# Is every basis for $\bigwedge^kV$ satisfying a “complementary” property a rescaling of a “standard” basis?

This question was inspired by this beautiful answer:

Let $$V$$ be a $$4$$-dimensional real vector space. Let $$\omega_{i_1,i_2}$$ ($$1 \le i_1 < \ldots < i_2 \le 4$$) be a basis for $$\bigwedge^2V$$, where each $$\omega_{i_1,i_2}$$ is decomposable. Suppose the following property holds: For every basis element $$\omega=\omega_{i_1,i_2}$$, there is exactly one basis element $$\tilde \omega=\omega_{j_1,j_2}$$ such that $$\omega \wedge \tilde \omega \neq 0$$.

Must $$\omega_{i_1,i_2}$$ be a rescalig of a "standard" basis for $$\bigwedge^2V$$ induced by a basis of $$V$$? i.e. does there exist a basis $$v_i$$ for $$V$$, such that $$\omega_{i_1,i_2}=\lambda_{i_1,i_2}v^{i_1} \wedge v^{i_2}$$?

Note that we must allow a possible rescaling of the original basis: The "complementary" property is scale-invariant, but being a standard basis is not:

Indeed, if $$\omega^{i_1,\ldots,i_k}$$ is a standard basis for $$\bigwedge^kV$$, and $$\lambda_{i_1,\ldots,i_k} \omega^{i_1,\ldots,i_k}$$ is also standard, then the $$\lambda_{i_1,\ldots,i_k}$$ must be the $$k$$-minors of some diagonal $$d \times d$$ matrix. In other words, we have $$\lambda_{i_1,\ldots,i_k}=\sigma_{i_1}\cdot \ldots\cdot\sigma_{i_k}$$ for some $$\sigma_1,\ldots,\sigma_d \in \mathbb{R}$$. This implies that the $$\lambda_{i_1,\ldots,i_k}$$ cannot be chosen freely; there are non-trivial relations.

Thus, the rescalings of "standard" bases which remain standard are restricted.

Comment: The question can be asked for any even $$d$$, and $$k=d/2$$. I thought it would be easier to start with the simplest case.

For the interested reader, here is a proof for the rigidity of standard bases:

We shall prove that $$\lambda_{i_1,\ldots,i_k}$$ must be the $$k$$-minors of some diagonal $$d \times d$$ matrix.

Suppose that $$\omega^{i_1,\ldots,i_k} =v^{i_1} \wedge \ldots \wedge v^{i_k}$$ and $$\lambda_{i_1,\ldots,i_k} \omega^{i_1,\ldots,i_k} =u^{i_1} \wedge \ldots \wedge u^{i_k}$$ for some bases $$u_i,v_i$$ of $$V$$. Then, we have $$\text{span}(v_{i_1},\dots,v_{i_k})=\text{span}(u_{i_1},\dots,u_{i_k})$$, for every $$1 \le i_1 < \ldots < i_k \le d$$. This implies that $$u_i \in \text{span}(v_i)$$: Indeed, by switching between $$i_k$$ and $$i_{k+1}$$ in $$\text{span}(v_{i_1},\dots,v_{i_{k-1}},v_{i_k})=\text{span}(u_{i_1},\dots,u_{i_{k-1}},u_{i_k}), \tag{1}$$ we obtain

$$\text{span}(v_{i_1},\dots,v_{i_{k-1}},v_{i_{k+1}})=\text{span}(u_{i_1},\dots,u_{i_{k-1}},u_{i_{k+1}}). \tag{2}$$

By intersecting (1) and (2), we deduce that

$$\text{span}(v_{i_1},\dots,v_{i_{k-1}})=\text{span}(u_{i_1},\dots,u_{i_{k-1}}). \tag{3}$$

In the passage from $$(1)$$ to $$(3)$$ we have "removed" the last vectors $$v_{i_k},u_{i_k}$$. Continuing in this way, we can remove all vectors, until we get to $$\text{span}(v_{i_1})=\text{span}(u_{i_1})$$ as required.