The BGG Resolution

For every simple root ${\alpha }_i$ one has $⟨\lambda ,{\alpha }_i^{\vee }⟩\in {\mathbb{Z}}_{\ge 0}$ by dominance. Applying $s_i$ to $2\rho ={\sum }_{\alpha \in {\mathrm{\Phi }}^{+}}\alpha$ and using that $s_i$ permutes the positive roots other than ${\alpha }_i$ while $s_i{\alpha }_i=-{\alpha }_i$ gives $2s_i\rho =2\rho -2{\alpha }_i$, whereas $s_i\rho =\rho -⟨\rho ,{\alpha }_i^{\vee }⟩{\alpha }_i$. Comparing the two expressions gives $⟨\rho ,{\alpha }_i^{\vee }⟩=1$, so $⟨\lambda +\rho ,{\alpha }_i^{\vee }⟩\in {\mathbb{Z}}_{>0}$.

Let $\beta \in {\mathrm{\Phi }}^{+}$ and write $\beta ={\sum }_i{n}_i{\alpha }_i$ with $n_i\in {\mathbb{Z}}_{\ge 0}$, not all zero. Put $c_i:=⟨{\omega }_i,{\beta }^{\vee }⟩=\frac{2({\omega }_i,\beta )}{(\beta ,\beta )}$. Then $c_i\in \mathbb{Z}$ because ${\omega }_i\in P$ and ${\beta }^{\vee }$ is the coroot of the root $\beta$, and $c_i\ge 0$ because $({\omega }_i,\beta )={\sum }_j{n}_j({\omega }_i,{\alpha }_j)=\frac{1}{2}{n}_i({\alpha }_i,{\alpha }_i)\ge 0$ and $(\beta ,\beta )>0$. Since the ${\omega }_j$ are a basis of $E$ and both sides have the same pairing with every ${\omega }_j$, this gives the identity ${\beta }^{\vee }={\sum }_i{c}_i{\alpha }_i^{\vee }$: every positive coroot is a nonnegative integral combination of the simple coroots, and for $\beta >0$ at least one $c_i$ is positive.

Combining the two previous steps, $⟨\lambda +\rho ,{\beta }^{\vee }⟩={\sum }_i{c}_i⟨\lambda +\rho ,{\alpha }_i^{\vee }⟩$ is a nonnegative integer combination in which at least one coefficient is positive and every paired simple coroot contributes at least $1$; hence it lies in ${\mathbb{Z}}_{>0}$. For a negative root $-\beta$ one has $⟨\lambda +\rho ,(-\beta {)}^{\vee }⟩=-⟨\lambda +\rho ,{\beta }^{\vee }⟩<0$, so $\lambda +\rho$ pairs nontrivially with the coroot of every root and is therefore regular: it lies on no root hyperplane.

A regular vector lies in some open chamber, and $W$ acts simply transitively on open chambers, so its stabilizer is trivial. If $w\circ \lambda =w^{\mathrm{\prime }}\circ \lambda$, then $w(\lambda +\rho )=w^{\mathrm{\prime }}(\lambda +\rho )$ and hence $(w^{\mathrm{\prime }-1}w)(\lambda +\rho )=\lambda +\rho$; triviality of the stabilizer gives $w^{\mathrm{\prime }-1}w=1$, that is $w=w^{\mathrm{\prime }}$.

For $w\in W$ and $\beta \in \mathrm{\Phi }$ one has $(w\beta {)}^{\vee }=w({\beta }^{\vee })$ because $w$ preserves the form, and pairing with $w(\lambda +\rho )$ against $w({\beta }^{\vee })$ equals pairing against ${\beta }^{\vee }$ because $w$ is an isometry. Hence $⟨w(\lambda +\rho ),(w\beta {)}^{\vee }⟩=⟨\lambda +\rho ,{\beta }^{\vee }⟩$, and since $w\circ \lambda +\rho =w(\lambda +\rho )$ the second displayed identity is the same statement; the nonnegative-integral-combination statement at a simple coroot, transported along $W$, is the "$W$-invariance" used throughout.

Let $y⊲x$. By [F1] with a one-step chain, $x=ty$ for a root reflection $t$; write $t=s_{\gamma }$ with $\gamma$ a root and replace $\gamma$ by $-\gamma$ if necessary so that $\gamma \in {\mathrm{\Phi }}^{+}$. Since $x=s_{\gamma }y$ and $\mathrm{\ell }(x)=\mathrm{\ell }(y)+1>\mathrm{\ell }(y)$, the criterion in [F2] applied to $w=y$ forbids $y^{-1}\gamma <0$; hence $\beta :=y^{-1}\gamma \in {\mathrm{\Phi }}^{+}$. Conjugation gives $x=y(y^{-1}{s}_{\gamma }y)=y{s}_{y^{-1}\gamma }=y{s}_{\beta }$ with $\beta =y^{-1}\gamma \in {\mathrm{\Phi }}^{+}$.

Suppose $x=y{s}_{\beta }=y{s}_{{\beta }^{\mathrm{\prime }}}$ with $\beta ,{\beta }^{\mathrm{\prime }}\in {\mathrm{\Phi }}^{+}$. Then $s_{\beta }=s_{{\beta }^{\mathrm{\prime }}}$, so ${\beta }^{\mathrm{\prime }}=\pm \beta$ by [F3], and positivity forces ${\beta }^{\mathrm{\prime }}=\beta$. Thus the positive root in step 1.1 is unique, and multiplying $x=y{s}_{\beta }$ on the left by $y^{-1}$ gives $y^{-1}x=s_{\beta }$, so the label is determined by the group elements.

Conversely let $\beta \in {\mathrm{\Phi }}^{+}$ and suppose $\mathrm{\ell }(y{s}_{\beta })=\mathrm{\ell }(y)+1$. Put $t:=y{s}_{\beta }{y}^{-1}=s_{y\beta }$ by conjugation, so $y{s}_{\beta }=ty$ is a one-step saturated reflection chain; by [F1], $y\le y{s}_{\beta }$. If $z$ satisfied $y<z<y{s}_{\beta }$, then by [F1] any saturated chain from $y$ to $y{s}_{\beta }$ through $z$ would have more than one step, so $\mathrm{\ell }(y{s}_{\beta })-\mathrm{\ell }(y)\ge 2$, contradicting the hypothesis. Hence $y{s}_{\beta }$ covers $y$.

Finally, if $\gamma \in {\mathrm{\Phi }}^{+}$ satisfies $y=x{s}_{\gamma }$ and $\mathrm{\ell }(x{s}_{\gamma })=\mathrm{\ell }(x)-1$, then $x=y{s}_{\gamma }$ by multiplying on the right by $s_{\gamma }$, and $\mathrm{\ell }(y{s}_{\gamma })=\mathrm{\ell }(x)=\mathrm{\ell }(y)+1$; step 1.1 applied to the cover $y⊲x$ (whose existence is the hypothesis $y=x{s}_{\gamma }$) gives $\gamma =\beta$. Step 2.1 supplied the uniqueness of $\beta$ from the pair $(x,y)$ alone, so the two characterisations of the label coincide.

We prove by induction on $\mathrm{\ell }(w)$ that $w\circ \lambda \uparrow \lambda$ and that $M(w\circ \lambda )$ embeds in $M(\lambda )$ with image generated by a singular vector of weight $w\circ \lambda$. For $w=1$ the chain is empty and the identity embeds $M(\lambda )$ in itself. If $w\mathrm{\ne }1$, choose a reduced word $w=s_{i}u$ with $\mathrm{\ell }(w)=\mathrm{\ell }(u)+1$; then $w\circ \lambda =s_i\circ (u\circ \lambda )$ and $u\circ \lambda \uparrow \lambda$ by induction. Since $\mathrm{\ell }(s_{i}u)=\mathrm{\ell }(u)+1$, step [F5] gives $u^{-1}{\alpha }_i>0$; hence $⟨u\circ \lambda +\rho ,{\alpha }_i^{\vee }⟩=⟨u(\lambda +\rho ),{\alpha }_i^{\vee }⟩=⟨\lambda +\rho ,u^{-1}{\alpha }_i^{\vee }⟩\in {\mathbb{Z}}_{>0}$ by [F4]. So [F2] provides the one-step chain $u\circ \lambda \succ s_i\circ (u\circ \lambda )=w\circ \lambda$, which concatenated with the inductive chain gives $w\circ \lambda \uparrow \lambda$, and [F1] gives an embedding $M(w\circ \lambda )\hookrightarrow M(u\circ \lambda )$ that we compose with $M(u\circ \lambda )\hookrightarrow M(\lambda )$.

Now let $w\ge w^{\mathrm{\prime }}$. By [F6] fix a saturated chain $w=v_0,\dots ,v_r=w^{\mathrm{\prime }}$ with $v_j=s_{{\alpha }_j}{v}_{j-1}$ and $\mathrm{\ell }(v_j)=\mathrm{\ell }(v_{j-1})-1$. At each step the pairing $⟨v_j\circ \lambda +\rho ,{\alpha }_j^{\vee }⟩=⟨\lambda +\rho ,v_j^{-1}{\alpha }_j^{\vee }⟩$ is a positive integer: $v_j^{-1}{\alpha }_j=(s_{{\alpha }_j}{v}_{j-1}{)}^{-1}{\alpha }_j=v_{j-1}^{-1}{s}_{{\alpha }_j}{\alpha }_j=-v_{j-1}^{-1}{\alpha }_j$, and $\mathrm{\ell }(s_{{\alpha }_j}{v}_{j-1})=\mathrm{\ell }(v_{j-1})-1$ gives $v_{j-1}^{-1}{\alpha }_j<0$ by [F5], so $v_j^{-1}{\alpha }_j>0$. Hence [F1] gives embeddings $M(v_{j-1}\circ \lambda )\hookrightarrow M(v_j\circ \lambda )$ for all $j$, whose composite embeds $M(w\circ \lambda )$ in $M(w^{\mathrm{\prime }}\circ \lambda )$; the pair $(w,w^{\mathrm{\prime }})$ is nonzero in $\mathrm{Hom}$, which is one dimensional by [F3].

By the embeddings constructed in step 1.1, ${\mathrm{Hom}}_{\mathfrak{g}}(M(w\circ \lambda ),M(\lambda ))\mathrm{\ne }0$, of dimension exactly $1$ by [F3]; its image is the submodule generated by the image of the highest weight vector, which by [F7] is the unique-up-to-scalar singular vector of weight $w\circ \lambda$ in $M(\lambda )$. This proves the first two assertions.

Both the composite $M(w\circ \lambda )\hookrightarrow M(w^{\mathrm{\prime }}\circ \lambda )\hookrightarrow M(\lambda )$ and the embedding of step 2.1 are nonzero elements of the one-dimensional space ${\mathrm{Hom}}_{\mathfrak{g}}(M(w\circ \lambda ),M(\lambda ))$; rescaling the chosen embedding $M(w\circ \lambda )\hookrightarrow M(w^{\mathrm{\prime }}\circ \lambda )$ by the reciprocal scalar makes the composite equal to the canonical embedding, so after this normalisation the image of $M(w\circ \lambda )$ lies inside the image of $M(w^{\mathrm{\prime }}\circ \lambda )$ inside $M(\lambda )$, i.e. $M(w\circ \lambda )\subseteq M(w^{\mathrm{\prime }}\circ \lambda )\subseteq M(\lambda )$ for the canonical singular-vector submodules. The stated uniqueness is exactly the one-dimensionality of [F3].

For every $w\in W$, fix an embedding $j_w\text{\hspace{0.17em}⁣}:M(w\circ \lambda )\hookrightarrow M(\lambda )$ with image $S_w$, taking $j_e$ to be the identity. There are only finitely many choices. For every comparable pair $u\ge v$, define ${\iota }_{u\to v}=j_v^{-1}\circ j_u$, where $j_v^{-1}$ is the inverse from $S_v$ to $M(v\circ \lambda )$; [F1] gives $S_u\subseteq S_v$, so this is well-defined and $j_v{\iota }_{u\to v}=j_u$. These are precisely the literal submodule inclusions transported to the abstract Verma copies. For a cover $x⊳y$, the map is nonzero and injective and spans the one-dimensional Hom space by [F2]. Its image is proper: otherwise the two Verma modules would have the same highest weight, contradicting $x\circ \lambda \mathrm{\ne }y\circ \lambda$ by [F3].

For $x⊳m⊳y$, the defining equations give $j_y{\iota }_{m\to y}{\iota }_{x\to m}=j_m{\iota }_{x\to m}=j_x=j_y{\iota }_{x\to y}$. Since $j_y$ is injective, ${\iota }_{m\to y}{\iota }_{x\to m}={\iota }_{x\to y}$. The same argument for $m^{\mathrm{\prime }}$ shows that the two diamond composites are equal as maps, rather than merely proportional.

More generally, for $x\ge y\ge z$ one has $j_z{\iota }_{y\to z}{\iota }_{x\to y}=j_y{\iota }_{x\to y}=j_x=j_z{\iota }_{x\to z}$, so injectivity of $j_z$ proves ${\iota }_{y\to z}{\iota }_{x\to y}={\iota }_{x\to z}$. Iterating this equality gives path independence, including the claimed length-gap-two case. The normalization depends on the chosen $j_w$ on abstract copies; the submodules $S_w$ and their literal inclusions are canonical. Arbitrarily rescaled cover maps need not have equal diamond composites.

Right lifting. Let $s$ be a simple reflection and $x\le y$ with $ys<y$ and $xs>x$. Then $x\le ys$ and $xs\le y$. Indeed, choose a reduced expression $y=s_1\cdots s_q$ ending in $s$, which exists because $ys<y$; by the subword characterisation in [F1], $x$ has a reduced subword expression inside $s_1\cdots s_q$. That subword cannot use the last letter: if it did, then $x=x^{\mathrm{\prime }}s$ with $\mathrm{\ell }(x)=\mathrm{\ell }(x^{\mathrm{\prime }})+1$ and thus $xs=x^{\mathrm{\prime }}$ has length $\mathrm{\ell }(x)-1$, contradicting $xs>x$. Hence $x$ is a reduced subword of $s_1\cdots s_{q-1}$, which is a reduced expression for $ys$, so $x\le ys$. Adjoining the last letter to a reduced subword for $x$ gives a reduced expression of length $\mathrm{\ell }(x)+1=\mathrm{\ell }(xs)$ for $xs$, so $xs\le y$.

Three consequences of step 1.1 are used below. (a) If $x\le y$ and $xs<x$, then $xs\le y$ since $xs<x\le y$. (b) If $x\le y$, $xs<x$ and $ys<y$, then $xs\le ys$: apply (a) to get $xs\le y$, then apply step 1.1 to the pair $xs\le y$, whose right products are $(xs)s=x>xs$ and $ys<y$. (c) If $x\le y$, $xs>x$ and $ys>y$, then $xs\le ys$, which is the order-preservation statement in [F1].

Main claim, case $ys>y$. Let $y<x$ with $\mathrm{\ell }(x)=\mathrm{\ell }(y)+2$. By [F2] choose a simple reflection $s$ with $xs<x$, and put $x_0=xs$, $y_0=ys$; then $\mathrm{\ell }(x_0)=\mathrm{\ell }(x)-1=\mathrm{\ell }(y)+1$ and $\mathrm{\ell }(y_0)=\mathrm{\ell }(y)\pm 1$. Assume first that $ys>y$, so $y_0>y$. Step 1.1 applied to $y\le x$ gives $y\le x_0$ and $y_0\le x$. Hence $y_0⊳y$ and $y_0\le x$ with $\mathrm{\ell }(x)=\mathrm{\ell }(y_0)+1$ give $x⊳y_0$; similarly $x_0⊳y$ and $x⊳x_0$. Thus $y_0$ and $x_0$ are two distinct elements between $y$ and $x$. Conversely, let $m$ be any element with $y<m<x$. If $ms>m$, step 1.1 applied to $m\le x$ gives $m\le x_0$, and $\mathrm{\ell }(m)=\mathrm{\ell }(y)+1=\mathrm{\ell }(x_0)$ forces $m=x_0$. If $ms<m$, step 1.1 applied to $y\le m$ gives $y_0\le m$, and $\mathrm{\ell }(y_0)=\mathrm{\ell }(y)+1=\mathrm{\ell }(m)$ forces $m=y_0$. Hence the interval is exactly $\{y,y_0,x_0,x\}$.

Main claim, case $ys<y$; reduction. Now assume $ys<y$, so $y_0=ys$. By consequence (b) of step 2.1 applied to $y\le x$, we get $y_0\le x_0$, and $\mathrm{\ell }(x_0)-\mathrm{\ell }(y_0)=\mathrm{\ell }(x)-\mathrm{\ell }(y)=2$; since $y_0\mathrm{\ne }{x}_0$ we may apply the induction hypothesis (strong induction on $\mathrm{\ell }(x)$) to the pair $y_0<x_0$: its interval has exactly two elements $n_1,n_2$, with $x_0⊳n_i⊳y_0$. This is the induction step: we analyse the elements between $y$ and $x$. Every such $m$ satisfies $ms\mathrm{\ne }m$; if $ms>m$, step 1.1 applied to $m\le x$ gives $m\le x_0$, and $\mathrm{\ell }(m)=\mathrm{\ell }(y)+1=\mathrm{\ell }(x_0)$ gives $m=x_0$. If $ms<m$, then $n:=ms$ satisfies $y_0\le n$ and $n\le x_0$ by the two applications of consequence (b) to $y\le m$ and $m\le x$, and $\mathrm{\ell }(n)=\mathrm{\ell }(m)-1=\mathrm{\ell }(y_0)+1$, so $n\in \{n_1,n_2\}$; moreover $ns=m>n$. Conversely, if $n\in \{n_1,n_2\}$ and $ns>n$, then $m:=ns$ satisfies $y\le m$ and $m\le x$ by the two applications of consequence (c) to $y_0\le n$ and $n\le x_0$, and $\mathrm{\ell }(m)=\mathrm{\ell }(n)+1=\mathrm{\ell }(y)+1$, so $x⊳m⊳y$. Thus the elements between $y$ and $x$ are exactly $x_0$ (if it lies between, i.e. if $y\le x_0$) together with the elements $ns$ for those $n\in \{n_1,n_2\}$ with $ns>n$.

Case $ys<y$ and $y\le x_0$. Then $x_0$ lies between $y$ and $x$. The element $y$ itself is a middle of the interval $y_0<x_0$: indeed $y⊳y_0$ (case hypothesis), $y\le x_0$ and $\mathrm{\ell }(x_0)=\mathrm{\ell }(y)+1$, so $x_0⊳y$. As $y=ys\cdot s=y_{0}s$ does not rise under $s$, at least one of $n_1,n_2$ fails to rise. If the other one, say $n$, also failed to rise, then step 1.1 applied to $y_0\le n$ would give $y=y_{0}s\le n$ with $\mathrm{\ell }(y)=\mathrm{\ell }(n)$, so $y=n$, contradiction. Hence exactly one of $n_1,n_2$ rises, and by step 3.1 the interval between $y$ and $x$ consists of $x_0$ and that one element $ns$: exactly two elements.

Case $ys<y$ and $y\nleqq x_0$. Then $x_0$ is not between $y$ and $x$. If some $n\in \{n_1,n_2\}$ failed to rise, then step 1.1 applied to $y_0\le n$ would give $y=y_{0}s\le n$ with $\mathrm{\ell }(y)=\mathrm{\ell }(n)$, hence $y=n$, and then $y\le x_0$ because $n\le x_0$—contrary to the case hypothesis. Therefore both $n_1,n_2$ rise, and step 3.1 exhibits exactly the two elements $n_{1}s,n_{2}s$ between $y$ and $x$.

The two cases of steps 2.2 and 3.1 (with the sub-cases resolved in steps 4.1 and 4.2) cover all possibilities for $y<x$ with $\mathrm{\ell }(x)=\mathrm{\ell }(y)+2$, and in each the set $\{m:y<m<x\}$ has exactly two elements. The base case of the induction is $\mathrm{\ell }(x)=2$, $\mathrm{\ell }(y)=0$, where $y=1$ and the hypothesis $ys>y$ of step 2.2 holds for every simple $s$, so step 2.2 applies; the induction step uses only the pair $y_0<x_0$ with $\mathrm{\ell }(x_0)=\mathrm{\ell }(x)-1<\mathrm{\ell }(x)$. Since every element strictly between $y$ and $x$ has length $\mathrm{\ell }(y)+1=\mathrm{\ell }(x)-1$ (the chain description of [F1] forces length to increase by one along any saturated chain), each such element is a cover of $y$ and is covered by $x$, and $[y,x]=\{y,m_1,m_2,x\}$.