# Orthogonal complements in exterior powers

I previously asked this on Mathematics Stack Exchange, to no result:

Consider the standard induced inner product structure on $$\wedge^k\mathbb{R}^d$$ given by defining $$\langle u_1\wedge \cdots \wedge u_k, v_1\wedge \cdots \wedge v_k\rangle:=\det ([\langle u_i, v_j\rangle]_{i,j=1}^d)$$ for simple $$k$$-vectors and extending to general $$k$$-vectors in the obvious fashion. If a linear subspace $$W$$ of $$\wedge^k \mathbb{R}^d$$ admits a basis of simple $$k$$-vectors, does its orthogonal complement have the same property?

The answer is no, there are counter-examples already with $$V:=\wedge^2\mathbb{R}^4\,(\cong \mathbb{R}^6)$$. In this case the simple bivectors form a smooth quadric $$Q\subset\mathbb{P}(V)$$. Take a line tangent to that quadric: the corresponding 2-plane $$P\subset V$$ contains only one simple bivector (up to a scalar), hence is not spanned by simple bivectors. Consider its orthogonal $$P^{\perp}$$. $$\mathbb{P}(P^{\perp})$$ intersects $$Q$$ along a quadric, which is at worst of rank 2, i.e. the union of 2 distinct planes. It is easy to find 4 points in this quadric such that no 3 of them are collinear; they correspond to 4 simple bivectors which form a basis of $$P^{\perp}$$. Hence $$W=P^{\perp}$$ is the required counter-example.
If you want an explicit example, take an orthonormal basis $$(e_1,\ldots ,e_4)$$ of $$\mathbb{R}^4$$, and take $$P=\langle e_1\wedge e_2, e_1\wedge e_3+e_2\wedge e_4 \rangle$$. It should be easy to find a basis of simple bivectors in $$P^{\perp}$$.