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I borrow notation from this answer. Let $X,Y\subseteq\mathbb{P}^{n}$ be two irreducible varieties over an algebraically closed field $k$. Consider $$ S^{0}_{X,Y}:=\{(x,y,z)\in X\times Y\times \mathbb{P}^{N}:x\neq y, z\in\langle x,y\rangle\}, $$ and $$ S_{X,Y}:=\overline{S^{0}_{X,Y}}\subseteq X\times Y\times\mathbb{P}^{N} $$ (where the line means Zariski closure). Let us consider the projections $$ \pi_{1}:S_{X,Y}\rightarrow X\times Y, $$
$$ \pi_{2}:S_{X,Y}\rightarrow \mathbb{P}^{N}. $$ The join of $X,Y$ is defined to be $$ S(X,Y):=\pi_{2}(S_{X,Y}). $$

Question: Is $\pi_2(S_{X,Y}^0)$ a quasi-projective variety?

Motivation: If you take $X=Y$, the join is equal to the secant variety of $X$. If $X$ is the Segre variety, then $\pi_2(S_{X,X}^0)$ is equal to the set of tensors of tensor rank at most $2$, and this can be extended to the set of tensors of tensor rank at most $r$ by taking more joins. My motivation is that I want to prove that the set of tensors of tensor rank at most $r$ is quasi-porjective. I have seen it mentioned that this is the case (see Landsberg; Tensors: Geometry and Applications, top of page 119), but I have been unable to prove it.

First attempt at a solution: It is clear that $S_{X,Y}^0$ is locally closed (i.e., quasi-projective). Since projection maps are open, $\pi_2$ is an open map. Since projective varieties are complete, $\pi_2$ is a closed map. The proof would be finished if $\pi_2$ sends locally closed sets (the intersection of an open set and a closed set) to locally closed sets (I would imagine such a map would be called locally closed, but I haven't found this definition anywhere).

Another idea is to show that $\pi_2|_{S_{X,Y}}:S_{X,Y} \rightarrow S(X,Y)$ is an open map. One way I know how to do this is by proving that it is flat, which in turn holds if the fibres are equidimensional and the input and output spaces are smooth. It would be nice not to have to assume smoothness here. (I don't think the join of two smooth varieties is smooth, in general).

Attempt 2: A completely different approach is to consider the abstract (or ruled) join of $X$ and $Y$ (see Flenner; Join Varieties and Intersection Theory, section 2):

$$ J_{X,Y}=\{[x:y]: \big((x \in X \text{ or } x=0) \text{ and } (y \in Y \text{ or } y=0)\big) \text{ and } \big(x,y \text{ not both zero}\big)\}\subseteq \mathbb{P}^{2N+1} $$

Then $\pi_2(S^0_{X,Y})=\pi (J_{X,Y} \setminus \Delta)$, where $$\pi : \mathbb{P}^{2N+1} \dashrightarrow \mathbb{P}^N$$ is given by $\pi [x: y] = [x-y]$, and $\Delta=\{[x:x] : x \in X \cap Y\}$.

This again expresses $\pi_2(S^0_{X,Y})$ as the image of a quasi-projective variety, which is in general only constructible, but maybe $\pi$ is ``locally closed" for some reason?

...

I am looking for completions of both of these attempts, but I would settle for one of the two!

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No.

The pre-closure of the join of two irreducible projective varieties is NOT necessarily quasi-projective.

Let $X$ be a smooth plane conic and let $Y$ be a single point of $X$. The pre-closure of the join of $X$ and $Y$ is the union of all the lines through points $x \in X$, $y \in Y$, $x \neq y$. Of course there is only one point in $Y$. So we get the union of all the lines through $Y$, except for the line through $Y$ which is tangent to the conic $X$. However we do get the point $Y$ itself, since it lies on all the other lines.

The union is the whole plane, minus one line, plus one point of that missing line. That's not quasi-projective.

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  • $\begingroup$ If that example is "weird", there are plenty of "normal" examples. For instance, let $X$ be the twisted cubic (a Veronese variety!). The pre-closure of the self-join of $X$ with itself is all of $\mathbb{P}^3$, minus the tangential surface of $X$, but then plus $X$ itself. So "pre-closures of secant varieties" are not necessarily locally closed. Specifically, this is the set of symmetric tensors of symmetric rank at most $2$. I would not hold out much hope that general tensors would behave any better. $\endgroup$ – Zach Teitler Apr 18 at 8:53
  • $\begingroup$ Or join a line to a point off it. Or two skew lines in $\mathbb{P}^3$. By the way, I’m going to speculate that some authors mix up (nonempty, irreducible) quasiprojective = open subset of irreducible variety = open dense subset “=“ dense subset. These preclosures are dense in their closures (tautologically), but not open in their closures. $\endgroup$ – Zach Teitler Apr 18 at 14:02
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Suppose first $X = Y = \mathbb{P}^n$. Let $V = \{x, y, L \,|\,x, y \in L \}$ where $L$ is a line in $\mathbb{P}^n$. Write $W = X \times Y = \mathbb{P}^n \times \mathbb{P}^n$ and $\Delta$ for the diagonal copy of $\mathbb{P}^n \subseteq W$. Then $V = Bl_\Delta W$ is the blowup of $W$ along the diagonal. The universal line bundle $\mathcal{O}(-1)$ has total space parametrizing points on the line, so contains your $S^0$ as the open subset away from the diagonal where $x \neq y$. If you want to allow the "point at infinity" on your line, you need to take the associated $\mathbb{P}^1$-bundle.

To restrict to $X,Y$ not necessarily equal to $\mathbb{P}^n$, pull back (take the preimage) of $X \times Y \subseteq \mathbb{P}^n \times \mathbb{P}^n$ along $V \to \mathbb{P}^n \times \mathbb{P}^n$.

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  • $\begingroup$ Thank you. This answer uses some unfamiliar topics to me, so it will probably take me some time to verify. Do you think that either of my solution attempts can be easily finished off? The first attempt seems so close -- we know that the projection is open and closed, and just need to prove that it preserves locally closed sets. $\endgroup$ – Ben Nov 19 '20 at 16:33
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    $\begingroup$ This answer is talking about $S^0$ being quasiprojective. The question was about whether the projection $\pi(S^0)$ is quasiprojective. $\endgroup$ – Zach Teitler Nov 20 '20 at 2:48

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