Simplest examples of nonisomorphic complex algebraic varieties with isomorphic analytifications - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T04:45:31Z http://mathoverflow.net/feeds/question/68421 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/68421/simplest-examples-of-nonisomorphic-complex-algebraic-varieties-with-isomorphic-an Simplest examples of nonisomorphic complex algebraic varieties with isomorphic analytifications Ravi Vakil 2011-06-21T18:46:09Z 2012-12-03T20:08:14Z <p>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. </p> <blockquote> <p>What are the simplest examples of nonisomorphic complex algebraic varieties with isomorphic analytificaitons?</p> </blockquote> <p>By "simplest", I mean by one of the following measures.</p> <ol> <li><p>(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.</p></li> <li><p>A known example that is simple to state, but may have a complicated proof. (Ideally there should be a reference.)</p></li> <li><p>An expected, folklore, or conjectured example. </p></li> </ol> http://mathoverflow.net/questions/68421/simplest-examples-of-nonisomorphic-complex-algebraic-varieties-with-isomorphic-an/68423#68423 Answer by ulrich for Simplest examples of nonisomorphic complex algebraic varieties with isomorphic analytifications ulrich 2011-06-21T19:09:09Z 2011-06-21T19:09:09Z <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/68421/simplest-examples-of-nonisomorphic-complex-algebraic-varieties-with-isomorphic-an/68425#68425 Answer by Georges Elencwajg for Simplest examples of nonisomorphic complex algebraic varieties with isomorphic analytifications Georges Elencwajg 2011-06-21T19:36:09Z 2011-06-22T11:39:08Z <p>Dear Ravi,<br> 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.</p> <p>1) $\mathbb G_m \times \mathbb G_m$<br> 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 $E$ : $$Pic (U)=Pic (E)$$</p> <p>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!</p> <p>Details can be found in Hartshorne's <em>Ample Subvarieties of Algebraic Varieties</em> Chapter VI, §3,p.232 (Springer, LNM 156). A link to an earlier discussion is <a href="http://mathoverflow.net/questions/634/a-complex-manifold-which-is-quasiprojective-in-two-different-ways/657#657" rel="nofollow">here</a> .</p> <p><strong>Edit:</strong> 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: $$Pic(\mathbb G_m \times \mathbb G_m)=0$$</p> <p>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.<br> 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 $V=\mathbb G_m \times \mathbb G_m$.</p> <p><strong>Second edit:</strong> Let us finally recall that the group of <em>analytic</em> line bundles on $\mathbb C^\ast\times \mathbb C^\ast$ is $\mathbb Z$, more precisely that the first Chern class is an isomorphism $$c_1:Pic_{an}(\mathbb C^\ast\times \mathbb C^\ast)\ =H^1(\mathbb C^\ast\times \mathbb C^\ast,\mathcal O^\ast)\stackrel {\sim}{\to} H^2(\mathbb C^\ast\times \mathbb C^\ast,\mathbb Z)=\mathbb Z$$<br> 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: $H^1(\mathbb C^\ast\times \mathbb C^\ast,\mathcal O)=H^2(\mathbb C^\ast\times \mathbb C^\ast,\mathcal O)=0$ </p> http://mathoverflow.net/questions/68421/simplest-examples-of-nonisomorphic-complex-algebraic-varieties-with-isomorphic-an/68444#68444 Answer by Minhyong Kim for Simplest examples of nonisomorphic complex algebraic varieties with isomorphic analytifications Minhyong Kim 2011-06-21T22:20:30Z 2011-06-22T15:55:44Z <p>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:</p> <hr> <p>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.</p> <hr> <p>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.</p> <hr> <p>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 $M$.</p> <p>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 </p> <p><em>continuum</em> of distinct algebraic structures </p> <p>on the analytic space $X\times A^1$.</p> http://mathoverflow.net/questions/68421/simplest-examples-of-nonisomorphic-complex-algebraic-varieties-with-isomorphic-an/68755#68755 Answer by Ravi Vakil for Simplest examples of nonisomorphic complex algebraic varieties with isomorphic analytifications Ravi Vakil 2011-06-24T18:39:42Z 2011-06-25T12:32:05Z <p>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 <a href="http://jones.math.unibas.ch/~kraft/Papers/Bourbaki.pdf" rel="nofollow">this wonderful paper of Hanspeter Kraft</a>, 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 $\mathbb{A}^3$).</p> <p>See also Ilya Nikokoshev's great question <a href="http://mathoverflow.net/questions/7603/topologically-contractible-algebraic-varieties" rel="nofollow">here</a> (which also links to Kraft's paper)</p> http://mathoverflow.net/questions/68421/simplest-examples-of-nonisomorphic-complex-algebraic-varieties-with-isomorphic-an/72831#72831 Answer by Isac Hedén for Simplest examples of nonisomorphic complex algebraic varieties with isomorphic analytifications Isac Hedén 2011-08-13T12:51:02Z 2011-08-14T05:08:37Z <p>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 <a href="http://arxiv.org/PS_cache/arxiv/pdf/0807/0807.4085v1.pdf" rel="nofollow">this</a> article that the Makar-Limanov invariant of its product with $\mathbb C$ is trivial. Maybe that helps.</p> http://mathoverflow.net/questions/68421/simplest-examples-of-nonisomorphic-complex-algebraic-varieties-with-isomorphic-an/115323#115323 Answer by Mohan for Simplest examples of nonisomorphic complex algebraic varieties with isomorphic analytifications Mohan 2012-12-03T18:45:48Z 2012-12-03T20:08:14Z <p>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.) </p>