In his recent post Bartek demonstrated an example of two equivariant sets with atoms that are related by a finitely supported bijection, but not by any equivariant bijection. We show that if the sets are orbit-finite then this situation is impossible.
Proposition. Let and
be two orbit-finite equivariant sets with atoms. If there exists a finitely supported bijection
then there exists an equivariant bijection
.
Let be a finite set of atoms. An
-orbit of an element
is the set
We introduce some terminology regarding -orbits. An
-support of an element
contains those elements of the least support of
that do not belong to
, i.e.,
. An element which least support has an empty intersection with
is called
-free.
Recall that the dimension of an element is the size of its least support. We define an
-dimension of
to be the size of its
-support. Each element of a single
-orbit has the same
-dimension. Therefore, we can also talk about an
-dimension of an
-orbit.
In equality atoms, every single-orbit set (i.e. a set that consists of one
-orbit) is a disjoint union of finitely many
-orbits. Exactly one of these
-orbits contains all
-free elements of
.
Example. The single-orbit set of pairs of distinct atoms is a union of three
-orbits. The first one consists of
-free pairs (it has
-dimension 2), the second one consists of pairs with the atom
on the first coordinate (it has
-dimension 1), and the third one consists of pairs with the atom
on the second coordinate (it also has
-dimension 1).
Proof. Let be a bijection supported by a finite set
of atoms. Note that the function
has the following properties:
it maps
-orbits of
to
-orbits of
,
it preserves the
-supports of elements (and hence also the
-dimension).
Let us first deal with the case when is single-orbit. Take an
-free element
of
. It is not difficult to see that
maps
to an
-free element of
of the same
-dimension. We define an equivariant function
by
, where
is any permutation of atoms.
The function is well defined. Indeed, suppose that
. Since
and
are disjoint sets of atoms, there exist permutations
and
which act as
and
on
but are equal on
, i.e.,
. The bijection
is supported by
, so
Observe that agrees with
on the
-free arguments, so it is injective on the set of
-free elements of
. Therefore,
is injective on the whole set
.
We have shown that is an equivariant bijection between
and
. Hence, it induces a bijection between the set of
-orbits of
and the set of
-orbits of
. But the number of
-orbits of
is finite and equal to the number of
-orbits of
, which proves that
.
Now suppose that has an arbitrary number of orbits. We will define the equivariant bijection
for each orbit of
separately, starting from the orbits of the biggest dimension.
Let denote the number of
-orbits of
with the
-dimension
. We have that
for every
. Let
be the biggest
such that
. In both
and
there are
-orbits of
-dimension
. It is not difficult to see that each of them is
-free and belongs to a different orbit of
or
. Hence, in both
and
there are
orbits of dimension
, denoted
and
, respectively. The function
induces a bijective correspondence between these sets. For every pair of corresponding orbits
and
we define an equivariant bijection
similarly as in the single-orbit case above.
We call two sets with atoms -isomorphic if they are related by some
-supported bijection. Let
and
. Both those sets are equivariant. Moreover,
and
are
-isomorphic. This follows from the following characterization: two orbit-finite sets with atoms are
-isomorphic if and only if there exists a bijection between the set of
-orbits of
and the set of
-orbits of
such that the corresponding
-orbits are
-isomorphic. Hence, there exists an an
-supported bijection
and the same construction can be repeated.