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.
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(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.
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A known example that is simple to state, but may have a complicated proof. (Ideally there should be a reference.)
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An expected, folklore, or conjectured example.
Best Answer
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 $X\times A^1$.