# Is the stack of stable curves with no rational component algebraic?

Let $$g\geq 2$$ be an integer and let $$\overline{\mathcal{M}}_g$$ be the (smooth proper Deligne-Mumford) algebraic stack of stable curves of genus $$g$$.

Let $$\mathcal{M}_g^{nr}$$ be the substack of stable curves $$C\to S$$ such that, for every geometric point $$\overline{s}$$ of $$S$$, the fibres $$C_{\overline{s}}$$ does not admit a morphism from $$\mathbb{P}^1_{\overline{s}}$$. (In other words, none of the irreducible components of $$C_{\overline{s}}$$ are rational.)

Is $$\mathcal{M}_g^{nr}$$ an algebraic substack of $$\overline{\mathcal{M}}_g$$?

I feel like $$\mathcal{M}_g^{nr}$$ is the complement of a some boundary divisor, but I'm not sure how to make this precise.

• Welcome new contributor. For every irreducible component of the boundary divisor of $\overline{\mathcal{M}}_g$, the generic point of that component parameterizes a stable curve for which each irreducible component has genus $\geq 1$. Thus, the open subset $\mathcal{M}_g^{\text{nr}}$ is not equal to the complement of a boundary divisor. – Jason Starr Oct 9 '18 at 16:52

I am adding a few details to my comment above. For notational convenience, in addition to the usual notation $$\overline{\mathcal{M}}_{g,n}$$ for genus-$$g$$, $$n$$-pointed curves, also for every finite set $$N$$ with $$n$$ elements, use the notation $$\overline{\mathcal{M}}_{g,N}$$ when the $$n$$ marked sections are explicitly indexed by $$N$$. Thus, there are boundary morphisms, $$\Delta_{(g',N'),(g'',N'')}:\overline{\mathcal{M}}_{g',N'\sqcup\{\ell'\}}\times \overline{\mathcal{M}}_{g'',N''\sqcup\{\ell''\}} \to \overline{\mathcal{M}}_{g'+g'',N'\sqcup N''},$$ $$\Delta_{g,N}:\overline{\mathcal{M}}_{g,N\sqcup\{\ell',\ell''\}}\to \overline{\mathcal{M}}_{g+1,N},$$ defined in the usual way.

Definition. An assignment to every integer $$g$$ and finite set $$N$$ satisfying $$2g-2+|N|>0$$ of a reduced substack $$Z_{g,N}\subset \overline{\mathcal{M}}_{g,N}$$ is $$\Delta$$-compatible if it is compatible with permutations of $$N$$, and for every boundary morphism $$\Delta_{(g',N'),(g'',N'')}$$, resp. $$\Delta_{g,N}$$, the inverse image of $$Z_{g'+g'',N'\sqcup N''}$$, resp. of $$Z_{g+1,N}$$, equals the union of the pullbacks of $$Z_{g',N'\sqcup\{\ell'\}}$$ and $$Z_{g'',N''\sqcup\{\ell''\}}$$, resp. the pullback of $$Z_{g,N\sqcup\{\ell',\ell''\}}$$. It is combinatorial if for every $$(g,N)$$ such that $$Z_{g,N}$$ does not equal $$\overline{\mathcal{M}}_{g,N}$$, then $$Z_{g,N}$$ is contained in the boundary divisor of $$\overline{\mathcal{M}}_{g,N}$$.

Proposition. Every combinatorial, $$\Delta$$-compatible assignment $$(Z_{g,N})_{g,N}$$ is a system of closed substacks $$Z_{g,N}$$ of $$\overline{\mathcal{M}}_{g,N}$$.

Proof. By hypothesis, for every $$(g,N)$$ such that the boundary of $$\overline{\mathcal{M}}_{g,N}$$ is empty, then $$Z_{g,N}$$ is either empty or else equal to all of $$\overline{\mathcal{M}}_{g,N}$$, both of which are closed substacks of $$\overline{\mathcal{M}}_{g,N}$$. Thus, by way of induction (on the integer $$2g-2+|N|$$, for example), assume that the boundary of $$\overline{\mathcal{M}}_{g,N}$$ is nonempty.

If $$Z_{g,N}$$ equals all of $$\overline{\mathcal{M}}_{g,N}$$, then it is a closed substack of $$\overline{\mathcal{M}}_{g,N}$$. Thus, assume that does not equal all of $$\overline{\mathcal{M}}_{g,N}$$. Since the system is combiantorial, the substack $$Z_{g,N}$$ is contained in the boundary. Since the boundary is a union of finitely many irreducible closed substacks of $$\overline{\mathcal{M}}_{g,N}$$, it suffices to check that the intersection with each of these irreducible closed substacks is closed.

Each of these irreducible closed substacks is the image of a boundary morphism. Since the system is $$\Delta$$-compatible, the inverse image of $$Z_{g,N}$$ under the boundary morphism is obtained from $$Z_{h,P}$$ with $$2h-2+|P| < 2g-2+|N|$$. Thus, by the induction hypothesis, the inverse image of $$Z_{g,N}$$ under each boundary morphism is a closed substack of the domain of the boundary morphism. Since the boundary morphism is finite, hence proper, the image of this closed substack of the domain is also a closed substack of $$\overline{\mathcal{M}}_{g,N}$$. Thus, $$Z_{g,N}$$ is a closed substack of $$\overline{\mathcal{M}}_{g,N}$$. By induction on $$2g-2+|N|$$, every $$Z_{g,N}$$ is a closed substack of $$\overline{\mathcal{M}}_{g,N}$$. QED

Notation. For every nonnegative integer $$h$$, denote by $$Z^{h}_{g,N} \subset \overline{\mathcal{M}}_{g,N}$$ the reduced substack parameterizing those genus-$$g$$, $$N$$-marked curves such that there exists a subcurve that is the image of a nonconstant morphism from a proper, connected, reduced, at-worst-nodeal curve of arithmetic genus $$\leq h$$.

Corollary. The system $$(Z^{h}_{g,N})$$ is $$\Delta$$-compatible and combinatorial. Thus, it is a system of closed substacks.

Proof. For a curve that is a cofiber coproduct of connected subcurves, the coproduct curve contains an irreducible component of geometric genus $$\leq h$$ if and only if one of the connected subcurves contains an irreducible component of geometric genus $$\leq h$$. Similarly, for a curve with a non-disconnecting node, the normalizations of the irreducible components all factor through the partial normalization of the non-disconnecting node. Thus, the nodal curve has an irreducible component of geometric genus $$\leq h$$ if and only if the partial normalization has an irreducible component of geometric genus $$\leq h$$. Thus, the system is $$\Delta$$-compatible.

If $$g\leq h$$, then $$Z^h_{g,N}$$ equals all of $$\overline{\mathcal{M}}_{g,N}$$. If $$h < g$$, then every curve parameterized by the complement of the boundary is irreducible of genus $$g>h$$. Thus, there is no irreducible component of this curve that has geometric genus $$\leq h$$. So, in this case, $$Z^h_{g,N}$$ is contained in the boundary divisor. In all cases, the substack $$Z^h_{g,N}$$ is combinatorial. QED

Finally, when $$h$$ equals $$0$$, the substack $$Z^0_{g,n}$$ is, by definition, the complement of the substack $$\mathcal{M}^{\text{nr}}_{g,n}$$. As the complement of a closed substack, the substack $$\mathcal{M}^{\text{nr}}_{g,n}$$ is an open substack. For every integer $$g>2$$, the closed substack $$Z^0_{g,0}$$ of $$\overline{\mathcal{M}}_{g,0}$$ contains no irreducible component of the boundary divisor of $$\overline{\mathcal{M}}_{g,0}$$.