\section{Borel-Weil Theorem and Theorem of highest weight (notes by Allechar Serrano Lopez)} 

Our main goal is to construct a highest weight $\lambda$ irreducible representation of $G$ for all $\lambda\in X^\bullet(T)_+$ and to show that these are all the irreducible representations of $G$. Our approach is via the Borel-Weil theorem. Recall that, given a representation $V$ of a connected reductive group $G$ (containing a fixed Borel subgroup $B$ and a fixed maximal torus $T \leq B$) and a choice of a highest weight vector $v \in V$, we get a $G$-equivariant map
$$
G/B \rightarrow \mathbb{P}(V)
$$
$$
v \mapsto (gv).
$$
The idea of the Borel-Weil theorem is to run this argument in reverse: we construct $G$-equivariant line bundles on $G/B$ and study the associated maps (see above) to projective spaces. We begin with some preliminaries on highest weights.

\begin{lem}\label{hwdom}
If $V$ is a highest weight $\lambda$ representation, then $\lambda$ is dominant. 
\end{lem}

\begin{proof}
Recall that $\text{ch}(V)$ is $W-$invariant. So if $V_\lambda\neq 0$, then $V_{s_\alpha(\lambda)}\neq 0$ for all $s_\alpha(\lambda)\in W$. By assumption, $\lambda\geqslant s_\alpha(\lambda)$, i.e., $\langle \lambda,  \alpha^\vee\rangle\alpha \in \mathbb{Z}_{\geqslant 0}\Phi^+$, so $\langle \lambda, \alpha^\vee\rangle\geqslant 0$ for all $\alpha\in\Phi^+$. 
\end{proof}

\begin{defn}
Let $V$ be a representation of $G$. Say that $v\in V$ is a highest weight vector of highest weight $\lambda$ if $Nv=v$ (or $\mathfrak{n}v=0$, where $\mathfrak{n}=\text{Lie}(N)$) and $T$ acts on $V$ by $\lambda$.
\end{defn}

\begin{exmpl}
Consider $G=\text{SL}_2$ and $V=\text{Sym}^3 \mathbb{C}^2\oplus \text{Sym}^4\mathbb{C}^2$, where $\langle e_1, e_2\rangle$ and $\langle f_1, f_2\rangle$ generate $\mathbb{C}$ in the first and second summands, respectively. We have that $e_1^3\in \text{Sym}^3 \mathbb{C}^2$ and $f_1^4\in \text{Sym}^4 \mathbb{C}^2$. Then $e_1^3$ and $f_1^4$ are the highest weight vectors of weights $ \bp t & 0 \\ 0 & t^{-1} \ep \mapsto t^3, t^4$, respectively; but, $V$ is not a highest weight representation in the sense just defined.
\end{exmpl}

\begin{lem}
If $V$ is a highest weight representation of weight $\lambda$, then any $v\in V_\lambda\smallsetminus 0$ is a highest weight vector of weight $\lambda$.
\end{lem}
\begin{proof}
We need to show that $\mathfrak{n}v=0$. More generally, if $\alpha\in\Phi$, and if $X_\alpha\in\mathfrak{g}_\alpha$, then $X_\alpha\cdot V_\lambda\subset V_{\lambda+\alpha}\cup 0$: for all $t\in T$
\[
t\cdot X_\alpha v=\text{Ad}(t)X_\alpha\cdot tv=\alpha(t)X_\alpha\lambda(t)v=(\alpha+\lambda)(t)X_{\alpha} v
\]
In particular, in our highest weight $\lambda$ representation $V$, $V_{\alpha+\lambda}=0$ for all $\alpha\in\Phi^+$, so $\mathfrak{n}v=0$, i.e., $v\in V_\lambda$ is a highest weight vector.
\end{proof}

\begin{lem}
If $V$ is any non-zero $G-$representation, the $V$ contains a highest weight vector with dominant weight.
\end{lem}
\begin{proof}
Let $\lambda$ be any weight in $V$ that is maximal for the partial order $\leq$, and let $v$ be any non-zero element of the weight space $V_{\lambda}$. Since $\lambda$ is maximal, the argument of the previous lemma shows $v$ is a highest weight vector. The argument of Lemma \ref{hwdom} (i.e., $W$-invariance of the character) then shows $\lambda$ is dominant.
\end{proof}
Combined with semi-simplicity of $G$-representations, this lemma implies:
\begin{cor}
If $\dim V^N\leqslant 1$, then $V$ is irreducible.
\end{cor}

That concludes the preliminaries. Now we turn to the construction of highest weight representations; the Borel-Weil theorem finds these in the cohomology of $G$-equivariant line bundles on the flag variety $G/B$, so we start by reviewing a few facts about linear systems.

Recall that for any variety $X$ over $k$, we have the following bijection:

\begin{align*}
\{\text{maps } X\xrightarrow{f}\mathbb{P}(V)\} &\longleftrightarrow 
\left\{
\begin{tabular}{c} 
\text{base-point free linear system}
$(\mathcal{L}, V, i:V^*\rightarrow\Gamma(X, \mathcal{L}))$
\end{tabular} 
\right\}/\cong \\
f&\longmapsto 
\left(
\begin{tabular}{c} 
$f^*(\mathcal{O}(1)), i:V^* \rightarrow \Gamma(X, \mathcal{L})$
\text{ with the property that for all }\\$x\in X$\text{ there exists }$\alpha\in V^*$ \text{such that }$i(\alpha)|_x\neq 0$
\end{tabular} 
\right)
\end{align*}

Recall that $\mathbb{P}(V)$ has a tautological line bundle $\mathcal{O}(-1)$ whose fiber over a line $l$ is just the line $l$; the dual line bundle $\mathcal{O}(1)$ has fiber $l^*= \mathrm{Hom}_k(l, k)$ over $l$, and $\Gamma(\mathbb{P}(V),\mathcal{O}(1))$ is canonically isomorphic to $V^*$, via the map sending $\alpha \in V^*$ to the section $l \mapsto \alpha|_{l}$.

We want to upgrade this correspondence to a $G-$equivariant version:

\begin{cor}
If $X$ is a space with $G-$action and $V$ is a $G-$representation, then there is a bijection
\begin{align*}
\{G-\text{equivariant maps } &X\rightarrow \mathbb{P}(V)\}\leftrightarrow \\
&\{G-\text{equivariant base point free linear systems $(\mathcal{L}, V, \iota)$ on }X\}/\cong
\end{align*}
\end{cor}
Here is the precise definition of a $G$-equivariant linear system:
\begin{defn}
A $G-$equivariant linear system on $X$ is $(\mathcal{L}, V, i:V^*\rightarrow\Gamma(X, \mathcal{L}))$ such that

\begin{enumerate}
\item[1)] $\mathcal{L}$ is a $G-$equivariant line bundle on X. This means that $\mathcal{L}$ is a line bundle on $X$ equipped with a $G-$action compatible with the action on $X$ under the projection $\mathcal{L} \to X$, and for all $x\in X$ and all $g\in G$, the map $g:\mathcal{L}_x\rightarrow \mathcal{L}_{g\cdot x}$ is linear.
\item[2)] $V$ is a $G-$representation.
\item[3)] $i$ is a map of $G-$representations (see below).
\end{enumerate}
\end{defn}

In part (3) of the above definition, we define the action of $G$ on $\Gamma(X, \mathcal{L})$ as follows. For a section $s:X \to \mathcal{L}$, and $g\in G$, define $g\cdot s\in \Gamma(X, \mathcal{L})$ by setting $(g\cdot s)(x)=gs(g^{-1}x)$ .

\begin{lem}
The correspondence

\begin{align*}
\left\{\begin{tabular}{c}
$G-$\text{equivariant line}\\
\text{bundles on }$G/B$
\end{tabular}
\right\}&\longleftrightarrow 
\left\{
\begin{tabular}{c} 
$1-$\text{dimensional}\\
$B-$\text{representations}
\end{tabular} 
\right\} \\
\mathcal{L}&\longmapsto \mathcal{L}_B
\end{align*}

induces a bijection

\[
\left\{G-\text{equivariant line}
\text{bundles on }G/B
\right\}/ \cong\xrightarrow{\sim}\text{Hom}(B,\mathbb{G}_m)
\]
\end{lem}

\begin{proof}
We will construct the inverse map. Given $\chi: B\rightarrow\mathbb{G}_m$, we form the associated line bundle

\[G\times^B\mathbb{A}^1=\{(g, x)\in G\times\mathbb{A}^1\}/(gb, x)\sim(g, \chi(b)x)\rightarrow G/B\]

where the equivalence relation is defined for all $g\in G, x\in \mathbb{A}^1, b\in B$. Analytically, this is clearly a line bundle. In the Zariski topology, we use the fact that the map $G\rightarrow G/B$ Zariski-locally has sections (this follows from the description of the big Bruhat cell $C(w_0)=N\times B$).

\end{proof}

\begin{lem}
$\text{Hom}(B, \mathbb{G}_m)=\text{Hom}(T, \mathbb{G}_m)=X^\bullet(T)$.
\end{lem}
\proof
Any homomorphism $B \rightarrow \mathbb{G}_m$ is trivial on $N$ (we have shown this before; it reduces to showing that $\op{Hom}(\mathbb{G}_a, \mathbb{G}_m)=0$), hence factors as a homomorphism
$$
T \longiso B/N \rightarrow \mathbb{G}_m.
$$
\endproof

\begin{defn}
For $\lambda\in X^\bullet(T)$, define $L(\lambda)=G\times^B k(-\lambda)$ to be the line bundle on $G/B$ associated to $-\lambda$.
\end{defn}

We can now state the main theorem:

\begin{thm}[Borel-Weil] Let $G$ be a connected, reductive group with $B\leq G$ a Borel subgroup and $T \leq B$ a maximal torus. Then:
\begin{itemize}
\item[1)] If $\lambda \in {\cgroup T}_+$, then $V(\lambda):=\op{H}^0(G/B, L(\lambda))^*$ is an irreducible $G$-representation with highest weight $\lambda$.
\item[2)] The $V(\lambda)$ constructed in this way exhaust all the irreducible representations of $G$ up to isomorphism.
\item[3)] If $\lambda \in \cgroup T \backslash {\cgroup T}_+$, then $\op{H}^0(G/B, L(\lambda))=0$.
\item[4)] $\dim V(\lambda)_\lambda =1$.
\end{itemize}
\end{thm}

\begin{exmpl}
This example will clarify why we defined $L(\lambda)$ via the character $-\lambda$. Let $G=SL_2(k)$, with $B=\begin{pmatrix} * & * \\ 0 & * \\ \end{pmatrix}=\op{Stab} \langle \begin{pmatrix} 1 \\ 0 \end{pmatrix} \rangle$. Then the tautological line bundle

\begin{center}
\begin{tikzcd}
{\mathcal{O}(-1)} \arrow{d} \\
SL_2(k)/B = \mathbb{P}^1
\end{tikzcd}
\end{center}
is $SL_2(k)$-equivariant. In our classification, this corresponds to the $B$-representation on $\mathcal{O}(-1)_B= \langle \begin{pmatrix} 1 \\ 0 \end{pmatrix} \rangle$. We have
$$
\begin{pmatrix} t & * \\ 0 &  t^{-1} \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 0 \end{pmatrix} = t \begin{pmatrix} 1 \\ 0 \end{pmatrix},
$$
but the tautological bundle doesn't give us global sections; we need its inverse. The bundles with sections are $\mathcal{O}(n)$ for $n\geq 0$: the corresponding characters in $\cgroup T$ are
$$
\begin{pmatrix} t & 0 \\ 0 & t^{-1} \end{pmatrix} \mapsto t^{-n}.
$$
\end{exmpl}

\begin{cor}
$R(G) \longiso R(T)^W.$
\end{cor}

\section{Proof of the Theorem (notes by Sean McAffee)}

We begin the proof of Borel-Weil with the following proposition:

\begin{prop}\text{}
\vspace{5mm}
\begin{itemize}
\item[1)] For any irreducible representation $V$ of $G$, there exists a $\lambda \in {\cgroup T}_+$ such that $V \cong V(\lambda)$.
\item[2)] For any $\lambda \in {\cgroup T}$, if $V(\lambda) \neq 0$, then $V(\lambda)$ is irreducible.
\end{itemize}

\end{prop}

\begin{proof}

Any irreducible representation $V$ has some highest weight vector $v$. As discussed at the beginning of this section, $v$ induces a $G$-equivariant map $f: G/B \rightarrow \mathbb{P}(V)$. This in turn induces a nonzero $G$-equivariant map
$$
i: V^* \rightarrow \Gamma(G/B, f^*(\mathcal{O}(1)).
$$
This must be injective, since $V$ is irreducible. Dualizing, then, we have $V(\lambda) \twoheadrightarrow V$, so 1) will follow from 2). 

To prove 2), first recall that a $G$-representation $W$ is irreducible if $\dim W^N \leq 1$. We will check this for $W=V(\lambda)$. 

\begin{lem}
We have that $\dim \Gamma(G/B, L(\lambda))^N \leq 1$, and if nonzero, then the $T$-action is given by the character $-w_0\cdot \lambda$ (here $w_0$ denotes the longest element of the Weyl group).
\end{lem}

\begin{proof}
The restriction map
$$
\Gamma(G/B, L(\lambda))^N \rightarrow \Gamma(C(w_0)/B,L(\lambda))^N\rightarrow L(\lambda)_{w_0}
$$
is injective, since $C(w_0)$ is dense in $G/B$ and $Nw_0 \longiso C(w_0)/B$. Thus $\Gamma(G/B,L(\lambda))^N$ injects into a line, i.e. its dimension is $\leq 1$.

To compute the $T$-action, note that this restriction map is $T$-equivariant and the following claim: for any $w \in W$, and any $\lambda \in \cgroup T$, $T$ acts on $L(\lambda_w)$ by $-w \cdot \lambda$. Indeed, write $L(\lambda)_w=\left\{ (\dot{w}, x) \vert x \in k \right\}$, where $\dot{w}$ is a fixed lift of $w$  (an element of $ W=N_G(T)/T)$) to $N_G(T)$. Then, $t$ acts by
$$
t \cdot (\dot{w},x)=(t\dot{w},x)=(\dot{w}\dot{w}^{-1}t\dot{w},x) \sim (\dot{w},-\lambda(\dot{w}^{-1}t\dot{w})x)=(-w\lambda)(t)\cdot x.
$$
This completes the proof of the lemma, and hence of part 2) of the proposition.


\end{proof}

\end{proof}

To complete the proof of Borel-Weil, we need to show that $\Gamma(G/B, L(\lambda))$ is nonzero if and only if $\lambda$ is dominant. First, we show that $\Gamma \neq 0$ implies $\lambda$ is dominant. We do this by analyzing the possible weights of $\Gamma$. We have seen  that restriction to $C(w_o)$ gives a $T$-equivariant injection
$$
\Gamma(G/B,L(\lambda)) \hookrightarrow \Gamma(Y(w_0),L(\lambda)).
$$

Let $\lambda^*$ denote $-w_0 \cdot \lambda$, and let $\mathfrak{n}$ denote $\op{Lie}(N)$. We have that $T$ acts on $\mathfrak{n}$ via $\op{Ad}$, hence $T$ acts on the dual $\mathfrak{n}^*$. 

\begin{lem}
$$
\Gamma(Y(w_0),L(\lambda)) \cong_T k(\lambda^*) \otimes \op{Sym}^\bullet(\mathfrak{n}^*).
$$

(Here $\cong_T$ denotes isomorphism as $T$-representations). In particular, this is a highest weight representation with highest weight $\lambda^*$.
\end{lem}

\begin{proof}
Recall the isomorphism
\begin{align*}
N & \longiso Y(w_0) \\
n & \mapsto n\dot{w}_0.
\end{align*}

Under this map, the left $T$-action on $Y(w_0)$ intertwines $\op{Ad}(T)$ on $N$:

$$
tn\dot{w}_0=tnt^{-1}t\dot{w}_0=tnt^{-1}\dot{w}_0t' \sim \op{Ad}(t)n\dot{w}_0.
$$

This extends to an isomorphism between $N \times L(\lambda)_{w_0}$ and $L(\lambda) \vert_{Y(w_0)}$:
\begin{center}

\begin{tikzcd}
(n,s) \arrow[r, maps to] & n \cdot s \\
{N \times L(\lambda)_{w_0}} \arrow{r}{\sim} \arrow{d}  & {L(\lambda) \vert_{Y(w_0)}} \arrow{d} \\
N \arrow{r}{\sim} & {Y(w_0)}.
\end{tikzcd}

\end{center}

This isomorphism is $T$-equivariant; to be precise, the left-hand side is given the adjoint representation (on the $N$-component) and left $T$-multiplication (on the fiber $L(\lambda)_{w_0}$), while the right-hand side has its given left $T$-multiplication. To check this equivariance claim, we let $s=(\dot{w}_0,x) \in \op{L}(\lambda)_{w_0}, n \in N, t \in T$; we have

\begin{align*}
t\cdot (n, s) & = t \cdot (n, (\dot{w}_0,x))\\
& = (t \cdot n,  t\cdot(\dot{w}_0,x))\\
& = (\op{Ad}(t)(n), (t\dot{w}_0,x)) \\
& = (\op{Ad}(t)(n), (\dot{w}_0(\dot{w}_0^{-1}t\dot{w}_0),x)\\
&= (\op{Ad}(t)(n), (\dot{w}_0, -\lambda(\dot{w}_0^{-1}t\dot{w}_0)x).
\end{align*}


Via this $T$-equivariant trivialization of $L(\lambda) \vert_{Y(w_0)}$, we can compute

$$
\Gamma(Y(w_0),L(\lambda)) \cong_T \Gamma(N,N \times L(\lambda)_{w_0}) \cong_T \mathcal{O}(N) \otimes k(\lambda^*) \cong \op{Sym}^\bullet (\mathfrak{n}^*) \otimes k(\lambda^*).
$$

This completes the proof of the lemma, and the proposition follows.

\end{proof}

The main step in the proof of Borel-Weil is to show that for any $\lambda \in {\cgroup T}_+$  we have $\op{H}^0(G/B,L(\lambda)) \neq 0$. To show this, our strategy will be to construct a highest weight vector of weight $\lambda^*$. We will do this in two steps:
\begin{itemize}
\item[Step 1:] Construct the highest weight vector over the ``big cell'' $Y(w_0)$.
\item[Step 2:] Show that this section extends to all of $G/B$.
\end{itemize}

First, we remark that a section $s$ on the line bundle 

\begin{center}
\begin{tikzcd}
{G \times_B k(-\lambda)} \arrow{d} \\
G/B
\end{tikzcd}
\end{center}

can be viewed as a map $gB \mapsto (g, f(g))$ for some function $f:G \rightarrow k$. Since $gB=gbB$, we must have $(g,f(g))=(gb,f(gb))$ for all $g \in G, b \in B$. Furthermore, the equivalence relation on $G \times_B k(-\lambda)$ requires
$$
(gb, f(gh)) \sim (g, -\lambda(b)f(gb)).
$$
Thus, we can think of sections as being determined by a function $f:G \rightarrow k$ such that
$$
f(gb)=\lambda(b)f(g), \forall g\in G, b \in B.
$$
Under this correspondence, the $G$-action on a section $s$ translates into the left regular action on a function $f$ with the above properties. 





We have that $N \times T \times N$ and $C(w_0)=Nw_0B$ are isomorphic as varieties under the map $(n_1,t,n_2) \mapsto n_1w_0tn_2$; we define a function $f$ on $C(w_0)$ by

$$
f(n_1w_0tn_2):=\lambda(t).
$$

We claim that $f$ is a highest weight vector of weight $\lambda^*:=-w_0\lambda$. Indeed, $N\cdot f =f$ since $f$ is left $N$-invariant by the left regular action, and we have
\begin{align*}
(t_1\cdot f)(n_1\dot{w}_0tn_2) &=f(t_1^{-1}n_1\dot{w}_0tn_2) \\
& = f((t_1^{-1}n_1t_1)t_1^{-1}\dot{w}_0tn_2) \\
& =f(t^{-1}_1\dot{w}_0tn_2) \\
& =f(\dot{w}_0(\dot{w}_0^{-1}t_1^{-1}\dot{w}_0)\cdot tn_2) \\
& =\lambda(\dot{w}_0^{-1}t_1^{-1}\dot{w}_0t) \\
& =(-w_0\lambda)(t_1)\lambda(t).
\end{align*}

This completes Step 1. Now, for the second (and final) step, we need to extend the section defined above to all of $G/B$. To do this, we first extend to a section on an open subset $U \subset G/B$ such that the complement of $U$ has codimension $\geq 2$. Then, we use the analytic continuation principle: for a line bundle $\mathcal{L}$ on an irreducible normal variety $X$, the restriction map $\Gamma(X, \mathcal{L}) \rightarrow \Gamma(U,\mathcal{L})$ is surjective when the codimension of $X-U$ is $\geq 2$. 

Given $f$ as above, we extend $f$ from $C(w_0)=N\dot{w}_0B$ to
$$
 C(w_0) \cup \bigcup\limits_{\alpha \in \Delta} s_\alpha N\dot{w}_0B.
$$
Note that each $s_\alpha N \dot{w}_0 B$ is open, but contains $C(s_\alpha\dot{w}_0)$, and in this way the union under consideration contains all Bruhat cells of codimension one.

We leave it as an exercise to check that any element of $s_{\alpha}N \dot{w}_0 B$ can be written in the form $n_1 s_\alpha u_\alpha(x)\dot{w}_0 t n_2$, where $n_1, n_2 \in N$ and $x \in \mathbb{G}_a$ (here $u_{\alpha}$ is the root group homomorphism associated to $\alpha$). We will use the following calculation in $SL_2$ to show that $f$ is defined on such an element with $x \neq 0 \in k$:

\begin{align*}
s_\alpha u_\alpha(x) & =\psi_\alpha \left( \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 1 & x \\ 0 & 1 \end{pmatrix} \right) \\
& =\begin{pmatrix} 1 & -x^{-1} \\ 0 & 1 \end{pmatrix} \begin{pmatrix} -x^{-1} & 0 \\ 0 & -x \end{pmatrix} \begin{pmatrix} 1 & 0 \\ x^{-1} & 1 \end{pmatrix} \\
& = u_\alpha (-x^{-1}) \alpha^\vee (-x^{-1}) u_{-\alpha} (x^{-1}).\\
\end{align*}

Also oberve the identity

\begin{align*}
u_{-\alpha}(x^{-1})\dot{w}_0t &= \begin{pmatrix} 1 & 0 \\ x^{-1} & 1 \end{pmatrix} \begin{pmatrix} t^{-1} & 0 \\ 0 & t \end{pmatrix} \\
& = \begin{pmatrix} t^{-1} & 0 \\ x^{-1}t ^{-1} & t \end{pmatrix} \\
& = \begin{pmatrix} t ^{-1} & 0 \\ 0 & t \end{pmatrix} \begin{pmatrix} 1 & 0 \\ x^{-1} t^{-2} & 1 \end{pmatrix} \\
& = \dot{w}_0 t u_{-\alpha}(x^{-1} t^{-2}).
\end{align*}

Thus, for $x \neq 0$, we have
\begin{align*}
n_1s_\alpha u_\alpha(x)\dot{w}_0tn_2 & = n_1u_\alpha(-x^{-1})\alpha^\vee (-x^{-1})u_{-\alpha}(x^{-1})\dot{w}_0tn_2 \\
&= n_1u_\alpha(-x^{-1})\alpha^\vee (-x^{-1})\dot{w}_0tu_{-\alpha}(x^{-1}t^{-2})n_2 \\
&= \underset{\in N}{\Big[ n_1u_\alpha(-x^{-1}) \Big]}\alpha^\vee (-x^{-1})(\dot{w}_0 t)\dot{w}_0 \underset{\in N}{\Big[ u_\alpha( x^{-1}t^{-2})n_2 \Big]}. \\
&= [ \in N ] \alpha^\vee (-x^{-1})(\dot{w}_0 t)\dot{w}_0 [\in N] \\
&= [ \in N ] \dot{w}_0 (\dot{w}_0 \alpha^\vee ) ( -x^{-1} ) t [\in N] \\
\end{align*}
We have that $f$ is already defined here for  $(\dot{w}_0 \alpha^\vee ) ( -x^{-1} ) t$ and takes value $\lambda(\dot{w}_0 \alpha^\vee (-x^{-1}) t)$, thus

\begin{align*}
 & =(-x^{-1})^{\langle \lambda, \dot{w}_0 \alpha^\vee \rangle} \lambda(t) \text{ for $x \neq 0$. }\\
\end{align*}

It is here and only here that we finally use the fact that $\lambda$ is dominant; this implies that $\langle \lambda, \dot{w}_0\alpha^\vee \rangle \leq 0$. Thus the expression above extends to $x=0$, since it involves a nonnegative power of $x$. So, we have shown we can extend $f$ to $s_\alpha N \dot{w}_0 B$ for any $\alpha \in \Delta$. Therefore, by the analytic continuation principle described earlier, we can extend $f$ to all of $G/B$. This completes the proof of Borel-Weil.