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

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?

• Why would it not be legitimate ? – Max Jul 20 '19 at 17:38
• @Max I don't know! In the books, Dixmier's lemma is given in the form "M - c Id is not invertibe", not in the simpler form M = c Id. Is this merely a tradition or is there some rationale behind it? – Michael_1812 Jul 20 '19 at 17:48
• Doesn't the conclusion that $M=c Id$ follow the given statement ? Stated e.g. as a corollary ? – Max Jul 20 '19 at 17:51
• @Max : In finite dimension, this is definitely so. In infinite dimensions, it is not that obvious -- please see what I wrote in my question – Michael_1812 Jul 20 '19 at 17:53
• No it obviously follows logically, I meant does it follow in the books ? (your argument is perfectly fine) – Max Jul 20 '19 at 18:53

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