Let $$ X, Y \subset \mathbb{P}^N$$ be two non singular algebraic varieties of dimensions $k$ and $l$ that intersect transversally. Is it true that the ``dimension'' of the variety $\overline{X} \cap \overline{Y} - X\cap Y$ is strictly less than $k+l-N$, which is the dimension of $X\cap Y$ as a complex manifold. What I am worried about is that when you take the closure and then take intersections you may add singular things of very high dimension to $X\cap Y$.

I think it is true that the dimension of $\overline{X\cap Y}- X \cap Y$ is strictly less than $k+l-N$.

  • $\begingroup$ Ritwik -- Why do your varieties $X$ and $Y$ intersect transversally? Are you using Bertini's theorem? If so, then you can apply Bertini's theorem to the closures of $X$ and $Y$. -- Jason $\endgroup$ – Jason Starr Oct 23 '11 at 18:50

There are already two answers pointing out why your statement cannot hold as stated, so let's see if we can fix it.

Let $X, Y\subseteq \mathbb P^N$ be two irreducible (quasi-projective) algebraic varieties of dimension $k$ and $l$ respectively. Then $\overline X,\overline Y\subseteq \mathbb P^N$ are two closed irreducible algebraic varieties of dimension $k$ and $l$ respectively. By the Projective Dimension Theorem you obtain that

Every irreducible component of the intersection $\overline X\cap\overline Y$ has dimension at least $k+l-N$.

This implies that if your initial $X$ and $Y$ are disjoint, then your desired statement cannot hold.

On the other hand since you assumed that $X$ and $Y$ intersect transversally, basically you only need to worry about the complements, that is, the interesting intersections are $\overline X\cap (\overline Y\setminus Y)$ and $(\overline X\setminus X)\cap \overline Y$.

If you know that these intersections are transversal, then I think what you want follows.

A perhaps interesting consequence of this is that if those intersections are transversal, then $X\cap Y\neq \emptyset$.

  • $\begingroup$ Your last statement implies that the part after "and" in your second-to-last statement is unnecessary, no? $\endgroup$ – Will Sawin Oct 24 '11 at 17:49
  • $\begingroup$ @Will: Yes, I meant to take that out after I realized this, but obviously forgot. Thanks for pointing it out. $\endgroup$ – Sándor Kovács Oct 24 '11 at 19:13

I am not sure I understand. If $\overline{X},\overline{Y}$ are two smooth irreducible huypersurfaces and $X=\overline{X}-\overline{X}\cap\overline{Y}$ and similarly for $Y$, then $X,Y$ are smooth with empty intersection and of dimension $N-1$. But the intersection of the closures is just $N-2$.


Moreover, one can use that technique to get very far beyond tat bound.

Let $Z$ be an $n$-dimensional subspace of $2N$-space. Let $\bar{X}$ be an $n+1$-dimensional subspace including $Z$, and let $\bar{Y}$ be another $n+1$-dimensional subspace including $Z$. Then apply Mohan's trick to create an $X$ and $Y$ that intersect transversely, or, rather, not at all. Then the formula fails severely, as $n$ is much larger than $2$.


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.