Let \(V\) be a sufficiently large finite-dimensional vector space over \(\mathbb{F}_q\), and let \(M \leq V\) be a fixed \(m\)-dimensional subspace. Given a \(\delta\)-coloring of the \(1\)-dimensional subspaces of \(V\), determine for which parameters \((m,k,l,q,\delta)\) one can always find a \(k\)-dimensional subspace \(K \leq V\) satisfying
\[
\dim(M\cap K)=l,
\]
such that all \(1\)-dimensional subspaces of \(K\setminus M\) receive the same color.

Remark.

The condition \(l=0\) is trivial, while counterexamples exist when the condition is \(0\leq l\leq k-2\). Thus the main question is to identify the correct geometric conditions under which such a subspace \(K\) must exist.

What are we concerned about?

Consider first the case of \(\delta=2\) and \(k=2\). Let \(S\) be a fixed subspace of codimension \(2\). In general, a \(2\)-dimensional subspace \(T\) disjoint from \(S\) need not exist. Indeed, there are exactly \(q+1\) hyperplanes containing \(S\), and every such \(T\) must intersect each of them. Coloring all points in \(x\) of these hyperplanes red and all points in the remaining \(q+1-x\) hyperplanes blue yields a coloring with no monochromatic \(T\).

Now suppose that \(S\) has codimension \(3\). Projecting onto a complementary \(3\)-space shows that the existence of a monochromatic \(2\)-space is equivalent to the inequality
\[
R_q(2,2)\leq 3,
\]
which holds precisely when \(q=2\). In that case, any complementary \(3\)-space \(R\) contains a monochromatic \(2\)-subspace \(T\). More generally, the same argument shows that one should require
\[
\operatorname{codim}(S)\geq R_q(2,2).
\]
An analogous projection argument applies when \(T\) is required to intersect \(S\) in a \(1\)-dimensional subspace: one considers a subspace \(R\) of dimension \(R_q(2,2)\) meeting \(S\) in a hyperplane and then applies the Ramsey property inside \(R\).