A line bundle that does not admit a G-linearisation - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T14:05:42Z http://mathoverflow.net/feeds/question/109310 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109310/a-line-bundle-that-does-not-admit-a-g-linearisation A line bundle that does not admit a G-linearisation George Melvin 2012-10-10T16:43:11Z 2012-10-10T19:35:46Z <p>I have been thinking about quotients lately and pondered the following:</p> <p>Let $G$ be a connected linear algebraic group and $X$ a $G$-variety acting via the morphism $\sigma:G\times X\rightarrow X$. Let $p:L\rightarrow X$ be a line bundle on $X$. </p> <p>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 <code>$L_{x}\rightarrow L_{g\cdot x}$</code> 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 <code>$p_{2}^{\ast}L\rightarrow \sigma^{\ast}L$</code> of bundles on $G\times X$, with $p_{2}$ the projection to $X$.</p> <p>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.</p> <p>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? . </p> <p>Question 1': If no such example can exist can someone point me towards the literature (if any) where this question is addressed?</p> <p>The existence result for normal $X$ relies on that fact that there is an exact sequence</p> <p>$0\rightarrow K \rightarrow Pic^{G}(X)\rightarrow Pic(X) \rightarrow Pic(G)$</p> <p>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 admitting a $G$-linearisation (or line $G$-bundles in Dolgachev's terminology).</p> <p>Question 2: Can we extend the exact sequence</p> <p>$0\rightarrow K \rightarrow Pic^{G}(X) \rightarrow Pic(X)$</p> <p>to the right for arbitrary $X$ and in a 'canonical' manner? (ie, is this exact sequence the tail of a canonical long exact sequence for any $G$-variety X?) </p> <p>Question 2': If so, what groups appear? Do they have any 'down-to-earth' interpretations? (eg, we have $Pic(G)$ appearing for normal $X$).</p> <p>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.</p> http://mathoverflow.net/questions/109310/a-line-bundle-that-does-not-admit-a-g-linearisation/109312#109312 Answer by Olivier Benoist for A line bundle that does not admit a G-linearisation Olivier Benoist 2012-10-10T17:34:04Z 2012-10-10T19:35:46Z <p>Let me give an example showing that the normality hypothesis is necessary.</p> <p>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.</p> <p>Note that it follows that there is no ample $G$-linearised line bundle on $X$.</p> <p>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). </p> <p>Moreover, as far as I could check, 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 : $$0\to K\to Pic^G(X)\to Pic(X)^G\to Pic(G).$$</p>