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Suppose we have a projective, nonsingular, convex variety $X$, $\beta \in H_2(X,\mathbb{Z})$ and a family $$ \begin{array}{ccc} \mathcal{C}& \to & X \cr \downarrow& & & \cr B & & \end{array}. $$ together with sections $\sigma_1, \ldots, \sigma_n: B \to \mathcal{C}$ making the diagram above a family of stable maps from $n$-pointed genus $g$ curves over $B$, inducing a map $B \to \overline M_{g,n}(X,\beta)$.

Assume $B$ is a smooth curve and all fibres $\mathcal{C}_b$ are smooth except for one fibre over a point $0 \in B$ which is a curve with one node.

Question: Is $\mathcal{C}$ being smooth a sufficient criterion for $B \to \overline M_{g,n}(X,\beta)$ being transverse to the boundary? If not, how could I try showing transversality (or computing the multiplicity of the intersection)?

If this helps: I only need the answer for $g=0$ and $X=\mathbb{P}^1 \times \mathbb{P}^1$.

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    $\begingroup$ For $g=0$, that is sufficient. For larger $g$, it depends on what you mean by "transverse to the boundary" in cases where "the boundary" is not actually a Cartier divisor in the moduli stack. $\endgroup$
    – Jason Starr
    Commented May 11, 2015 at 10:22
  • $\begingroup$ Ah, ok, but that would be sufficient for me. Do you happen to know a reference for this? $\endgroup$
    – JoS
    Commented May 11, 2015 at 10:28
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    $\begingroup$ I will try to find you a reference. Personally, I would handle it in two steps. First, the induced (1-)morphism from $B$ to the Artin stack $\mathfrak{M}_{0,n}$ of all $n$-pointed, at-worst-nodal curves pulls back the boundary divisor $\Delta$ to be $\underline{0}$, with multiplicity $1$. Second, because of convexity, the $1$-morphism $\overline{M}_{g,n}(X,\beta)\to \mathfrak{M}_{0,n}$ is smooth, so that the pullback of $\Delta$ is a Cartier divisor -- the usual boundary divisor. $\endgroup$
    – Jason Starr
    Commented May 11, 2015 at 12:27
  • $\begingroup$ Thank you very much for your help. If you find a reference please give it as an answer, so I can accept it. $\endgroup$
    – JoS
    Commented May 11, 2015 at 12:43

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The following argument should answer the original question. However it still seems that the right way to do it is via the Artin stack of $n$-pointed, at-worst-nodal curves as proposed by Jason Starr.

My argument why $\mathcal{C}$ being smooth suffices for obtaining transversality goes as follows:

  • Over the locus $\Delta_1 \subset \overline M_{0,n}(X, \beta)$ of $q=(f:C \to X; p_1, \ldots, p_n)$ with $C$ having exactly one node, the universal family $\mathcal{U} \to \overline M_{0,n}(X, \beta)$ has a section corresponding to this node (where the fibre over $q$ is identified with $C$). This can be shown using the inductive boundary structure via gluing maps.

  • Locally around $0 \in B$ the family $\mathcal{C} \to B$ is the pullback of this universal family (if $0$ has no automorphisms).

  • If $B$ was tangent to the boundary at $0$ then every tangent vector $v$ of $0$ in $B$ would be in the tangent space of $\Delta_1$.

  • Via pushforward by the section of the universal family over $\Delta_1$ we obtain a tangent vector in $\mathcal{C}$ based at the node $N$ of $\mathcal{C}_0$ which maps to $v$ under $\pi$.

  • This is a contradiction: the derivative of $\pi$ at $N$ must vanish because otherwise, by the implicit function theorem, $\mathcal{C}_0 = \pi^{-1}(0)$ would be smooth around $N$.

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