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A fellow grad student asked me this, I have been playing for a while but have not come up with anything. Note that $\mathbb{C}$ is homeomorphic to $\mathbb{C} - \{0\}$ in the Zariski topology - just take any bijection and the closed sets (finite sets) will biject as well. Concocting a similar thing for the plane is harder though.

I think I can show that the rational plane and the rational plane minus the origin are homeomorphic by enumerating the irreducible curves and using a back and forth argument, but I have not written it all up formally to see if I am missing something yet.

I know the question isn't natural from the point of view of algebraic geometry, because one of the objects isn't even a variety. I think it is still interesting just to see how weird the zariski topology really is.

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    $\begingroup$ Actually $\mathbb{C}^2 - (0,0)$ is a perfectly fine object in algebraic geometry--it's just a quasiprojective variety. The non-natural bit is that merely continuous maps really play no role in algebraic geometry. Not sure on the answer to your question though. $\endgroup$
    – Jack Huizenga
    Commented Oct 21, 2011 at 16:04
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    $\begingroup$ As an easier example, $\mathbb{C}^2$ is not homeomorphic to $\mathbb{CP}^2$, because any two 1-dimensional closed sets have nonempty intersection in the latter. You might be able to devise a more complicated intersection problem that obstructs homeomorphism in the question under consideration. $\endgroup$
    – S. Carnahan ♦
    Commented Oct 21, 2011 at 16:48
  • $\begingroup$ Can you cook up some way to distinguish the generic points of lines from those of general curves by generic intersection numbers under Bezout's theorem? Then you could say that before removing the point, "either intersecting trivially or being identical" is an equivalence relation on lines (cutting them into equivalence classes of parallel lines), whereas after removing the origin it is not (since $(y=0) \simeq (x=0) \simeq (x=1)$). $\endgroup$
    – Tyler Lawson
    Commented Oct 21, 2011 at 17:06
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    $\begingroup$ @Tyler I thought so or hoped so at first, but i's not that simple because there are lots of algebraic automorphisms of the affine plane, for instance $(x,y) \mapsto (x,y+p(x))$. $\endgroup$
    – Greg Kuperberg
    Commented Oct 21, 2011 at 17:12
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    $\begingroup$ I believe the question is equivalent to this: consider the simplicial complexes whose vertices are the (irreducible) curves in $\mathbb{C}^2$ (resp. $\mathbb{C}^2\setminus\{(0,0)\}$) with a face connecting certain curves when they have a nonempty intersection — are these two structures isomorphic? It may not be simpler that way, but it puts the emphasis on curves, and also suggests looking at it from the model-theoretic point of view: the back-and-forth argument can be phrased by saying in terms of an Ehrenfeucht-Fraïssé game between these structures. (And I'm running out of space.) $\endgroup$
    – Gro-Tsen
    Commented Oct 21, 2011 at 21:53

2 Answers 2

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This question was analyzed by Roger Wiegand in the context of algebraic surfaces over fields $k$ which are algebraic closures of finite fields.

[1] R. Wiegand, Homeomorphisms of affine surfaces over a finite field, J. London Math. Soc. (2) 18 (1978), no. 1, 28–32.

and in the followup paper:

[2] R. Wiegand, W. Krauter, Projective surfaces over a finite field. Proc. Amer. Math. Soc. 83 (1981), no. 2, 233–237.

In [1] he proves that the affine plane $A^2_k$ (over $k$) is Zariski-homeomorphic to any open nonempty subset of $A^2_k$ (Corollary 7). In particular, $A^2_k$ is homeomorphic to $A^2_k -\{(0,0)\}$. In particular, Greg's argument (as written) cannot be complete (as zero characteristic is not used anywhere in his sketch).

After proving the corollary Wiegand asks if the same holds for $k={\mathbb R}$ and ${\mathbb C}$.

In [2] it is proven (amon other things) that every proper nonempty open subset of $P^2_k$ is homeomorphic to either $P^2_k - \{point\}$, or to the affine plane $A^2_k$.

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  • $\begingroup$ Nice, Moishe. I assume it is also proved in [2] that $\mathbb{C}^2$ and $\mathbb{P}^2(\mathbb{C})-\text{point}$ are not homeomorphic. Could the method help deciding which one is hemoe to $\mathbb{C}^2-\text{point}$ (I cannot access the paper currently)? $\endgroup$
    – Uri Bader
    Commented Mar 4, 2017 at 8:05
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    $\begingroup$ @UriBader: $\mathbb{P}^2$, $\mathbb{P}^2-\{P\}$, $\mathbb{A}^2$ are pairwise nonhomeomorphic because, as argued in the penultimate paragraph of [2], the latter two spaces contain disconnected (field other than $\mathbb{C}$!) irreducible curves while $\mathbb{P}^2$ does not; and $\mathbb{P}^2-\{P\}$ contains curves that meet every other curve (e.g., any curve not through $P$) while every curve in $\mathbb{A}^2$, say $f=0$, is disjoint from some other curve, such as $f=1$. I don't see how it helps for $\mathbb{A}^2-\{P\}$. $\endgroup$
    – Zach Teitler
    Commented Mar 4, 2017 at 14:41
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$\newcommand\C{\mathbb{C}}$ In the Zariski topology on any quasiprojective variety over $\C$, curves are a distinguished class. They are minimal elements of the set of infinite closed subsets. (In fact, by this type of construction, you can define closed subvarieties in general, and recognize the dimension of a closed subvariety.) Then you can look at fibrations of a quasiprojective variety by curves, in particular fibrations of either $C^2$ or $\C^2 \setminus \{0\}$ by curves. In $\C^2$, there are many pairs of fibrations with the property that every pair of fibers meets in exactly one point. For instance, you can take the fibrations by any two families of parallel lines. The question is delicate because there are many algebraic automorphisms of $\C^2$, and therefore many Zariski self-homeomorphisms. (In fact the Jacobian conjecture is about algebraic automorphisms of $\C^2$.)

I don't think that this is possible in $\C^2 \setminus \{0\}$. I'm going to handwave some, but I think that it works. First of all I think that a fibration by curves has to be an algebraic family, only with its parameterization erased. Suppose that you have two fibrations $F$ and $G$ of $\C^2 \setminus \{0\}$ such that every pairs of fibers meets at one point. Then on each side, 0 is in the closure of every fiber. It's either that or it's in the closure of finitely many fibers. If it were in the closure of finitely many fibers, you would get two families of curves $F'$ and $G'$ in $\C^2$ such that the intersections leap in isolated places from 1 to 2, which is not possible. (The intersection cardinality is lower semicontinuous where it is finite.) On the other hand if every fiber of $F$ approaches 0, then I think that $F$ is a projective family of affine curves, I guess a Riemann sphere of curves. So then a fiber in $G$, which is not projective, would have an algebraic bijection with the projective parameter space of $F$, which is also not possible.

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