14
$\begingroup$

I am having some trouble finding and/or understanding a general definition of subvarieties intersecting transversally. Assume that $Z_1,\ldots,Z_k$ are closed, irreducible subvarieties of a nonsingular algebraic variety $Y$.

Intuitively, I would say that these subvarieties intersect transversally if all varieties have pairwise intersection multiplicity one, i.e. $i(W,Z_i\cdot Z_j; Y)=1$ along any component $W$ of $Z_i\cap Z_j$ for $i\ne j$.

In our scenario, we should have $i(W,Z_i\cdot Z_j;Y)=\mathop{\mathrm{length}}_{\mathscr{O}_{Y,W}}\left(\mathscr{O}_{Z_i\cap Z_j, W}\right)$, if I am not mistaken.

Another intuitive definition would be that the tangent sheaves $\mathscr{T}_{Z_i}$ form a direct sum inside $\mathscr{T}_{Y}$. This seems to agree with the definition in this paper (in 5.1.2), but the author gives an equivalent definition which looks interesting:

For any $y\in Y$, there exists

  1. a system of parameters $x_1,\ldots,x_n$ on $Y$ at $y$ that are regular on an affine neighborhood $U$ of $y$ such that $y$ is defined by the maximal ideal $(x_1,\ldots,x_n)$ as well as
  2. integers $0=r_0 \le r_1 \le \cdots \le r_k \le n$ such that the subvariety $Z_i$ is defined by the ideal $I_i=\left(x_{r_{i-1}+1},\ldots,x_{r_i}\right)$ for all $1\le i\le k$.

Just for the record, what precisely does "defined by" mean here? I assume it means that $U\cap Z_i = Z(I_i)$ ... or do we actually get that $I_i=I(Z_i\cap U)$?

I would like to know if and how these three definitions are equivalent - there is no (general) treatment of this in Hartshorne or even in Fulton's book on Intersection Theory, which befuddled me greatly.

$\endgroup$
2
  • $\begingroup$ "defined by" must be scheme-theoretic, ie, $I_i=I(Z_i\cap U)$ $\endgroup$
    – quim
    May 20, 2011 at 9:03
  • $\begingroup$ Well, that is at least something. Maybe I will just work with that, it's so comfortably local. $\endgroup$ May 23, 2011 at 15:18

1 Answer 1

5
$\begingroup$

This naive answer is strictly intended to provoke a knowledgeable answer from someone. Transversal intersection at p should mean that all subvarieties are smooth at p, all contain p, and the codimension of the intersection of the tangent spaces equals the sum of the codimensions of the individual tangent spaces. ????

$\endgroup$
3
  • $\begingroup$ The only correction I'd make is: the codimension of the intersection of the Zariski tangent spaces equals the sum of the codimensions of the subvarieties at the point (so, intersections at the singular points won't be allowed). $\endgroup$
    – quim
    May 20, 2011 at 9:01
  • $\begingroup$ I thought my requirement that the points be smooth includes this? $\endgroup$
    – roy smith
    May 22, 2011 at 3:38
  • $\begingroup$ Well. Yea, this is yet another intuitive thing you would want a transversal intersection to satisfy, but as you already pointed out, it's not the answer to my question. I am somewhat startled that noone posted a follow-up. Thanks anyway! $\endgroup$ May 23, 2011 at 15:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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