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Let $\mathfrak{g}$ be the nonabelian $2$-dimensional complex Lie algebra. It can be generated by two independent vectors $e_1,e_2$ such that $[e_1,e_2]=e_1$. Thus, $\mathfrak{g}$ is solvable and it follows that its irreducible representations are $1$-dimensional (I guess this is Lie's theorem). Also we may look at the adjoint representation: it is $2$-dimensional but it can be easily shown that it is indecomposable, however. So, is it true that for each $n\in \mathbb{N}$ there is a indecomposable representation of $\mathfrak{g}$ of dimension $n$? I will be grateful if you share the examples.

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Start with the $n$ dimensional irrep of $\mathfrak{sl}_2$ and restrict it to its Borel subalgebra of upper triangular matrices (a non-abelian $2$ dimensional Lie algebra). It is indecomposable because any proper submodule does not contain the highest weight vector. This produces one indecomposable in every dimension.

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Supplementing Stephen's excellent answer (+1) with concrete matrices to cater for the possibility that this exercise was encountered prior to the theory of representations of $\mathfrak{sl}_2$. Let us fix a dimension $n$. Let $A$ be the diagonal matrix with entries $-(n-1)/2$, $-(n-3)/2$, $\ldots$, $(n-3)/2$,$(n-1)/2$ in an arithmetic progression from $-(n-1)/2$ to $(n-1)/2$ (so step $=1$). Let $B=(b_{ij})$ be the $n\times n$-matrix with $b_{i,i+1}=1$ for all $i=1,2,\ldots,n-1$, and zeros elsewhere. Then it is not too hard to check that $BA-AB=B$, so by mapping $e_1\mapsto B$ and $e_2\mapsto A$ we get an $n$-dimensional representation $V$ of your Lie algebra. This is indecomposable, because no proper submodule contains the vector $v=(0,0,0,\ldots,0,1)^T$. For if a submodule does contain $v$, then it also contains $Bv=(0,0,\ldots,0,1,0)^T$, $B^2v$ and all the way down to $B^{n-1}v=(1,0,0,\ldots,0)$. These vectors span all of $V$, so the submodule wasn't a proper one. In light of Stephen's answer the reason for this is immediate. The element $e_1$ (or the matrix $B$) acts as a ladder operator, and this is just the ladder in action - $n$ rungs to it this time.

As an example consider the case $n=5$. Then the matrices are $$ e_1\mapsto\left(\begin{array}{ccccc}0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&1\\ 0&0&0&0&0\end{array}\right)\quad\text{and}\quad e_2\mapsto\left(\begin{array}{rrccc}-2&0&0&0&0\\0&-1&0&0&0\\0&0&0&0&0\\0&0&0&1&0\\ 0&0&0&0&2\end{array}\right). $$

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  • $\begingroup$ Thanks, Jyrki (+1 back at you). $\endgroup$ – Stephen Sep 16 '14 at 19:45
  • $\begingroup$ There were two typos. Now corrected. $\endgroup$ – Jyrki Lahtonen Sep 16 '14 at 19:47
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    $\begingroup$ Sorry for not looking so carefully at the matrices! These are the same representations my answer produces. I don't know off the top of my head how many more there are (I kind of suspect there are lots and lots). $\endgroup$ – Stephen Sep 16 '14 at 19:51
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Here is an answer derived from the answer of Stephen.

First, write the equivalent relation $[ - e_2, e_1]= e_1$ and denote $e = - e_2$, $f= e_1$ to get $[e,f] = f$.

For $\lambda$ a parameter ( take it $\text{mod}\ \mathbb{Z}$) consider the infinite dimensional vector space $V_\lambda$ spanned by $v_i$, $i \in \mathbb{Z}$. It is easy to check that the formulas below give an action of our Lie algebra on $V_{\lambda}$:

\begin{eqnarray} e (v_i) &=& (\lambda + i ) \cdot v_i \\ f( v_i) &=& v_{i+1} \end{eqnarray}

For $n$ in $\mathbb{Z}$ the subspace $V_{\lambda,N}$ spanned by $v_i$, $i \ge N$ is subrepresentation of $V_{\lambda}$. Let's show that there are no other subrepresentations. Indeed, let $W\ne 0$ an invariant subspace of $V_{\lambda}$. Let $v \in W$, $v = \sum_{i \ge i_0} a_i v_i$, $a_{i_0} \ne 0$. Since all the $v_i$ are eigenvectors of $e$ for distinct eigenvalues and $W$ is invariant under $e$ it follows that $v_{i_0} \in W$. Now raise the indexes to conclude $W \supset V_{\lambda, i_0}$.

The quotient representations $V_{\lambda, N} / V_{\lambda, N + n}$ ( a subquotient of $V_{\lambda}$ ) has dimension $n$ and all its subrepresentations are $V_{\lambda, N+m} / V_{\lambda, N + n}$ for $0\le m \le n$. Hence the subrepresentations of $V_{\lambda, N} / V_{\lambda, N + n}$ form a complete flag and hence no two can decompose the space $V_{\lambda, N} / V_{\lambda, N + n}$ in a nontrivial way.

These $V_{\lambda, N} / V_{\lambda, N + n}$, let's denote them by $W_{\lambda+n-1,n}$ should give you all the finite dimensional representations of your algebra. The dimension is of $W_{\mu, n}$ is $n$ and the highest weight of $e$ is $\mu$.

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