Standard-costandard Hom and Ext-one orthogonality

By [F1] a homomorphism $\mathrm{\Delta }(\mu )\to \mathrm{\nabla }(\nu )$ corresponds to an ${\mathfrak{n}}^{+}$-fixed vector of weight $\mu$ in $\mathrm{\nabla }(\nu )$, and by [F2] the space $\mathrm{\nabla }(\nu {)}_{\mu }$ is $(M(\nu {)}_{\mu }{)}^{\ast }$, a functional being extended by zero off weight $\mu$, with $(x\psi )(m)=\psi (\tau (x)m)$; the fixed condition therefore says exactly that $\psi$ annihilates $\tau ({\mathfrak{n}}^{+})M(\nu )={\mathfrak{n}}^{-}M(\nu )$. By [F1] one has $M(\nu )={\mathfrak{n}}^{-}M(\nu )\oplus \mathbb{C}{v}_{\nu }$, so a functional supported in weight $\mu$ and vanishing on ${\mathfrak{n}}^{-}M(\nu )$ is zero when $\mu \mathrm{\ne }\nu$ (its weight space lies in ${\mathfrak{n}}^{-}M(\nu )$) and is determined by an arbitrary value on $\mathbb{C}{v}_{\nu }$ when $\mu =\nu$; hence the Hom space has dimension $1$ for $\mu =\nu$ and $0$ otherwise.

Let $0\to \mathrm{\nabla }(\nu )\stackrel{}{\to }N\stackrel{}{\to }\mathrm{\Delta }(\mu )\to 0$ be an extension and assume $\nu -\mu \mathrm{\notin }{Q}^{+}\setminus \{0\}$; pull the sequence back along the $\mathfrak{b}$-linear map ${\mathbb{C}}_{\mu }\to \mathrm{\Delta }(\mu )$, $1\mapsto v_{\mu }$, to obtain the $\mathfrak{b}$-exact sequence $0\to \mathrm{\nabla }(\nu )\to N^{\mathrm{\prime }}\to {\mathbb{C}}_{\mu }\to 0$ with $N^{\mathrm{\prime }}=N{\times }_{\mathrm{\Delta }(\mu )}{\mathbb{C}}_{\mu }$. A $\mathfrak{g}$-splitting of the original sequence restricts to a $\mathfrak{b}$-splitting of the pulled-back sequence, and conversely a $\mathfrak{b}$-splitting ${\mathbb{C}}_{\mu }\to N^{\mathrm{\prime }}$, composed with $N^{\mathrm{\prime }}\to N$, is a $\mathfrak{b}$-map ${\mathbb{C}}_{\mu }\to N$ whose image is an ${\mathfrak{n}}^{+}$-fixed vector of weight $\mu$, so it extends to a $\mathfrak{g}$-map $\mathrm{\Delta }(\mu )\to N$ by [F1], and the composite $\mathrm{\Delta }(\mu )\to N\to \mathrm{\Delta }(\mu )$ is a $\mathfrak{g}$-endomorphism of $\mathrm{\Delta }(\mu )$ sending $v_{\mu }$ to $v_{\mu }$, hence the identity; so the original sequence splits exactly when the pulled-back one does. The weights of $N^{\mathrm{\prime }}$ are those of $\mathrm{\nabla }(\nu )$, namely $\nu -Q^{+}$, together with $\mu$; if a weight $\gamma$ of $\mathrm{\nabla }(\nu )$ were strictly above $\mu$, then $\mu <\gamma \le \nu$, so $\mu <\nu$ by [F4], contrary to the case assumption, and no weight of $N^{\mathrm{\prime }}$ is strictly above $\mu$. The quotient map $N^{\mathrm{\prime }}\to {\mathbb{C}}_{\mu }$ is surjective in weight $\mu$, so choose a lift $v$ of its basis vector that is a $\mu$-weight vector. For $x\in {\mathfrak{n}}^{+}$ nonzero of weight $\alpha \in Q^{+}\setminus \{0\}$ the vector $xv$, if nonzero, would be a weight vector of weight $\mu +\alpha >\mu$ in $N^{\mathrm{\prime }}$, which is impossible; hence $v$ is ${\mathfrak{n}}^{+}$-fixed and the pulled-back sequence splits, so the original extension splits.

It remains to treat the case $\nu -\mu \in Q^{+}\setminus \{0\}$, i.e. $\mu <\nu$. Applying the exact contravariant involution $D$ of [F2] to the extension $0\to \mathrm{\nabla }(\nu )\to N\to \mathrm{\Delta }(\mu )\to 0$ gives the extension $0\to D(\mathrm{\Delta }(\mu ))=\mathrm{\nabla }(\mu )\to D(N)\to D(\mathrm{\nabla }(\nu ))=\mathrm{\Delta }(\nu )\to 0$, in which the pair of weights is $(\nu ,\mu )$; since the strict order is transitive and $\mu <\nu$, antisymmetry gives $\mu -\nu \mathrm{\notin }{Q}^{+}\setminus \{0\}$ for the reversed pair, so step 1.2 shows that the dual extension splits. Applying the involution $D$ again, and using $D^2\cong \mathrm{id}$ and exactness, the original extension splits.

Every pair of weights satisfies $\nu -\mu \mathrm{\notin }{Q}^{+}\setminus \{0\}$ or $\mu <\nu$, so steps 1.2 and 2.1 show that every extension of $\mathrm{\Delta }(\mu )$ by $\mathrm{\nabla }(\nu )$ splits; by the Yoneda identification of [F3] this is exactly ${\mathrm{Ext}}_{\mathcal{O}}^1(\mathrm{\Delta }(\mu ),\mathrm{\nabla }(\nu ))=0$. Together with the Hom computation of step 1.1 this proves both assertions of the statement.