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.

Definition 1. A $k$-idempotented algebra $(H, I)$ is a $k$-algebra $H$ together with the set idempotents $I$ satisfying

(*) For any finite family $(x_i)$, there exists an idempotent $e \in I$ such that $ex_ie = x_i$ for all $i$.

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.

Lemma 2. Let $e, f \in I$. The following are equivalent.

(1) $eHe \subset fHf$
(2) $e \in fHf$
(3) $e = fef$

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 $h = ehe \in eHe$.

Definition 3. An $H$-module $M$ is non-degenerate if $HM=M$.

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

Proposition 4. An $H$-module $M$ is non-degenerate if and only if for any $m \in M$, there exists an $e \in I$ such that $em = m$.

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.

Definition 5 (Topology on $H$). The topology $\tau$ on $H$ is given by a fundamental system of neighborhoods $\{H(1-e): e \in I\}$ at $0 \in H$, where $H(1-e) = \{h-he: h \in H\},$ is the set of left annihilators of $e$.

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 $\begin{equation*} \begin{split} \psi(hx) &= \mathbf{1}_{gK} = (gkg^{-1}x) = \mathbf{1}_{K}(kg^{-1}x) \\ &= \mathbf{1}_{K}(g^{-1}x) = \mathbf{1}_{gK}(x) = \psi(x). \end{split} \end{equation*}$

Lemma 9. Under the above notations, we have $e_H \ast \psi = \psi \ast e_H = \psi$. In particular, $e_H \ast \psi \ast e_H = \psi$.

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 $I_C$: $\widehat{H(G)} = \varprojlim_{e \in I} H(G) \ast e = \varprojlim_{e \in I_C} H(G) \ast e.$