0
$\begingroup$

I am remembering this from something I read a while ago, but I'm not sure how accurate this is and I would like clarification and appreciate explanations if possible.

Is the following correct? :

The reading defined a cone as the "conical hull" of finitely many points in Euclidean space, and if the points had rational coordinates the cone was called rational.

To such a rational cone, the lattice points contained in it form a monoid that is in fact finitely generated. If the cone was pointed, then the monoid had a unique minimal generating set, I think they called it a Hilbert basis.

Then we associated a monomial to each lattice point, so that the addition of lattice points corresponded to the multiplication of the corresponding monomials.

Do we consider sums of these monomials too or are they meaningless? What structure do these monomials have (are they a multiplicative monoid or something more)?

I think I remember the book considering the ideal generated by these monomials, and then considering the variety defined by this collection of polynomials. Would this have been something that the book might have done? What is the purpose of this?

$\endgroup$
2
$\begingroup$

This is called the theory of toric varieties. Canonical references are

Fulton: Introduction to Toric Varieties (very concise)

Cox, Little, Schenck: Toric Varieties (very detailed)

The point of this theory is that describing these varieties in terms of combinatorial data, i.e. cones, makes it straightforward to compute a huge number of things that typically in algebraic geometry are difficult or even intractable --- for example, cohomology groups of sheaves.

$\endgroup$
  • $\begingroup$ Nice to see you posting again! :) $\endgroup$ – André 3000 Jan 30 at 19:36
  • $\begingroup$ @André3000: thanks! $\endgroup$ – Asal Beag Dubh Jan 31 at 13:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.