Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

When speaking about blow-ups, I've seen that everybody says "let $Y \subset X$ be a smooth closed subvariety of codimension $r$..."

What happens when one want to blow-up subvarieties which are not of pure dimension?

You can always decompose $Y$ into a union of pure dimension subvarieties and blow-up them successively. Does the order matter? Or the assumption of $Y$ being smooth mean that these components cannot intersect?

share|improve this question

2 Answers 2

Blowing up commutes with passing to open sets in $X$. Therefore, if two components don't intersect the blowups commute. And yes, $Y$ smooth would imply the components are smooth and don't intersect. (According to some authors "smooth" should also imply that $Y$ is of pure dimension!)

Even if $Y$ is of pure dimension, but has intersecting components, the order can matter. Consider the parable of the orthant $\mathbb R_+^3$. If we use a planing device to shave off edge #1, then (less of) what remains of edge #2, we get a combinatorially inequivalent polyhedron than if we do #2 first, then #1. This polytopal picture illustrates (via the toric variety dictionary) what happens if we blow up one coordinate line, then the proper transform of another, in $\mathbb A^3$.

share|improve this answer

Note that the blow-up of the union of irreducible is not the same as the consecutive blow-ups of the components (except if they are disjoint, as Allen Knutson explained). For example the blow-up of two lines intersecting in the projective space is singular (check in coordinates) and the two ways of blowing-up the lines one after the other give two smooth threefolds.

share|improve this answer
    
Right, blowing up both together corresponds to planing off those two edges of the orthant the same amount, rather than the second one less than the first. The resulting 3-d polyhedron has a vertex of degree 4. –  Allen Knutson Feb 28 at 7:23

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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