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

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?

• i think that one main reason that your question was not asked on MathSE is (a) the lack of relevant tags (I added operator-theory and representation-theory tags a few minutes ago, before you cross-posted), (b) its lack of conciseness. I also didn't try to answer it because it does not really ask whether a mathematical statement is true, but whether some proof that doesn't work really doesn't work. This "proof verification" style is more appropriate on MathSE than here anyway. – YCor Jul 19 '19 at 23:16
• Dear @YCor , I am not in fact sure if my formulation is correct. What is correct is that (M - cI) is not invertible - that is for sure. It, however, remains questionable if I had the right to interpret it in a stronger way, that M = cI . – Michael_1812 Jul 19 '19 at 23:22
• @YCor : Also, thank you for adding the tags. – Michael_1812 Jul 19 '19 at 23:23

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.

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."

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 $$k(X)$$).

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

• Allowing myself a bit of nitpicking, you should not write "for countable dimensions only", but "for dimension less than continuum only", as explained in my answer. Moreover, your answer here is essentially the content of my last line (in parentheses). – Uri Bader Sep 13 '19 at 8:16
• @UriBader Thank you for your comment! You are certainly right: less than continuum would be more exact. The problem is not with your answer, but with us physicists who lack the ability to understand the language of pure math. In the beginning of your answer, you say: "we fix a filed k". What is "filed k"? – Michael_1812 Sep 13 '19 at 16:26
• Michael, I meant to say a field $k$ and $D$ meant to denote a generic $k$-division algebra. Using $k$ and $D$ in these context are to some (but apparently not to all) as hintful as using $\epsilon$ and $N$ when regarding the convergence of a sequence. The use of the letter $k$ for a field is maybe more standard than the use of $F$. I believe it comes from the German term "Körper" introduced by Dedekind. The spelling mistake is mine alone. Also, I should have explained better what $D$ meant - I will edit my answer. – Uri Bader Sep 14 '19 at 7:03