Dixmier's lemma as a generalisation of Schur's first lemma

Question

I thought that this question is simple, and asked it at Stackexchange. To my surprise, no one was able to answer it there. Now have to elevate it to Overflow.

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What mathematicians call Schur's lemma is known to physicists as Schur's second lemma:

• An intertwiner of two irreducible representations of a group is either zero or isomorphism.

It is valid for all dimensionalities -- finite, countable, uncountable.

The following statement is referred to in physics books as Schur's first lemma:

• An intertwiner from an irreducible representation to itself is a scalar times the identity operator.

In finite dimensions, the latter lemma easily follows from the former one:

• Let $\,{\mathbb{A}}\,$ be the said irreducible representation, with an element $\,g\,$ of the group $\,G\,$ mapped to an operator $\,{\mathbb{A}}_g\,$. If $\,{\mathbb{M}}\,$ is an intertwiner, i.e., if $~{\mathbb{M}}\,{\mathbb{A}}_g\,=\,{\mathbb{A}}_g\,{\mathbb{M}}~$ for $\,\forall\, g\in G\,$, then $~({\mathbb{M}}\,-\,\lambda\,{\mathbb{I}})\,{\mathbb{A}}_g\,=\,{\mathbb{A}}_g\,({\mathbb{M}}\,-\,\lambda\,{\mathbb{I}})\,$, where $\,\lambda\,$ is any eigenvalue of $\,{\mathbb{M}}\,$, while $\,{\mathbb{I}}\,$ is the identity matrix. Schur's Second Lemma says that the matrix $\,({\mathbb{M}}\,-\,\lambda\,{\mathbb{I}})\,$ is either zero or nonsingular. The latter option, however, is excluded because the eigenvector corresponding to $\,\lambda\,$ is mapped by the operator $~({\mathbb{M}}\,-\,\lambda\,{\mathbb{I}})\,{\mathbb{A}}_g\,$ to zero. So this is a zero operator, and $\,{\mathbb{M}}\,=\,\lambda\,{\mathbb{I}}\,$. $\left.\qquad\right.$

${\mathbb{QED}}$

This proof works only for finite dimensions, because it requires a nonzero $\,{\mathbb{M}}\,$ to possess at least one nonzero eigenvalue.

A generalisation of Schur's first lemma to countable dimensions is Dixmier's lemma.

I present its formulation for group representations, because this is the language understandable to a physicist.

• Suppose that $\,V\,$ is a countable-dimension vector space over $\,{\mathbb{C}}\,$ and that $\,{\mathbb{A}}\,$ is a group representation acting irreducibly on $\,V\,$. If the intertwiner $\,{\mathbb{M}}\in\,$Hom$\,_C(V, V )\,$ commutes with the action of $\,{\mathbb{A}}\,$, then there exists a number $\,c\in{\mathbb{C}}\,$ for which $\,{\mathbb{M}}\,-\,c\,{\mathbb{I}}\,$ is not invertible on the space $\,V\,$.

Proof

To employ reductio ad absurdum, start with an assumption that the map $\,{\mathbb{M}}\,-\,c\,{\mathbb{I}}\,$ is invertible for $\,\forall c\in {\mathbb{C}}\,$. Then, for any non-zero polynomial $$\,P(x)\,=\,(x-p_1)\,(x-p_2)\,.\,.\,.\,(x-p_N)\,\;,$$ invertible is the map $$\,P({\mathbb{M}})\,=\,({\mathbb{M}}\,-\,p_1\,{\mathbb{I}})\,({\mathbb{M}}\,-\,p_2\,{\mathbb{I}})\,.\,.\,.\,({\mathbb{M}}\,-\,p_N\,{\mathbb{I}})\,\;,$$ because the composition of invertible maps is invertible.

Consider all rational functions $\,R(x)\,=\,P(x)/Q(x)\,$, with $\,P(x)\,$ and $\,Q(x)\,$ complex-valued polynomials in a complex variable $\,x\,$. Defined on $\,{\mathbb{C}}\,$ except an unspecified finite subset (allowed to vary with each function), they constitute a space $\,{\mathbb{C}}(x)\,$ over $\,{\mathbb{C}}\,$. While the space $\,{\mathbb{C}}[x]\,$ of polynomials is of countable dimensions over $\,{\mathbb{C}}\,$, the space $\,{\mathbb{C}}(x)\,$ of rational functions is of uncountable dimensions.

For any $\,R(x)\,=\,P(x)/Q(x)\,$, there exists a map $\,R({\mathbb{M}})\,=\,P({\mathbb{M}})/Q({\mathbb{M}})\,$. Hence a map $$ {\mathbb{C}}(x)\,\longrightarrow\,\mbox{Hom}_C\,(V,\,V)\,\;. $$

As we saw above, our initial assumption that the map $\,{\mathbb{M}}\,-\,c\,{\mathbb{I}}\,$ is invertible for $\,\forall c\in {\mathbb{C}}\,$ implies that all nonzero polynomials in $\,{\mathbb{M}}\,$ are invertible. For an invertible polynomial $\,Q({\mathbb{M}})\,$, invertible is the map $\,1/Q({\mathbb{M}})\,$. So the maps $\,R({\mathbb{M}})\,=\,\left(\,Q({\mathbb{M}})\,\right)^{-1}\,P({\mathbb{M}})\,$ are compositions of invertible transformations, and thus are invertible. Stated alternatively, if $\,v\in V\,$ is non-zero, then $\,R({\mathbb{M}})\, v\,=\,0\,$ necessitates $\,P({\mathbb{M}})v\,=\,0\,$.

This, in its turn, can be true only if $\,P\,$ is the zero polynomial: $\;P(x)\,=\,0\;$ and, therefore, $\,R\,$ is the zero function, $\,R(x)\,=\,0\,$. In other words, only one element of the space $\,{\mathbb{C}}(x)\,$, the function $\,R(x)\,=\,0\,$, is mapped to the zero element $\,R({\mathbb{M}})\,=\,0\,$ of the space $\,\mbox{Hom}_C\,(V,\,V)\,$. Hence the map $\,{\mathbb{C}}(x)\,\longrightarrow\,\mbox{Hom}_C\,(V,\,V)\,$ is injection -- which implies that the dimensionality of $\,\mbox{Hom}_C\,(V,\,V)\,$ is uncountable, because such is the dimensionality of ${\mathbb{C}}(x)\,$. This, however, is incompatible with the assumption that $\,V\,$ is of countable dimensions.

${\mathbb{QED}}$

Now, my question.

We have proven that, for some $\,c\in{\mathbb{C}}\,$, the operator $\,{\mathbb{M}}\,-\,c\,{\mathbb{I}}\,$ is not invertible.

Can we now use Schur's second lemma, to state that $\,{\mathbb{M}}\,$ is a scalar multiple of the identity operator?

In finite dimensions, the noninvertibility of $\,{\mathbb{M}}\,-\,c\,{\mathbb{I}}\,$ is equivalent to $\,c\,$ being an eigenvalue of the matrix $\,{\mathbb{M}}\,$. However, in infinite dimensions this is not necessarily so. When $\,{\mathbb{M}}\,-\,c\,{\mathbb{I}}\,$ is noninvertible (while the linear operator $\,{\mathbb{M}}\,$ is bounded), $\,c\,$ is said to belong to the spectrum of $\,{\mathbb{M}}\,$ -- which does not necessitate it being an eigenvalue. An operator on an infinite-dimensional space may have a nonempty spectrum and, at the same time, lack eigenvalues.

Despite this circumstance, will it be legitimate to say that, if

• $\exists\,c\in{\mathbb{C}}\,$ for which $\,{\mathbb{M}}\,-\,c\,{\mathbb{I}}\,$ is not invertible,

• $\,{\mathbb{M}}\,-\,c\,{\mathbb{I}}\,$ is an intertwiner of an irreducible representation to itself,

• Schur's second lemma works in all dimensions,

then $\,{\mathbb{M}}\,-\,c\,{\mathbb{I}}\,=\,0\,$ and $\,{\mathbb{M}}\,$ is proportional to the identity operator?

Answer 1

Yes, it indeed is correct that Dixmier's lemma and Schur's lemma, together, entail the stronger statement that $\,{\mathbb{M}}\,-\,c\,{\mathbb{I}}\,=\,0\,$ and, therefore, $\,{\mathbb{M}}\,$ is proportional to the identity operator.

See Lemma 99 in https://arxiv.org/abs/1212.2578 , where that story follows N. R. Wallach's "Real Reductive Groups. 1."

Answer 2

If $M-c$ is not invertible, then its kernel is non-zero or its image is smaller than $A$. But $A$ is irreducible. This means any proper submodule is zero. This forces $M-c=0$.

BTW, for finite-dimensional Lie algebras it works over $\overline{\mathbb{Q}}$ as well. It is known as Quillen Lemma. A good discussion when it works can be found in Noncommutative Noetherian Rings by McConnell and Robson.

Answer 3

I'm writing this just to give context, as others already answered the question. In the discussion below we fix a field $k$ of cardinality $|k|$.

The following version of Schur's lemma is very general, also trivial to prove: For an irreducible $k$-representation of a group $G$ the $k$-algebra of intertwiners is a $k$-division algebra.

By a $k$-division algebra we mean, as usual, a unital $k$-algebra in which every non-zero element is invertible and $k$ is central in it.

In fact, every $k$-division algebra $D$ could be seen as the algebra of intertwiners of an irreducible $k$-representation for some group $G$ - just take $G=D^*$ and consider its left regular representation on $D$.

Note that $\dim k(x)\geq |k|$, as the set $\{(x-\alpha)^{-1}\mid \alpha\in k\}$ is linearly independent (in fact an equality holds for infinite $k$). By fixing $d\in D-k$ and considering the evaluation map $k(x)\to D$, $x\mapsto d$ we get

Lemma: If $k$ is algebraically closed and $D$ is a non-trivial $k$-division algebra then $\dim(D)\geq \dim k(x) \geq |k|$.

If $G\to \text{GL}(V)$ is an irreducible representation with $D=\text{End}(V)^G$ then, fixing $0\neq v\in V$, the map $D\to V$, $d\mapsto d(v)$ is clearly injective, thus $\dim(D)\leq\dim(V)$.

It follows from the discussion above that if $k$ is algebraically closed and $\dim(V)<|k|$ then $\text{End}(V)^G=k$ (and the cardinality condition is strict by considering the regular representation of $k(X)$).

Answer 4

As an interesting aside, it should be added that Dixmier's generalisation of Schur's First Lemma works for countable dimensions only.

In uncountable dimensions, intertwiners from a representation to itself are not warranted to be scalar multiples of the identity. This can be illustrated by a counterexample kindly offered to me by Jacob Tsimerman.

Let the uncountable-dimensional space $\,V\,$ be implemented by the field $\,{\mathbb{C}}(x)\,$ of rational functions in $\,x\in{\mathbb{C}}\,$. Let the role of a group representation $\,{\mathbb{A}}\,$ acting on this space be played by the same field $\,{\mathbb{C}}(x)\,$. A rational function acts on a rational function by multiplication, to render another rational function. Then any nonzero element of $\,V\,$ generates all of $\,V\,$, wherefore $\,V\,$ is irreducible. Now, take $\,{\mathbb{M}}\,$ to be any non-zero integer power of $\,x\,$, e.g., $\,{\mathbb{M}}\,=\,x\,$. Not being a scalar multiple of the identity, this map commutes with the action of any operator $\,{\mathbb{A}}\;$.