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The order dimension of a poset $(P,\leq)$ is the least number of linear extensions of $(P,\leq)$ such that the intersection of these extensions is $(P,\leq)$. The wikipedia entry provides some examples.

I know that there is quite a bit of research about this, but I haven't found anything concerning the following question:

Assume that $(P_1,\leq_1),\ldots,(P_n,\leq_n)$ are all partial orders and subspaces of $(P,\leq)$ such that they form a weak partition of $P$, that is, we have $\bigcup_i^n P_i = P$, but the posets are not necessarily pairwise disjoint.

As a variant of this, let us also consider the case in which we additionally require that $\leq$ is the smallest order-relation on $P$ that contains $\leq_1,\ldots,\leq_n$.

Assume that $P$ can be written as the weak partition of $n$ posets, where each of these posets has order dimension at most $k$. Does this tell us anything about the order dimension of $(P,\leq)$? Does it, perhaps, yield an upper bound? What if we take the variant?

Both cases are easy if all $n$ posets are pairwise disjoint or if $k =1$ (in which case it is just a covering of $(P,\leq)$ by chains). But it doesn't seem very easy if they intersect and we have $k \geq 2$, so I was wondering if anybody could point me towards some research that was done in this direction.

The case $k=2$ alone seems to be very interesting.

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The link to Wikipedia should be fixed. –  Samuele Giraudo Apr 15 '13 at 11:35
    
Thanks for pointing that out and sorry for that being necessary. I have fixed the link. –  Niemi Apr 15 '13 at 11:41
    
Related question: mathoverflow.net/questions/29169/… –  François G. Dorais Apr 15 '13 at 19:24

1 Answer 1

Here is what I think is a partial answer to the problem.

Assuming that $(P_1,\leq_1),…,(P_n,\leq_n)$ are all subspaces of $(P,\leq)$ and $P_1,\ldots,P_n$ form a weak partition of $P$. If all $n$ spaces have order dimension at most $k$, this does not yield a bound on the order dimension of $(P,\leq)$ in general.

Of course, it does yield an upper bound if we have $k=1$. Then, $P_1,\ldots,P_n$ are chains and this is known to cause that the order dimension of $(P,\leq)$ is at most $k$ (in fact, this is one of the most fundemental facts about the order dimension).

However, as soon as we have $k=2$, this is not true anymore, even if we require that $(P,\leq)$ is connected and has top and bottom. Take, for instance, $P(n) = \{0,a_1,\ldots,a_n,b_1,\ldots,b_n,1\}$ and define $\leq$ to be the order given by $x \leq y$ for $x \neq y$ if and only $y = 1$ or $x=0$ or $x = a_i$ and $y = b_j$ for some $i \neq j$. Then, for each $n \in \mathbb{N}$, $P(n)$ can be covered by three subspaces $(P_1,\leq_1),(P_2,\leq_2),(P_3,\leq_3)$, each of which is a tree. Trees have order dimension $2$, so $P(n)$ can always be covered by three subspaces of order dimension $2$. But now, $(P(n),\leq)$ has order dimension $n$ (in fact, it is the poset that is obtained by adding a greatest and least element to what is often called the standard example of a poset of order dimension $n$).

This, however, does not answer the version with the variant in which the transitive closure of $\bigcup \leq_i$ has to be $\leq$.

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