Hi: I have a problem as follows. Consider the category $\mathcal{O}$ of $\mathfrak{g}: = \mathfrak{sl}_2(\mathbb{C})$.
Let $r\in \mathbb{C}$ but $r\notin\mathbb{Z}$. Let $s_\alpha$ be the simple reflection of the Weyl group of $\mathfrak{g}$ and so $s_{\alpha} \cdot r = -r -2$. How can we prove that there is no non-trivial extension between simple modules $L(r)$ and $L(s_{\alpha}\cdot r)$ in the category $\mathcal{O}$ of $\mathfrak{g}$? I will glad to see a elementary proof (e.g., don't need to know $L(r)$ is projective in $\mathcal{O}$... etc.) . Thanks very much!
For general semisimple ${\mathfrak g}$, if ${\mathbb Z}\Phi$ is the root lattice, then any ${\mathfrak g}$-module $X$ in ${\mathcal O}$ decomposes as $X = \bigoplus\limits_{C\in {\mathfrak h}^{\ast}/{\mathbb Z}\Phi} X_C$, where $X_C := \bigoplus\limits_{\lambda\in C} X_\lambda$. As vector spaces, this is ok since it is only nested way of writing the weight space decomposition $X=\bigoplus\limits_{\lambda\in{\mathfrak h}^{\ast}} X_\lambda$ (which exists by definition of ${\mathcal O}$), and it is even a decomposition of ${\mathfrak g}$-modules since ${\mathfrak g}={\mathfrak h}\oplus\bigoplus\limits_{\alpha\in\Phi} {\mathfrak g}_{\alpha}$ and $[{\mathfrak g}_{\alpha},X_\lambda]\subset X_{\lambda+\alpha}$.
In case ${\mathfrak g}={\mathfrak s}{\mathfrak l}_2({\mathbb C})$, identifying ${\mathfrak h}\cong{\mathbb C}$ via $\lambda\leftrightarrow\lambda(h_\alpha)$, you have $\lambda - s_\alpha.\lambda = 2(\lambda + 1)$ belong to the same class modulo ${\mathbb Z}\Phi = 2{\mathbb Z}$ if and only if $\lambda\in{\mathbb Z}$. Hence, if $\lambda\notin{\mathbb Z}$, then any module $X$ in an extension $0\to L(s_\alpha.\lambda)\to X\to L(\lambda)\to 0$ will decompose as $X = X_{[\lambda]}\oplus X_{[s_\alpha.\lambda]}$ with $X_{[\lambda]}$ mapping isomorphically to $L(\lambda)$ and $L(s_\alpha.\lambda)$ mapping isomorphically into $X_{[s_\alpha.\lambda]}$.