Any non-singular projective variety over $\mathbb{C}$ is easily seen to be a smooth manifold. Presumably the same is not true for algebraic varieties - one would not expect varieties with singular points to have a smooth structure. But do there exist non-singular varieties that are not smooth manifolds?
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$\begingroup$ The first sentence: seriously? How do you see that? (I may be just too dumb) $\endgroup$– user717Commented Dec 1, 2009 at 15:08
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$\begingroup$ Moreover, if X is a variety over C and X^{an} is the corresponding (reduced) analytic space, then X^{an} should be smooth if and only if X is smooth. Or am I absolutely wrong? (Sorry, if this is the case!) $\endgroup$– user717Commented Dec 1, 2009 at 15:15
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$\begingroup$ For an $n$-dimensional space, just take $U_i$ - the subset of points of the variety for which the $i$-th coordinates are non-zero - to be your atlas. It's explained in wikipedia - but when I think about it one may need non-singularity in this case also. $\endgroup$– Aston SmytheCommented Dec 1, 2009 at 15:22
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$\begingroup$ I don't understand your example. But in general: how do you check smoothness? This should work in both cases (algebraic and analytic) by taking an affine cover and then apply the Jacobi criterion. This is in both cases identical and so a point on the algebraic variety should be smooth if and only if the point in the analytic space is smooth. But maybe, I'm just missing something obvious... $\endgroup$– user717Commented Dec 1, 2009 at 15:28
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$\begingroup$ @Arminius: I edited the first sentence of this question by adding the phrase "non-singular". Otherwise, the first sentence was false. $\endgroup$– Daniel ErmanCommented Dec 1, 2009 at 15:30
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3 Answers
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Every non-singular algebraic variety over $\mathbb C$ is a smooth manifold. See for instance: http://en.wikipedia.org/wiki/Manifold under "Generalizations of Manifolds".
In fact, Arminius' suggested answer in the comments seems to give a proof of this fact, and I'll attempt to flesh it out a small amount. Every algebraic variety is locally a quasi-affine variety. So we may take an open cover $U_i$ of the variety, where each $U_i$ is a closed subset of an open subset of affine n-space. We may then check smoothness at each point of $U_i$ via the Jacobian criterion. The same procedure illustrates that each $U_i$ is a complex manifold. Since the gluing maps are algebraic, they are smooth, and hence our non-singular variety is also a smooth manifold.
answered Dec 1, 2009 at 15:48
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$\begingroup$ Why do you use "quasi-affine" instead of "affine"? $\endgroup$– user717Commented Dec 1, 2009 at 15:56
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$\begingroup$ I am not sure what terminology is standard here. I learned that "quasi-affine variety" refers to an open subvariety of an affine variety. For instance, I would refer to the affine plane minus the origin as a quasi-affine variety which is not affine. If I remember correctly, I also learned that an abstract variety is one that is locally quasi-affine. $\endgroup$ Commented Dec 1, 2009 at 16:47
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$\begingroup$ Yes, this is Hartshorne terminology, so it's standard :) My question was more, why you don't simply use affine opens. $\endgroup$– user717Commented Dec 1, 2009 at 16:55
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$\begingroup$ Daniel, better: quasi-affines are locally affine, so an abstract variety is locally affine. To see this, take a quasi-affine variety $U$, it's open in some affine $X$. Now, the complement of $U$, say, $V$, is defined by $f_1=\ldots=f_k=0$. Now look at the open sets $U_{f_i}$ in $X$. Their union will be the set where none are zero, and thus precisely the $U$ that you started with. For instance, if $U=\mathbb{A}^2\setminus\{(0,0)\}$, we can write it as $U_x\cup U_y$. So every abstract variety by your definition is locally affine. $\endgroup$ Commented Dec 1, 2009 at 17:46
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Every nonsingular variety over $\mathbb{C}$ is a smooth manifold, period. Take any affine open cover $X=\cup U_i$. Then each $U_i$ is a smooth manifold, and the transition maps are algebraic, so in particular, smooth. Thus, manifold.
answered Dec 1, 2009 at 15:49
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$\begingroup$ Are they also complex manifolds? $\endgroup$ Commented Dec 1, 2009 at 15:59
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$\begingroup$ Of course, everything is holomorphic! $\endgroup$– user717Commented Dec 1, 2009 at 16:02
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$\begingroup$ Yeah, I only said smooth because smooth was what was mentioned in the question. $\endgroup$ Commented Dec 1, 2009 at 16:05
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If $k$ is a complete valued field (e.g. $\mathbb{Q}_p$, etc.), one may define analytic manifolds over $k$ in the natural way. Precisely, these are topological spaces that locally look like open balls in $k^n$ and the transition functions must be analytic. Then the $k$-points of a smooth variety over $k$ is an analytic manifold (over $k$); Charlie's reasoning for the case $k = \mathbb{C}$ works for any $k$ as above.
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$\begingroup$ +1 for considering more general fields that e.g. number theorists care about $\endgroup$ Commented May 22, 2013 at 2:24