Suppose $X$, $Y$, and $Z$ are smooth irreducible schemes [EDIT: of finite type over an algebraically closed field of characteristic zero], and $X \to Z$ and $Y \to Z$ are dominant maps.

I have a certain point $(x, y)$ in the fiber product $X \times_Z Y$; and I'd like to know that it lies in some component of the fiber product which dominates $Z$.

What condition on the differentials $df|_x : T_x X \to T_z Z$ and $df|_y : T_y Y \to T_z Z$ will allow me to verify this?

It's fairly easy to see that the surjectivity of $df|_x$ (or $dg|_y$) alone is enough to guarantee the desired conclusion --- I'm hoping here for something weaker that would still suffice. I'd by happy if the surjectivity of $df|_x + dg|_y$ suffices, for example; but if it does not, I would be interested in any other weaker condition too.

  • $\begingroup$ Smoothness is a property of a morphism, not of a scheme. $\endgroup$ – Question Mark Aug 7 '14 at 3:01
  • $\begingroup$ Is this question being posed for schemes of finite type over an (unspecified) algebraically closed field? Characteristic 0? Please clarify the hypotheses so that a useful answer can be given. $\endgroup$ – user27920 Aug 7 '14 at 3:32
  • $\begingroup$ A weaker condition than tangential surjectivity of a map is the dimension formula $\dim(X_z) = \dim(X) - \dim(Z)$, which in this setting implies flatness (weaker than smoothness of the morphism) and so is sufficient. But presumably that formula does not hold (or may not hold) in your setting or you wouldn't have raised the question. Can you say something more about the motivation? $\endgroup$ – user27920 Aug 7 '14 at 15:45
  • $\begingroup$ I don't know how to compute the dimension of $X_z$ in the case I care about; but I have some control over the differential. (My $X$ and $Y$ are certain Kontsevich spaces of stable $n$-pointed maps; and $Z$ is the Hilbert scheme of $n$ points. The differentials are partially-computable since they are induced maps on cohomology groups of the normal bundle.) $\endgroup$ – Eric Larson Aug 7 '14 at 17:24

[Observation of David Yang.]

The condition $df|_x(T_x) + dg|_y(T_y) = T_z$ is not sufficient; more generally, the only closed condition depending on the images of $df|_x(T_x)$ and $dg|_y(T_y)$ that could be sufficient is the surjectivity of either $df|_x$ or $dg|_y$.

To see this, it suffices to construct a counterexample where $df|_x(T_x)$ and $dg|_y(T_y)$ are codimension one subspaces in general position. For this, we can take $Z = \mathbb{P}^n$, and $X = Y = \text{Bl}_\Lambda \mathbb{P}^n$, where $\Lambda$ is a subspace of codimension two.


Edit: I gave a proof of a false statement earlier. It can be salvaged to the folowing much weaker statement: the point $(x,y)$ is in the dominant irreducible component if we have $df|_x(T_x)+dg|_y(T_y)=T_z,$ together with the following assumption:

Let $r=\operatorname{rank}(df|_x).$ Then there is a local coordinate system $z_1,\ldots, z_n$ with origin $z\in Z$ such that

(1) the first $r$ coordinates $z_1,\dots, z_r$ define a full-rank map $(f_1,\dots, f_r):X\to A^r$ and

(2) for fixed $(z_1,\dots, z_r)$, let $X_{(z_1,\dots, z_r)}=(f_1,\dots, f_r)^{-1}(\{(z_1,\dots, z_r)\})$ be the fiber. I ask that the map $(f_{r+1},\dots, f_n):X_{(z_1,\dots, z_r)}\to A^{n-r}$ be dominant for any tuple $(z_1,\dots, z_r)$ in a neighborhood of 0.

Don't know how useful this is. But in this case one can see using the implicit function theorem that the intersection $f(D_x)\cap g(D_y)$ of small disks near $x,y$ has nonempty interior, and the statement follows.

  • 1
    $\begingroup$ The error/gap in that this argument is that it doesn't make any contact with the topology of $X \times_Z Y$ around $(x,y)$, only give a conclusion on some open piece of $X \times Y$ that may well be "far" from a neighborhood of $(x,y)$ in the fiber product. $\endgroup$ – user27920 Aug 7 '14 at 15:48
  • $\begingroup$ Sorry! The fact that $f|_{I^n_X}$ is the identity on the first $r$ coordinates doesn't imply that its image is $I^r\times U$. (@user52824, the incorrect arguments I gave are in fact local). $\endgroup$ – Dmitry Vaintrob Aug 7 '14 at 19:50

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