Is there an algebraic geometry analogue of the closed graph theorem?

2012-11-19T19:14:30Z

In functional analysis, the closed graph theorem asserts that if a linear map $T: X \to Y$ between two Banach spaces $X, Y$ has a closed graph $S := \{ (x,Tx): x \in X \}$, then the map is continuous. Thus, it gives a criterion for regularity of a map in terms of regularity of the graph of that map.

I am curious as to whether there is any analogous statement in algebraic geometry. The naive formulation would be: if $T: X \to Y$ was a function (in the set-theoretic sense) between algebraic varieties $X, Y$ over an algebraically closed field $k$ whose graph $S := \{ (x,Tx): x \in X \}$ was also an algebraic variety, then $T$ would be a regular map. (Here I will be vague as to whether I want varieties to be affine, projective, quasiprojective, or abstract.) But this is false, even in characteristic zero: for instance, the coordinate function $(t^2,t^3) \mapsto t$ from the cuspidal curve $\{ (t^2,t^3): t \in k \}$ to $k$ has a graph which is an algebraic variety, but is not a regular map (it is not given by a rational function in a neighbourhood of the origin). In characteristic $p$, the inverse of the Frobenius map $x \mapsto x^p$ provides another counterexample. Somehow the difficulty is that regular functions in $S$ need not come from pullback from regular functions in $X$, even though the vertical line test suggests that such maps should be "degree 1" in some sense.

Still, I feel like there should be some positive statement to be made here, though I was not able to find one after searching through a few algebraic geometry texts. For instance, if one demands that $X, Y, S$ are all smooth and that the field has characteristic zero, does the claim now hold? Ideally, I would like to only have conditions on the varieties $X,Y,S$ and not on the various maps between these varieties; for instance, I would prefer not to have to assume that the projection map from $S$ to $X$ is finite (though perhaps this is automatic?).

Answer by Piotr Achinger for Is there an algebraic geometry analogue of the closed graph theorem?

2012-11-19T19:29:27Z

You might be rediscovering Zariski's Main Theorem, which implies your statement in case $X$ is normal (or just weakly normal) and the projection from the graph $\Gamma$ to $X$ is proper and separable. What you really need is the map $\Gamma\to X$ to be an isomorphism, so the question is equivalent to "when is a bijective map an isomorphism"?

You already explained why we need (weak) normality (example with the cuspidal curve) and separability (the Frobenius map). To see why properness is also necessary, look at the map $\mathbb{A}^1\to\mathbb{A}^1$ sending $x$ to $1/x$ for $x\neq 0$ and sending $0\to 0$, whose graph is a union of $(0,0)$ and a hyperbola $xy=1$.

Is there an algebraic geometry analogue of the closed graph theorem? - MathOverflow

Is there an algebraic geometry analogue of the closed graph theorem?

Answer by Piotr Achinger for Is there an algebraic geometry analogue of the closed graph theorem?