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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??

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  • $\begingroup$ The answer depends on your convention on what $\mathbb{P}V$ is --- it can be either the moduli space of 1-dimensional subspaces in $V$, or the moduli space of 1-dimensional quotient spaces. Both conventions are used, so it is better to specify what do you mean here. $\endgroup$
    – Sasha
    Commented Jul 21, 2014 at 20:33
  • $\begingroup$ I know, but I think the definition of \mathcal{O}(1) changes accordingly. Isn't it?? You should get in both cases the same answer $\endgroup$
    – Giulio
    Commented Jul 21, 2014 at 20:48
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    $\begingroup$ No! In the first case the answer is $V^\vee$, while in the second it is $V$. $\endgroup$
    – Sasha
    Commented Jul 21, 2014 at 21:35
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    $\begingroup$ It is a bit confusing. $\mathbb{P}V$ and $\mathbb{P}V^{\vee}$ are isomorphic; even as polarized varieties with polarization $\mathcal{O}(1)$. But they are NOT isomorphic as $SL(V)$-varieties?? (Moreover, $SL(V)$ does not have characters, so the linearization is unique) $\endgroup$
    – Giulio
    Commented Jul 22, 2014 at 11:28
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    $\begingroup$ As varieties, they are both isomorphic to $\mathbb{P}^n$. I think the group acting on $Proj \, Sym \,V$ is $SL(V^{\vee})$ rather than $SL(V)$. $\endgroup$
    – Giulio
    Commented Jul 22, 2014 at 12:43

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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 $V^\vee$.

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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 $\textrm{SL}(V)$.

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