional $R$-module $T(\mathcal{X}) \cdot \mathcal{I} = (T(X) \cdot \mathcal{I}, \mathcal{I} \cdot T(X'), \tilde{g})$.

*Proof.* We can represent elements of $\mathcal{L}_R(T(X))$ by (infinite) matrices by giving the action of the operator on each subspace $X^{\otimes n}$. In particular

$$T_x = \begin{pmatrix} 0 & 0 & 0 & 0 & \cdots \\ x & 0 & 0 & 0 & \cdots \\ 0 & x & 0 & 0 & \cdots \\ 0 & 0 & x & 0 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix}, T_{\phi} = \begin{pmatrix} 0 & \phi & 0 & 0 & \cdots \\ 0 & 0 & \phi & 0 & \cdots \\ 0 & 0 & 0 & \phi & \cdots \\ 0 & 0 & 0 & 0 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix}, \pi_0 \cdot i = \begin{pmatrix} i & 0 & 0 & 0 & \cdots \\ 0 & 0 & 0 & 0 & \cdots \\ 0 & 0 & 0 & 0 & \cdots \\ 0 & 0 & 0 & 0 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix}$$
Where the symbol $x$ (resp. $\phi$) in the above matrix is to be understood as the operator $T_x$ (resp. $T_\phi$) restricted to the corresponding subspace $X^{\otimes n}$. Moreover $\mathcal{K}_R(T(\mathcal{X}) \cdot \mathcal{I}) = T(X) \cdot \mathcal{I} \otimes_R \mathcal{I} \cdot T(X') = T(X) \cdot \mathcal{I} \otimes_R T(X')$ because $\mathcal{I}$ has local units. Distributing the tensor product we get

$$ \mathcal{K}_R(T(\mathcal{X} \cdot \mathcal{I})) = \bigoplus_{(n,m) \in \mathbb{N}^2} X^{\otimes n} \otimes \mathcal{I} \otimes (X')^{\otimes m} $$

which is an ideal of $\mathcal{L}_R(T(X))$. We have $J_{\mathcal{X},\mathcal{I}} \subset \mathcal{K}_R(T(\mathcal{X}) \cdot \mathcal{I})$ because $P_0 \in \mathcal{K}_R(T(\mathcal{X} \cdot \mathcal{I}))$ and any matrix of the form $T_\phi$ or $T_x$ multiplied by a finite matrix is still a finite matrix. In addition, for any $i \in \mathcal{I}$, $p = x_1 \otimes \dots \otimes x_n \in X^{\otimes n}$, $\psi = \phi_1 \otimes \dots \otimes \phi_m \in (X')^{\otimes m}$:

$$ \left( \begin{array}{cccccc} 0 & \dots & 0 & 0 & \dots \\ \dots & \dots & 0 & 0 & \dots \\ 0 & (p \cdot i) \otimes \psi & 0 & 0 & \dots \\ 0 & 0 & 0 & 0 & \dots \\ \dots & & & & & \end{array} \right) = T_{x_1} \dots T_{x_n} (i \cdot P_0) T_{\phi_1} \dots T_{\phi_m} \in \mathcal{T}_{\mathcal{X}} $$

□

**Example 2.17.** Let $\mathcal{X} = (M, M^*, \text{ev})$ with $M$ an $R$-$R$ bimodule that is finitely generated and projective on the right. We have

$$ \mathcal{O}_{\mathcal{X}} \simeq \lim_{k \in \mathbb{N}} \bigoplus_{d \in \mathbb{Z}} \operatorname{Hom}_{-,R}(M^{\otimes k}, M^{\otimes k+d}) $$

as rings. We take the convention that $M^{\otimes n}$ is the zero module for negative $n$. Here the colimit is taken over tensoring by the identity of $M$ and the product is given by the composition. Indeed, because $M_R$ is finitely generated and projective we have an isomorphism (see [AÁM87]) $\operatorname{Hom}_{-,R}(M^{\otimes k}, M^{\otimes l}) \simeq M^{\otimes l} \otimes (M^*)^{\otimes k}$. If we let $O$ be the ring obtained as the above direct sum of colimits then there is a ring homomorphism $\mathcal{T}_{\mathcal{X}} \to O$ obtained by restriction to the modules $M^{\otimes k}$ for big enough $k$. The kernel of this map is clearly $J_{\mathcal{X}} = T(M) \otimes T(M^*)$.

**Example 2.18.** Let $\mathcal{X} = (R, R, \mu)$, $\mathcal{I} = R$ and $R$ acts on $R$ by multiplication then $T(X) = R^{(\mathbb{N})}$. Each $T_x$ acts on $T(X)$ by multiplication by $x$ in $R$ and by adding one to the degree, each $T_\phi$ multiplies by $\phi$ and subtracts one to the degree. Moreover $J_{\mathcal{X}} = \mathcal{K}_R(T(\mathcal{X})) = M_\infty(R)$ and

$$ \mathcal{T}_{\mathcal{X}} \simeq R\langle x, y \rangle/(xy - 1), \quad \mathcal{O}_{\mathcal{X}} \simeq R[x, x^{-1}]. $$

More generally, if $\alpha$ is a ring automorphism of $R$ and each $r \in R$ acts on $R$ on the left by multiplication by $r$ and on the right by multiplication by $\alpha(r)$ then $\mathcal{O}_{\mathcal{X}} \simeq R \rtimes_{\alpha} \mathbb{Z}$ is the crossed product ring of $R$ by $\mathbb{Z}$ with $\alpha$.

The Toeplitz ring has a universal characterization in terms of representations analogous to the one for $C^*$-algebras.

**Definition 2.19.** [CO11, Definition 1.2] A covariant representation of $(R, \mathcal{X})$ is a quadruple $(S, T, \sigma, D)$ where

(1) $D$ is a ring

(2) $\sigma: R \to D$ is a ring homomorphism

(3) $S: X' \to D$ and $T: X \to D$ are $R$-bimodule homomorphisms with respect to the bimodule structure induced by multiplication and by $\sigma$

(4) $\forall \phi \in X', \forall x \in X, \sigma(g(\phi \otimes x)) = S(\phi)T(x)$

If $j: R \to \mathcal{T}_{\mathcal{X}}$ is the natural inclusion, $T_{\mathcal{X}}, S_{\mathcal{X}}: X, X' \to \mathcal{T}_{\mathcal{X}}$ map respectively $x$ to $T_x$ and $\phi$ to $T_\phi$ then $(S, T, j, \mathcal{T}_{\mathcal{X}})$ is a covariant representation. It is injective by non-degeneracy and surjective by definition. It is universal in the following sense, which characterizes the Toeplitz ring:

**Theorem 2.20.** Let $R$ be a ring with local units and let $\mathcal{X}$ be an $R$-$R$ correspondence. If $(S, T, \sigma, D)$ is a covariant representation of $(R, \mathcal{X})$, then there exists a unique ring homomorphism

$$ \eta: \mathcal{T}_{\mathcal{X}} \to D $$
such that $\eta \circ j = \sigma$, $\eta \circ T_X = T$, and $\eta \circ S_X = S$.

*Proof.* See [CO11, Theorem 1.7 and Proposition 4.2]. □

The relative algebraic Cuntz-Pimsner ring $\mathcal{O}_{X, \mathcal{I}}$ also has a universal property [CO11, Theorem 3.18] but we will not use it in this article.

### 3. PROOFS OF MAIN RESULTS

Let $\mathcal{X} = (X, X', g)$ be a correspondence over a ring with local units $R$, let $\mathcal{I}$ be a two-sided ideal of $R$ that has local units and acts on $X$ on the left by compact operators. Let $E: \textbf{Rings} \to \textbf{Ab}$ be a homotopy invariant, split-exact, $M$-stable functor (we refer to [CMR07] and [CT06] for the definitions of these notions and classical computations using quasihomomorphisms). There is a short exact sequence of rings

$$0 \to \mathcal{K}_R(T(\mathcal{X} \cdot \mathcal{I})) \to \mathcal{T}_X \to \mathcal{O}_{\mathcal{X}, \mathcal{I}} \to 0$$

Write $T(\mathcal{X} \cdot \mathcal{I}) = \mathcal{I} \oplus T^1(\mathcal{X} \cdot \mathcal{I})$ where $T^1(\mathcal{X} \cdot \mathcal{I}) = \bigoplus_{n>1} (\mathcal{X} \cdot \mathcal{I})^{\otimes n}$. By Proposition 2.11 and $M$-stability of $E$ the upper left corner embedding $e: \mathcal{I} \to \bar{\mathcal{K}}_R(T(\mathcal{X} \cdot \mathcal{I}))$ induces an isomorphism

$$E(\mathcal{K}_R(T(\mathcal{X} \cdot \mathcal{I}))) \simeq E(\mathcal{I})$$

The main difficulty is thus to compute the value of $E$ on $\mathcal{T}_X$. We have a map $j: R \to \mathcal{T}_X$ by definition of the Toeplitz ring. We want to build a map that is the inverse to $j$ once we apply $E$. We define a quasi-homomorphism

$$\pi = \mathcal{T}_X \xrightarrow{\pi_0} \mathcal{L}_R(T(\mathcal{X})) \triangleright \mathcal{K}_R(T(\mathcal{X})).$$

by letting $\pi_0$ be the natural inclusion and by defining $\pi_1$ to be such that for all pure tensors $p \in T(X)$,

$$\pi_1(T_x)(p) = \begin{cases} 0 & \text{if } \deg(p) = 0 \\ x \otimes p & \text{otherwise} \end{cases} \qquad \pi_1(T_\phi)(p) = \begin{cases} 0 & \text{if } \deg(p) \le 1 \\ \phi(p_1) \otimes p_2 \cdots \otimes p_n & \text{otherwise} \end{cases} \\ \\ \pi_1(a)(p) = \begin{cases} 0 & \text{if } \deg(p) = 0 \\ a \cdot p & \text{otherwise} \end{cases}$$

One can check that both $\pi_0$ and $\pi_1$ preserve the pairing and thus define a homomorphism on $\mathcal{T}_X$ according to Theorem 2.22.

**Lemma 3.1.** For all $\tau \in \mathcal{T}_X$,

$$\pi_0(\tau) - \pi_1(\tau) \in \mathcal{K}_R(T(X))$$

*Proof.* Every element of the Toeplitz ring can be written as a finite sum of elements of the form $\tau = T_{p_1} \cdots T_{p_k} T_{\phi_1} \cdots T_{\phi_l}$ for $k, l \ge 0$, $p_1, \cdots, p_k \in X$ and $\phi_1, \cdots, \phi_l \in X'$. $\pi_0(\tau) - \pi_1(\tau)$ vanishes on homogeneous tensors of degree $\ge l+1$ and of degree $\le l-1$. Let $q = q_1 \otimes \cdots \otimes q_l$ be a pure tensor of degree $l$, we have:

$$(\pi_0(\tau) - \pi_1(\tau))(q) = (p_1 \otimes \cdots \otimes p_k \cdot (\phi_1(\cdots \phi_l(q_1) \cdots)q_l))$$

Hence $\pi_0(\tau) - \pi_1(\tau)$ belongs to $\mathcal{K}_R(T(X)) = \bigoplus_{n,m \in \mathbb{N}} X^{\otimes n} \otimes_R (X')^{\otimes m}$. □

**Theorem 3.2.** Let $\mathcal{X}$ be an $R$-correspondence over a ring with local units $R$. Every homotopy invariant, split-exact, $M$-stable functor $E: \textbf{Rings} \to \textbf{Ab}$ sends the inclusion $j: R \to \mathcal{T}_X$ to an isomorphism $E(\mathcal{T}_X) \simeq E(R)$
Proof. First we compute $E(\pi \circ j)$. Let $\iota = \iota_{R,T(X)}: R \to K_R(T(X))$ be the inclusion in the upper left corner $(\iota(r)(p) = 0 \text{ if deg}(p) \ge 1 \text{ and } \iota(r)(p) = r \cdot p \text{ otherwise})$. $E(\iota)$ is an isomorphism by Proposition 2.11.

$$E(\pi \circ j) = E(j, \pi_1 \circ j) = E(\iota + \pi_1 \circ j, \pi_1 \circ j) = E(\iota)$$

as $\pi_1 \circ j$ and $\iota$ are orthogonal quasi-homomorphisms (see [CMR07, Proposition 3.3]).

Now we compute $E(j) \circ E(\pi)$. Write $\mathcal{T}_X = \mathcal{T}$ for simplicity. The map $j: R \to \mathcal{T}$ makes $(\mathcal{T}, \mathcal{T}, \mu)$ an $R$-$\mathcal{T}$ correspondence. Consider the following diagram :

$$ \begin{tikzcd} \mathcal{T} \arrow[r, "\pi_0"'] \arrow[d, "\pi_1"] & \mathcal{L}_R(T(\mathcal{X})) \triangleright \mathcal{K}_R(T(\mathcal{X})) \arrow[d, "[-\otimes id_{\mathcal{T}}]"'] \\ & \mathcal{L}_{\mathcal{T}}(T(\mathcal{X}) \otimes_R \mathcal{T}) \triangleright \mathcal{K}_{\mathcal{T}}(T(\mathcal{X}) \otimes_R \mathcal{T}) \arrow[d, "[-\otimes id_{\mathcal{T}}]"'] \arrow[r, "\pi_1"'] \arrow[d, "\pi_0"'] \end{tikzcd} $$

Let us first verify that $[- \otimes id_{\mathcal{T}}]$ maps the compact operators to the compact operators. $\mathcal{K}_{\mathcal{T}}(T(\mathcal{X}) \otimes_R \mathcal{T}) = T(X) \otimes_R \mathcal{T} \otimes_{\mathcal{T}} \mathcal{T} \otimes_R T(X') \simeq T(X) \otimes_R \mathcal{T} \otimes_R T(X')$ and $\mathcal{K}_R(T(\mathcal{X})) = T(X) \otimes_T T(X') \simeq T(X) \otimes R \otimes T(X')$

Let $p \in X^{\otimes n}, q \in X'^{\otimes m}$ and $r \in R$. The compact operator $p \otimes r \otimes q$ acting on $T(\mathcal{X}) \otimes_R \mathcal{T}$ is equal to $p \otimes j(r) \otimes q \in T(X) \otimes_R \mathcal{T} \otimes_R T(X')$. Hence $[- \otimes id_{\mathcal{T}}]$ restricted to $\mathcal{K} = \mathcal{K}_R(T(\mathcal{X}))$ is the map induced by $j$ on each coordinate. We write $[- \otimes id_{\mathcal{T}}]: \mathcal{L}_R(T(\mathcal{X})) \to \mathcal{L}_{\mathcal{T}}(T(\mathcal{X}) \otimes_R \mathcal{T})$ as $\tilde{j}$. $E(\tilde{j}|_{\mathcal{K}})$ equals $E(j)$ up to stabilization isomorphism :

$$E(j) = E(\iota_{\mathcal{T},T(\mathcal{X})\otimes_R \mathcal{T}})^{-1} E(\tilde{j}|_{\mathcal{K}}) E(\iota)$$

$\iota_{\mathcal{T},T(\mathcal{X})\otimes_R \mathcal{T}}: \mathcal{T} \to \mathcal{K}_{\mathcal{T}}(T(\mathcal{X}) \otimes_R \mathcal{T})$ being the upper left corner embedding.

We now define $\lambda_1, \lambda_0: \mathcal{T} \to \mathcal{L}_{\mathcal{T}}(T(\mathcal{X}) \otimes_R \mathcal{T})$ two ring homomorphisms. We let $\lambda_1(\tau)$ be zero on $p \otimes \tau$ with deg$(p) \ge 1$ and the operator of left multiplication by $\tau$ on $R \otimes_R \mathcal{T} \simeq \mathcal{T}$ (this isomorphism holds because we assume the left action to be non-degenerate). Let $\lambda_0(T_x)$ send $p \otimes \tau$ to zero if deg$(p) \ge 1$ and to $x \otimes p \otimes \tau$ otherwise. We let $\lambda_0(T_\phi)$ send $p \otimes \tau$ to zero if deg$(p) \ge 2$ or if deg$(p) = 0$ and to $\phi(p) \cdot \tau$ if deg$(p) = 1$. Finally we let $\lambda_0(r)$ be zero on tensors of degree $\ge 1$ and send $\tau \in \mathcal{T}$ to $r \otimes \tau$. One easily checks that $\lambda_0$ preserves the pairing of $\mathcal{X}$. We can write elements of $\mathcal{L}_{\mathcal{T}}(T(\mathcal{X}) \otimes_R \mathcal{T})$ as infinite matrices by giving their action on each subspace of the form $X^{\otimes n} \otimes \mathcal{T}$. We have

$$ \lambda_0(T_x) = \begin{pmatrix} 0 & 0 & 0 & \dots \\ T_x & 0 & 0 & \dots \\ 0 & 0 & 0 & \dots \\ \dots \end{pmatrix}, \lambda_0(T_\phi) = \begin{pmatrix} 0 & T_\phi & 0 & \dots \\ 0 & 0 & 0 & \dots \\ 0 & 0 & 0 & \dots \\ \dots \end{pmatrix}, \lambda_0(r) = \begin{pmatrix} r & 0 & 0 & \dots \\ 0 & 0 & 0 & \dots \\ 0 & 0 & 0 & \dots \\ \dots \end{pmatrix} $$

The last step of the proof is to build a polynomial homotopy

$$H: \pi_0 \otimes id \sim \lambda_1 + \pi_1 \otimes id$$

We define, using a “rotational” homotopy similar to that of [CT06]

$$ \begin{align*} H(T_x) &= (1 - t^2)\lambda_0(T_x) + (2t - t^3)\lambda_1(T_x) + (\pi_1 \otimes id)(T_x) \\ H(T_\phi) &= (1 - t^2)\lambda_0(T_\phi) + t\lambda_1(T_\phi) + (\pi_1 \otimes id)(T_\phi) \\ H(r) &= r \cdot id \end{align*} $$

In other words

$$ H(T_x) = \begin{pmatrix} (2t - t^3)T_x & 0 & 0 & 0 & \dots \\ (1 - t^2)T_x & 0 & 0 & 0 & \dots \\ 0 & T_x & 0 & 0 & \dots \\ 0 & 0 & T_x & 0 & \dots \\ \dots & & & & \end{pmatrix} $$
$$H(T_{\phi}) = \begin{pmatrix} tT_{\phi} & (1-t^2)T_{\phi} & 0 & 0 & \dots \\ 0 & 0 & T_{\phi} & 0 & \dots \\ 0 & 0 & 0 & T_{\phi} & \dots \\ 0 & 0 & 0 & 0 & \dots \\ \dots \end{pmatrix}$$

A direct computation shows that $H$ indeed preserves the pairing and hence defines a ring homomorphism

$$H: \mathcal{T} \to \mathcal{L}_{\mathcal{T}}(T(\mathcal{X}) \otimes_{\mathcal{R}} \mathcal{T})[t]$$

such that $H(0) = \pi_0 \otimes \mathrm{id}$ and $H(1) = \lambda_1 + \pi_1 \otimes \mathrm{id}$. Hence, using the homotopy invariance of $E$

$$E(\tilde{j} \circ \pi) = E(\pi_0 \otimes \mathrm{id}, \pi_1 \otimes \mathrm{id}) = E(\lambda_1 + \pi_1 \otimes \mathrm{id}, \pi_1 \otimes \mathrm{id}) = E(\lambda_1)$$

as $\lambda_1$ and $\pi_1 \otimes \mathrm{id}$ are orthogonal homomorphisms. But $\iota_{\mathcal{T},T(\mathcal{X})\otimes_{\mathcal{R}}\mathcal{T}}$ is the upper left corner embedding of $\mathcal{T}$ in $\mathcal{K}_{\mathcal{T}}(T(\mathcal{X}) \otimes_{\mathcal{R}} \mathcal{T})$. Thus $E(\lambda_1)$ is an isomorphism by Proposition 2.11 and M-stability of $E$.

$E(\tilde{j} \circ \pi) = E(\tilde{j}|_{\mathcal{K}}) \circ E(\pi)$, this finishes the proof because we now have two isomorphisms which are inverses of each other :

$$E(j)E(\iota)^{-1}E(\pi) = \mathrm{id}_{\mathcal{T}}$$

$$E(\pi)E(j)E(\iota)^{-1} = \mathrm{id}_{\mathcal{R}}$$

□

**Lemma 3.3.** Let $E$ be a split-exact and M-stable functor. Let $\iota_{R,T(\mathcal{X})}$ be the upper left corner embedding $R \to \mathcal{K}_R(T(\mathcal{X}))$. We write $i: \mathcal{I} \to R$, and $i': \mathcal{K}_R(T(\mathcal{X}) \cdot \mathcal{I}) \to \mathcal{T}_{\mathcal{X}}$ for the inclusion maps. We have

$$E(\pi) \circ E(i' \circ \iota_{\mathcal{I},T(\mathcal{X})\cdot\mathcal{I}}) = E(\iota_{R,T(\mathcal{X})}) \circ (E(i) - E(\mathcal{X}))$$

where $E(\mathcal{X})$ is defined to be the map induced by the structure of $\mathcal{I}$-R correspondence on $R$.

*Proof.* Recall that $E(\mathcal{X}) = E(\iota_{R,R\oplus\mathcal{X}})^{-1} \circ E(\iota_{\mathcal{X},R\oplus\mathcal{X}}) \circ E(\Delta)$ where $\Delta: \mathcal{I} \to \mathcal{K}_R(\mathcal{X})$ is the left action, $\iota_{\mathcal{X},R\oplus\mathcal{X}}: \mathcal{K}_R(\mathcal{X}) \to \mathcal{K}_R(R \oplus \mathcal{X})$ is the lower right corner embedding and $\iota_{R,R\oplus\mathcal{X}}: R \to \mathcal{K}_R(R \oplus \mathcal{X})$ is the upper left corner embedding. There is another natural inclusion $\iota_{\mathcal{X},T(\mathcal{X})}: \mathcal{K}_R(\mathcal{X}) \to \mathcal{K}_R(T(\mathcal{X}))$.

$$E(\pi) \circ E(i' \circ \iota_{\mathcal{I},T(\mathcal{X})\cdot\mathcal{I}}) = E(i' \circ \iota_{\mathcal{I},T(\mathcal{X})\cdot\mathcal{I}}, \pi_1 \circ i' \circ \iota_{\mathcal{I},T(\mathcal{X})\cdot\mathcal{I}})$$

For all write $a \in \mathcal{I}: \Delta(a) = \sum_k x_k \otimes \phi_k$. We have

$$i'(\iota_{\mathcal{I},T(\mathcal{X})\cdot\mathcal{I}}(a)) = a \cdot p_0 = a \cdot \mathrm{id}_{T(\mathcal{X})} - \sum_k T_x T_{\phi_k}$$

By using the definition of $\pi_1$ we get $\pi_1(i'(\iota_{\mathcal{I},T(\mathcal{X})\cdot\mathcal{I}}(a))) = \iota_{\mathcal{X},T(\mathcal{X})}(\Delta(a))$. This is orthogonal to $i'(\iota_{\mathcal{I},T(\mathcal{X})\cdot\mathcal{I}}(a))$ and also maps in the compact operators. Hence

$$E(\pi) \circ E(i' \circ \iota_{\mathcal{I},T(\mathcal{X})\cdot\mathcal{I}}) = E(\iota_{R,T(\mathcal{X})} \circ i) - E(\iota_{\mathcal{X},T(\mathcal{X})}) \circ E(\Delta)$$

We now just have to show that $E(\iota_{R,T(\mathcal{X})}) \circ E(\iota_{R,R\oplus\mathcal{X}})^{-1} \circ E(\iota_{\mathcal{X},R\oplus\mathcal{X}}) = E(\iota_{\mathcal{X},T(\mathcal{X})})$. This is a direct consequence of the commutativity of the following diagram

<figure><img src="image_3.png" /></figure>

□
**Theorem 3.4.** Let $(E_n)_{n \in \mathbb{Z}}$: **Rings** $\rightarrow$ **Ab** be a sequence of homotopy invariant and M-stable functors satisfying excision. Let R be a ring with local units. Let $\mathcal{X}$ an R and $\mathcal{I} \subsetneq R$ a two-sided ideal that has local units and acts on X on the left by compact operators. There is a long exact sequence

$$ \dots \longrightarrow E_n(\mathcal{I}) \xrightarrow{E_n(i)-E_n(\mathcal{X})} E_n(R) \xrightarrow{E_n(j)} E_n(\mathcal{O}_{\mathcal{X},\mathcal{I}}) \longrightarrow E_{n-1}(\mathcal{I}) \longrightarrow \dots $$

Here $j: R \rightarrow \mathcal{O}_{\mathcal{X},\mathcal{I}}$ and $i: \mathcal{I} \rightarrow R$ are the natural inclusions.

We now show how some long exact sequences of KH/HP-groups of some specific classes of rings can be obtained directly when realizing these specific rings as algebraic Cuntz-Pimsner rings. We fix $(E_n)_{n \in \mathbb{Z}}$: **Rings** $\rightarrow$ **Ab** a sequence of homotopy invariant, stable, functors satisfying excision. As a direct corollary of Theorem B we get (see [CO11, Example 5.5] for details)

**Theorem 3.5.** (Pimsner-Voiculescu)

Let R be a ring with local units, let $\alpha: R \rightarrow R$ be a ring automorphism and $R \times_{\alpha} \mathbb{Z}$ be the crossed product ring of R by $\alpha$. There is a long exact sequence :

$$ \dots \longrightarrow E_n(R) \xrightarrow{1-E_n(\alpha)} E_n(R) \xrightarrow{E_n(j)} E_n(R \times_{\alpha} \mathbb{Z}) \longrightarrow E_{n-1}(R) \longrightarrow \dots $$

Let $Q = (Q^0, Q^1, r, s)$ be a quiver. Recall that a vertex $v \in Q^0$ is regular if $0 < |s^{-1}(v)| < \infty$, and let $\rho(Q) \subseteq Q^0$ denote the set of regular vertices. It is known (see [CO11, Example 5.8]) that the Leavitt path algebra $L_k(Q)$ is the algebraic Cuntz-Pimsner ring associated to the ring R, the ideal $\mathcal{I}$, and the R-correspondence $\mathcal{X} = (X, X', (\cdot, \cdot))$ defined by

$$ R = \bigoplus_{v \in Q^0} k \cdot 1_v, \quad \mathcal{I} = \bigoplus_{v \in \rho(Q)} k \cdot 1_v, \quad X = \bigoplus_{e \in Q^1} k \cdot 1_e, \quad X' = \bigoplus_{e \in Q^1} k \cdot 1_{e^*}. $$

The evaluation map is given, for $e, f \in Q^1$, by

$$ \langle 1_{e^*}, 1_f \rangle = \delta_{e,f} 1_{r(e)} $$

and the bimodule structures are determined on generators by

$$ \begin{align*} 1_e \cdot 1_v &= \delta_{r(e),v} 1_e, & 1_v \cdot 1_e &= \delta_{s(e),v} 1_e \\ 1_{e^*} \cdot 1_v &= \delta_{s(e),v} 1_{e^*}, & 1_v \cdot 1_{e^*} &= \delta_{r(e),v} 1_{e^*} \end{align*} $$

The functional homomorphism of the correspondence is given by the maps $U: X \rightarrow R^{(Q^1)}$ and $V: X' \rightarrow R^{(Q^1)}$ defined by $U(1_e) = (\delta_{e,f} 1_{r(e)})_{f \in Q^1}$ and $V(1_{e^*}) = (\delta_{e,f} 1_{r(e^*)})_{f \in Q^1}$ for all $e \in Q^1$.

Let $N'_Q = (n_{x,y})_{x,y \in Q^0}$ be the adjacency matrix of $Q$, where $n_{x,y}$ is the number of arrows from $x$ to $y$. Let $N_Q$ be the matrix obtained from $N'_Q$ by removing the columns indexed by elements of $Q^0 \setminus \rho(Q)$.

**Theorem 3.6.** Let $k$ be a ring with local units, there is a long exact sequence ($n \in \mathbb{Z}$)

$$ \dots \longrightarrow E_n(k)^{\rho(Q)} \xrightarrow{E_n(i)-N_Q} E_n(k)^{Q^0} \xrightarrow{E_n(j)} E_n(L_k(Q)) \longrightarrow \dots $$

*Proof.* We just have to prove that the map

$$ E_n(\mathcal{X}): E_n(\mathcal{I}) = E_n(k)^{\rho(Q)} \rightarrow E_n(R) = E_n(k)^{Q^0} $$

equals $N_Q$. $K_R(\mathcal{X})$ is the set of $k$-linear combinations of elements of the form $1_e \otimes 1_{f^*}$ for $e, f \in Q^1$. The functional homomorphism of $\mathcal{X}$ described above induces the ring homomorphism $\rho: K_R(\mathcal{X}) \rightarrow M_{Q^1}(R)$ which sends every element of the form $1_e \otimes 1_{f^*}$ to the matrix $(\delta_{e,e'} 1_{r(e)} \otimes \delta_{f,f'} 1_{r(f)})_{e',f' \in Q^1} = \delta_{e,f} 1_{r(e)}$. By Proposition 2.13, $E_n(\mathcal{X})$ equals the following map

$$ E_n(\mathcal{I}) \xrightarrow{E_n(\Delta)} E_n(K_R(\mathcal{X})) \xrightarrow{E_n(\rho)} E_n(M_{Q^1}(R)) \xrightarrow{\sim} E_n(R) $$

Here $\Delta: \mathcal{I} \rightarrow K_R(\mathcal{X})$ is the left action. For all $v \in \rho(Q)$,

$$ \rho(\Delta(1_v)) = \rho(\sum_{e \in s^{-1}(v)} 1_e \otimes 1_{e^*}) = \sum_{e \in s^{-1}(v)} 1_{r(e)} $$
<figure><img src="image_4.png" /></figure>

Nekrashevych algebras were introduced originally for $C^*$-algebras as Cuntz-Pimsner algebras naturally associated to self-similar groups [Nek04]. Analogues in the discrete setting were recently found [SS23].

Let $X$ be a finite set with $|X| \ge 2$, and let $X^*$ be the free monoid generated by $X$. A self-similar group over $X$ is a group $G$ together with a faithful action on $X^*$ by length-preserving permutations such that, for every $g \in G$ and every $x \in X$, there exists an element $g|_x \in G$ satisfying

$$g(xw) = g(x)g|_x(w) \quad \text{for all } w \in X^*.$$

Let $k$ be a field, let $R = kG$ be the convolution algebra of $G$ with coefficient in $k$, let

$$R = \bigoplus_{g \in G} k \cdot g = k^{(G)}, X = R^{(X)} = k^{(G \times X)}, X' = R^{(X)}$$

we define the $k$-bilinear scalar product

$$\langle \sum_{x \in X} \lambda_x x \cdot g_x, \sum_{x \in X} \mu_x x \cdot h_x \rangle = \sum_{x \in X} \lambda_x \mu_x g_x^{-1} h_x$$

$X$ and $X'$ are respectively right and left free $R$-modules, but as a left $R$-module we define $X$ by the relation $g \cdot x = g(x) \cdot g|_x$ for all $g \in G, x \in X$ (we have $g|_x \in G$) using the self-similarity condition. The Nekrashevych algebra associated to the self similar group $G$ (and the set $X$) with coefficients in $k$, is the algebraic Cuntz-Pimsner ring associated to the ring $R$ and the correspondence $\mathcal{X} = (X, X', \langle, \rangle)$

$$\mathcal{N}_k(G, X) = \mathcal{O}_{\mathcal{X}}$$

As $X$ is finite, the left action obviously maps into the compact operators. The ring of compact operators is just $M_d(kG)$ where $d = |X|$.

**Theorem 3.7.** For any self similar group $(G, X)$ there is a long exact sequence $(n \in \mathbb{Z})$

$$\dots \longrightarrow E_n(kG) \xrightarrow{1-E_n(\mathcal{X})} E_n(kG) \xrightarrow{E_n(j)} E_n(\mathcal{N}_k(G, X)) \longrightarrow \dots$$

## ACKNOWLEDGEMENT

The author would like to thank Professor Ralf Meyer for insightful discussions and valuable comments on earlier versions of this paper.

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