2. Relation bimodules

In what follows $K$ is a commutative ring, and $A$ is an associative $K$ -algebra. Usually, $A$ will be projective as a $K$ -module. By an $A$ -bimodule, we mean a left $A\otimes _K A^{op}$ -module. We say that $A$ is bi- $\mathrm {FP}_n$ over $K$ if $A$ has a projective resolution as an $A$ -bimodule that is finitely generated through degree $n$ . We record here some elementary facts.

The free monoid on a set $X$ is denoted by $X^*$ , and $KX^*$ is then the free $K$ -algebra generated by $X$ . Following Ivanov [Reference Ivanov5], if $A$ is $X$ -generated and $J=\ker (KX^*\to A)$ , then $J/J^2$ is an $A$ -bimodule, termed the relation bimodule. Note that if $J=(R)$ , then $J/J^2$ is generated by the $r+J^2$ with $r\in R$ as an $A$ -bimodule. The relation bimodule was first studied by Bergman and Dicks [Reference Bergman and Dicks1], but not under that name.

Let $I=\ker\!(KX^*\otimes _K KX^*\xrightarrow {\mu } KX^*)$ be the kernel of the multiplication map. It is well known that $I$ is a free $KX^*$ -bimodule with basis the elements $x\otimes 1-1\otimes x$ with $x\in X$ . We give a quick sketch of the proof for completeness.

Proof. First we show that $I$ is generated by these elements. There is an isomorphism $K[X^*\times X^*]\to KX^*\otimes _K KX^*$ given by $(u,v)\mapsto u\otimes v$ for $u,v\in X^*$ . Composing this with $\mu$ is the $K$ -linear extension of the multiplication map $X^*\times X^*\to X^*$ , which clearly has kernel with $K$ -basis all differences $(u,v)-(1,uv)$ with $u,v\in X^*$ and $u\neq 1$ . It follows that $I$ has $K$ -basis all elements of the form $u\otimes v-1\otimes uv$ with $u\neq 1$ . But $u\otimes v-1\otimes uv=(u\otimes 1-1\otimes u)v$ , and so $I$ is generated by all elements of the form $w\otimes 1-1\otimes w$ with $1\neq w\in X^*$ as a $KX^*$ -bimodule. We show by induction on $|w|$ that $w\otimes 1-1\otimes w$ is in the sub-bimodule generated by the $x\otimes 1-1\otimes x$ with $x\in X$ with the case $|w|=1$ trivial. If $w=xu$ with $x\in X$ , then $w\otimes 1-1\otimes w=xu\otimes 1-1\otimes xu = x(u\otimes 1-1\otimes u)+(x\otimes 1-1\otimes x)u$ and the claim follows by induction.

Suppose now that $B$ is a $KX^*$ -bimodule and $f\colon X\to B$ is any map. Then

\begin{equation*}A=\left \{\begin{bmatrix} a & b\\ 0 & a\end{bmatrix} \mid a\in KX^*, b\in B\right \}\end{equation*}

is a $K$ -algebra, and we can define a homomorphism $\psi \colon KX^*\to A$ such that

\begin{equation*}\psi (x) = \left[\begin{array}{c@{\quad}c} x & f(x)\\ 0 &x\end{array}\right], \quad \psi (a) = \left[\begin{array}{c@{\quad}c} a & d(a)\\ 0 &a\end{array}\right]\end{equation*}

for $x\in X$ and $a\in KX^*$ . One easily verifies that $d$ is a derivation, meaning that $d(aa^{\prime}) = ad(a^{\prime})+d(a)a^{\prime}$ , and that $d(x)=f(x)$ . Since $(a,a^{\prime})\mapsto d(a)a^{\prime}$ is clearly $K$ -bilinear, we can define a $K$ -linear map $KX^*\otimes _K KX^*\to B$ by $\alpha (a\otimes a^{\prime}) = d(a)a^{\prime}$ . We claim that $\alpha \colon I\to B$ is a $KX^*$ -bimodule homomorphism with $\alpha (x\otimes 1-1\otimes x)=f(x)$ for $x\in X$ . This will complete the proof. Indeed, if $\sum u_i\otimes v_i\in I$ , then $\sum u_iv_i=0$ . Trivially, $\alpha (\sum u_i\otimes v_iw) = \sum d(u_i)v_iw=d(\sum u_i\otimes v_i)w$ . On the other hand, $\alpha (\sum wu_i\otimes v_i) = \sum d(wu_i)v_i = w\sum d(u_i)v_i+ d(w)\sum u_iv_i = wd(\sum u_i\otimes v_i)$ . Finally, $\alpha (x\otimes 1-1\otimes x) = d(x)-d(1)x= f(x)$ , as $d(1)=0$ .

The following result is due to Bergman and Dicks [Reference Bergman and Dicks1]. We give a proof for completeness.

Theorem 2.3. Let $A$ be a $K$ -algebra generated by $X$ that is projective over $K$ . Then there is an exact sequence

\begin{equation*}0\to J/J^2\xrightarrow {d_2} A\otimes _K KX\otimes _K A\xrightarrow {d_1} A\otimes _A A\xrightarrow {\mu } A\to 0\end{equation*}

where $\mu$ is the multiplication map, $d_1(1\otimes x\otimes 1) = x\otimes 1-1\otimes x$ and $d_2$ is induced by the derivation $KA^*\to A\otimes _K KX\otimes _K A$ given by $x_1\cdots x_m\mapsto \sum _{i=1}^m (x_1\cdots x_{i-1}+J)\otimes x_i\otimes (x_{i+1}\cdots x_m+J)$ .

Proof. Since $I$ is freely generated as a $KX^*$ -bimodule by the elements of the form $x\otimes 1-1\otimes x$ with $x\in X$ , we have an exact sequence

(2.1) \begin{equation} 0\to KX^*\otimes _K KX\otimes _K KX^*\to KX^*\otimes _K KX^*\to KX^*\to 0 \end{equation}

where the first map takes $1\otimes x\otimes 1$ to $x\otimes 1-1\otimes x$ and the second map is the multiplication map. Since $KX^*$ is free as a right $KX^*$ -module, this sequence splits over $(KX^*)^{op}$ , and so tensoring on the right with $A$ yields the exact sequence $0\to KX^*\otimes _K KX\otimes _K A\to KX^*\otimes _K A\to A\to 0$ . Since $A$ is projective over $K$ , it follows that $KX^*\otimes _K KX\otimes _K A$ and $KX^*\otimes _K A$ are projective left $KX^*$ -modules. Thus, tensoring on the left with $A=KX^*/J$ and recalling that $\operatorname {\mathrm {Tor}}\nolimits _1^{KX^*}(KX^*/J,KX^*/J)\cong J/J^2$ (cf. [Reference Cartan and Eilenberg2, Exercise 19, p. 126]), yields the exact sequence

\begin{equation*}0\to J/J^2\to A\otimes _K KX\otimes _K A\to A\otimes _K A\to A\to 0.\end{equation*}

It is a straightforward exercise to check that the connecting map is given by $d_2$ above (or see [Reference Bergman and Dicks1]).

In particular, if $X$ is finite, then $A$ is bi- $\mathrm {FP}_2$ over $K$ if and only if the relation bimodule is finitely generated. Next we shall need a result I learned of from Webb [Reference Webb9].

As a corollary, we can deduce that $J$ is projective as both a left and right $KX^*$ -module.

The next proposition will be used to show that $J^m$ is projective as both a left and right module.

3. The homotopy bimodule

A presentation of a $K$ -algebra is a gadget $\mathcal P=\langle X\mid R\rangle$ where $X$ is a set and $R\subseteq KX^*$ . The presentation is finite if $X$ and $R$ are finite. A $K$ -algebra $A$ is presented by $\mathcal P$ if $A\cong KX^*/(R)$ . As usual, we put $J=(R)$ . We shall assume throughout this section that $A$ is projective over $K$ , and hence $J$ is projective as both a left and right $KX^*$ -module by Corollary 2.5. The relation bimodule $J/J^2$ is generated by the cosets $r+J$ with $r\in R$ , whence $A$ is bi- $\mathrm {FP}_2$ if the presentation is finite. It follows there is an exact sequence $A\otimes _K KR\otimes _K A\to J/J^2\to 0$ , and we aim to compute the kernel. This was done in the case that $A$ is an integral monoid ring and $\mathcal P$ comes from a monoid presentation in the work of Pride [Reference Pride8] and of Guba–Sapir [Reference Guba and Sapir4], where it was shown that the kernel is the first homology of the Squier complex of the presentation. Here we construct the analog of the first homology of the Squier complex for any presentation of a $K$ -algebra $A$ that is projective over $K$ . We then prove the analog of the Pride–Guba–Sapir exact sequence. In the next section, we shall derive anew the Pride–Guba–Sapir sequence in the monoid case from our more general result.

Since $J=(R)$ , we have an exact sequence of $KX^*$ -bimodules

(3.1) \begin{equation} 0\to H\to KX^*\otimes _K KR\otimes _K KX^*\xrightarrow {\alpha } J\to 0. \end{equation}

To keep $R$ and its image in $KX^*$ formally distinct, we write $ {\overline {r}}$ for the basis element of $KR$ corresponding to $r\in R\subseteq KX^*$ . Then $\alpha (u\otimes {\overline {r}}\otimes v)=urv$ . Note that since $J$ is projective as both a left and right $KX^*$ -module by Corollary 2.5, the sequence (3.1) splits when viewed as either a sequence of left or of right $KX^*$ -modules. Moreover, since $KX^*\otimes _K KR\otimes _K KX^*$ is free as both a left and right $KX^*$ -module, we deduce that $H$ is projective as both a left and right $KX^*$ -module.

If $r_i,r_j\in R$ and $u\in X^*$ , then we put $[r_i,u,r_j] = 1\otimes {\overline {r}}_i\otimes ur_j-r_iu\otimes {\overline {r}}_j\otimes 1\in KX^*\otimes _K KR\otimes _K KX^*$ . Note that $\alpha ([r_i,u,r_j]) = r_iur_j-r_iur_j=0$ , and so $[r_i,u,r_j]\in H$ . Let $D$ be the sub- $KX^*$ -bimodule of $H$ generated by $\{[r_i,u,r_j]\mid r_i,r_j\in R, u\in X^*\}$ . Note that $D$ is spanned as a $K$ -module by the elements $v[r_i,u,r_j]w$ with $v,w\in X^*$ .

The $A$ -bimodule $H/D$ will be denoted $\pi _2(\mathcal P)$ in analogy to the corresponding notation used by Pride in the case of integral monoid rings and monoid presentations.

The crucial idea is now the following. The exact sequence (3.1) splits over $(KX^*)^{op}$ , and so tensoring on the right with $A=KX^*/J$ yields an exact sequence $0\to H/HJ\to KX^*\otimes _K KR\otimes _K A\to J/J^2\to 0$ . Now tensoring on the left with $A$ and observing that $KX^*\otimes _K KR\otimes _K A$ is a projective $KX^*$ -module, because $A$ is projective over $K$ , yields an exact sequence $0\to \operatorname {\mathrm {Tor}}\nolimits _1^{KX^*}(KX^*/J,J/J^2)\to H/(JH+HJ)\to A\otimes _K KR\otimes _K A\to J/J^2\to 0$ . Note that $0\to J^2\to J\to J/J^2\to 0$ is a projective resolution of $J/J^2$ as a left $KX^*$ -module by Corollary 2.7, and so $\operatorname {\mathrm {Tor}}\nolimits _1^{KX^*}(KX^*/J,J/J^2) = \ker\!(J^2/J^3\to J/J^2)=J^2/J^3$ . We will compute the connecting map $J^2/J^3\to H/(JH+HJ)$ and see that its image is $D/(JH+HJ)$ . This will then yield the exact sequence $0\to \pi _2(\mathcal P)\to A\otimes _K KR\otimes _K A\to J/J^2\to 0$ , that we desire.

Figure 1.

A commutative diagram.

Figure 2.

A commutative diagram.

Splicing together the exact sequences in Theorems2.3 and 3.2 yields the following result.

Corollary 3.3. Let $A$ be a $K$ -algebra that is projective over $K$ with presentation $\mathcal P=\langle X\mid R\rangle$ . Then there is an exact sequence of $A$ -bimodules

\begin{equation*}0\to \pi _2(\mathcal P)\to A\otimes _K KR\otimes _K A\to A\otimes _K KX\otimes _K A\to A\otimes _K A\to A\to 0.\end{equation*}

In particular, if $\mathcal P$ is a finite presentation, then $A$ is of type bi- $\mathrm {FP}_3$ over $K$ if and only if $\pi _2(\mathcal P)$ is finitely generated.

4. Monoid presentations

We now consider the case of a monoid presentation and apply Theorem3.2 to recover a result of Pride [Reference Pride8, Theorem 4.2] and Guba–Sapir [Reference Guba and Sapir4, Theorem 11.3] (see also [Reference Kobayashi and Otto6]). The proof of Theorem3.2 is more elementary than the proofs found in [Reference Guba and Sapir4] or [Reference Kobayashi and Otto6] as we avoid Knuth–Bendix completions.

Let $\mathcal P = \langle X\mid R\rangle$ be a monoid presentation of a monoid $M$ . We view $R\subseteq X^*\times X^*$ . The derivation graph $\Gamma (\mathcal P)$ has vertex set $X^*$ and directed edge set $X^*\times R\times X^*$ where $(u,(\ell, r),v)\colon u\ell v\to urv$ . It is evident that two vertices are in the same connected component of $\Gamma (\mathcal P)$ if and only if they represent the same element of $M$ . The Squier complex $D(\mathcal P)$ has $1$ -skeleton $\Gamma (\mathcal P)$ and has set of $2$ -cells $X^*\times R\times X^*\times R\times X^*$ . The cell corresponding to $(u,(\ell, r),v,(\ell ^{\prime},r^{\prime}),w)$ has boundary path

\begin{equation*}(u,(\ell, r),v\ell ^{\prime} w)(urv,(\ell ^{\prime},r^{\prime}),w)[(u\ell v,(\ell ^{\prime},r^{\prime}),w)(u,(\ell, r),vr^{\prime}w)]^{-1}.\end{equation*}

Roughly speaking, paths correspond to monoid pictures of a derivation from one word to another, and two paths are homotopic if and only if they can be obtained from each other by isotopy and removing or inserting dipoles. See [Reference Guba and Sapir4] for details, as we will not need monoid pictures.

Notice that $X^*$ has commuting left and right actions on $D(\mathcal P)$ by regular cellular maps induced by multiplication.

The monoid algebra $\mathbb ZM$ has presentation as a $\mathbb Z$ -algebra $\mathcal P^{\prime}=\langle X\mid {\widetilde {R}}\rangle$ where $ {\widetilde {R}}= \{r-\ell \mid (\ell, r)\in R\}$ . Moreover, $\mathbb ZM$ is free over $\mathbb Z$ so the above theory applies. Put $J=( {\widetilde {R}})$ as usual. Then $J/J^2$ is the relation bimodule considered by Ivanov [Reference Ivanov5]. Notice that if we look at the cellular chain complex of $\Gamma (\mathcal P)$ , we get the exact sequence $0\to H_1(\Gamma (\mathcal P))\to \mathbb Z[X^*\times R\times X^*]\xrightarrow {d_1} \mathbb ZX^*$ , and the image of $d_1$ is $J$ as $d_1((1,(\ell, r),1)) = r-\ell$ . Identifying $\mathbb Z[X^*\times R\times X^*]$ with $\mathbb ZX^*\otimes _{\mathbb Z} \mathbb Z {\widetilde {R}}\otimes _{\mathbb Z} \mathbb ZX^*$ via $(u,(\ell, r),v)\mapsto u\otimes {\overline {r-\ell }}\otimes v$ then identifies $H_1(\Gamma (\mathcal P))$ with $H$ from (3.1).

On the other hand, if $d_2\colon \mathbb Z[X^*\times R\times X^*\times R\times X^*]\to \mathbb Z[X^*\times R\times X^*]$ is the boundary map in the cellular chain complex of $D(\mathcal P)$ , then $d_2(u,(\ell, r),v,(\ell ^{\prime},r^{\prime}),w) = (u,(\ell, r),v\ell ^{\prime}w)-(u,(\ell, r),vr^{\prime}w)+(urv,(\ell ^{\prime},r^{\prime}),w)-(u\ell v,(\ell ^{\prime},r^{\prime}),w)$ . Under our identification of $\mathbb Z[X^*\times R\times X^*]$ with $\mathbb ZX^*\otimes _{\mathbb Z} \mathbb ZR\otimes _{\mathbb Z} \mathbb ZX^*$ this maps to $-u\otimes {\overline {r-\ell }}\otimes v(r^{\prime}-\ell ^{\prime}) w+u(r-\ell )v\otimes {\overline {r^{\prime}-\ell ^{\prime}}}\otimes w= -u[(r-\ell, v,r^{\prime}-\ell ^{\prime})]w$ . It follows that $H_1(D(\mathcal P))\cong H/D=\pi _2(\mathcal P^{\prime})$ as $\mathbb ZX^*$ -bimodules, and hence $H_1(D(\mathcal P))$ is a $\mathbb ZM$ -bimodule (as was already observed by Pride). Pride puts $\pi _2^{(b)}(\mathcal P)=H_1(D(\mathcal P))$ . Putting together these observations with Theorem3.2 recovers the exact sequence of Pride [Reference Pride8] and Guba–Sapir [Reference Guba and Sapir4].

Theorem 4.1 (Pride/Guba–Sapir). Let $\mathcal P=\langle X\mid R\rangle$ be a monoid presentation. Then there is an exact sequence of $\mathbb ZM$ -bimodules

\begin{equation*}0\to \pi _2^{(b)}(\mathcal P)\to \mathbb ZM\otimes _{\mathbb Z} \mathbb ZR\otimes _{\mathbb Z} \mathbb ZM\to J/J^2\to 0\end{equation*}

where $J/J^2$ is the relation bimodule of this presentation.

This theorem implies that the abelianization of a diagram group is free abelian [Reference Guba and Sapir4, Theorem 11.2].

There is an augmentation homomorphism $\varepsilon \colon \mathbb ZM\to \mathbb Z$ with $\varepsilon (m)=1$ for all $m\in M$ . The left (respectively, right) relation module, studied by Ivanov [Reference Ivanov5], is $J/J^2\otimes _{\mathbb ZM} \mathbb Z$ (respectively, $\mathbb Z\otimes _{\mathbb ZM}J/J^2$ ). Pride defined $\pi _2^{(\ell )}(\mathcal P) = \pi _2^{(b)}(\mathcal P)\otimes _{\mathbb ZM}\mathbb Z$ and $\pi _2^{(r)}(\mathcal P)=\mathbb Z\otimes _{\mathbb ZM} \pi _2^{(b)}(\mathcal P)$ . The following exact sequences are known [Reference Kobayashi and Otto6, Reference Pride8], but our proof is different.

Corollary 4.2. Let $\mathcal P=\langle X\mid R\rangle$ be a presentation of a monoid $M$ . Let $J=\ker\!(\mathbb ZX^*\to \mathbb ZM)$ . Then there are exact sequences

\begin{gather*} 0\to J/J^2\otimes _{\mathbb ZM}\mathbb Z\to \mathbb ZM\otimes _{\mathbb Z}\mathbb ZX\to \mathbb ZM\to \mathbb Z\to 0\\ 0\to \pi _2^{(\ell )}(\mathcal P)\to \mathbb ZM\otimes _{\mathbb Z}\mathbb ZR\to J/J^2\otimes _{\mathbb ZM}\mathbb Z\to 0 \end{gather*}

and dually for the right relation module and $\pi _2^{(r)}(\mathcal P)$ .

If we splice the two exact sequences in Corollary 4.2 together, we get the exact sequence $0\to \pi _2^{(\ell )}(\mathcal P)\to \mathbb Z[M\times R]\to \mathbb Z[M\times X]\to \mathbb ZM\to \mathbb Z\to 0$ . Recall that the (left) Cayley complex $C^{(\ell )}(\mathcal P)$ of $M$ with respect to the presentation $\mathcal P$ is the free $M$ -CW complex with vertex set $M$ , edge set $M\times X$ and set of $2$ -cells $M\times R$ [Reference Gray and Steinberg3]. Here $(m,x)\colon m\to mx$ and the boundary path of $(m,(\ell, r))$ is the path from $m$ to $m\ell =mr$ labeled by $r$ followed by the inverse of the path from $m$ to $m\ell =mr$ labeled by $\ell$ . This complex is simply connected and the above exact sequence identifies $\pi _2^{(\ell )}(\mathcal P)$ as a left $\mathbb ZM$ -module with $H_2(C^{(\ell )}(\mathcal P))\cong \pi _2(C^{(\ell )}(\mathcal P))$ .

We observe that Theorem4.1 provides a similar result for the two-sided Cayley complex $C^{(b)}(\mathcal P)$ [Reference Gray and Steinberg3]. The vertex set of $C^{(b)}(\mathcal P)$ is $M\times M$ , the edge set is $M\times X\times M$ and the set of $2$ -cells is $M\times R\times M$ . The edge $(m,x,m^{\prime})$ goes from $(m,xm^{\prime})\to (mx,m^{\prime})$ . If $(\ell, r)\in R$ and $(m,m^{\prime})\in M\times M$ , then $m\ell =mr$ , $\ell m^{\prime}=rm^{\prime}$ , and it is not difficult to see that there are paths labeled by $\ell$ and by $r$ from $(m,\ell m^{\prime})\to (m\ell, m^{\prime})$ . The boundary path of the cell $(m,(\ell, r),m^{\prime})$ is the aforementioned path labeled by $r$ followed by the inverse of the aforementioned path labeled by $\ell$ . There are natural commuting left and right cellular actions of $M$ on $C^{(b)}$ induced by multiplication. The set of path components of $C^{(b)}(\mathcal P)$ is in bijection with $M$ via $(m,m^{\prime})\mapsto mm^{\prime}$ on vertices and each path component is simply connected. Moreover, one has $H_0(C^{(b)}(\mathcal P))\cong \mathbb ZM$ as a $\mathbb ZM$ -bimodule.

The exact sequence in Theorem4.1 can be spliced with that in Theorem2.3 to yield an exact sequence $0\to \pi _2^{(b)}(\mathcal P)\to \mathbb ZM\otimes _{\mathbb Z}\mathbb ZR\otimes _{\mathbb Z}\mathbb ZM\to \mathbb ZM\otimes _{\mathbb Z}\mathbb ZX\otimes _{\mathbb Z}ZM\to \mathbb ZM\otimes _{\mathbb Z}\mathbb ZM\to \mathbb ZM\to 0$ . It is not difficult to check using Theorem2.3 that this can be identified with the exact sequence $0\to \pi _2^{(b)}(\mathcal P)\to \mathbb Z[M\times R\times M]\to \mathbb Z[M\times X\times M]\to \mathbb Z[M\times M]\to H_0(C^{(b)}(\mathcal P))\to 0$ , where the maps come from the cellular chain complex of $C^{(b)}(\mathcal P)$ . Therefore, we can identify $\pi _2^{(b)}(\mathcal P)$ with $H_2(C^{(b)}(\mathcal P))$ as a $\mathbb ZM$ -bimodule. Since each path component of $C^{(b)}(\mathcal P)$ is simply connected, as an abelian group this is the direct sum of the $\pi _2$ of the path components of $C^{(b)}(\mathcal P)$ , justifying the notation. It follows from the above exact sequence that if the presentation is finite, then $M$ is bi- $\mathrm {FP}_3$ if and only if $\pi _2^{(b)}(\mathcal P)$ is finitely generated, as was observed by Kobayashi and Otto [Reference Kobayashi and Otto7].