# Idempotented algebra

This post is a reading note to Bernstein’s Le “centre” de Bernstein, on the part of idempotented algebras (section 1.1 to 1.7, ibid.). Let \(R\) be a ring, an element \(e\) is called *an idempotent* if \(e^2=e\). Let \(k\) be a field.

The \(k\)-structures of \(H\) is not involved in the discussion of this post, so we will ignore it. Let \((H, I)\) be an idempotented algebra for the rest of the post.

*Pf.* If \(eHe \subset fHf\), then in particular, \(e = eee \in fHf\). If \(e \in fHf\), then \(e = fhf\) for some \(h \in H\), so
\[fef = f(fhf)f = fhf = e.\]
If \(e = fef\), then \(eHe = fefHfef \subset fHf\).
\(\vartriangleleft\)

For \(e, f \in I\), we say \(e \le f\) if the equivalent conditions in Lemma 2 hold. This gives an order on the set \(I\). Definition 1 of idempotented algebra forces \(I\) to be directed under this order: For any \(e, f \in I\), there exists by definition a \(g \in I\) such that \(e = geg\) and \(f = gfg\), i.e., \(e \le g\) and \(f \le g\). Moreover, \(H\) is the union of \(eHe\), for \(e\) runs over \(I\): for any \(h \in H\), there exists an \(e \in I\) such that \(h = ehe \in eHe\).

Below is a very useful characterization of non-degenerate modules we shall use later:

*Proof.* “if” part is obvious.

“only if” part: For any \(h \in H\), there exists an \(e \in I\) such that \(h = ehe\). Therefore, \(eh = e(ehe) = ehe = h\). Now for any \(m \in M\), there exist \(h \in H\) and \(n \in M\) such that \(hn = m\), since \(HM=M\). We can find an \(e \in I\) with \(eh = h\), then \(em = ehn = hn = m\). \(\square\)

In particular, we can see from the proof that \(H\) itself is non-degenerate as an \(H\)-module. We denote \(\mathcal{M}\) the category of all non-degenerate \(H\)-modules.

The next thing we would like to do is to put a topology on \(H\), then consider the completion \(\hat{H}\) of \(H\) under this topology. It turns out that \(\hat{H}\) is the endomorphism ring of the forgetful functor \(\omega: \mathcal{M} \to \mathcal{A}\), where \(\mathcal{A}\) is the category of abelian groups. We shall see the details of these in the following.

Since \(H\) has a group structure, the topology \(\tau\) above is well-defined. More specifically, for any \(h \in H\), there associates to it a fundamental system of neighborhoods \(\{h+H(1-e): e \in I\}\). A *Cauchy sequence* in \(H\) is a sequence \(\{x_n\}_{n \in \N} \subset H\) that, for any \(e \in I\), there exits an \(N_e > 0\) such that for any \(n, m \ge N_e\), \(x_n - x_m \in H(1-e)\). Given two Cauchy sequences \(\{x_n\}\) and \(\{y_n\}\), we say that \(\{x_n\} \sim \{y_n\}\) if for any \(e \in I\), there exists an \(N_e > 0\) such that for any \(n \ge N_e\), \(x_n - y_n \in H(1-e)\). One can easily check that \(\sim\) is an equivalent relation for these Cauchy sequences. We define the *completion* of \(H\) under \(\tau\) to be
\[\hat{H} = \{\text{Cauchy sequences in }H\}/\sim.\]

**Theorem 6.**\(\displaystyle \hat{H} = \varprojlim_{e \in I} He\), where \(\varprojlim\) is the inverse limit, taking over the inverse system \(x \mapsto xe: Hf \to He\) for \(e \le f\), of all the \(He\) with discrete topology.

*Proof.* Let \(\tilde{H} = \varprojlim_{e \in I} He\). \(\tilde{H}\) is a topological space, with initial topology induced from the projections \(\{\pi_e: \tilde{H} \to He, e \in I\}\). \(H\) is embedded into \(\tilde{H}\) via diagonal embedding, i.e., \(h \mapsto (he)_{e \in I}\). We view \(H\) is a subgroup in \(\tilde{H}\) under this embedding. The proof contains two steps: (i) the subspace topology of \(H\) in \(\tilde{H}\) is the same of the one define in Definition 5; (ii) \(H\) is dense in \(\tilde{H}\).

Since \(\tilde{H}\) is also a group, we just need to study the topology at \(0\). The initial topology of \(\tilde{H}\) at \(0\) is generated by \(\{\pi_e^{-1}(0)\}\). Since \(I\) is directed, \(\{\pi_e^{-1}(0)\}\) is actually a fundamental system of neighborhoods. But \(\pi_e^{-1}(0) \cap H = H(1-e)\), so (i) is done.

Let \((h_e)_{e \in I} \in \tilde{H}\). We need to show that \(((h_e)+\pi_f^{-1}(0)) \cap H\) is not empty, for any \(f \in I\). We take \(x = h_f \in H\), then \[\pi_f((h_e) - (xe)) = h_f - xf = h_f - h_f f=0.\] Therefore, \(x \in ((h_e)+\pi_f^{-1}(0)) \cap H\). This proves (ii). \(\square\)

Let \(M \in \mathcal{M}\) be a non-degenerate \(H\)-module. If we endow \(M\) with discrete topology, the action \(H \times M \to M\) is continuous: for any \(x \in M\), let \(U = \{(h, m): hm = x\}\), we need to show \(U\) is open in \(H \times M\). For any \((h, m) \in U\), since \(M\) is non-degenerate, there exists an \(e \in I\) such that \(em = m\). Consider \(V = (h + H(1-e)) \times \{m\}\). Then \(V\) is open in \(H \times M\), and a simple calculation shows \(V \subset U\). Thus, \(U\) is indeed open. By continuity, we can extend the action of \(H\) on \(M\) to that of \(\hat{H}\). By Theorem 6, we can write elements in \(\hat{H}\) as \((h_e)_{e \in I}\). For any \(m \in M\), there exists \(e \in I\) such that \(em = m\). Then \((h_e) \cdot m\) is defined to be \(h_e \cdot m\). It’s easy to check that the action of \(\hat{H}\) on \(M\) is well-defined.

Recall that \(\omega: \mathcal{M} \to \mathcal{A}\) is the forgetful functor, from the category of non-degenerate \(H\)-modules to the category of abelian groups. Here is another characterization of \(\hat{H}\).

**Theorem 7.**\(\displaystyle \hat{H} \cong \mathrm{End}(\omega)\).

Before the proof, let’s recall the notion of endomorphisms of a functor. Let \(F: \mathcal{C} \to \mathcal{D}\) be a functor. Then an *endomorphism* \(\varphi\) of \(F\) is a collection of morphisms \(\{\varphi_C \in \mathrm{Hom}_\mathcal{D}(FC, FC): C \in \mathcal{C}\}\), such that, for any morphism \(f: C_1 \to C_2\) in \(\mathcal{C}\), we have the following commutative diagram:
\[\begin{equation*}
\require{AMScd}
\begin{CD}
FC_1 @>{\varphi_{C_1}}>> FC_1 \\
@V{Ff}VV @VV{Ff}V \\
FC_2 @>>{\varphi_{C_2}}> FC_2
\end{CD}
\end{equation*}\]

*Proof.* [Proof of Theorem 7]
Since every non-degenerate \(H\)-module \(M\) is a quotient of \(H^{\oplus S}\) for some \(S\), it’s enough to show that \(\hat{H}\) is exactly those morphisms in \(\mathrm{End}_{\Z}(H)\) that commute with \(\mathrm{End}_H(H)\). Let’s set
\[A = \{\varphi \in \mathrm{End}_{\Z}(H): \varphi(f(x)) = f(\varphi(x)), f \in \mathrm{End}_H(H), x \in H\}.\]

Let’s show \(\hat{H} \subseteq A\) first. Let \(\hat{h} = (h_e)_{e \in I} \in \hat{H}\). \(H\) is itself a non-degenerate \(H\)-module, so for any \(x \in H\), there exists \(e \in I\) such that \(ex = x\). Then \(\hat{h} \cdot x := h_e x\). It’s easy to check that this action is well-defined and respects abelian structures, so \(\hat{h} \in \mathrm{End}_{\Z}(H)\). Moreover, if \(\hat{h_1} \ne \hat{h_2}\), then they maps differently into \(\mathrm{End}_{\Z}(H)\): there must be an \(e \in I\) so that \(h_{1, e} \ne h_{2, e}\), so \(\hat{h_1} \cdot e \ne \hat{h_2} \cdot e\).

Take any \(f \in \mathrm{End}_H(H)\), and any \(x \in H\). Then there exists by Definition 1 an \(e \in I\) such that \(ex = x\) and \(ef(x) = f(x)\). Hence, \[f(\hat{h} \cdot x) = f(h_e x) = h_ef(x) = \hat{h} \cdot f(x).\] This proves \(\hat{h} \in A\), so \(\hat{H} \subseteq A\).

Next we would like to show \(A \subseteq \hat{H}\). Let \(\varphi \in A\). Define \(\hat{h} = (h_e)\) by setting \(h_e = \varphi(e)\). The claim is that \(\varphi(x) = \hat{h} \cdot x\) for \(x \in H\). For any element \(y \in H\), right multiplication by \(y\) is an element in \(\mathrm{End}_H(H)\), so \(\varphi\) commutes with right multiplication in \(H\). For any \(x \in H\), let \(e \in I\) be such that \(ex = x\). Then, \[\varphi(x) = \varphi(ex) = \varphi(e)x = h_e x = \hat{h} \cdot x.\] Thus, \(A \subseteq \hat{H}\). \(\square\)

**Remark.** In the original proof, Bernstein actually considered also the topology on \(H\) and \(\hat{H}\). The topology of \(H\) (see Definition 5) induces a topology of pointwise convergence on \(\mathrm{End}_{\Z}(H)\). Under these topologies, \(\hat{H}\) is embedded in \(\mathrm{End}_{\Z}(H)\) as a closure of \(H\), where \(H\) is viewed a subset in \(\mathrm{End}_{\Z}(H)\) via left multiplication.
Lastly, we would like to discuss the center of \(\mathcal{M}\). Let \(\mathcal{C}\) be an abelian category, and \(\mathrm{Id}_{\mathcal{C}}: \mathcal{C} \to \mathcal{C}\) be the identity functor. The *center* of \(\mathcal{C}\) is defined to be the endomorphism ring \(\mathrm{End}(\mathrm{Id}_{\mathcal{C}})\).

**Theorem 8.**\(\displaystyle Z(\hat{H}) = \mathrm{End}(\mathrm{Id}_{\mathcal{M}}) = \varprojlim_{e \in I} Z(eHe)\), where \(Z(R)\) is the center of a ring \(R\).

*Proof.* By Theorem 7, we know already that \(\hat{H} = \mathrm{End}(\omega)\). For \(\hat{h} \in \hat{H}\), \(\hat{h} \in \mathrm{End}(\mathrm{Id}_{\mathcal{M}})\) if and only if \(\hat{h}\) also commutes with left multiplication by elements in \(H\). \(\hat{h}\) commutes with left multiplication by elements in \(H\), is exactly the same as \(\hat{h} \in Z(\hat{H})\). Therefore, we have \(Z(\hat{H}) = \mathrm{End}(\mathrm{Id}_{\mathcal{M}})\).

Let \(C = \varprojlim_{e \in I} Z(eHe)\). Let’s see that \(C\) is actually well-defined, i.e., \(\{Z(eHe): e \in I\}\) forms an inverse system. For \(f \ge e\), let \(fzf \in Z(fHf)\), we need to show \(eze \in Z(eHe)\). Let \(exe \in eHe\), we note that \(ef = fe = e\) and \(ez = ze = eze\), then \[exe \cdot eze = efxfze = ezfxfe = eze \cdot exe.\] Therefore, \(eze \in Z(eHe)\), so \(C\) is well-defined.

Let \(\hat{c} = (c_e) \in C\), then \(\hat{c} \in \hat{H}\). Let \(x \in H\), for any \(y \in H\), then there exists an \(e \in I\) such that \(exe = x\) and \(eye = y\). Thus, \(exye = xy\) and \[x\hat{c}(y) = xc_ey = c_e xy = \hat{c}(xy).\] This proves that \(\hat{c}\) commutes with left multiplication in \(H\), so \(\hat{c} \in Z(\hat{H})\).

Now let \(\hat{h} = (h_e) \in Z(\hat{H})\). Then \(\hat{h}\) commutes with left multiplication in \(H\). In particular, \(e\hat{h}(e) = \hat{h}(e)\), i.e., \(eh_ee = h_ee = h_e \in eHe\) (last equality uses the fact that \(h_e \in He\)). Let \(exe \in eHe\), we use again the fact that \(\hat{h}\) commutes with left multiplication: \[exeh_e = exe \hat{h}(e) = \hat{exe} = h_e exe.\] This shows that \(h_e \in Z(eHe)\), so \(\hat{h} \in C\). \(\square\)

### An example of idempotented algebras

Let \(G\) be a locally compact, totally disconnected group. Let \(H(G)\) be the set of locally constant functions from \(G\) to \(\C\) with compact open support. The addition in \(H(G)\) is just pointwise addition: for \(f, g\in H(G)\), \((f+g)(s) = f(s) + g(s)\), for \(s \in G\). The multiplication in \(H(G)\) is given by convolution: for \(f, g \in H(G)\), \[f \ast g(t) = \int_G f(s)g(s^{-1}t)d\mu(s),\] where \(\mu\) is a left Haar measure on \(G\). The above addition and multiplication give a ring structure on \(H(G)\). There are natural idempotents in \(H(G)\). Let \(K\) be an open compact subgroup in \(G\), and let \(e_k\) be the normalized characteristic function on \(K\), i.e., \[\begin{equation*} e_K(s) = \left\{ \begin{array}{ll} \frac{1}{\mu(K)}, & s \in K; \\ 0, & s \not\in K. \end{array} \right. \end{equation*}\] Here, \(\mu(K) = \int_Kd\mu(s)\) is the measure of \(K\). In a simpler form, we can write \(e_K = \frac{1}{\mu(K)} \mathbf{1}_K\), where \(\mathbf{1}_K\) is the indicator function on \(K\). A direct computation shows \(e_K \ast e_K = e_K\), so \(e_K\) is an idempotent. Let \(I\) be a set of idempotents in \(H(G)\). We will check that \((H(G), I)\) is an idempotented algebra.

Let \(K \subset H\) be two open compact supgroups in \(G\), then \[\begin{equation*} \begin{split} e_K \ast e_H(t) &= \int_G e_K(s)e_H(s^{-1}t) d\mu(s) \\ &= \frac{1}{\mu(K)}\int_K e_H(s^{-1}t)d\mu(s) \\ &= \frac{1}{\mu(K)\mu(H)} \mathbf{1}_H(t) \int_K d\mu(s) \\ &= \frac{1}{\mu(H)} \mathbf{1}_H(t) = e_H. \end{split} \end{equation*}\] Similarly, \(e_H \ast e_K = e_H\). Therefore, \(e_K \ast e_H \ast e_K = e_H\), i.e., \(e_K \ge e_H\). Now consider \(\varphi = \mathbf{1}_C\) for some compact open set \(C\) in \(G\). We can then cover \(C\) with open sets \[\{gK \subset C: g \in G, K \subset G \text{ is a compact open subgroup}\}.\] This can be done because \(\{K \subset G \text{ is a compact open subgroup}\}\) is a basis for the topology at the identity of \(G\) and \(C\) is open. Since \(C\) is compact, we can then select finite many \(g_iK_i\) that cover \(C\). Since \(K_i\) are also open, we can further make \(g_iK_i\) disjoint, so we can rewrite \(\varphi\) as \[\varphi = \sum_i \mathbf{1}_{g_iK_i}.\] For simplicity, let’s consider for now \(\psi = \mathbf{1}_{gK}\), for some \(g \in G\) and \(K \subset G\) some compact open subgroup. Let \(H = K \cap gKg^{-1}\). It’s easy to see that \(\psi\) is right invariant under multiplication by elements in \(H\). In fact, \(\psi(xk) = \psi(x)\) for any \(x \in G\) and \(k \in K\). Moreover, \(\psi\) is left invariant under multiplication by elements in \(H\), i.e., \(\psi(hx) = \psi(x)\), for any \(x \in G\) and \(h \in H\): write \(h = gkg^{-1}\) for some \(k \in K\), then \[\begin{equation*} \begin{split} \psi(hx) &= \mathbf{1}_{gK} = (gkg^{-1}x) = \mathbf{1}_{K}(kg^{-1}x) \\ &= \mathbf{1}_{K}(g^{-1}x) = \mathbf{1}_{gK}(x) = \psi(x). \end{split} \end{equation*}\]

*Pf.* We need to use the fact that \(\psi\) is left and right invariant under multiplication by elements of \(H\). Here, only calculation for \(e_H \ast \psi = \psi\) is given. \(\psi \ast e_H = \psi\) can be shown using a similar calculation.

\[\begin{equation*} \begin{split} e_H \ast \psi(t) &= \int_G e_H(s)\psi(s^{-1}t) d\mu(s) \\ &= \frac{1}{\mu(H)} \int_H \psi(s^{-1}t) d\mu(s) \\ &= \frac{\psi(t)}{\mu(H)}\int_H d\mu(s) = \psi(t). \end{split} \end{equation*}\] Note that we need the fact that \(\psi\) is left invariant under multiplication by elements of \(H\) in the above calculation. \(\vartriangleleft\)

Come back to our \(\varphi = \sum_i \mathbf{1}_{g_iK_i}\). Then for each \(\mathbf{1}_{g_iK_i}\), we can find by Lemma 9 a compact open subgroup \(H_i\) such that \[\begin{equation} e_{H_i} \ast \mathbf{1}_{g_iK_i} \ast e_{H_i} = \mathbf{1}_{g_iK_i}. \label{20161004-1} \end{equation}\]

Let \(H = \cap_i H_i\), then \(H\) is also a compact open subgroup. \(H_i \subset H\), so \(e_H \ge e_{H_i}\), so \[\begin{equation} e_{H} \ast e_{H_i} \ast e_{H} = e_{H_i}, \label{20161004-2} \end{equation}\]

for each \(i\). Combining \(\eqref{20161004-1}\) and \(\eqref{20161004-2}\), we have \[e_{H} \ast \mathbf{1}_{g_iK_i} \ast e_{H} = \mathbf{1}_{g_iK_i},\] for all \(i\). This implies that \(e_H \ast \varphi \ast e_H = \varphi\). To summarize, we have the following theorem:

**Theorem 10.**Let \(C\) be a compact open subset in \(G\) and let \(\varphi = \mathbf{1}_C\), then there exist a compact open subgroup \(K\) such that \[e_K \ast \varphi \ast e_K = \varphi.\]

Basically, we have already shown that \((H(G), I)\) is an idempotented algebra by Theorem 10. Given a finite set of elements \(\{\varphi_i\}\) in \(H(G)\), we can assume \(\varphi_i = \mathbf{1}_{C_i}\) for some compact open subset. The assumption is harmless, since elements in \(H(G)\) are of form \[\sum_i \alpha_i \mathbf{1}_{C_i},\] where \(\alpha_i\) are constants in \(\C\) and \(C_i\) are compact open subsets in \(G\). Then by Theorem 10, there exists a compact open \(K_i\) so that \[e_{K_i} \ast \varphi_i \ast e_{K_i} = \varphi_i,\] for each \(i\). Now take \(K = \cap_i K_i\), then \(e_K \ast \varphi_i \ast e_K = \varphi_i\), for all \(i\). Therefore, we conclude that \((H(G), I)\) is an idempotent algebra.

**Remark.** Let \(I_C = \{e_K: K \text{ is a compact open subgroup in }G\} \subset I\). Then the above argument actually shows that we can find for each finite set \(\{\varphi_i\} \subset H(G)\) an idempotent \(e \in I_C\) such that \(e \ast \varphi_i \ast e = \varphi_i\). Therefore, \(I_C\) is cofinal in \(I\). Instead of taking inverse limit over \(I\), we can take it over \(I_C\):
\[\widehat{H(G)} = \varprojlim_{e \in I} H(G) \ast e = \varprojlim_{e \in I_C} H(G) \ast e.\]