# Are algebraic varieties strictly more general than (differentiable) manifolds?

I have read that every non-singular algebraic variety is a smooth manifold. However, I was wondering if every smooth manifold can be expressed as a non-singular algebraic variety, or even just a general algebraic variety; such that algebraic varieties are a strict generalization of manifolds.

If not, does not restricting algebraic varieties to be defined in terms of polynomial equations allow for a generalization?

(Any references on the relation of varieties and manifolds, and any practical use of such a relation, would be appreciated, as I'm new to algebraic geometry. Thanks!)

• – KCd
Jul 27, 2020 at 19:10

The classes of varieties and manifolds have nontrivial intersection, but neither one contains the other. A complex projective variety can have singularities which would disqualify it from being a complex manifold. If it is nonsingular, then using the Jacobian criterion we see that we get a complex manifold. For the converse, if we can holomorphically embed our manifold in $$\Bbb{P}^n$$ as a closed subset then by Chow's Lemma it can be given the structure of an algebraic variety.

Of course, there are complex manifolds that are not algebraic. For instance the Hopf Surfaces $$(\Bbb{C}^2\setminus \{0\})/\Gamma$$ with $$\Gamma$$ a discrete group acting freely admit no Kähler metric and cannot be projective.

Actually, Serre's GAGA paper gives a way to translate between analytic geometry and algebraic geometry in the complex case. However, the suitable "analytic" notion of algebraic variety is an "analytic space" which is like a complex manifold but allowing for singularities.

Once we stop working over $$\Bbb{C}$$, we notice that algebraic varieties make sense over any field $$k$$ whereas the notion of manifold pretty much needs one to use $$\Bbb{R}$$ or $$\Bbb{C}$$. (Apparently there are analogues of manifolds over $$\Bbb{Q}_p$$ but I don't know much about that.) So, the notion of variety is much more general in that being cut out by polynomial equations makes sense over any ring let alone a field; varieties are more "algebraically general" whereas manifolds are more general in that over $$\Bbb{R}$$ and $$\Bbb{C}$$ there are many functions that are smooth/holomorphic but not polynomial.

• For your variety to make sense, your base field should still be an algebraically closed field, no? If it's not a field wouldn't you use a scheme instead? Jul 27, 2020 at 20:28
• You're right that things get more complicated, but nothing stops you from defining varieties over non-algebraically closed fields. Jul 27, 2020 at 21:01
• Yes, I guess that's true in the sense that you can define them. Jul 28, 2020 at 5:36

Another useful piece of the story is the Kähler group problem: which finitely presented groups arise as fundamental groups of Kähler manifolds? Smooth projective varieties are Kähler manifolds, and I believe there are no known properties of fundamental groups that distinguish these classes. On the other hand, any finitely presented group arises as the fundamental group of a compact complex manifold, so the Kähler condition is crucial.