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## The double commutant theorem

Let $A$ be an abelian group and $T = \{ T_i : A \to A \}$ be a collection of endomorphisms of $A$. The commutant $T'$ of $T$ is the set of all endomorphisms of $A$ commuting with every element of $T$; symbolically,

$\displaystyle T' = \{ S \in \text{End}(A) : TS = ST \}$.

The commutant of $T$ is equal to the commutant of the subring of $\text{End}(A)$ generated by the $T_i$, so we may assume without loss of generality that $T$ is already such a subring. In that case, $T'$ is just the ring of endomorphisms of $A$ as a left $T$-module. The use of the term commutant instead can be thought of as emphasizing the role of $A$ and de-emphasizing the role of $T$.

The assignment $T \mapsto T'$ is a contravariant Galois connection on the lattice of subsets of $\text{End}(A)$, so the double commutant $T \mapsto T''$ may be thought of as a closure operator. Today we will prove a basic but important theorem about this operator.

Warmup: multiplicities

If $G$ is a finite group and $V$ a finite-dimensional complex representation of $G$, then $V$ breaks up into a direct sum

$\displaystyle V = \bigoplus_i n_i V_i$

of irreducible representations $V_i$ with some multiplicities $n_i$. However, this direct sum decomposition is not canonical if the multiplicities $n_i$ are greater than $1$. In the worst case, $G$ may act trivially on $V$, and then $V$ is a direct sum of $\dim V$ copies of the trivial representation. Actually choosing such a direct sum decomposition is equivalent to choosing a basis of $V$.

However, there is an alternate and completely canonical way of describing a representation in terms of its irreducible subrepresentations without choosing a direct sum decomposition as above. As a first hint, note that

$\displaystyle n_i = \dim \text{Hom}_G(V_i, V)$.

This suggests that it might be useful to replace $n_i$ with the vector space $\text{Hom}(V_i, V)$. And, in fact, this turns out to be a great idea: there is a canonical evaluation map

$\displaystyle V_i \otimes \text{Hom}_G(V_i, V) \to V$

whose image is precisely the $V_i$-isotypic component of $V$, and this gives an alternate canonical decomposition of $V$ as

$\displaystyle V = \bigoplus V_i \otimes \text{Hom}_G(V_i, V)$

which does not require making any choices. One can think of $\text{Hom}_G(V_i, V)$ as the multiplicity space associated to $V_i$, the correct canonical replacement for the multiplicity $n_i$.

The idea of the double commutant theorem is to think about what kind of structure multiplicity spaces have. So far we have been using them only as vector spaces, but in fact they are $\text{End}_G(V)$-modules. Note that $\text{End}_G(V)$ is precisely the commutant of the image of $\mathbb{C}[G]$ in $\text{End}(V)$.

Basic properties of commutants

Now that our warmup is done, we list some basic properties of the commutant operation

$\displaystyle \text{End}(A) \supseteq T \mapsto T' \subseteq \text{End}(A)$.

1. $T'$ is a subring of $\text{End}(A)$.
2. $S \subseteq T$ implies $T' \subseteq S'$.
3. $S \subseteq T'$ if and only if $T \subseteq S'$.
4. $S \subseteq S''$ (by 3).
5. $S''' \subseteq S'$ (by 2 and 4).
6. $S' = S'''$ (by 4 and 5).

The second and third properties assert that the commutant establishes a special type of Galois connection. In the language of category theory, the second and third properties assert that the commutant is a contravariant functor from the poset of subsets of $\text{End}(V)$ to itself which happens to be its own adjoint. The remaining properties verify something slightly stronger than the statement that the double commutant is a closure operator: they also verify that the subsets of $\text{End}(A)$ which are their own double commutant are precisely the commutants of other subsets of $\text{End}(A)$.

The double commutant theorem

Theorem (double commutant): Let $A$ be an abelian group and let $T \subseteq \text{End}(A)$ be a subring of $\text{End}(A)$ such that

1. $T$ is a semisimple ring, and
2. $A$ is a finite direct sum of simple $T$-modules.

Then $T = T''$ is its own double commutant. Moreover, $T'$ is also semisimple, and as a $T \otimes T'$-module, $A$ decomposes as a direct sum

$\displaystyle A = \bigoplus_i M_i \otimes_{D_i} N_i$

where $M_i$ is a complete list of the simple $T$-modules, $N_i$ is a complete list of the simple $T'$-modules, and $D_i = \text{End}_T(M_i) = \text{End}_T(N_i)^{op}$. In particular, there is a canonical bijection between simple $T$-modules and simple $T'$-modules.

Proof. Choose a finite direct sum decomposition

$\displaystyle A \cong \bigoplus n_i M_i$

where the $M_i$ are the simple $T$-modules. Since $T$ acts faithfully on $A$, it follows (for example by Artin-Wedderburn) that the multiplicities $n_i$ are all positive. By Schur’s lemma,

$\displaystyle \text{End}_T(A) = T' \cong \prod_i \mathcal{M}_{n_i}(D_i)$

where $D_i = \text{End}_T(M_i)$ are division rings; in particular, $T'$ is semisimple. Now, $T'$ acts on the multiplicity spaces $\text{Hom}(M_i, A)$, and by inspection of the two decompositions above these are precisely the simple $T'$-modules. More precisely, $N_i = \text{Hom}_T(M_i, A)$ is the unique simple
$T'$-module on which the factor $\mathcal{M}_{n_i}(D_i)$ acts nontrivially, and it is in particular an $n_i$-dimensional $D_i^{op}$-vector space (since $D_i$ acts on $M_i$ on the left, it acts on $N_i$ on the right). Hence, as in the finite group case above, the natural map

$\displaystyle \bigoplus_i M_i \otimes_{D_i} \text{Hom}_T(M_i, A) \to A$

is an isomorphism. Writing $N_i = \text{Hom}_T(M_i, A)$, we may now think of the $M_i$ as the multiplicity spaces of the decomposition of $A$ as a $T'$-module, we conclude that $A$ is also the finite direct sum of simple $T'$-modules (with multiplicities given by the dimensions of the $M_i$ as $D_i$-vector spaces), and it follows from here that $T'' = T$ by Artin-Wedderburn. $\Box$

If you don’t like division rings, feel free to assume that $A$ is a finite-dimensional vector space over an algebraically closed field $k$, which case everything above is a $k$-vector space.

Example. Let $G$ be a finite group and $K$ a subgroup, and consider the representation $V = \mathbb{C}[G/K] = \text{Ind}_K^G(1)$ of $G$. The double commutant theorem tells us that $V$ decomposes into a direct sum as

$\displaystyle V = \bigoplus_i V_i \otimes W_i$

where the $V_i$ are irreducible representations of $G$ and the $W_i$ are a complete list of the simple $\text{End}_G(V)$-modules. Understanding $\text{End}_G(V)$ thus gives us a information about the decomposition of $V$ as a $G$-module.

$\text{End}_G(V)$ is one definition of the Hecke algebra $H(G, K)$. It may be described explicitly as spanned by double cosets $KgK, g \in G$, which have a well-defined product by a left coset on the left as a left coset. This construction is morally responsible for many of the Hecke algebras appearing in mathematics by making particular choices of $G$ and $K$ (usually $G$ and $K$ are not finite groups and so one passes from $\mathbb{C}[G/K]$ to a suitable space of functions on $G/K$, but the idea is the same).

### 4 Responses

1. I am having trouble making the final connection in your proof: how do we use A-W with the information we have to deduce that T” = T? I would greatly appreciate a careful explanation.

2. […] of Schur-Weyl duality asserts that the answer is yes. The double commutant theorem plays an important role in the proof and also highlights an important corollary, namely that […]

3. Nice! I like that listing of properties of the commutant, and how 3 follow from contravariant functoriality

$S \subseteq T \implies T' \subseteq S'$

$S \subseteq T' \iff T \subseteq S'$
I guess it’s worth reminding people that these are familiar properties of ‘negation’ of propositions, and ‘complement’ of subsets, even in intuitionistic contexts where $S'' \ne S$.