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

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 $$\text{span}(v_{i_1})=\text{span}(u_{i_1})$$.

• Consider splitting your basis of $\Lambda^2 V$ into 3+3, where every pair taken from each set of three have vanishing wedge product...? – AlexArvanitakis Jan 31 at 17:06
• 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? – მამუკა ჯიბლაძე Jan 31 at 19:27

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.