# Linear relations among wedge products.

Let $v_1,v_2,v_3,v_4$ be $4$ vectors in a two dimensional space $V$. Then one can work out by hand that: $$(v_1\wedge v_2)(v_3\wedge v_4) + (v_1\wedge v_3)(v_4\wedge v_2) + (v_1\wedge v_4)(v_2\wedge v_3) = 0$$ where multiplication is identified with the dot product since $\Lambda^2(V)$ is a one dimensional space.

Is there a conceptual reason for why this is true? Note that there is a lot of symmetry. For instance, the indices correspond to the three permutations: $$(1,2,3,4), (1,3,4,2), (1,4,2,3)$$ which form a $3$ cycle in $S_4$. One might also flip one of the indices and then with alternating signs, it seems to be the determinant of some matrix - but which one?

There should also probably be a generalization to arbitrary $n$-dimensional vector spaces with $m$ vectors and certain (cyclic?) subgroups of $S_m$...

• I have a question about notations used in your post. What do you mean when you write two 2-forms in parentheses next to each other? For example $$\left(v_1\wedge v_2\right)\left(v_3\wedge v_4\right) \stackrel{?}{=}\left\langle v_1\wedge v_2,\,v_3\wedge v_4 \right\rangle$$ From the post it seems to me that this stands for inner product of two forms. However, I am not clear on how the inner product is defined on $\Lambda^2\left(V\right)$. – Vlad Feb 29 '16 at 13:32
• Oops, that should be a $\lambda^2(V)$. This is a one dimensional space and you just identify it with the base field. – Asvin Feb 29 '16 at 13:33
• I am still confused. Is $\left(v_1\wedge v_2\right)\left(v_3\wedge v_4\right) =\left(v_1\wedge v_2\right)\wedge\left(v_3\wedge v_4\right)$? – Vlad Feb 29 '16 at 13:35
• So if $v_i = a_ie_1 + b_1e_2$, then $v_i\wedge v_j = (a_ib_j-a_jb_i)e_1\wedge e_2$. To multiply two such things, you just forget about the $e_1\wedge e_2$. – Asvin Feb 29 '16 at 13:36

Here is my interpretation of the equality, although it may be short in "conceptual" part.

According to this Wikipedia article inner product for 2-forms can be defined as determinant:

\begin{align} \big(a\wedge b\big)\big(c\wedge d\big) &= \big\langle a\wedge b,c\wedge d \big\rangle = \det \begin{pmatrix} \left\langle a, c \right\rangle & \left\langle a, d \right\rangle \\ \left\langle b, c \right\rangle & \left\langle b, d \right\rangle \end{pmatrix} \end{align}

so that expression $$\,\left(v_1\wedge v_2\right)\left(v_3\wedge v_4\right) + \left(v_1\wedge v_3\right)\left(v_4\wedge v_2\right) + \left(v_1\wedge v_4\right)\left(v_2\wedge v_3\right) = 0\,$$ can be rewritten as

\begin{align} \begin{vmatrix}w_{13} & w_{14} \\ w_{23} & w_{24}\end{vmatrix} + \begin{vmatrix}w_{14} & w_{12} \\ w_{34} & w_{23}\end{vmatrix} + \begin{vmatrix}w_{12} & w_{13} \\ w_{24} & w_{34}\end{vmatrix} = 0 \end{align}

where $\,w_{ij} := \left\langle v_i, v_j \right\rangle = \left\langle v_j, v_i \right\rangle$.

One can see that in the last expression the main diagonal of every matrix coincides with the anti diagonal of the following matrix, therefore all the summands cancel out. More specifically,

\begin{multline} \begin{vmatrix}w_{13} & w_{14} \\ w_{23} & w_{24}\end{vmatrix} + \begin{vmatrix}w_{14} & w_{12} \\ w_{34} & w_{23}\end{vmatrix} + \begin{vmatrix}w_{12} & w_{13} \\ w_{24} & w_{34}\end{vmatrix} = \\ \big(\color{#e50000}{w_{13}w_{24}} \color{#0Dee08}{ - w_{14}w_{23}}\big) + \big(\color{#0Dee08}{w_{14}w_{23}} \color{blue}{ - w_{12}w_{34}}\big) + \big(\color{blue}{w_{12}w_{34}} \color{#e50000}{ - w_{13}w_{24}}\big) = 0 \end{multline}

I understand that this explanation is not too different from explicitly computing expression by hands, but I believe the sum of three determinants can be viewed as row/column expansion of determinant of a larger matrix of specific form. I could not come up with such a matrix on a spot, but I also could not prove that such a matrix does not exist.

• This is more or less the explicit computation I had in mind. I tried finding a determinant that would give the result too but with no success. – Asvin Feb 29 '16 at 15:25