Name of the order of vectors

Question

I define the following partial order $\succ$ of vectors: if $\mathbf{x}=(x_1,\dots,x_n)$ and $\mathbf{y}=(y_1,\dots,y_n)$ such that for some $i$ and $j$, $y_i y_j > x_i x_j$ and $y_i+y_j=x_i+x_j$, but for all $k \neq i,j$ we have $y_k=x_k$, then $\mathbf{y} \succ \mathbf{x}$.

In my application, vectors are of non-negative integers. So, that for example $(1,1,1) \succ (2,0,1) \succ (0,0,3)$, but $(2,2,2,0)$ and $(3,1,1,1)$ are not comparable.

Is there a name for such order? It comes out as a natural property of a proof in a larger project I'm working on and I do not want to reinvent a wheel if it is already known in some literature.

Answer 1

It looks like there won't be a definite answer to this question. Just to wrap up my current thinking on this: let $S(\mathbf{x}) = \sum_{i \neq j} x_i x_j$ and let $\mathbf{x} \succ_S \mathbf{y}$ iff $S(\mathbf{x}) > S(\mathbf{y})$. Then $\succ_S$ is a total order that expands $\succ$ for obvious reasons. It is also equivalent to many other reasonable orders, such as the one that minimizes the Euclidean distance from $\mathbf{0}$ or from $\mathbf{1}$, etc. But unfortunately, it is not the answer I was looking for as the ordering of elements that are not pairwise comparable (as in my definition) is not the one I'm looking for.

Answer 2

For vectors $(x_1,x_2)$ with $x_1+x_2=\text{const}$ and $x_1,x_2\geq 0$, the quantity $x_1x_2$ is maximized when $x_1=x_2$. Thus the smaller elements of this order are the ones where the components are more uneven.

In general, for vectors with non-negative coefficients, this order is in a sense measuring entropy. By entropy, I mean here how far the vector is from having all equal components. Another way to measure this notion of entropy is to look at the energy $$\left|(x_1,\dots,x_n)\right|^2 = x_1^2+\dots+x_n^2,$$ which is simply the square of the Euclidean norm. In fact, $y\succ x$ implies that $\left|y\right|< \left|x\right|$ and $\sum y_k=\sum x_k$.

To see this implication, consider $y=(y_1,\dots,y_n)\succ (x_1,\dots,x_n)=x$ such that there are no intermediate vectors in the definition of the partial order. To simplify the notation a bit, suppose $i=1$ and $j=2$, so $y_1y_2 > x_1x_2$, $y_1+y_2=x_1+x_2$ and $y_k=x_k$ for $k\geq 3$. Then $$y_1^2+y_2^2 = (y_1+y_2)^2-2y_1y_2 < (x_1+x_2)^2-2x_1x_2 = x_1^2+x_2^2,$$ which further implies that $$\left|y\right|^2= \sum_{k=1}^ny_k^2 = y_1^2+y_2^2+\sum_{k=3}^ny_k^2 < x_1^2+x_2^2+\sum_{k=3}^nx_k^2=\left|x\right|^2$$ and hence $\left|y\right|<\left|x\right|$. The condition $\sum y_k=\sum x_k$ is necessary because two comparable vectors under $\succ$ will by definition necessarily have the same sums.

So the partial order defined by $y\succ x$ is simply a restriction of the partial order $\left|y\right|<\left|x\right|$.