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I hear people use these words relatively interchangeably. I'd believe that any differentiable manifold can also be made into a variety (which data, if I understand correctly, implicitly includes an ambient space?), but it's unclear to me whether the only non-varietable manifolds should be those that don't admit smooth structures. I'd hope there's more to it than that.

I've heard too that affine schemes are to schemes local coordinates are to manifolds, so maybe my question should be about schemes instead -- I don't even know enough to know...

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Oh, and clearly the topologies are vastly different. So, beyond that, I guess. (Although of course that's going to fundamentally change the nature of the games you play. But my question is just at the very lowest level.) –  Aaron Mazel-Gee Nov 5 '10 at 2:59
Who on earth do you know (who is not French) who uses the words "variety" and "manifold" interchangeably? –  Qiaochu Yuan Nov 5 '10 at 3:22
@Qiaochu, in Spanish we also use the same word, variedad. –  Mariano Suárez-Alvarez Nov 5 '10 at 3:45
My professor today used them interchangeably. He is French, which maybe explains something. Also, I said relatively interchangeably -- I realize they're not actually the same thing! But, e.g., people will use both "Grassmann manifold" and "Grassmann variety" to mean the same thing (on the level of points, I suppose). –  Aaron Mazel-Gee Nov 5 '10 at 5:49
It's the same in Hebrew, although one usually adds "algebraic" or "smooth" to avoid confusion. –  Pandora Jul 8 '14 at 10:53

4 Answers 4

up vote 15 down vote accepted

In English (as opposed to French, in which language variety and manifold are synonyms) the word variety is short for algebraic variety. The main differences, then, between (algebraic) varieties and (smooth) manifolds are that:

(i) Varieties are cut out in their ambient (affine or projective) space as the zero loci of polynomial functions, rather than simply as the zero loci of smooth functions. This gives them a more rigid structure. (Here I am thinking just of quasi-projective varieties; there are objects that people call varieties which can't be immersed into projective space, but there is no need to think about them when you are just learning the subject. Also, a manifold need not be regarded as lying in an ambient Eulcidean space, but can always be immersed into one, and can then be thought of as being cut out as the zero locus of smooth functions.)

The rigidity of varieties is reflected in the definition of isomorphism: we define two varieties to be isomorphic if we can find polynomial maps giving rise to mutually inverse bijections from one to the other, while two manifolds are isomorphic (i.e. diffeomorphic) if we can find smooth maps giving rise to mutually inverse bijections between them. It turns out, for example, that the only diffeomorphism invariant of a compact connected surface is its genus $g$, while if we look at smooth connected projective curves over the complex numbers (which, when we forget the variety structure and just think of them as manifolds, are compact connected surfaces --- note that one complex dimension gives two real dimensions) then the genus is not a complete invariant. For a fixed genus $g \geq 2$, there is a $6g-6$-dimensional family of non-isomorphic curves of genus $g$. (When $g = 1$ there is a $2$-dimensional family, and when $g = 0$, the variety structure is actually uniquely determined. Also, by "dimension" here I mean real dimension; but these families have their own natural algebraic variety structures, of half the dimension --- i.e. there is a $3 g - 3$ dimensional variety parameterizing isomorphism classes of genus $g$ curves when $g \geq 2$. Again the halving of dimension reflects the difference between real and complex dimension.)

(ii) Varieties can admit singularities, whereas we stipulate that manifolds be non-singular (i.e. locally Euclidean). Here it is useful to think about the fact that the critical locus of a (collection of) smooth function(s) can be pretty nasty, and so if we consider the zero loci of smooth functions and allow singularities, we will allow extremely nasty singularities. On the other hand, the critical locus of a (collection) of polynomial(s) is not so bad (e.g. it is always of codimension at least one in the zero locus), and so allowing singularities in the theory turns out to be okay (and in fact to be more than okay; it turns out to be one of the more powerful features of algebraic geometry).

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It really isn't necessary to refer to an ambient affine or projective space, is it? At least for affine varieties over an algebraically closed field k one can simply take the opposite category of the category of reduced finitely-generated integral k-algebras. (The point being that a choice of embedding into affine space is equivalent to a choice of generators, which one need not make.) –  Qiaochu Yuan Nov 5 '10 at 3:58
@Matt E: Isn't it also okay to allow singularities in the case of complex-analytic manifolds as well? –  user126 Nov 5 '10 at 4:00
@Qiaochu: Dear Qiaochu, When someone asks about the difference between varieties and manifolds, I don't think introducing the opposite category of finite type integral $k$-algebras is the best approach. If they are able to understand why this is a geometric notion at all, then they probably already understand precisely what varieties and manifold are, and what their similarities and differences are. In any case, I personally always prefer to begin by thinking of varieties as embedded in an ambient space. (E.g. this is the viewpoint I'm taking in my algebraic geometry course this quarter.) –  Matt E Nov 5 '10 at 4:08
@97832123: Dear 978, Complex analytic manifold would always mean "smooth" (in a holomorphic sense). But one can enlarge the category of complex analytic manifolds to complex analytic spaces, which can have singularities (or even nilpotents in the structure sheaf, if you want); these are (by definition) obtained by gluing zero loci of holomorphic functions. Again, though, the singularities of a collection of holomorphic functions are much better behaved than of random smooth functions, and so this is a workable notion. –  Matt E Nov 5 '10 at 4:12
@LeonLampret: Dear Leon, My somewhat vague memory is that there are theorems of Nash to this effect. This wikipedia entry seems to confirm my memory. (You need some hypotheses --- compact seems to be good enough --- and you maybe can't be sure that you have an affine algebraic set, but just a union of connected component of such.) Regards, –  Matt E May 7 '12 at 3:02

If you are French, a variety is a manifold.

If not, the connection is a bit more subtle. The correct analogue of a manifold is a scheme of finite type over a field. That is, it has a compatible covering by ring spectra (which is similar to the requirement that a manifold have a covering by compatible copies of euclidean space), and it lives over a field (the base space is an honest "point"). Here, I am being intentionally vague about "compatibility", since this would lead me into a discussion of sheaves, which I would rather not discuss unless you feel it's necessary.

The classical notion of a variety much more resembles the classical notion of a manifold (see, for instance, Milnor's book "Topology from a differentiable viewpoint"). The classical notion of a manifold is simply a locally closed subset of $\mathbf{P}^n(\mathbf{R})$ (or just $\mathbf{R}^n$, although looking at locally-closed subsets of projective space makes the analogy that much stronger), and the notion of a variety (quasi-projective, I guess) comes from considering locally closed subsets of a projective space.

Lastly, there is a theorem of Serre called GAGA (Géométrie algébrique et géométrie analytique), which says that in nice cases, we can "replace" complex varieties with complex-analytic manifolds to "do" cohomology and get a better picture of the real underlying space. (Here, by complex variety, we simply mean finite-type schemes over $\mathbf{C}$)

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in your last parenthetical sentence you probably want finite type, separated, irreducible, reduced (at least!) scheme over $C$, in general. –  Mariano Suárez-Alvarez Nov 5 '10 at 3:47
@Mariano: For the sake of argument, I refer to M. Hakim's thesis and her notion of topos-theoretic GAGA. –  user126 Nov 5 '10 at 3:51
You might also notice that I cheated in my "classical definition of a manifold", since the union of the two axes in $\mathbf{R}^2$ is not really a manifold, even though it is locally closed. However, the OP is asking "low-level" questions, and I'm not going to bother with telling the whole story if it's just going to fall on deaf ears. –  user126 Nov 5 '10 at 3:57
Thanks, this is roughly what I'm looking for. In fact, after looking at the wikipedia entry to which a search for GAGA takes me ("Algebraic geometry and analytic geometry"), I think that GAGA and Chow's theorem were really the point of my question to begin with. –  Aaron Mazel-Gee Nov 5 '10 at 6:02
And what does "sober" mean, btw? –  Aaron Mazel-Gee Nov 5 '10 at 6:30

A variety is usually defined to be the zero set of some polynomials. I don't know much about this, but I think people usually call manifolds that can be realized as varieties "algebraic manifolds". Take for example the polynomial $x^2+y^2-1$, then the zero set of this is $x^2+y^2=1$, or the circle $S^1$. So that is an algebraic manifold.

There can be varieties that are not manifolds, for instance, $y^2-x^2(x+1)=0$ is a "nodal cubic" and so it has a singularity at $(0,0)$. It can't be a manifold because it looks like "X" a cross at the origin so is not homeomorphic locally to $\mathbb{R}$.

There is a very useful theorem, sometimes called the Regular Level Set Theorem, which says that if you have a smooth map $M\to N$, then the level set of a regular value is a smooth manifold. So in this particular case, our polynomial $f:\mathbb{R}^n \to \mathbb{R}$ is a smooth map, and we care about whether or not zero is a regular value.

This amounts to checking whether or not the matrix $(\frac{\partial f}{\partial x_1}(c), \ldots , \frac{\partial f}{\partial x_n}(c))$ has rank 1 at all $c$ such that $f(c)=0$. i.e. if at least one of the partial derivatives is non-zero. This can easily be extended to work if your variety is defined by a bunch of polynomials $\mathbb{R}^n\to\mathbb{R}^m$.

This has a name, called the Jacobian criterion, and for varieties it says that if the Jacobian has full rank, then the variety is non-singular. But these two things are equivalent (Jacobian criterion for non-singular variety and the level set theorem to be a smooth manifold). Thus every non-singular variety is a smooth manifold (Just to be clear, this was considering "variety" to mean a zero set of a collection of polynomials over $\mathbb{R}$)

So to reiterate, some varieties are manifolds (if the defining polynomials satisfy a certain condition on partial derivatives) and some are not. Is every smooth manifold a variety? I think not, but it seems harder. A quick Google search turns up a result that seems to say every smooth compact manifold is algebraic.

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Dear Matt, This is a nice answer. When thinking about the question of whether every smooth manifold is (the real points of) a variety, it is helpful to remember that diffeomorphism of the real points is much weaker than isomorphism of varieties, so that the variety structure won't be uniquely determined by the manifold structure. This is what people have in mind when they talk about algebraic geometry being more rigid than differential topology. (And this rigidity is why it gets to be called a branch of geometry, rather than of topology.) –  Matt E Nov 5 '10 at 4:21

This is answer to 978...answer (I have not enough reputation to comment). He writes "Also, we can be jerks and do something silly: Every compact manifold admits the structure of an affine scheme, by Hochster's Theorem" This is not true because Hochster theorem writes that quasicompact opens sets must be base of opens sets and that is completely false for usual manifolds.

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fixed!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!‌​!!!!111111111111111111111111111111111111111111111111111111111111111111111 –  user126 Nov 5 '10 at 15:03
Your rep is now greater than 50, and you should now be able to comment. –  Guess who it is. Nov 5 '10 at 15:07
First comment is : thank you very much –  evgeniamerkulova Nov 5 '10 at 18:22

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