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Suppose we have a family of objects $\Xi \to S$ over a base smooth projective scheme $S$. Take a closed point $p\in S$ and consider the tangent space to $S$ at $p$. Can one construct an "induced family" over this tangent space (or probably rather over its projectivized space) starting from $ \Xi$ ? what interpretation could one give to this? a second-order deformation?

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Would you settle for having a family over the tangent cone to $p\in S$? Which one could instead call the normal cone. Then the normal cone to $\Xi_p \subseteq \Xi$ will map to the normal cone to $p \in S$. I don't have much interpretation for its fibers, though. – Allen Knutson Feb 8 '13 at 12:30
Thank you, Allen. I don't fully understand what you mean by $\Xi_p$, is it the full fiber over $p$, right? – IMeasy Feb 10 '13 at 17:59
Yes, that's what I meant. – Allen Knutson Feb 15 '13 at 4:15
up vote 2 down vote accepted

Imagine that $S$ is an open subset of some fine moduli space which does not contain any rational curves. This does happen although I don't know any examples. Then any family over an affine space will have to be constant, so the answer would be no.

On the other hand, of course there is formal family over the completion of $S$ at the given point $p$ ($\hat{S}_p = \lim Spec (\mathcal{O}_{S, p}/\mathfrak{m}_{S, p}^n)$), which looks like the infinitesimal neighborhood of $0$ in the tangent space ($\hat{T}_0 = Spf(k[[t_1, \ldots, t_s]])$ where the $t_i$ form a basis of the dual of $T= T_{p} S$). I think morally this might play the role of what you want.

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For an example, take the modular curve $X(n)$ (with $n$ not too small). – Dan Petersen Feb 6 '13 at 17:53
thank you two! @Dan: could you develop your example a little more? – IMeasy Feb 6 '13 at 18:19
What I think Dan meant is: if the families you are considering are families of elliptic curves with fixed basis of $n$-torsion, the moduli space will be the modular curve $X(n)$ (or rather its open subset $Y(n)$). After removing finitely many points, it will be a fine moduli space. As computed here , the genus of $X(n)$ is nonzero for most $n$. Since $X(n)$ is smooth, for such $n$ there are no non-constant maps $\mathbb{A}^1\to X(n)$, that is, every family of the considered type over $\mathbb{A}^1$ has to be constant. – Piotr Achinger Feb 7 '13 at 23:05

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