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Let us consider a Lie algebra $\mathfrak g$ whose dimension is (at most) countable, i.e. it admits a countable set of generators $T_n$ with $n\in\mathbb N$, and let the linear space $V$ be an irreducible representation of $\mathfrak g$.

I would like to prove that if $\phi:V\to V$ is a homomorphism of representations of $\mathfrak g$, then $\phi = \lambda\, \mathrm{id}_V$ for some $\lambda \in \mathbb C$.

To this end, I was suggested to first show that $V$ may attain at most countable dimension. This first step can be carried out as follows: since $V$ is irreducible, given a reference vector $0\neq v \in V$, we may write any element $w\in V$ as a linear combination of vectors of the type $T_{i_1} T_{i_2}\cdots T_{i_n}v$; the latter form a countable basis, implying that $V$ has countable dimension as well.

At this point, I consider $\psi_\lambda \equiv \phi - \lambda\, \mathrm{id}_V$. Since $\phi$ is a homomorphism, so is $\psi_\lambda$, therefore both its kernel and its image are subrepresentations of $V$. But $V$ is irreducible, so $\psi_\lambda$ is either invertible (zero kernel and image equal to $V$) or identically zero (zero image, kernel equal to $V$) and in the latter case $\phi=\lambda\,\mathrm{id}_V$. The thesis is obtained once we conclude that $\psi_\lambda$ cannot be invertible for all $\lambda\in\mathbb C$.

I was shown a proof of this last step where one considers the polynomials in $\phi$ (for instance, see this Math.SE post). If one assumes that all such nonzero polynomials are invertible, then one coucludes that, for any set of $\lambda_k\in\mathbb C$, $$ \sum_{k} \frac{c_k}{\phi-\lambda_k\mathrm{id}_V}=0 $$ has only the trivial solution $c_k=0$, hence that all the $(\phi-\lambda\,\mathrm{id}_V)^{-1}$ are linearly independent; but this in contradiction with the fact that $V$ has at most countable dimension.

However, I was trying a more direct approach. Let us consider a fixed $0\neq v\in V$ and the set $\{\psi_\lambda(v): \lambda \in \mathbb C\}$. Since $V$ is countable dimensional, there are (infinitely many) nonzero solutions $c_k\neq 0$ to the equation $$ \sum_{k} c_k \psi_{\lambda_k}(v)=0\,, $$ that is $$ \sum_{k} c_k \phi(v) = \sum_{k}c_k \lambda_k v\,. $$ Now, if I were able to argue that the $c_k$ may be chosen such that $\sum c_k\neq 0$, I could conclude that $$ \phi(v) = \frac{\sum_{k}c_k \lambda_k}{\sum_{k} c_k} v\,. $$ This would indeed prove the theorem with $\lambda = (\sum c_k \lambda_k)/(\sum c_k)$. However every attempt at showing that $\sum c_k\neq 0$, for some choice, has proven unsuccessful until now.

Is there a way to make this type of argument right?

Brightsun
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  • Are you able to explain the last step in the proof you linked? I don't see how the fact that the $(\phi - \lambda \mathrm{id}_V)^{-1}$ being linearly independent is in contradiction with the fact that $V$ has countable dimension. If $V$ has countable dimension, won't $\mathrm{End}(V)$ be of uncountable dimension? – Joppy Feb 20 '18 at 11:36
  • Let me check if I understand where your claim that there are non-zero solutions to $\sum_k c_k \psi_{\lambda_k} =0$ comes from. Suppose that 3 is my favorite number. I consider all three element subsets $K := {k_1, k_2, k_3} \subset \mathbb{C}$. For each $K$ we get a three element subset $V_K := {\psi_{\lambda_{k_1}}(v), \psi_{\lambda_{k_2}}(v), \psi_{\lambda_{k_3}}(v) }$ of $V$. Now the span of the union of all the $V_K$ has dimension equal to at least the number of sets $K$ for which the elements of $V_K$ are linearly independent (??), so this latter number is at most countable (ctd...) – Vincent Feb 20 '18 at 12:25
  • ...however the total number of choices for the sets $K$ is uncountable so uncountably many $K$s give rise to sets $V_K$ that are linearly dependent and hence give a non-all-zero solution to your equation. Is that what is happening? As you can see from my notation I am not really sure about the obviousness of the step I marked "??". But perhaps it is obvious? Or you had something different in mind? – Vincent Feb 20 '18 at 12:27
  • @Vincent The claim is simply that, since that set of vectors is uncountable, but the whole space is countable, some of them must be linearly dependent. This is also why I left the summation range unspecified. – Brightsun Feb 20 '18 at 12:45
  • @Joppy If $V$ has a countable basis $e_n$, then any endomorphism $A$ is specified by giving its action on such a basis $A e_n = \sum a_{nm} e_m$. A basis of the $a_{nm}$ is given by $e_{ij}$, whose element with $n=i$ and $m=j$ is 1 and the rest zero. This is also a countable basis, so $End(V)$ is countable. – Brightsun Feb 20 '18 at 12:55
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    @Brightsun But you cannot represent the identity map $V \to V$ with a finite sum of those $e_{ij}$, so they cannot form a basis. – Joppy Feb 20 '18 at 13:04
  • @Joppy Mmh, ok if we are not allowed to take countable linear combinations maybe we can argue as follows in this specific case. We have a set of linearly independent homomorphisms $A_k$ (the inverse of $\phi - \lambda_k$) and we want to prove that the $A_k v$ are linearly independent vectors for nonzero $v$. Suppose by contradiction $\sum c_k A_k v=0$ with some nonzero coefficients. Then, by the homomorphism property, either $\sum c_k A_k$ is zero, but this is not the case by linear independence, or $v=0$ which we assumed was not the case. Please tell me what you think. – Brightsun Feb 20 '18 at 13:32
  • Yes, my concern is similar to that of Joppy: the range of $k$ maybe unspecified, but it should still be finite. – Vincent Feb 20 '18 at 13:34
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    @Brightsun I could take an example $V$ being two-dimensional, and the linearly independent collection of operators $A= \begin{pmatrix}1&0\0&0\end{pmatrix}$, $B= \begin{pmatrix}1&1\0&0\end{pmatrix}$. Then if $v = \begin{pmatrix}1 \ 0\end{pmatrix}$, $Av = Bv$, so I don't think your argument works. The whole thing still seems like a reasonable approach, I just don't see how it implies the result. – Joppy Feb 20 '18 at 13:38
  • @Joppy I agree, this is not true in general. In my last comment I gave a proof, I think, which uses the homomorphism property. So, in this respect, you couterexample is relevant only if $A$ and $B$ are homomorphisms of a Lie algebra representation on $\mathbb R^2$. I think this cannot be the case. – Brightsun Feb 20 '18 at 13:42
  • Also I would like to stress the fact that the approach using ${\psi_\lambda (v), \lambda \in \mathbb C}$ and the one using the inverse operators are, as far as I see, to some extent unrelated. – Brightsun Feb 20 '18 at 13:43
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    Right, I deleted my wrong answer, but the process of writing it and discovering what was wrong with it clarified some matters for me. The crux is in your line 'we conclude that $\phi_\lambda$ cannot be invertible for all $\lambda \in \mathbb{C}$.' This is what you want to show. One would expect that you would explore a world where alle the $\psi_\lambda$ are invertible and derive a contradiction, but in the subsequent text the (non)-invertibility of the $\psi_\lambda$ never comes into play, making it unlikely that it will work – Vincent Feb 20 '18 at 15:44
  • The first $\phi_\lambda$ in my previous comment must be $\psi_\lambda$. – Vincent Feb 20 '18 at 15:56
  • @Vincent I agree, this approach is doomed to fail: without a more refined strategy, it could be applied to any endomorphism $f:V\to V$ of a countable dimensional vector space...! Indeed, the $f_\lambda(v) = f(v)-\lambda v$ cannot be linearly independent, hence $\sum_k c_k f(v)=\sum_k c_k \lambda_k v$ with some nonzero $c_k$. But we should not be able to conclude that $f$ is always $\lambda$ id${}_V$! – Brightsun Feb 21 '18 at 13:22
  • Yes indeed! Btw I like the approach you sketch in the question, where you appeal to the other answer: first use the other answer to show that $\phi$ has at least one eigenvector $v$ (using no more info about $\phi$ than that it is a linear map), THEN use the fact that $\phi$ is an intertwiner and $V$ is irreducible to show that every vector in $V$ is then an eigenvector at the same eigenvalue as $v$ was. – Vincent Feb 21 '18 at 13:37

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