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Recall that an associative algebra $A$ is called idempotented provided that is the filtered union of subalgebras $eAe$ for $e \in A$ idempotent. I think sometimes people say that $A$ has approximate identity.

It is well-known that for any idempotent $e \in A$, the functor $M \mapsto eM$ induces a bijection from simple $A$-modules $M$ satisfying $eM \neq 0$ to simple $eAe$-modules. My question is whether this bijection can be deduced from some equivalence of categories. Here's my guess: the aforementioned functor restricts to an equivalence from $A$-modules $M$ satisfying $AeM = M$ to $eAe$-modules. Note that this functor has a left adjoint, namely $N \mapsto Ae \otimes_{eAe} N$, which must be the inverse if my guess is true.

Here's the tricky point: if $M$ is a $A$-module satisfying $AeM = M$, we have to show that $Ae \otimes_{eAe} eM \to M$ is injective. This is not at all clear to me.

Is my guess correct? If not, can it be modified into a true statement? I really only care about the case where $A$ is the Hecke algebra of a $p$-adic group and $e$ is the indicator function of a compact open subgroup, so if there is a counterexample of this kind I'd like to see it.

Edit: As Benjamin Steinberg's answer shows, this is false in the stated generality. So let me reiterate that I'm interested in Hecke algebras of $p$-adic groups, and ask more directly: is my guess correct in this setting?

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    $\begingroup$ If your left adjoint is a quasi inverse, then it preserves simplicity. If M is simple, then $Ae\otimes_{eAe}M$ will be simple if and ony if no submodule is annihilated by e. Does this happen in your context? $\endgroup$ Commented Sep 17, 2013 at 19:00

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The map $Ae\otimes_{eAe}eM\to M$ is usually not injective when M is simple. The kernel is the largest submodule killed by e. This is part of Green's theory.

Added. Here is an example. Take A to be the algebra over any field $\Bbbk$ of the monoid $\{1,a,b\}$ where $ax=a,bx=b$ for all x and 1 is the identity. Then $aAa\cong \Bbbk$. Let M be the trivial module ($\Bbbk$ with trivial action of the monoid). It is simple and $aM\cong \Bbbk$. But $Aa\otimes_{aAa}aM\cong Aa$ which is two dimensional so not isomorphic to $M$. The kernel of your counit is the submodule generated by $a-b$.

We use this construction here to study simple modules for finite monoids. In that context your map is very rarely injective.

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  • $\begingroup$ Can you give an example? Green's theory of what? $\endgroup$ Commented Sep 17, 2013 at 18:19
  • $\begingroup$ I added an example. Look at chapter 6 of Green's book on polynomial reps of GL(n). $\endgroup$ Commented Sep 17, 2013 at 18:23
  • $\begingroup$ Oh dear, I shouldn't have tried for so much generality. $\endgroup$ Commented Sep 17, 2013 at 18:28
  • $\begingroup$ If eAe is a semisimple algebra then I think you are ok as long as no non-trivial submodule of Ae is annihilated by e. $\endgroup$ Commented Sep 17, 2013 at 18:30
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Here is a counterexample in the world of Hecke algebras of p-adic groups to the claim that there is an equivalence of categories.

Let G=GL_2(F) (F a nonarchimedean local field, O_F its valuation ring), B=Borel, K=GL_2(O_F). Let A be the convolution algebra of locally constant compactly supported functions on G. and let e=characteristic function of K (an idempotent, for the appropriate choice of Haar measure). Call a module M e-nice if M=AeM (equivalently, M is generated by its K-fixed points).

Let triv_G be the trivial representation of G. It can be written as a quotient of a principal series representation $I:=Ind_B^G \delta_B^{-1}$. Then both I and triv_G are e-nice. Applying the functor M to eM from e-nice representations to eAe-modules, the quotient map I to triv_G gets sent to an isomorphism. But the quotient from I to triv_G isn't an isomorphism so we can't have an equivalence of categories.

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In complement to Peter McNamara's answer:

Assume that $A$ is the Hecke algebra of a connected reductive $p$-adic group $G$. Take a pair $(K,\lambda )$ formed of a compact open subgroup $K$ and of an irreducible smooth representation $\lambda$ of $K$. It gives rise to an idempotent $e$ of $A$, so that for each $A$-module $M$, $eM$ is the $\lambda$-isotypic component of $M$. Then your functor is not in general an equivalence of categories. It is an equivalence exactly when $(K,\lambda )$ is a type in the sense of Bushnell and Kutzko:

Bushnell, Colin J.; Kutzko, Philip C. Smooth representations of reductive $p$-adic groups: structure theory via types. Proc. London Math. Soc. (3) 77 (1998), no. 3, 582–634.

For instance, if $K$ is a parahoric subgroup and $\lambda$ the trivial character, then $(K,\lambda )$ is a type if and only if $K$ is an Iwahori subgroup.

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