Example of linearization for GIT Take a vector space $V$ (finite dimensional, over the complex numbers), let $G=SL(V)$. The group $G$ acts on $\mathbb{P}V$ and we can linearize its action to an action on the line bundle $\mathcal{O}(1)$. This gives an action on $H^0(\mathbb{P}V,\mathcal{O}(1))$. My question is
as $G$-module, is $H^0(\mathbb{P}V,\mathcal{O}(1))$ isomorphic to $V$ or to  $V^{\vee}$?? and why??
 A: By definition $\mathbb{P} V = \mathrm{Proj}\ \mathrm{Sym}(V^\vee)$. Quasi-coherent sheaves on the $\mathrm{Proj}$ are identified with graded modules over the graded ring (modulo torsion). Under this correspondence the sheaf $\mathcal{O}(1)$ goes to the graded module $\mathrm{Sym}(V^\vee)[1]$ (the ring shifted down by 1). The functor of global sections corresponds to taking the degree 0 part of the module, which is just $V^\vee$.
A: The vector space $H^0(\mathbb{P}V,\mathcal{O}(1))$ can be identified with $V^{\vee}$. This is because elements of $H^0(\mathbb{P}V,\mathcal{O}(1))$ can be identified with linear functionals on $V$ (just think of the case where $V=\mathbb{C}^{n+1}$, where $H^0(\mathbb{P}V,\mathcal{O}(1))$ is the collection of linear homogeneous polynomials on $\mathbb{C}^{n+1}$). The action of $\textrm{SL}(V)$ on $H^0(\mathbb{P}V,\mathcal{O}(1))$ is just the natural action on $V^{\vee}$.
More generally, the linearization on $\mathcal{O}(1)$ gives a linearization on $\mathcal{O}(d)$ for any $d$. We may identify $H^0(\mathbb{P}V,\mathcal{O}(d))$ with the symmetric power $S^d(V^{\vee})$ together with its natural action of $\textrm{SL}(V)$.
