Let $X$ be a smooth projective variety with an action of $\mathbb{C}^{\*}$. Let us suppose that the set $X^{\mathbb{C}^{\*}}$ is finite. For $x \in X^{\mathbb{C}^{\*}}$, let $A\_{x}$ denote the attractor (under our $\mathbb{C}^{\*}$-action) to $x$ and let $R\_{x}$ denote the repellent. Take $x\_{1}, x\_{2} \in X^{\mathbb{C}^{\*}}$. Suppose that $x\_{2} \in \bar{A}\_{x\_{1}}$, where $\bar{A}\_{x\_{1}}$ is the closure of $A\_{x\_{1}}$.


Is that true that $x\_{1} \in \bar{R}\_{x\_{2}}$?



About the connection between repellents and attractors under a $\mathbb{C}^{*}$ action on a projective variety
Counterexample: Take the action of $\mathbb C^\*$ on $\mathbf P^2(\mathbb C)$ given by
$$
t\cdot[a\_0:a\_1:a\_2]=[a\_0:ta\_1:t^2a\_2].
$$
There are three fixed points $p\_0$, $p\_1$, $p\_2$ where $p\_0$ is a sink and $p\_2$ is a source. There are two orbits having $p\_1$ in its closure, namely the lines $L\_0$, $L\_2$ through $p\_0,p\_1$ and $p\_1,p\_2$, respectively.


Now let $X$ be the blow up of $\mathbf P^2(\mathbb C)$ in $p\_1$ with exceptional divisor $E$. Then $X$ has four fixed points $p\_0,p\_1',p\_1'',p\_2$ where $L\_0\cap E=\{p\_1'\}$ and $L\_2\cap E=\{p\_1''\}$. Moreover, $L\_0,E,L\_2$ are the only orbit closures not containing both $p\_0$ and $p\_2$. They are organized in a chain.


The fixed point $x\_1:=p\_0$ is still a sink. So, the closure of its attracting set is all of $X$. In particular, it contains $x\_2:=p\_1''$. On the other side, the closure of the repelling set of $p\_1''$ is $E$ which does not contain $x\_1$.


