Let $f:X\rightarrow Y$ be a morphism of schemes with smooth curves as fibers. Let $g:X\rightarrow Z$ be a family of smooth or nodal curves with $Z$ a regular scheme. Does the push-out $Z\coprod_X Y$ exist (at least as algebraic stack)? I saw existence of the push-out as algebraic space when one has $f$ a closed immersion and $g$ a finite morphism (fibers of dimension zero). I am wondering what is going on if the fibers have dimension one.
4
-
1$\begingroup$ I do not see any "stacky" nature to such a pushout. How are nontrivial automorphisms supposed to arise? You can see the basic difficulty if you let $Y$ be $\mathbb{P}V \cong \mathbb{P}^2$, let $Z$ be $\mathbb{P}(V^\vee) \cong (\mathbb{P}^2)^\vee$, the dual projective plane of lines in $\mathbb{P}V$, and let $X$ be the flag variety $\text{Flag}(1,2;V)$ of pointed lines in $\mathbb{P}V$, resp. in $\mathbb{P}(V^\vee)$, with its natural projections to $Y$ and $Z$. The pushout (presheaf) contracts $Y$ and $Z$ with $1$-dimensional fibers, but there is no such contraction of the projective plane. $\endgroup$– Jason StarrDec 29, 2014 at 14:03
-
$\begingroup$ Regarding my last example, one problem is lack of "associativity". To deal with this, some authors (cf. Koll'ar's "Rational Curves on Algebraic Varieties") work with "chains", e.g., replace $X$ by the disjoint union over all possible fiber products of copies of $X$ over both $Y$ and $Z$. When you do that with the previous example, eventually $X$ with its morphisms to $Y$ and $Z$ dominates $Y\times Z$, so that the coproduct is just a point. For more general families of curves, Koll'ar's book discusses quotients by chains of curves. $\endgroup$– Jason StarrDec 29, 2014 at 14:40
-
$\begingroup$ @JasonStarr thanks a lot! I don't understand well your claim in the example. Are you saying that such "contraction" cannot exist even as algebraic stack? $\endgroup$– ecceJan 6, 2015 at 17:06
-
$\begingroup$ "I don't understand well your claim in the example. Are you saying that such 'contraction' cannot exist even as algebraic stack?" Unfortunately, I do not understand your question. In what sense is the quotient a stack (algebraic or not), rather than a (set-valued) presheaf, sheaf, etc.? If you tell us the stack structure, MO users can try to help you determine whether or not your stack is algebraic. However, I see no natural stack structure beyond the obvious structure of a (set-valued) presheaf, or maybe a sheafification of that presheaf. $\endgroup$– Jason StarrJan 6, 2015 at 17:56
Add a comment
|