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?


1 Answer 1


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 $P^{\perp}$.


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