If you take the closure of two smooth varieties and then take their intersections, is the singular locus still small?

Question

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 $k+l-N$.

Answer 1

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 $X\cap Y\neq \emptyset$.

Answer 2

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 $N-2$.

Answer 3

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 $2$.