MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

3 edited tags
2 added 53 characters in body

There's a fair amount of work on valuations on (modular) lattices, by which I mean functions $v : \mathcal{L} \rightarrow R$ that satisfy the modular expression $$v(x) + v(y) = v(x \wedge y) + v(x \vee y)$$

I'm wondering if there's been any work on vector-valued valuations (where the range of v is $R^k$ and the same relation holds) ?

In addition, I'm also interested in lower valuations (I'm not sure if this name is standard) that satisfy the submodular inequality $$v(x) + v(y) \ge v(x \wedge y) + v(x \vee y)$$ and possibly the generalization to $R^k$ where we replace the above by $$v(x) + v(y) \succeq v(x \wedge y) + v(x \vee y)$$ ($\succeq$ being the coordinate-wise partial order)

This is a reference request, for the most part.

1

Vector-valued valuations on lattices

There's a fair amount of work on valuations on (modular) lattices, by which I mean functions $v : \mathcal{L} \rightarrow R$ that satisfy the modular expression $$v(x) + v(y) = v(x \wedge y) + v(x \vee y)$$

I'm wondering if there's been any work on vector-valued valuations (where the range of v is $R^k$ and the same relation holds) ?

In addition, I'm also interested in lower valuations (I'm not sure if this name is standard) that satisfy the submodular inequality $$v(x) + v(y) \ge v(x \wedge y) + v(x \vee y)$$ and possibly the generalization to $R^k$ where we replace the above by $$v(x) + v(y) \succeq v(x \wedge y) + v(x \vee y)$$

This is a reference request, for the most part.