Representations of the Lie algebra $\mathfrak{sl}(2,\mathbb{C})$ In Corollary 7.2 of http://math.uchicago.edu/~may/REU2012/REUPapers/Bosshardt.pdf, why is the set of weights an unbroken string? I understand we get a finite number of weights by looking at the invariant subspace $$W = \bigoplus\limits_{k \in \mathbb{Z}} V_{\lambda+2k}$$ where $\lambda$ is a fixed weight, but I don't understand why it is unbroken.
I mean, why can't we have something like:
\begin{align*} 
   V_{\lambda-2m_1} & \neq 0 \\
   V_{\lambda} &= 0 \\
   V_{\lambda+2m_2} & \neq 0
\end{align*}
where $m_1,m_2 \geq 1$?
 A: As Mariano says, we assume that the representation is irreducible.  If $V_\lambda = 0$ then $V = A \oplus B$ where
$$A = \sum_{\lambda_1 > 0}V_{\lambda - 2\lambda_1}$$
and
$$B = \sum_{\lambda_1 > 0}V_{\lambda + 2\lambda_1}.$$
Observe that this is certainly a direct sum of vector spaces.  Each space is a sum of weight spaces so it's invariant under the action of $h$.  The actions of $x$ and $y$ act on weight vectors by sending them to weight vectors of weight exactly $2$ higher (in the case of $x$) or $2$ lower (in the case of $y$).  Thus it's obvious that $A$ is closed under the action of $y$ and $B$ is closed under the action of $x$.  Finally note that $A$ is closed under the action of $x$ because $xV_{\lambda - 2\lambda_1} = V_{\lambda - 2(\lambda_1 - 1)}$ and either $\lambda_1 - 1 > 0$, making $V_{\lambda - 2(\lambda_1 - 1)} \subseteq A$ or $\lambda_1 - 1 = 0$, making $V_{\lambda - 2(\lambda_1 - 1)} = V_\lambda = 0$.  The same argument shows that $B$ is closed under the action of $y$.
But now, this is worse than irreducible, this is decomposable.
