Is the flip graph of every point set in $\mathbb R^3$ connected? If not, is there a set with an isolated node?
Def: For a point set $S$, the flip graph of $S$ is a graph whose nodes are the set of triangulations of $S$. Two nodes $T_1$ and $T_2$ of the flip graph are connected by an arc if one diagonal of $T_1$ can be flipped to obtain $T_2$.
Any idea would be appreciate.
These tags not available for me : flip-graph
According to this page at the Open Problems Project and other sources, the problem of whether the flip graph is connected is open for $\Bbb R^3$ and $\Bbb R^4$, and it is not even known whether it can have an isolated point.
C.L. Lawson showed in ‘Transforming triangulations’, Discrete Mathematics $3(1972)$, $365$-$372$ that the flip graph in $\Bbb R^2$ is connected, and Francisco Santos showed in ‘A point configuration whose space of triangulations is disconnected’, Journal of the American Mathematical Society, $13(2000)$, $611$-$637$, that in $\Bbb R^5$ it need not be connected.