"Algebraic geometry is more rigid than differential geometry or topology"

Question

I am reading Robin Hartshorne's Algebraic Geometry. In chapter $1$, section $4$, he says:

Since an open subset of a variety is dense, this already carries a lot of information. In this respect algebraic geometry is more "rigid" than differential geometry or topology. In particular, the concept of birational equivalence is unique.

What does he mean? I did not understand what does he mean when he says that "algebraic geometry is more "rigid" than differential geometry or topology".

Answer 1

At the simplest level, one can say that differential geometry is more "flabby" because it admits things like partitions of unity, which are necessarily defined by nonanalytic functions. An algebraic geometry the defining transition functions are by definition at least rational and therefore analytic. An analytic function is completely determined by how it behaves on a dense set in a neighborhood of a point.

Answer 2

This is not a deep answer, but this is how I explain this claim to myself. Algebraic geometry is study of zeroes of polynomials and a polynomial is a very rigid object. For example, in the case of one variable if you know value of the polynomial on a finite set (of cardinality equal to degree) you know it everywhere.

In complex algebraic geometry we study zeroes of holomorphic functions. Holomorphic function is still rigid, again in one dimensional case it completely determined by its values on a set with a limit point.

In topology or differential geometry we work with continuous or differentiable functions and here we have a lot of freedom to change functions in some neighbourhood without affecting its behavior in the complement.