TITLE: Difference between rationally connected and $\mathrm{CH}_0\simeq \mathbb Z$
QUESTION [12 upvotes]: For a projective variety $X$ over an algebraically closed field of characteristic $0$, why the condition $\mathrm{CH}_0(X)\simeq \mathbb Z$ does not imply that $X$ is rationally (chain) connected ?

REPLY [19 votes]: I hope that my answer will make it clearer. If $\mathrm{CH}_0X=\mathbb{Z}$, not all points are necesserily connected by rational curves. For instance, fake projective planes satisfy conditions of Bloch's conjecture, but they don't contain rational curve at all (they are quotients of the complex ball). However, points should be "connected by rational curves" in some symmetric power of $X$. I mean that for every two points $P,Q$ of $X$ there are an effective cycle $T$ of degree $n$ and a map $f:\mathbb{P}^1 \to X^{(n+1)}$, such that $f(0)=T+P$ and $f(\infty)=T+Q$. (See Mumford "Rational equivalence of 0-cycles
on surfaces" and Fulton "Intersection theory").   
Let me give a simple example of a non-rationally connected surface with trivial Chow group. I am going to prove Bloch's conjecture for Enriques surfaces (it will be little different from the original proof by Bloch, Kas, Lieberman in "Zero cycles on surfaces with $p_g = 0$"). Let $X$ be an Enriques surface. It is well-known that $X$ admits an elliptic fibration of index 2 and its Jacobian fibration $J$ is a rational surface [See e.g. Beaville's book on algebraic surfaces].  There is a map $\pi:X^{(2)}_{\mathbb{P}^1} \to J$  from the second relative symmetric power to the relative Jacobian with fibers $\mathbb{P}^1$. Thus, by Graber-Harris-Starr theorem $X^{(2)}_{\mathbb{P}^1}$ is rationaly connected. We can map it to $X^{(2)}$ ( non-relative symmetric power) and we can embed $X$ to $X^{(2)}$ as the diagonal. The variety $X^{(2)}_{\mathbb{P}^1}$ is a rationaly connected divisor in $X^{(2)}$ which contains the diagonal. So, points of $X$ are connected by rational curves in the second symmetric power. Precisely, for every $P,Q \in X$ we  have $2P \sim 2Q$. By Roitman's theorem (proof in Bloch's lectures on cycles) $\mathrm{CH}_0X$ is torsion free. Hence, $P \sim Q$ and $\mathrm{CH}_0X= \mathbb{Z}$.