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An order isomorphism is any bijective map $f:X\rightarrow Y$ between posets such that $f(x)\leq f(y)$ if and only if $x\leq y$.

Let $[0,1]^n$ be a poset with the standard order: $(x_i)\leq (y_i)$ if and only if $x_i\leq y_i$ for all $i$.

If $\sigma\in S^n$ is a permutation then the map $g_\sigma:[0,1]^n \rightarrow [0,1]^n$ defined by $g_\sigma((x_i)) = (x_{\sigma i})$ is obviously an order isomorphism.

If we have $n$ strictly increasing bijective maps $f_i:[0,1]\rightarrow [0,1]$ then the product map $f=(f_i):[0,1]^n\rightarrow [0,1]^n$ is also an order isomorphism.

My question is, can any order isomorphism of $[0,1]^n$ be written as a composition of these types of order isomorphisms (this is obviously true for $n=1$)? (I'm not sure if we need to assume continuity as well or if that automatically follows from the map being an order isomorphism.)

It is not too hard to prove that if the order isomorphism is linear that it must be given by some $\sigma\in S^n$, so if for any order isomorphism we can find a product map such that the composition is linear we would be done, but I'm not sure if this can be done.

Any more general references on where I can find answers to questions similar to these would also be appreciated.

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My question is, can any order isomorphism of $[0,1]^n$ be written as a composition of these types of order isomorphisms?

Yes.

Call the first type of map a coordinate permutation and the second type a product automorphism.

Let $f\colon [0,1]^n\to [0,1]^n$ be a poset automorphism. It must also be a lattice automorphism, since the lattice operations are definable order-theoretically. The key fact you need about the lattice $[0,1]^n$ is this:

The nonzero join irreducible elements of $[0,1]^n$ are the tuples that are nonzero in exactly one coordinate.

In particular, the tuple ${\bf e}_i$ that is $1$ in the $i$th coordinate and zero in all others is a maximal join irreducible and the interval $[{\bf 0},{\bf e}_i]$ is a maximal chain of join irreducibles. Every tuple $(r_1,\ldots,r_n)$ is a join of at most $n$ join irreducibles, namely $(r_1,0,\ldots,0), (0, r_2,\ldots,0), \ldots$.

Since the finite set $E=\{{\bf e}_1, \ldots, {\bf e}_n\}$ is definable order theoretically as the set of maximal join irreducibles, it must be invariant under $f$. There is a coordinate permutation $\sigma$ that agrees with $f$ on $E$, hence $g:=\sigma^{-1}\circ f$ is the identity on $E$. The endpoints of the intervals $$ [{\bf 0},{\bf e}_i] = \{0\}\times \cdots\times [0,1]\times \cdots \times \{0\} $$ are fixed by the automorphism $g$, so $g$ restricts to an automorphism of each of these intervals. Let $g_i\colon [0,1]\to [0,1]$ be taken so that $g(0,\ldots,x,\ldots,0)=(0,\ldots,g_i(x),\ldots,0)$. Then $G = \prod g_i$ is a product automorphism which agrees with $g$ on any interval of the form $[{\bf 0},{\bf e}_i]$. That is, $G^{-1}\circ g = G^{-1}\circ \sigma^{-1}\circ f$ fixes every elements of every interval of the form $[{\bf 0},{\bf e}_i]$. But now the order automorphism $G^{-1}\circ \sigma^{-1}\circ f$ fixes every element of $[0,1]^n$, since every element of this lattice is a join of fixed points of $G^{-1}\circ \sigma^{-1}\circ f$. This shows that $G^{-1}\circ \sigma^{-1}\circ f(x) = x$, or $f(x) = (\sigma\circ G)(x) = (\sigma\circ (\prod g_i))(x)$.

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