What is the difference between a variety and a manifold?

Question

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...

Answer 1

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).

Answer 2

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.

Answer 3

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 $\mathbf{C}$)