## additive functions on a lattice

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Given a lattice $L$. Can one classify all functions $f:L\rightarrow \mathbb{R}$, that satisfy

$f(a \wedge b)+f(a\vee b) = f(a)+f(b)$.

Some examples are the the set of all finite subsets of a given set $S$. Then every such function is uniquely determined by the element $(f(\{s\}))_{s\in S}\in \prod_S\mathbb{R}$ plus the value on the empty set. Indeed this gives a vector space isomorphism from the set of all such functions to $\mathbb{R}^{|S|+1}$.

For other lattices there are also other additive functions arising naturally. For example if one considers the set of all natural numbers (without $0$) ordered by divisibility and assigns to a natural number (for a fixed prime $p$) the biggest $n$, so that $p^n$ divides this number. The logarithm is another example for such a function.

More general a ultrafilter on the underlying poset can also be viewed as such a function. So there are a lot of interesting examples. There are even more interesting examples like lattices of measurable sets and their measures, Euler-characteristic of subcomplexes and so on. So my question is: Can one classify the set of all those functions, probably in terms of filters / ultrafilters on the underlying poset?

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One example to keep in mind is the case when $L$ is a total order, e.g., $\mathbf{Z}$ with join and meet given by max and min. Then your condition on $f$ is empty, i.e. any function $f\colon \mathbf{Z} \to \mathbf{R}$ is additive. – Guntram Feb 15 2011 at 16:21
One characterization of modular lattices is that a graded lattice is modular if and only if its rank function is additive in your sense. Both of your examples arise in this way: the boolean lattice and $\mathbf N$ under divisibility are both modular, and the additive functions you have are obtained by altering the values of the rank function on the atoms of the lattice. This might give you some more interesting examples. – Dan Petersen Feb 16 2011 at 7:54

## 2 Answers

An additive function with the additional property that $f(a) < f(b)$ whenever $a < b$ (i.e., strict monotonicity) can exist only when the lattice is modular. A partial converse was mentioned in a comment above: If the lattice is modular and has finite height, so that it has an integer-valued rank function, then this rank function is additive. But modular lattices of infinite height need not support any strictly monotone additive function. Any additive function on a non-modular lattice factors through the projection to a modular quotient lattice. I believe all this information is in Birkhoff's classic book "Lattice Theory".

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Update: My answer is incorrect. See comments below.

This answer may be obvious, and it probably isn't what you were looking for, but I realized that for any lattice $L$, there is a subset $M$, in which any $f:M\rightarrow \mathbb{R}$ extends uniquely to an additive $f$ on $L$.

I can't think of any great characterization of this set, except to just pick it using the axiom of choice, i.e. well-order $L$ and build both the set $M$ of generators and $N$ of determined points. For each $x \in L$, add it to $M$ if $M\cup N$ doesn't contain a three elements of $a,b,a\wedge b,a \vee b$ where $x$ is the fourth. Else the value $f(x)$ of any additive $f$ is determined by the values of $f$ on the elements in $M\cup N$, which are in turn determined by the values on $M$. So add $x$ to $N$.

In the linear case, $M$ is just all of $L$. In the case of a finite Boolean algebra with atoms, then $M$ can be set of the atoms plus the bottom element.

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I don't buy it yet. For example if I consider a lattice having a smallest element 0, then 3 noncomparable elements a,b,c and a largest element d. It is a lattice. My well ordering is $a,b,c,0, d$. Then the set $M$ should be $\\{a,b,c,0\\}$. But not any function $M\rightarrow \mathbb{R}$ extends to a additive function. Note that every additive function on that lattice sends $a,b,c$ to the same value. – HenrikRüping Feb 16 2011 at 10:56
But the weaker claim does seem to hold, that every additive function is determined by its values on $M$. – Joel David Hamkins Feb 16 2011 at 11:44
@Hendrik - Yes, I seemed to have neglected the possibility of over-specification. Worst, consider the lattice $0,1,a,b,c,d$ where $a,b,c,d$ are incompatible and $0,1$ are the smallest, largest elements. This has no subset $M$ on which every $f:M \rightarrow R$ extends uniquely to an additive $f$ on the lattice. – Jason Rute Feb 16 2011 at 14:34
Actually, my above comment is incorrect as well. Knowing $a,0$ determines $b,c,d,1$. We must have $a=b=c=d$, and then $0+1=2a$. So maybe there is something salvageable about my argument. – Jason Rute Feb 16 2011 at 14:48
OK, I have it for the finite case at least. Your problem can be reduced to a system of linear equations with one variable for each node $a$ representing the value $f(a)$. Then, if $L$ is finite, so is the system, and we can pick a (possibly empty) set of variables, with corresponding nodes $a_1,\ldots,a_n$, that are free and the rest are determined from them. Hence if $M= \\{ a_1,\ldots,a_n \\}$, then any $f:M\rightarrow \mathbb{R}$ extends uniquely to an additive function $f$ on $L$. I don't know if this property is true of infinite systems of finite linear equations. – Jason Rute Feb 16 2011 at 17:05
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