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.


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