flatness in complex analytic geometry

Question

It is always a pain to move back and forth between definitions in algebraic geometry and complex analytic geometry. Dictionary is much easier when are working with (family of) smooth varieties but the pain grows exponentially when we include singular varieties.

Here is some of that:

Suppose $f: X\rightarrow Y$ is a map of possibly singular complex analytic varieties, then is there a simpler definition of flatness for $f$ ? this one should be hard to answer but how about following,

---Let $f:X\rightarrow Y$ be a family of curves. there might be multiple fibers, non-reduced fibers, nodal curves, cusp curves,... in the family, but fibers are connected.

---When this fibration is flat? What kind of bad fibers mentioned above are allowed in a flat family?

---Same question for higher dimensional family of analytic varieties?

For simplicity you may assume that the base of fibration is smooth.

Answer 1

Instead of trying to say what flatness in analytic geometry means I'll give you some street-fighting tricks for recognizing whether a morphism of analytic spaces ( not necessarily reduced) $f:X\to Y $ is , or has a chance to be, flat.

a) A flat map is always open. . Hence, contraspositely, the embedding $\lbrace 0\rbrace \hookrightarrow \mathbb C$ is not flat. More generally, the embedding of a closed (and not open !) subspace $X \hookrightarrow Y$ is never flat.

b) An open map need not be flat: think of the open map $ Spec(\mathbb C) \to Spec(\mathbb C[\epsilon ]) $ from the reduced point to the double point, which is open but not flat.
[It is not flat because the $\mathbb C[\epsilon]$- algebra $\mathbb C=\mathbb C[\epsilon]/(\epsilon)$ is not flat : recall that a quotient ring $A/I$ can only be flat over $A$ if $I=I^2$ and here $I=(\epsilon) \neq I^2=(\epsilon)^2=(0)$]

An example with both spaces reduced is the normalization [see g) below] $f:X^{nor} \to X $ of the cusp $X\subset \mathbb C^2$ given by the equation $y^2=x^3$ .That normalization is a homeomorphism and so certainly open, but it is not flat : this results either from g) or from h) below.

c) Given the morphism $f:X\to Y $ , consider the following property:
$\forall x \in X, \quad dim_x(X)=dim_{f(x)} (Y) +dim_x(f^{-1}(f(x)))$ $\quad (DIM) $
We then have: $ f \; \text {flat} \Rightarrow f \;\text { satisfies } (DIM)$
For example a (non-trivial) blowup is not flat.

d) For a morphism $f:X\to Y $ between connected holomorphic manifolds we have:

$$f \text { is flat} \iff f \text { is open} \quad \iff (DIM) \;\text {holds}$$ For example a submersion is flat, since it is open.

e) Given two complex spaces $X,Y$ the projection $X\times Y\to X$ is flat ( Not trivial: recall that open doesn't imply flat!).

f) flatness is preserved by base change.

g) The normalization $f:X^{nor} \to X $ of a non-normal space is never flat.
For example if $X\subset \mathbb C^2$ is the cusp $y^2=x^3$, the normalization morphism $\mathbb C\to X: t\mapsto (t^2,t^3)$ is not flat.

h) Given a finite morphism $f:X\to Y $, each $y\in Y$ has a fiber $X(y) \subset X$ and for $x\in X$ we can define $\mu (x)= dim_{\mathbb C} (\mathbb C \otimes_{\mathcal O_{Y,y}} \mathcal O_{X,x})$ . Now for $ y\in Y$ we put $\; \nu (y)= \Sigma_{x\in X(y)} \mu(x)$ and we obtain :

$$f \; \text{ flat} \iff \nu :Y\to \mathbb N \text { locally constant}$$

[Of course for connected $Y$, locally constant = constant]

Bibliography:
A. Douady, Flatness and Privilege, L'Enseignement Mathématique, Vol.14 (1968)
G.Fischer, Complex Analytic Geometry, Springer LNM 538, 1976

Answer 2

Mohammad,

flatness is not simple, so you are not going to get a simple overall definition. On the other hand, in the example you mention in your comment, there is a simple criterion:

Let $f:X\to Y$ be a morphism of schemes such that $X$ is reduced and irreducible (most likely satisfied in the cases you are interested in, at least for now) and $Y$ is a smooth curve. Then if $X$ dominates $Y$ (i.e., the image is dense in $Y$) then $f$ is flat.

In particular $f:\mathbb C^2\to \mathbb C$ given by $(x,y)\mapsto xy$ is flat.

Over higher dimensional basis it is a little more complicated, but there is one simple criterion you can check:

If $f$ is flat, then the fibers have the same dimension.

For example a blowup is the typical not flat map (think of the meaning of "flat").

On the other hand, the fibers being equidimensional is certainly not enough for a morphism to be flat. An example is provided by taking $X$ to be the union of two planes meeting in a single point mapping to just one copy of the plane in the obvious way. All fibers are $0$-dimensional, but the morphism is not flat.

However, there is a criterion that is pretty useful and says that if both $X$ and $Y$ are relatively nice, then flatness is equivalent to that.

Let $f:X\to Y$ be a dominant morphism of irreducible varieties such that $X$ is Cohen-Macaulay and $Y$ is regular (smooth if you prefer), then $f$ is flat if and only if the fibers are equidimensional.

Finally, my suggestion is that in order to get a better feeling for flatness, try looking at finite morphisms as in the above example. If you understand those, you essentially have understood all flat maps.

Answer 3

Well, I definitely know nothing on the subject. However, there is something called Hironaka division, which gives some way to check flatness explicitly. Check Section 4.2 of these notes.