Are curves with `fractional points' uniquely determined by their residual gerbes? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T01:59:22Z http://mathoverflow.net/feeds/question/1467 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/1467/are-curves-with-fractional-points-uniquely-determined-by-their-residual-gerbes Are curves with `fractional points' uniquely determined by their residual gerbes? David Zureick-Brown 2009-10-20T18:23:20Z 2009-10-29T03:05:42Z <p>One makes precise the vague notion of "curve with a fractional point removed" (see for instance <a href="http://www-math.mit.edu/~poonen/slides/campana%5Fs.pdf" rel="nofollow">these slides</a>) using stacks -- one should really consider Deligne-Mumford stacks whose coarse spaces are curves, and the "fractional points" correspond to the residual gerbes at the stacky points.</p> <p>One example: let a,b,c > 1 be coprime integers and let S be the affine surface given by the equation x^a + y^b + z^c = 0. Then there is a weighted Gm action on S (t sends (x,y,z) to (t^bc x, t^ac y, t^ab z) and one can check that the stack quotient [S-{0}/Gm] has coarse space P^1 and, that since the action is free away from xyz = 0 but has stabilizers at those 3 points, one gets a stacky curve with three non-trivial residual gerbes.</p> <blockquote> <p><strong>Question</strong>: Are two 1-dimensional DM stacks with isomorphic coarse spaces and residual gerbes themselves isomorphic?</p> </blockquote> <p>I have an idea for how to prove this when the coarse space is P^1, but in general don't know what I expect the answer to be. Also, one may have to restrict to the case when the coarse space is a smooth curve.</p> <p>This might be analogous to the statement that the `angle' of a node of a rational nodal curve with one node doesn't affect the isomorphism class of the curve.</p> <p>Also, the recent papers of Abromivich, Olsson, and Vistoli (on stacky GW theory) may be relevant.</p> http://mathoverflow.net/questions/1467/are-curves-with-fractional-points-uniquely-determined-by-their-residual-gerbes/1715#1715 Answer by Anton Geraschenko for Are curves with `fractional points' uniquely determined by their residual gerbes? Anton Geraschenko 2009-10-21T20:31:54Z 2009-10-25T05:08:45Z <p>Here's an example of two non-isomorphic Deligne-Mumford stacks whose coarse spaces are <b>A</b><sup>1</sup>, and the only non-trivial residual gerbe in each case is B(<b>Z</b>/2) at the origin.</p> <p>First, take the <b>Z</b>/2 action on <b>A</b><sup>1</sup> given by reflection around 0, given by x&rarr;-x. The stack quotient [<b>A</b><sup>1</sup>/(<b>Z</b>/2)] has coarse space <b>A</b><sup>1</sup> and there's a B(<b>Z</b>/2) gerbe at the origin. Note that this stack is smooth since it has an etale cover by something smooth.</p> <p>On the other hand, you can take the stack I defined in <a href="http://mathoverflow.net/questions/1565/can-a-singular-deligne-mumford-stack-have-a-smooth-coarse-space/1584#1584" rel="nofollow">this answer</a>: the stack quotient of the coordinate axes in <b>A</b><sup>2</sup> by the <b>Z</b>/2 action which switches the two axes. The coarse space is again <b>A</b><sup>1</sup> and the stack has a B(<b>Z</b>/2) gerbe at the origin. Note that this stack is <em>not</em> smooth since it has an etale cover by something singular, so it cannot be isomorphic to the previous stack.</p> <h2>An example where both stacks are smooth</h2> <p>The first stack will be the same [<b>A</b><sup>1</sup>/(<b>Z</b>/2)] I used above.</p> <p>Let G be the affine line with a doubled origin, regarded as a group over <b>A</b><sup>1</sup> (most of the fibers are trivial groups, but the fiber over the origin is <b>Z</b>/2). This G has the trivial action on <b>A</b><sup>1</sup>. Consider the quotient stack [<b>A</b><sup>1</sup>/G]=B<sub><b>A</b><sup>1</sup></sub> G. This is a DM stack whose coarse space is <b>A</b><sup>1</sup>, and there's a B(<b>Z</b>/2) gerbe at the origin. Note that this stack is smooth since it has an etale cover by a smooth scheme (namely, <b>A</b><sup>1</sup>). Note that this stack has non-separated diagonal (since the pullback G&rarr;<b>A</b><sup>1</sup> is non-separated), but the diagonal of [<b>A</b><sup>1</sup>/(<b>Z</b>/2)] is separated, so the two stacks are non-isomorphic.</p> http://mathoverflow.net/questions/1467/are-curves-with-fractional-points-uniquely-determined-by-their-residual-gerbes/3198#3198 Answer by Anton Geraschenko for Are curves with `fractional points' uniquely determined by their residual gerbes? Anton Geraschenko 2009-10-29T03:05:42Z 2009-10-29T03:05:42Z <p>I love this question so much that I can't stop thinking about it. Since this is a separate line of thought, I'll post it as a separate answer. Here's a general way to distinguish between DM stack whose coarse spaces are isomorphic with all the same residual gerbes.</p> <p>Suppose a DM stack X contains a BG. Then the tangent space to the stack at that point is a representation of G. To get this representation, pull the sheaf of differentials of X back to BG. This is a coherent sheaf on BG, which is the same thing as a representation of G. Alternatively, there is a structure theorem for DM stacks (see Theorem 38.20 of <a href="http://math.berkeley.edu/~anton/written/Stacks/Stacks.pdf" rel="nofollow">my notes from Martin Olsson's course</a>) that says that if G is the stabilizer of a geometric point x&isin;X, then there is an etale cover U&rarr;X of a neighborhood of x with a G action so that the neighborhood of x is the stack quotient [U/G]. I'm pretty sure that the action of G on the tangent space of the point over x is the same representation as in the other construction (or maybe it will be the dual representation).</p> <p>Now the idea is that if two DM stacks have the same coarse space and the same residual gerbes, then you might still be able to tell them apart if they have different "residual tangent representations". In the three examples I gave in my other answer, the residual representations were (1) the sign representation of <b>Z</b>/2, (2) a two-dimensional representation (sign plus trivial) of <b>Z</b>/2, and (3) the trivial representation of <b>Z</b>/2. However, there is the annoying bit that you only recover G up to isomorphism, so if two representations differ by an automorphism of G, you can't conclude that the stacks are different.</p> <p><hr /></p> <p>Using this kind of thinking, we can construct two DM stacks which are both smooth, separated, and have any other kind of nice properties you can imagine, and have isomorphic coarse spaces with the same residual gerbes, but are non-isomorphic. Unfortunately, the coarse space is a surface.</p> <p>Consider the action of <b>Z</b>/6 on <b>A</b><sup>2</sup> given by 1&sdot;(x,y)=(x,&zeta;y), where &zeta; is a primative 6-th root of unity. This action is generated by pseudo-reflections, so by the <a href="http://en.wikipedia.org/wiki/Chevalley%2DShephard%2DTodd%5Ftheorem" rel="nofollow">Chevalley-Shephard-Todd theorem</a> (the easy half of it), the coarse space of the quotient stack [<b>A</b><sup>2</sup>/(<b>Z</b>/6)] is <b>A</b><sup>2</sup>, with a residual B(<b>Z</b>/6) at the origin.</p> <p>On the other hand, consider the action of <b>Z</b>/6 on <b>A</b><sup>2</sup> given by 1&sdot;(x,y)=(&zeta;<sup>2</sup>x,&zeta;<sup>3</sup>y). Again, the action is generated by pseudo-reflections, so the coarse space of the quotient stack is <b>A</b><sup>2</sup>, with a residual B(<b>Z</b>/6) at the origin.</p> <p>But the two residual representations of <b>Z</b>/6 are non-isomorphic, even if you twist by automorphisms of <b>Z</b>/6. The first representation has a vector invariant under the action (namely (1,0)), but the second does not.</p>