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This is a cross-post.

Let $V$ be a $4$-dimensional real vector space. Let $\omega_{i_1,i_2}$ be a basis for $\bigwedge^2V$, where each $\omega_{i_1,i_2}$ is decomposable. Suppose that for every $\omega_{i_1,i_2}$, there is exactly one other basis element $\omega_{j_1,j_2}$ such that $\omega_{i_1,i_2} \wedge \omega_{j_1,j_2} \neq 0$.

Is $\omega_{i_1,i_2}$ necessarily a rescaling of a basis that is induced by a basis of $V$? i.e. do there exist a basis $v_i$ for $V$ and $\lambda_{i_1,i_2} \in \mathbb R$, such that $\lambda_{i_1,i_2}\omega_{i_1,i_2}=v_{i_1} \wedge v_{i_2}$?

We must allow a possible rescaling of the $\omega_{i_1,i_2}$: The "complementary" property is scale-invariant, but being an "induced basis" is not invariant:

Indeed, if $\omega^{i_1,\ldots,i_k}$ is an "induced basis" for $\bigwedge^kV$, and $\lambda_{i_1,\ldots,i_k} \omega^{i_1,\ldots,i_k} $ is also induced, 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 induced bases which remain induced are restricted.

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.


If you are interested, here is a proof for the rigidity of induced 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 see 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 reach $\text{span}(v_{i_1})=\text{span}(u_{i_1})$.

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  • $\begingroup$ Consider splitting your basis of $\Lambda^2 V$ into 3+3, where every pair taken from each set of three have vanishing wedge product...? $\endgroup$ – AlexArvanitakis Jan 31 at 17:06
  • 2
    $\begingroup$ Am I right that the following is a reformulation: given 6 lines in (projective) 3-space such that each of them intersects all others except exactly one, are they then edges of a tetrahedron? $\endgroup$ – მამუკა ჯიბლაძე Jan 31 at 19:27
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The $\omega_{i_1,i_2}$ give six distinct two-dimensional spaces $L_1,L'_1,L_2,L'_2,L_3,L'_3$ such that each pair has a one-dimensional intersection except for $L_1\cap L'_1=L_2\cap L'_2=L_3\cap L'_3=0.$

Let $A=L_1\cap L_2,$ $B=L_1\cap L'_2,$ $C=L'_1\cap L_2,$ and $D=L'_1\cap L'_2.$ Then $L_1$ contains $A$ and $B,$ but these are distinct because $L_2\cap L'_2=0.$ So $L_1=A\oplus B$ (direct sum of one-dimensional spaces). Similarly $L'_1=C\oplus D,$ $L_2=A\oplus C,$ and $L'_2=B\oplus D.$ The decomposition $V=A\oplus B\oplus C\oplus D$ will provide the required basis.

Suppose that $L_3$ does not contain $A.$ Then $L_3\cap L_1$ and $L_3\cap L_2$ are distinct, because $L_1\cap L_2=A.$ So $L_3=(L_3\cap L_1)\oplus (L_3\cap L_2)\subset A\oplus B\oplus C.$ Therefore $L_3\cap L'_1\subset (A\oplus B\oplus C)\cap (C\oplus D)=C.$

We have shown that $A\not\subset L_3$ implies $C\subset L_3.$ Swapping $L_1,L'_1,L_2,L'_2$ with $L'_1,L_1,L'_2,L_2$ respectively in this argument swaps $A,B,C,D$ with $D,C,B,A$ respectively. Swapping $L_1,L'_1,L_2,L'_2$ with $L_2,L'_2,L_1,L'_1$ respectively swaps $A,B,C,D$ with $A,C,B,D$ respectively. Swapping $L_1,L'_1,L_2,L'_2$ with $L'_2,L_2,L'_1,L_1$ respectively swaps $A,B,C,D$ with $D,B,C,A$ respectively. So either of $A\not\subset L_3$ or $D\not\subset L_3$ has the consequence that $B,C\subset L_3.$ This means $L_3=A\oplus D$ or $L_3=B\oplus C.$ And we can swap $L_3$ with $L'_3$ in the argument to get the same conclusion for $L'_3.$ This proves that $L_3$ and $L'_3$ are $A\oplus D$ and $B\oplus C$ in some order. This shows that the $\omega_{i_1,i_2}$ are a rescaling of the basis induced by elements of $A,B,C,D$ respectively.

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