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An important fact in the theory of central simple algebras is the double centralizer theorem, which says: if $k$ is a field, $A$ is a $k$-algebra, $V$ is a faithful semisimple $A$-algebra, then $C(C(A)) = A$, where $C(A)$ is the centralizer of $A$ in $End_k(V)$.

Taking the centralizer is an inclusion-reversing operation. If we consider the algebra $M_n(k)$ of $n\times n$ matrices over a field $k$, identfying $k$ with the scalar matrices and $\Delta$ with the diagonal matrices, we get $C(k)$ = $M_n(k)$, $C(M_n(k)) = k$, and $C(\Delta) = \Delta$.

(1) This suggests that the centralizer operation is something like a duality between $k$-subalgebras. Is there any truth to this?

(2) Will there always be a subalgebra like $\Delta$ which is its own centralizer?

I'm not exactly sure what conditions to put on $k, A$, and $V$, so I'm open to flexibility on that.

(3) Does a double centralizer result hold in the context of groups? (i.e. $C_G(C_G(H)) = H$?). If not in general, for what groups does this hold?

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1) The connection you're looking for is that $C(-)$ is an anti-tone Galois connection (with itself.) This is indeed a "type of duality."

2) I'm not sure in general. The approach using anti-tone Galois connections certainly doesn't always work that way. For example, if you take the operation to be set complementation within a nonempty set $Y$, then $Y\setminus X = C(X)\neq X$ for any $X$.

On the other hand, if $A$ is the annihilator operator on the lattice of ideals of a commutative ring, then the lattice of all $A(I)$ might or might not have a fixed point. Example: $\mathbb R[x]/(x^2)$ it has the fixed point $A((x))=(x)$, but in $\mathbb R[x]/(x^3)$ it does not. This leads me to believe that the answer to your question depends somewhat on the algebra. You've picked a particularly nice type of algebra (a matrix algebra over a field.) If you chose, instead, $M_n(\mathbb H)$, then $\Delta \nsubseteq C(\Delta)$. But of course there might still be another subring that does the trick.

3) Because of the properties of $CC(-)$ considered as a closure operator (as described in the link above.) In fact for any nonempty set $X$, $C(X)=C(C(C(X))$. Proof: We know that $C^2(X)\supseteq X$ for any $X$. Applying $C$ on both sides you get $C(C^2(X))\subseteq C(X)$. On the other hand $C(X)\subseteq C^2(C(X))$. $C^3=C$ holds with any Galois closure operator, but in general you may have $C^2(X)\supsetneq X$ for some subsets. Anyhow, we've learned that $C$ acts like an involution on the set of all $C(X)$, whenever $C$ is a closure operator like this.

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  • $\begingroup$ Ok, so the centralizer operator is an anti-tone Galois connection both in the case of algebras over a field and in the case of groups? But I don't see how that statement for (2) answers the question of a fixed point of the operator. $\endgroup$ Apr 11, 2017 at 18:55
  • $\begingroup$ @festering Sorry, I think you're right: I was thinking of $C(X)$ being a fixed point of $C^2(X)$ for every $X$. In that case, $C$ is acting like an involution on the sets of the form $C(X)$. I will see if I can adjust that. $\endgroup$
    – rschwieb
    Apr 11, 2017 at 19:59
  • $\begingroup$ @festering In other words, my head was more in the realm of 3). I adjusted things, and you can let me know what else might be interesting to change. $\endgroup$
    – rschwieb
    Apr 11, 2017 at 20:13
  • $\begingroup$ Thanks, this mostly answers my questions. If you know where I can get any more information on fixed points of such a Galois connection, that would be of interest. $\endgroup$ Apr 12, 2017 at 11:49
  • $\begingroup$ @festering I will keep an eye out. I'd also like to know $\endgroup$
    – rschwieb
    Apr 12, 2017 at 12:57

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