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If $\{a,b,c\}$ is a basis for the Lie algebra $A_1$ with $[b,a]=2a, [b,c]=-2c$ and $[a,c]=b$. Suppose $V$ is a 13-dimensional $A_1$ module and that $\dim\{v\in V\mid bv=4v\}=\dim\{v\in V\mid bv=5v\}=1$. I want to show that $\dim\{v\in V\mid bv=8v\}=0$.

Taking $v\in \{v\in V\mid bv=8v\}$ then $b(cv)=[b,c]v+c(bv)=-2cv+c8v=6cv$. By the same considerations then $b(c(cv))=4(c(cv))$, so we may conclude $c(cv)\in \{v\in V\mid bv=4v\}$. Now I'm thinking, that if I somehow could conclude then $c(cv)$ is zero in $\{v\in V\mid bv=4v\}$, then $c(cv)=0$ giving that $v=0$, since $c$ is a basis element? However I'm kinda stuck right now, and can't get any further.

It seems pretty easy to solve, by using that $A_1$ can be classified with $sl(2)$ - then the dimension equals the corresponding nullspace. However I'm still curious to solve it without this classification.

njlieta
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    This is what I would do. Suppose that there exists $u\ne 0$ s.t. $bu=8u$. Let $U$ be the submodule of $V$ generated by $u$. Let $W$ denote the submodule of $V$ generated by $w\ne 0$ s.t. $bw=5w$. Show that $\dim U\ge 9$ and $\dim W\ge 6$, but $U\cap W=0$. However, $U+W\subseteq V$ and $\dim V=13<9+6$. – Batominovski Feb 25 '20 at 12:55
  • So you get a contradiction with the dimensions resulting in $u=0$, since the set of $v$ st. $bv=5v$ have dimension 1, so $w$ can surely be choosen nonzero, meaning $u=0$? – njlieta Feb 25 '20 at 13:34
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    Yes, that's the idea. – Batominovski Feb 25 '20 at 13:38
  • What is the difference of $W$ and ${v\in V | bv=5v}$? Why would'nt they have the same dimension? – njlieta Feb 25 '20 at 13:54
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    $W$ necessarily contains ${v\in V|bv=5v}$. However, since $W$ is a submodule (not just a subspace), it may contain other elements (and indeed it does). – Batominovski Feb 25 '20 at 13:58

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This follows from the representation theory of $\mathfrak{sl}_2$, covered in any book on Lie algebras and representation theory; cf. https://math.stackexchange.com/a/3344350/96384 and https://math.stackexchange.com/a/3288482/96384. The element you call $b$ is often called $h$ in those sources.

Namely, since the weight spaces to the weights $4$ and $5$ are not zero, they generate submodules of dimension at least $5$ resp. $6$ respectively; if there were a non-zero weight vector to the weight $8$, then there would also be weight spaces to the weights $-8$ as well as $\pm 6$, i.e. at least four more dimensions. But $4+11 > 13$.

Torsten Schoeneberg
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