I have been thinking about quotients lately and pondered the following:
Let $G$ be a connected linear algebraic group and $X$ a $G$-variety where the action is the morphism $\sigma:G\times X\rightarrow X$. Let $p:L\rightarrow X$ be a line bundle on $X$.
A $G$-linearisation of $L$ is an action of $G$ on $L$ such that $p(g\cdot l)= g\cdot p(l)$, for $l\in L, g\in G$, and which restricts to a linear isomorphism $L_{x}\rightarrow L_{g\cdot x}$ on the fibres. This last condition can be expressed as saying that there is an isomorphism $L\rightarrow g^{\ast}L$, for each $g\in G$ (here $g^{\ast}L$ is the pullback bundle by the automorphism $g$ of $X$). In fact, since $G$ is connected, a $G$-linearisation of $L$ exists if and only if there is an isomorphism $p_{2}^{\ast}L\rightarrow \sigma^{\ast}L$ of bundles on $G\times X$, with $p_{2}$ the projection to $X$.
It is known (Corollary 7.2, p.109, 'Lectures on Invariant Theory' – Dolgachev) that if $X$ is normal then for any $L$ there is some power of $L$ that admits a $G$-linearisation.
Question 1: Can someone provide an example of a non-normal $G$-variety $X$ and a line bundle $L$ for which no power $L^{n}$ admits a $G$-linearisation?
Question 1': If no such example can exist can someone point me towards the literature (if any) where this question is addressed?
The existence result for normal $X$ relies on that fact that there is an exact sequence
$ 0\rightarrow K \rightarrow Pic^{G}(X)\rightarrow Pic(X) \rightarrow Pic(G)$
and that $Pic(G)$ is finite. Here $K$ is the group of rational characters of $G$ and $Pic^{G}(X)$ is the group of line bundles with a $G$-linearisation (or line $G$-bundles in Dolgachev's terminology).
Question 2: Can we extend the exact sequence
$0\rightarrow K \rightarrow Pic^{G}(X) \rightarrow Pic(X)$
to the right for arbitrary $X$ and in a 'canonical' manner? (i.e., is this exact sequence the tail of a canonical long exact sequence for any $G$-variety X?)
Question 2': If so, what groups appear? Do they have any 'down-to-earth' interpretations? (e.g., we have $Pic(G)$ appearing for normal $X$).
Thanks in advance and apologies if this is standard material in GIT – I only have a copy of Dolgachev's notes at hand and these questions are not addressed.
Best Answer
Let me give an example showing that the normality hypothesis is necessary.
Let $Y=\mathbb{P}^1$ with natural $G=\mathbb{G}_m$-action. Let $X$ be the $G$-variety obtained by glueing transversally the two fixed points $0$ and $\infty$. Consider the line bundle $\mathcal{O}(l)$ with $l\neq 0$ on $Y$ and glue the fibers over $0$ and $\infty$ using any linear isomorphism to obtain a line bundle $\mathcal{L}$ on $X$. Suppose that $\mathcal{L}$ has a $G$-linearisation. Pulling it back to $Y$, we obtain a $G$-linearisation of $\mathcal{O}(l)$ on $Y$ such that $G$ acts on the fibers over $0$ and $\infty$ with the same character. However, the description of the $G$-linearisations of $\mathcal{O}(l)$ when $l\neq 0$ shows that this is not possible (more precisely, for any $G$-linearisation of $\mathcal{O}(l)$, the characters through which $G$ acts on the fibers over $0$ and $\infty$ differ by the character $t\mapsto t^l$). This argument shows moreover that no multiple of $\mathcal{L}$ has a $G$-linearisation.
Note that it follows that there is no ample $G$-linearised line bundle on $X$.
As for the second question, the natural map to study is more likely to be $Pic^G(X)\to Pic(X)^G$ where $Pic(X)^G$ denotes the group of line bundles whose class in $Pic(X)$ is $G$-invariant. When $X$ is normal and proper, its Picard group is an extension of a discrete group by an abelian variety so that if $G$ is linear connected, $G$ acts necessarily trivially on $Pic(X)$ and $Pic(X)^G=Pic(X)$. However, when $X$ is not normal, this is not the case anymore (for instance in the above example).
Moreover, the arguments in Dolgachev's notes extend to show that, if $X$ is an integral proper variety over an algebraically closed field endowed with an action of a connected linear algebraic group $G$, there is an exact sequence : $$0\to K\to Pic^G(X)\to Pic(X)^G\to Pic(G).$$
A more general statement, that does not assume $X$ proper or $G$ affine, may be found in [Brion, On linearization of line bundles, arXiv:1312.6267, Proposition 2.10].