Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?).

share|improve this question
It has nothing to do with $Y$, right? The projection $S\to X$ is a regular map that is one to one and onto, and the question is: when can you say that the inverse is also regular? Isn't it basically when $X$ is normal? –  Tom Goodwillie Nov 19 '12 at 19:30
There is a closed graph theorem in real algebraic geometry, or more generally in an $o$-minimal setting. See Exercise 7, Sec. 6.1, page 97 in L. Van den Dries' book Tame topology and $o$-minimal structures, London Math. Soc. Lecture Notes Series 248 Cambridge Univ. Press, 1998. –  Liviu Nicolaescu Nov 19 '12 at 20:11
add comment

1 Answer

up vote 16 down vote accepted

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 $xy=1$.

share|improve this answer
Ah, yes, that makes sense. So, in particular, if we insist on $X$ and the graph being smooth projective varieties over characteristic zero, the analogue of the closed graph theorem should hold, as the hypotheses of smoothness, projectiveness, and characteristic zero should guarantee normality, properness, and separability respectively? (I guess in retrospect it seems reasonable to have "smooth projective variety over characteristic zero" be the analogue to "complete normed vector space over the reals".) –  Terry Tao Nov 19 '12 at 19:56
Or perhaps this formulation is even closer to the functional analysis closed graph theorem: if $X,Y$ are smooth projective varieties over characteristic zero, then $f: X \to Y$ is regular iff its graph is Zariski closed. –  Terry Tao Nov 19 '12 at 20:02
Yes, that's a good point. I like your analogy :). –  Piotr Achinger Nov 19 '12 at 20:02
this reminds me that once years ago when i substituted one day for a colleague teaching real analysis, i recall i thought of zariski's main theorem as motivation for the closed graph theorem. –  roy smith Nov 19 '12 at 20:25
There was a question mathoverflow.net/questions/78696/… with much discussion of different ways of thinking about Zariski's Main Theorem. –  Tom Goodwillie Nov 20 '12 at 0:18
show 3 more comments

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.