Is there an algebraic geometry analogue of the closed graph theorem? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T07:50:32Z http://mathoverflow.net/feeds/question/113858 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/113858/is-there-an-algebraic-geometry-analogue-of-the-closed-graph-theorem Is there an algebraic geometry analogue of the closed graph theorem? Terry Tao 2012-11-19T19:14:30Z 2012-11-19T19:29:27Z <p>In functional analysis, the <a href="http://en.wikipedia.org/wiki/Closed_graph_theorem" rel="nofollow">closed graph theorem</a> 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.</p> <p>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 <a href="http://en.wikipedia.org/wiki/Regular_map_%28algebraic_geometry%29" rel="nofollow">regular map</a>. (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.</p> <p>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?). </p> http://mathoverflow.net/questions/113858/is-there-an-algebraic-geometry-analogue-of-the-closed-graph-theorem/113860#113860 Answer by Piotr Achinger for Is there an algebraic geometry analogue of the closed graph theorem? Piotr Achinger 2012-11-19T19:29:27Z 2012-11-19T19:29:27Z <p>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"?</p> <p>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 $xy=1$.</p>