You are right, even when the foliation does exist, achronicity demands it must be unique, so there is no freedom at all. The proof is as follows:
Consider a causal graph and suppose that a foliation with the given properties can be constructed. We construct the sequence of hypersurfaces (
$\Sigma_1$,
$\Sigma_2$,
$\Sigma_3$...). Now consider an event
$x \in \Sigma_i$.
If the graph is acyclic, there exists a longest future directed path connecting the initial event(s) with event
$x$. Let's say this path is long
$n-1$, meaning that it is composed of
$n$ nodes and
$n-1$ edges. Achronicity imposes that if two nodes are causally related, they must be in different foliations, therefore, since the number of nodes in the causal path is
$n$, there must be at least
$n$ hypersurfaces until the event
$x$. In other words
$i \geq n$.
There must also be a shortest future directed path connecting the initial event(s) with event
$x$. Let say this path is long
$m$, with
$m \leq n$. Now, the property 4 states that the Cauchy development of every hypersurface is the whole set of events. Therefore, every path connecting the initial event(s) with
$x$ must intersect every hypersurface
$\Sigma_j$ with
$j \leq i$. Since the shortest path contains
$m$ nodes, there must be at most
$m$ hypersurfaces until event
$x$. In other words,
$i \leq m$.
Therefore:
- If
$n \neq m$, the foliation cannot be constructed, contradicting the hypotheses.
- If
$i=n=m$, than the foliation is uniquely determined
Incidentally, this proof provides also an algorithm to obtain the foliation (or prove that it doesn't exit):
For every node in the causal graph, determine the length of the shortest and longest path from the initial state(s). If they are equal (i.e.
$i=n=m$), assign the node to
$\Sigma_i$. If they are different (i.e.
$n \neq m$), terminate the program: the foliation doesn't exist.
