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 <title>Borel-Bott-Weil theorem</title>
 <name>BorelBottWeilTheorem</name>
 <created>2003-08-14 16:48:00</created>
 <modified>2004-03-20 13:17:58</modified>
 <type>Theorem</type>
 <creator id="2727" name="mathcam"/>
 <author id="2760" name="yark"/>
 <author id="988" name="bwebste"/>
 <classification>
	<category scheme="msc" code="14M15"/>
 </classification>
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 <content>\PMlinkescapeword{states}

Let $G$ be a semisimple Lie group, and $\lambda$ be an integral weight for that group. $\lambda$ naturally defines a one-dimensional representation $C_\lambda$ of the Borel subgroup $B$ of $G$, by simply pulling back the representation on the maximal torus $T=B/U$ where $U$ is the unipotent radical of $G$. Since we can think of the projection map $\pi\colon G\to G/B$ as a \PMlinkname{principle $B$-bundle}{PrincipleBundle}, to each $C_\lambda$, we get an associated fiber bundle $\L_\lambda$ on $G/B$, which is obviously a line bundle. Identifying $\L_\lambda$ with its sheaf of holomorphic sections, we consider the sheaf cohomology groups $H^i(\L_\lambda)$. Realizing $\g$, the Lie algebra of $G$, as vector fields on $G/B$, we see that $\g$ acts on the sections of $\L_\lambda$ over any open set, and so we get an action on cohomology groups. This integrates to an action of $G$, which on $H^0(\L_\lambda)$ is simply the obvious action of the group.

The Borel-Bott-Weil theorem states the following: if $(\lambda+\rho,\alpha)=0$ for any simple root $\alpha$ of $\g$, then $$H^i(\L_\lambda)=0$$ for all $i$, where $\rho$ is half the sum of all the positive roots. Otherwise, let $w\in W$, the Weyl group of $G$, be the unique element such that $w(\lambda+\rho)$ is dominant (i.e. $(w(\lambda+\rho),\alpha)&gt;0$ for all simple roots $\alpha$).
Then $$H^{\ell(w)}(\L_\lambda)\cong V_\lambda$$ where $V_\lambda$ is the unique irreducible representation of highest weight $\lambda$, and $H^i(\L_\lambda)=0$ for all other $i$. In particular, if $\lambda$ is already dominant, then $\Gamma(\L_\lambda)\cong V_\lambda$, and the higher cohomology of $\L_\lambda$ vanishes.

If $\lambda$ is dominant, than $\L_\lambda$ is generated by global sections, and thus determines a map $$m_\lambda\colon G/B\to\mathbb{P}\left(\Gamma(\L_\lambda)\right).$$ This map is an obvious one, which takes the coset of $B$ to the highest weight vector $v_0$ of $V_\lambda$. This can be extended by equivariance since $B$ fixes $v_0$. This provides an alternate description of $\L_\lambda$.

For example, consider $G=\mathrm{SL}_2\C$. $G/B$ is $\C P^1$, the Riemann sphere, and an integral weight is specified simply by an integer $n$, and $\rho=1$. The line bundle $\L_n$ is simply $\O(n)$, whose sections are the homogeneous polynomials of degree $n$. This gives us in one stroke the representation theory of $\mathrm{SL}_2\C$: $\Gamma(\O(1))$ is the standard representation, and $\Gamma(\O(n))$ is its $n$th symmetric power. We even have
a unified decription of the action of the Lie algebra, derived from its realization as vector fields on $\C P^1$: if $H,X,Y$ are the standard generators of $\mathfrak{sl}_2\C$, then
\begin{align*}
H&amp;=x\frac{d}{dx}-y\frac{d}{dy}\\
X&amp;=x\frac{d}{dy}\\
Y&amp;=y\frac{d}{dx}\\
\end{align*}</content>
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