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.
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?
Best Answer
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.