# Prescribing the dimension of intersections of sub-vector spaces

Let $K$ be a field and $n$ a positive integer. To a finite dimensional $K$-vector space $V$, equipped with a family $V_1,\dots,V_n$ of subspaces, we attach the map $d:\mathcal P(\{1,\dots,n\})\to\mathbb N$ defined by $$d(S)=\dim\left(\bigcap_{s\in S}V_s\right).$$ [Here $\mathcal P(\{1,\dots,n\})$ denotes the set of subsets of $\{1,\dots,n\}$.] Then we have

$(1)\quad S\subset T\implies d(S)\ge d(T)$,

$(2)\quad d(S\cap T)\ge d(S)+d(T)-d(S\cup T)$

for all $S,T\in\mathcal P(\{1,\dots,n\})$. [Here $S\subset T$ means that each element of $S$ belongs to $T$.]

Conversely, given $d:\mathcal P(\{1,\dots,n\})\to\mathbb N$ satisfying $(1)$ and $(2)$, is there a family $(V,V_1,\dots,V_n)$ as above inducing $d$ in the way just described?

The answer is No in general, a counterexample being given as follows: $K=\mathbb F_2, n=4$, $$d(\varnothing)=2=\dim V,\quad d(\{i\})=1=\dim V_i,$$ and $d(S)=0$ if $S$ has at least two elements. [We use the fact that there are only three lines through the origin in $\mathbb F_2^2$.] There is an obvious analog for any finite field.

In view of the above observations, it seems natural to ask:

Assume that our field $K$ is infinite, and that $d:\mathcal P(\{1,\dots,n\})\to\mathbb N$ satisfies $(1)$ and $(2)$. Is there a family $(V,V_1,\dots,V_n)$ as above such that $$d(S)=\dim\left(\bigcap_{s\in S}V_s\right)$$ for all $S\ ?$

I find this easier to think about if we dualize everything: replace $d(S)$ with $\dim V - d(S)$ and each subspace with its annihilator in $V'$. Recall that the annihilator of a sum is the intersection of the annihilators. Thus your question is the following:
Assume $d:\mathcal{P}(\{1,\dots,n\}) \to \mathbb{N}$ is increasing and submodular (this is just the cool name for the condition $d(S \cup T) + d(S\cap T) \leq d(S) + d(T)$) and satisfies $d(\emptyset)=0$ and $d(\{1,\dots,n\}) \leq \dim V$. Then does there exists a collection of subspaces $V_i$ of $V$ such that $\dim \sum_{i \in S} V_i = d(S)$ for each $S \subset \{1,\dots,n\}$?
If we further assume that $d(\{x\}) \leq 1$ for each $x$ then this is exactly one of the equivalent definitions of a matroid, and your question is whether every matroid is linear. Apparently the answer is no.