Projective GIT from Linearized Line Bundles — Examples

The trivial linearization. Let $L_0$ be the trivial linearization of $L={\mathcal{O}}_X$, so that $G$ acts trivially on the total space $\mathbb{C}$. Then every global section of every $L_0^{\otimes n}$ is invariant and equal to a constant, and the section $1\in \mathrm{\Gamma }(X,{\mathcal{O}}_X)$ is nonzero at $p$; hence $p$ is semistable and $X^{ss}(L_0)=X$. The stabilizer $G_p=G$ is not finite, so $p$ is not stable and $X^s(L_0)=\mathrm{\varnothing }$. The invariant section ring is $R(X,L_0{)}^G=\mathbb{C}[t]$, so $X\mathrm{/}\text{\hspace{0.17em}⁣}{\mathrm{/}}_{L_0}G=\mathrm{Proj}\mathbb{C}[t]=X$.

The twisted linearization. Let $L_{\chi }$ be the twist of $L_0$ by the character $\chi (t)=t$. By [F2] the induced action on the one-dimensional space $\mathrm{\Gamma }(X,L^{\otimes n})=\mathbb{C}$ is multiplication by ${\chi }^n$, so a nonzero invariant section of $L_{\chi }^{\otimes n}$ would require ${\chi }^n=1$, which fails for every $n\ge 1$ since $\chi (2{)}^n=2^n\mathrm{\ne }1$. Hence there is no nonzero invariant section of positive degree, $X^{ss}(L_{\chi })=\mathrm{\varnothing }$, and the invariant section ring is $\mathbb{C}$ in degree zero, so $X\mathrm{/}\text{\hspace{0.17em}⁣}{\mathrm{/}}_{L_{\chi }}G=\mathrm{Proj}\mathbb{C}=\mathrm{\varnothing }$.

Conclusion. The two linearizations $L_0$ and $L_{\chi }$ have the same underlying ample invertible sheaf ${\mathcal{O}}_X$ up to isomorphism, the same group $G$ and the same projective action on the same variety $X$, yet they give the nonempty semistable locus $X$ and the empty semistable locus $\mathrm{\varnothing }$, and the one-point quotient $X$ and the empty quotient $\mathrm{\varnothing }$; in particular at least one of the two linearizations is not the other (their invariant rings and quotient loci differ). So semistability depends on the linearization and not only on the isomorphism class of the sheaf, and the assertion displayed in the Statement refuted is false.

The standard linearization (i). By [F1] the weight of $e_0^i{e}_1^{n-i}$ in $L_0^{\otimes n}$ is $2i-n$, which vanishes exactly when $i=n\mathrm{/}2$; hence the invariant sections of positive degree are the multiples of $(e_0{e}_1{)}^{n\mathrm{/}2}$ for even $n$, and the invariant section ring is $\mathbb{C}[e_0{e}_1]$. Therefore $X^{ss}(L_0)=\{e_0{e}_1\mathrm{\ne }0\}$, the complement of the two fixed points, and $X\mathrm{/}\text{\hspace{0.17em}⁣}{\mathrm{/}}_{L_0}G=\mathrm{Proj}\mathbb{C}[e_0{e}_1]$, a one-point scheme. On the locus $e_0{e}_1\mathrm{\ne }0$ write $z=e_0\mathrm{/}{e}_1$; then $z\mapsto t^{-2}z$, so the action is transitive onto ${\mathbb{C}}^{\times }$ with stabilizer ${\mu }_2$ at every point, and the unique orbit is the whole locus and is therefore closed there. Hence $X^s(L_0)=X^{ss}(L_0)\cong {\mathbf{G}}_m$, and by [F2] the good quotient restricts to a geometric quotient onto the point.

The twist by $\chi$ (ii). By [F1] the weight of $e_0^i{e}_1^{n-i}$ in $L_{+}^{\otimes n}$ is $2i$, so the invariant sections of positive degree are the multiples of $e_1^n$ and the invariant ring is $\mathbb{C}[e_1]$; hence $X^{ss}(L_{+})=\{e_1\mathrm{\ne }0\}\cong {\mathbb{A}}^1$ and $X\mathrm{/}\text{\hspace{0.17em}⁣}{\mathrm{/}}_{L_{+}}G=\mathrm{Proj}\mathbb{C}[e_1]$ is a point, so the quotient collapses the affine line to a point. The point $[0:1]$ (that is, $z=0$ in the coordinate $z=e_0\mathrm{/}{e}_1$) is fixed with stabilizer $G$, and every other orbit $z\mathrm{\ne }0$ has closure containing $0$, so no point of $X^{ss}(L_{+})$ is stable: $X^s(L_{+})=\mathrm{\varnothing }$ and the quotient is not geometric.

The opposite twist and the higher twists (iii). For $L_{-}=L_0\otimes {\chi }^{-1}$ the weight is $2i-2n$, vanishing exactly for $i=n$; the invariant sections are the multiples of $e_0^n$, the invariant ring is $\mathbb{C}[e_0]$, and $X^{ss}(L_{-})=\{e_0\mathrm{\ne }0\}\cong {\mathbb{A}}^1$ with $X^s(L_{-})=\mathrm{\varnothing }$ by the same fixed-point argument as in step 1.2. For $k\ge 2$ the weight is $2i-n+kn$, which vanishes only for $i=n(1-k)\mathrm{/}2<0$, impossible for a monomial; for $k\le -2$ it vanishes only for $i=n(1-k)\mathrm{/}2>n$, also impossible. Hence for $\mathrm{\mid }k\mathrm{\mid }\ge 2$ there is no nonzero invariant section of positive degree, $X^{ss}(L_0\otimes {\chi }^k)=\mathrm{\varnothing }$, and the quotient is $\mathrm{Proj}\mathbb{C}=\mathrm{\varnothing }$.

Conclusion (iv). All the linearizations considered have the same underlying ample sheaf $\mathcal{O}(1)$ and the same action on $X$: the standard linearization gives the nonempty stable locus ${\mathbf{G}}_m$ with quotient a point, the two single twists give the two affine lines with non-geometric one-point quotients and empty stable loci, and the remaining twists give the empty quotient. This exhibits the dependence of the GIT data on the linearization alone and matches the computations in Newstead's Lecture 4 example (stated there for $n\ge 2$, with the same argument at $n=1$) and in the corresponding example of Hoskins' notes.