Understanding weight spaces of weight module from its composition factors? If you have a weight module $V$, and you understand the weight spaces of the composition factors of $V$, does this determine the structure of the weight spaces of $V$ itself?
I'm confused, because I'm reading a passage where $V$ is a weight $\mathfrak{sl}_2(\mathbb{C})$-module with only two distinct composition factors (which may have repeated multiplicity) such that in both composition factors, the action of $f\in\mathfrak{sl}_2$ induces an isomorphism of the weight spaces with weight $\lambda+2i$, $\lambda\in\mathbb{C}$, $i\in\mathbb{N}$, with the weight space of weight $\lambda$. It then says that the same is true for the weight spaces of $V$ itself. 
I don't understand how one can lift such properties of the weight spaces of the composition factors back to a property of the weight spaces of the original module. 
 A: Let $\mathfrak{g}$ be a semisimple Lie algebra with Cartan subalgebra $\mathfrak{h}$. Recall that if $V=\bigoplus_{\lambda\in\mathfrak{h}^*}V_\lambda$ is a weight module, define the formal character of $V$ to be
$$
\mathrm{ch}V=\sum_{\lambda\in \mathfrak{h}^*}(\dim V_\lambda)e^{\lambda},
$$
where $e^\lambda$ is a formal symbol that obeys exponent laws. Since $\mathfrak{h}$ acts semi-simply on any representation, one can show that
$$\mathrm{ch}(V/W)=\mathrm{ch}V-\mathrm{ch}W$$
for any submodule $W$ (indeed, just show that $W_\mu=V_\mu\cap W$).
Now, given any chain complex
$$
0=V_0\to V_1\to\cdots\to V_n=V
$$
it follows that 
$$
\mathrm{ch}V=\sum_{i=1}^n\mathrm{ch}(V_i/V_{i-1})$$
(hint: expand a telescoping sum).
This proves that you can recover the weight spaces of $V$ from its composition factors.
Edit: Assume $V$ has two composition factors $X$ and $Y$ with multiplicity $m$ and $n$, respectively. Then
$$\mathrm{ch}V = m\,\mathrm{ch}X + n\,\mathrm{ch}Y.$$
Now, assuming that $f^i$ induced an isomorphism $X_\lambda\cong X_{\lambda+2i}$ and $Y_\lambda\cong Y_{\lambda+2i}$. Then
$$
\dim X_\lambda= \dim X_{\lambda+2i}\;\;\; \mbox{and} \;\;\; \dim Y_\lambda=\dim Y_{\lambda+2i}
$$
and, therefore 
$$\dim V_\lambda=m\dim X_\lambda+n\dim Y_\lambda=m\dim X_{\lambda+2i}+n\dim Y_{\lambda+2i}=\dim V_{\lambda+2i}.$$
Finally, we will show that the multiplication map $$f^i:V_\lambda\to V_{\lambda+2i}$$ is injective. Indeed, suppose $v\in V_\lambda$ and $f^iv=0$. 
Let $$V=W_0\supset W_1\supset W_2\supset\cdots\supset W_n=0$$ be a composition series for $V$. Suppose we hav shown that $v\in W_k$ (this is certainly true when $k=0$). Then, since $W_k/W_{k+1}$ is simple, $$f^i:(W_k/W_{k+1})_\lambda\to (W_k/W_{k+1})_{\lambda+2i}$$ is an isomorphism. Since
$$f^i(v+W_{k+1})=(f^iv)+W_{k+1}=0+W_{k+1}$$
we deduce that $v\in W_{k+1}$. It now follows that $v\in W_n=0$, so $v=0$.
