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 $\,{\mathbb{M}}\,=\,\lambda\,{\mathbb{I}}\,$. $\left.\qquad\right.$


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 $\,V\,$.


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.


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?

  • $\begingroup$ Why would it not be legitimate ? $\endgroup$ Jul 20, 2019 at 17:38
  • $\begingroup$ @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? $\endgroup$ Jul 20, 2019 at 17:48
  • $\begingroup$ Doesn't the conclusion that $M=c Id$ follow the given statement ? Stated e.g. as a corollary ? $\endgroup$ Jul 20, 2019 at 17:51
  • $\begingroup$ @Max : In finite dimension, this is definitely so. In infinite dimensions, it is not that obvious -- please see what I wrote in my question $\endgroup$ Jul 20, 2019 at 17:53
  • 1
    $\begingroup$ No it obviously follows logically, I meant does it follow in the books ? (your argument is perfectly fine) $\endgroup$ Jul 20, 2019 at 18:53

2 Answers 2


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.

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

Thanks to @Max who kindly drew my attention to that lemma.


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 $\,{\mathbb{A}}\;$.


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