# Is an open subscheme of a rationally connected variety, rationally connected?

Let $X$ be a projective, irreducible variety over an algebraically closed field (of characteristic zero) which is rationally connected. Is it true that any open dense subvariety of $X$ is rationally connected? If so, can we say that same for rationally chain connectedness?

By rationally (chain) connected we do not assume properness (similar to the definition in Kollar's "Rational curves on algebraic varieties"). We simply ask that two general points in the variety are connected by a (chain of) rational curve.

Any idea/reference is most welcome.

• When you write "rational curve" do you mean a dense open of $\mathbb{P}^1$? (If not, $\mathbb{P}^1 - S$ is not rationally connected when $S\neq \emptyset$.) – Ariyan Javanpeykar Sep 18 '18 at 11:43
• Can you please clarify your question? Affine space is an open dense subvariety of projective space, and there are no nonconstant morphisms from a projective variety (like a rational curve) to affine space. If the complement of the open is codimension $2$ and contained in the smooth locus, the result is straightforward (with your definition of rational connectedness). If the complement intersects the singular locus, things are tricky, but there are some results in this case as well (Keel-McKernan, Chenyang Xu, Zhiyu Tian, ...). – Jason Starr Sep 18 '18 at 11:48

If I have understood the OP's definitions correctly, the answer is no for rational chain connectedness. Let $E$ be a smooth genus $1$ curve in $\mathbb{P}^2$ and let $X \subset \mathbb{P}^3$ be the cone on $E$, with vertex $x_0$. For any $x_1$ and $x_2$ in $X$, the lines joining $x_1$ to $x_0$ and $x_0$ to $x_2$ are in $X$, so $x_1$ and $x_2$ are connected by a chain of rational curves.
However, let $Y = X \setminus \{ x_0 \}$. Let $\pi : Y \to E$ be the projection. For any rational curve $C \subset Y$, the map $\pi : C \to E$ maps a genus $0$ curve to a genus $1$ curve, so it is constant. Thus, all rational curves of $Y$ lie in fibers of $\pi$, and points in different fibers of $\pi$ cannot be connected by a chain of rational curves.
• If the singularities are klt then (over $\mathbb C$) rationally chain connected and rationally connected are the same see Corollary 1.8 of arxiv.org/pdf/math/0504330.pdf – Hacon Sep 18 '18 at 20:07