Let $X$ be a smooth and proper variety over $\mathbb{C}$. Let $F$ be an $\mathbb{A}^1$ family of $\mathbb{G}_m$ gerbes over $X$. Suppose the fibers over every point away from 0 in $\mathbb{A}^1$ are all isomorphic. Must the fiber over zero be isomorphic to the general fiber $F_t$ ? This is certainly true for line bundles. I could think of stronger results that might be true, but my general ignorance of the subject leads me to pose the question I am interested in. Sorry also if this question belongs on math.stackexchange.
$\begingroup$
$\endgroup$
asked Nov 12, 2012 at 16:13
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
3
By a family of gerbes you mean, I suppose, a gerbe over $X \times \mathbb A^{1}$. In any case, it has a class in $\mathrm H^0(\mathbb A^1, \mathrm R^2 \mathrm{pr}_{2*}\mathbb G_{\rm m})$. Since $\mathrm R^2 \mathrm{pr}_{2*}\mathbb G_{\rm m}$ is a torsion sheaf, this class comes from $\mathrm H^0(\mathbb A^1, \mathrm R^2 \mathrm{pr}_{2*}\mu_{n})$ for some $n$; and $ \mathrm R^2 \mathrm{pr}_{2*}\mu_{n}$ is a constant sheaf on $\mathbb A^1$. Hence any family of gerbes is in fact constant along the fibers.
And I don't think that your question, or any question about gerbes, belongs in https://math.stackexchange.com/.
-
$\begingroup$ Thank you professor Vistoli. One point of clarification, of all of the random hypotheses I threw into the bag for good luck, the only one you need is that we are working in characteristic zero? $\endgroup$ Commented Nov 13, 2012 at 0:39
-
$\begingroup$ Also you use smoothness of X to conclude that the sheaf $R^2 pr_*G_m$ is torsion. $\endgroup$ Commented Nov 13, 2012 at 7:21
-
$\begingroup$ I think that the statement should also be true in positive characteristic, but requires a little care. $\endgroup$– AngeloCommented Nov 13, 2012 at 7:51