I thought that this question is simple, and [asked it at Stackexchange](https://math.stackexchange.com/questions/3297283/dixmiers-lemma-as-a-generalisation-of-schurs-first-lemma). 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?



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


