If they are not proper, two complex algebraic varieties can be nonisomorphic yet have isomorphic analytifications. I've heard informal examples (often involving moduli spaces), but am not sure of the references.

What are the simplest examples of nonisomorphic complex algebraic varieties with isomorphic analytificaitons?

By "simplest", I mean by one of the following measures.

  1. (best) An example whose proof is as elementary as possible, and ideally short. This of course requires proof that the complex algebraic varieties are nonisomorphic, and that the analytifications are isomorphic.

  2. A known example that is simple to state, but may have a complicated proof. (Ideally there should be a reference.)

  3. An expected, folklore, or conjectured example.


I believe the following is an elementary example: Let $X$ be an affine smooth curve of geometric genus at least one. Let $L$ be a non-trivial algebraic line bundle on $X$ (easy to produce such things). Then $L$ is analytically trivial because $X$ is a Stein space ($H^1(X, O_X)=0$) with trivial integral $H^2$. Hence, there is an analytic isomorphism $L\simeq X\times A^1$. We see that there is no algebraic isomorphism by noting:

Suppose there is an isomorphism $$f: L\simeq X\times A^1$$ of algebraic varieties. Then $L$ and $X\times A^1$ are isomorphic as algebraic line bundles.

Proof: For any fiber $L_y$ of $L$, if we consider the composite $$p\circ f: L_y\rightarrow X\times A^1 \stackrel{p}{\rightarrow} X$$ of $f$ with the projection, it must be constant, since $X$ is not rational. Hence, we have $$f(L_y)\subset z\times A^1$$ for some point $z$. Since the map $L_y\rightarrow A^1$ thus obtained is injective, it must be of the form $ax+b$ for non-zero $a$, that is, $f$ induces an isomorphism $$L_y\simeq z\times A^1.$$ Also by injectivity, we see that $y\neq y'$ implies $f(L_y)=z\times A^1$ and $f(L_{y'})=z'\times A^1$ for $z\neq z'$. Now let $s:X\rightarrow L$ be the zero section. Then $$\phi=p\circ f\circ s: X\rightarrow X$$ is an injective map, and hence, an automorphism. Thus, $$(\phi^{-1}\times 1)\circ f:L\rightarrow X\times A^1 \stackrel{\phi^{-1}\times 1}{\rightarrow} X\times A^1$$ is a map preserving the fibers of the projections to $X$. Now let $$q:X\times A^1 \rightarrow A^1$$ be the other projection, and $$h=q\circ (\phi^{-1}\times 1)\circ f\circ s.$$ So $h(y)$ is the image in $A^1$ under $(\phi^{-1}\times 1)\circ f$ of the origin of $L_y$. We use this function to get a fiber-preserving isomorphism $g: X\times A^1\simeq X\times A^1$ that sends $(y,\lambda )$ to $(y, \lambda-h(y))$. So finally, $$g\circ (\phi^{-1}\times 1)\circ f: L\simeq X\times A^1$$ is a fiber-preserving isomorphism of varieties that furthermore preserves the origins of each fiber. It must then be fiber-wise an isomorphism of vector spaces. Thus, it is an isomorphism of line bundles.

Added: The argument above can be easily modified to show that if $X$ is an irrational smooth curve and $L$ and $M$ are line bundles on $X$, then any isomorphism of algebraic varieties $$f:L\simeq M$$ is of the form $$f=T_s\circ \tilde{\phi} \circ g$$ where $$\tilde{\phi}:\phi^*M\rightarrow M$$ is the base-change map for an automorphism $\phi$ of $X$, $$g:L\simeq \phi^*M$$ is an isomorphism of line bundles, and $$T_s:M\rightarrow M$$ is translation by a section $s:X\rightarrow M$ of $M$.

Since the algebraic automorphism group of an affine irrational curve is finite, we see, by varying $L$, that for $X$ as above, there is in fact a

continuum of distinct algebraic structures

on the analytic space $X\times A^1$.

  • $\begingroup$ Very slick. And I'd never heard this before, and it is simple, and the one I will remember most easily. The remark that there are a continuum of distinct algebraic structures (and that the idea for this is so simple) is beautiful. $\endgroup$ – Ravi Vakil Jun 24 '11 at 18:32

Dear Ravi,
maybe the simplest example is one by Serre: the holomorphic Stein surface $\mathbb C^\ast\times \mathbb C^\ast $ underlies two non-isomorphic smooth complex algebraic varieties.

1) $\mathbb G_m \times \mathbb G_m$
2) An open subset $U\subset L$ of a $\mathbb P^1$-bundle $L$ on an an elliptic (complete!) curve $E$, obtained by deleting a section $S$ of said bundle: $U=L\setminus S$. That variety $U$ is not affine and has a huge Picard group, namely that of the elliptic curve $E$ : $$Pic (U)=Pic (E) $$

So you can use two concepts to prove that $U$ and $\mathbb G_m \times \mathbb G_m$ are not algebraically isomorphic: affineness and Picard. Actually you can use a third concept: just regular functions! Indeed $U$ has the strange property that its regular functions are constant:, just as if it were projective: $\Gamma(U, \mathcal O_U)=\mathbb C$ . But it is far, far from projective since its analytification is Stein!

Details can be found in Hartshorne's Ample Subvarieties of Algebraic Varieties Chapter VI, §3,p.232 (Springer, LNM 156). A link to an earlier discussion is here .

Edit: I forgot to say (but of course you know it better than I!) that $\mathbb G_m \times \mathbb G_m$ has trivial Picard group: $$Pic(\mathbb G_m \times \mathbb G_m)=0$$

The way I see it is that $\mathbb G_m \times \mathbb G_m=Spec (A)$ where $A=S^{-1}\mathbb C[X,Y]$ with $S$ the multiplicative monoid consisting of the $X^iY^j$'s. So $A$ is a UFD (since it is a ring of fractions of a UFD) and its spectrum thus has trivial Picard group.
A slightly more geometric formulation is that we have a surjective group morphism $Pic(S) \to Pic(V) \to 0$ valid for every open subset $V\subset S$ of a locally factorial scheme $S$ [Hartshorne, Algebraic Geometry, page 133]. Apply to $S=\mathbb A^2$ which has trivial Picard group and to $V=\mathbb G_m \times \mathbb G_m$.

Second edit: Let us finally recall that the group of analytic line bundles on $\mathbb C^\ast\times \mathbb C^\ast $ is $\mathbb Z$, more precisely that the first Chern class is an isomorphism $$c_1:Pic_{an}(\mathbb C^\ast\times \mathbb C^\ast)\ =H^1(\mathbb C^\ast\times \mathbb C^\ast,\mathcal O^\ast)\stackrel {\sim}{\to} H^2(\mathbb C^\ast\times \mathbb C^\ast,\mathbb Z)=\mathbb Z $$
This follows as usual from the long cohomology exact sequence associated to the exponential exact sequence $0\to\mathbb Z\to \mathcal O \to \mathcal O^\ast \to 0$ and from the vanishing of the cohomology groups of the coherent sheaf $\mathcal O$ due to Steinness of $ \mathbb C^\ast\times \mathbb C^\ast$, namely: $H^1(\mathbb C^\ast\times \mathbb C^\ast,\mathcal O)=H^2(\mathbb C^\ast\times \mathbb C^\ast,\mathcal O)=0$

  • $\begingroup$ Sorry, what's the isomorphism between $U$ and $\mathbb{G}_m \times \mathbb{G}_m$? $\endgroup$ – Dima Sustretov Jun 21 '11 at 19:58
  • $\begingroup$ The indefinite article in "a $\mathbb P^1$-bundle" is a little bit ambiguous as you can not just take any $\mathbb P^1$-bundle with any section, you have to choose them so that the complement is a non-trivial $\mathbb A^1$-bundle as in ulrich's answer (see my comment to it to get an explicit isomorphism). $\endgroup$ – Torsten Ekedahl Jun 21 '11 at 20:03
  • $\begingroup$ Dear Torsten, yes the indefinite article is ambiguous : here it doesn't mean "any" but "a certain bundle that I'm not describing in detail". As I wrote, details are to be found by following the link given, but ultimately I am responsible for any lack of clarity in my answer: so I both apologize and thank you for your constructive comment. $\endgroup$ – Georges Elencwajg Jun 21 '11 at 20:20
  • $\begingroup$ Dear Dmitry, the (analytic !) isomorphism is proved by pulling back the projective $\mathbb P^1$ bundle $L\to E$ to the analytic covering space $\mathbb C \to E$ where it becomes trivial.Then you see that the pull back of $U$ is just $\mathbb C \times \mathbb C$ and $U$ is the quotient of that $\mathbb C \times \mathbb C$ by a suitable action of $\mathbb Z \times \mathbb Z$. The quotient is $\mathbb G_m \times \mathbb G_m$ and this is why $U$ is isomorphic to $\mathbb G_m \times \mathbb G_m$ $\endgroup$ – Georges Elencwajg Jun 22 '11 at 7:43
  • $\begingroup$ As of the time I write this, there are a number of different answers (in two families) that have given me a great deal of enlightenment. Thank you Georges for this one! $\endgroup$ – Ravi Vakil Jun 24 '11 at 18:28

Let $E$ be an elliptic curve. The moduli space $M_E$ of line bundles with a connection on $E$ is an $\mathbb{A}^1$ bundle over $Pic^0(E) \cong E$. In particular, $E$ can be recoved from $M_E$ as the Albanese variety (of any compactification). As an analytic variety $M_E \cong(\mathbb{C}^{\times})^2$ since a complex line bundle with connection is exactly the same as a character of the fundamental group (which is $\mathbb{Z}^2$ in this case). So for all $E$ the analytifications are isomorphic whereas the moduli spaces are not isomorphic as algebraic varieties.

A similar construction works for higher genus curves as well and also for higher rank vector bundles. I believe this observation is originally due to Serre, but I do not know the precise reference.

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    $\begingroup$ This is clearly the most conceptual way to present this example but measure 1 is increased if one notices that we are dealing with the quotient of $\mathbb C^2$ by the group of translations generated by $(1,0)$ and $(\tau,1)$. On the one hand the quotient is an $\mathbb A^1$-torsor over $\mathbb C/(\mathbb Z 1+\mathbb Z\tau$ (and thus algebraic), on the other hand a $2$-dimensional complex vector space divided by the subgroup generated by a basis, i.e., $(\mathbb C^\times)^2$ which itself is algebraic. $\endgroup$ – Torsten Ekedahl Jun 21 '11 at 19:51
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    $\begingroup$ In fact, this example is the same as Georges's one! Indeed, the universal vector extension of an elliptic curve (used there) can be interpreted as the moduli space of line bundles with integrable connections on the (dual) elliptic curve. $\endgroup$ – ACL Jun 23 '11 at 7:27
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    $\begingroup$ @ACL. I was aware of this; my answer was posted before his... $\endgroup$ – ulrich Jun 23 '11 at 8:07
  • $\begingroup$ I found both this example enlightening, and also Torsten Ekedahl's conceptual explanation --- thanks! $\endgroup$ – Ravi Vakil Jun 24 '11 at 18:29
  • $\begingroup$ I should have added: Torsten's comment reminds me that I forgot another valuable metric: one with a simple conceptual underpinning. $\endgroup$ – Ravi Vakil Jun 24 '11 at 18:31

By one of those coincidences so common in mathematics, two days after I asked this question, something unrelated I was thinking about led me to this wonderful paper of Hanspeter Kraft, which points out (among many other things) that $\mathbb{C}^3$ has other algebraic structures, for example the hypersurface in $\mathbb{A}^4$ given by $x + x^2 y + z^3 + t^2 = 0$ (Peter Russell showed it was analytically $\mathbb{C}^3$, and Makar-Limanov showed that it is not algebraically isomorphic to $\mathbb{A}^3$).

See also Ilya Nikokoshev's great question here (which also links to Kraft's paper)

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    $\begingroup$ Leonid Makar-Limanov is a single person, math.wayne.edu/~lml $\endgroup$ – Victor Protsak Jun 24 '11 at 19:25
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    $\begingroup$ [comment replaced 2 days later to fix latex] Isac Hedén pointed out to me that what I wrote is wrong: Peter Russell showed that it is diffeomorphic to $\mathbb{C}^3$, not analytically isomorphic. It is not known if it is analytically isomorphic to $\mathbb{C}^3$. Isac is new to Mathoverflow, so if he adds an additional comment below, please upvote him as a welcome (should you wish), not this comment! $\endgroup$ – Ravi Vakil Aug 15 '11 at 16:42

I've been asking around a little, and nobody seems to be able to tell whether or not Russell's hypersurface is analytically $\mathbb C^3$. Adrien Dubouloz shows in this article that the Makar-Limanov invariant of its product with $\mathbb C$ is trivial. Maybe that helps.


If you look at local rings, it is easy to construct such examples. For example, if $X,Y$ are smooth of same dimension, then for any point $x\in X,y\in Y$, the completions of $O_{X,x}, O_{Y,y}$ are isomorphic, but of course the algebraic local rings are not necessarily isomorphic (for example, if $X,Y$ are not birational, then even their fraction fields are not isomorphic.)

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    $\begingroup$ @Mohan: this is a nice idea, but this does not give you varieties, only schemes (of not finite type). $\endgroup$ – Sándor Kovács Dec 3 '12 at 20:11

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