# Multilinear Algebra, finding $z \wedge z.$

Define $A^k(V)$ to be the set of all alternating multilinear functions from $V^k \to \mathbb{R}$. Consider the space $A^2(\mathbb{R}^4)$, does there exists $z\in A^2(\mathbb{R}^4)$ such that $z \wedge z \neq 0$?

I played around with $$z\wedge z (v_1,\dots,v_4) = 4z(v_1,v_2)z(v_3,v_4) + 4z(v_1,v_3)z(v_2,v_4) + 4z(v_4,v_1)z(v_3,v_2).$$

It didn't provide me much insight to the answer. I mean you could let $v_i = v_j$, but that still doesn't lead me anywhere. So after playing around with some algebra, I concluded this map works.

$$z(v_1,v_2) = v_1^1v_2^2 - v_1^2v_2^1$$

It looks like some variation of the determinant $$\det\begin{bmatrix} v_1^1 &v_2^1 \\ v_1^2 & v_2^2 \end{bmatrix}.$$

I verified that $z(v_1,v_2) = -z(v_1,v_2)$, but is that all or am I missing something big?

Denote by $\{e_i\}$ a basis of $R^4$, and by $\{e^i\}$ the dual basis, $e^j(e_i)=\delta^j_i$. Then you can check that e.g. for $z=e^1\wedge e^2+e^3\wedge e^4$ we have $z\wedge z= 2 e^1\wedge e^2\wedge e^3 \wedge e^4\neq0$.

EDIT: So let me spell this out in more detail. First note that I have not used any particular properties of the $e^i$s, and since for any base of $V$ (finite dimensional) I can find a basis of $V^*$ defined by the relation above, so the example I gave is valid for any basis. Let us get the sign right. By definition of exterior product and apart an arbitrary conventional overall normalization factor we have $$(z\wedge z)(v_1,v_2,v_3,v_4)=\sum_{\sigma}(-1)^\sigma z(v_{\sigma_1},v_{\sigma_2})z(v_{\sigma_3},v_{\sigma_4}),$$ where $\{\sigma_1,\sigma_2,\sigma_3,\sigma_4 \}$ is a permutation of $\{1,2,3,4\}$, $(-1)^\sigma$ denotes the parity of the permutation and the sum is over all the permutations of $\{1,2,3,4\}$. So a priori we expect $4!=24$ terms. By the antisymmetry property of forms, we can multiply by 4 and sum over the permutations with $\sigma_1<\sigma_2$, $\sigma_3<\sigma_4$, thus getting only $4\cdot3/2=6$ terms. You can write down them explicitly or note that taking e.g. $(v_1,v_2)$ as the argument of the first $z$ and $(v_3,v_4)$ as the argument of the second is the same taking first $(v_3,v_4)$ and then $(v_1,v_2)$ (this only holds because you are taking the product of $z$ with itself). Thus we van multiply by another factor of 2 and get only 3 terms: $$(z\wedge z)(v_1,v_2,v_3,v_4)=8[(-1)^{\sigma} z(v_1,v_2)z(v_3,v_4)+(-1)^\tau z(v_1,v_3)z(v_2,v_4)+(-1)^\rho z(v_1,v_4)z(v_2,v_3)],$$ where $\sigma$ is the permutation $\{1,2,3,4\} \rightarrow \{1,2,3,4\}$ which is clearly even, $\tau$ is the permutation $\{1,2,3,4\}\rightarrow \{1,3,2,4\}$ which is odd and $\rho$ is the permutation $\{1,2,3,4\}\rightarrow \{1,4,2,3\}$ which is even. Therefore we finally have $$(z\wedge z)(v_1,v_2,v_3,v_4)=8[z(v_1,v_2)z(v_3,v_4)- z(v_1,v_3)z(v_2,v_4)+ z(v_1,v_4)z(v_2,v_3)].$$

More generally, since $e^i\wedge e^i=0$, if you want a form wedge itself to be non-zero, it has to be a sum of more than one (non-zero) term (that is not sufficient, e. g. if $z=e^1\wedge e^2+ e^1\wedge e^3$ then $z\wedge z=0$). Actually it also needs to be of even degree as an odd form wedge itself is necessarily zero, can you see why?

• That's an answer I found as well, but am I wrong? Because my map is really simple…I mean your example suggests a generalization, which is nice as well. – Hawk Sep 22 '14 at 22:05
• I think that with my notation your map is $z=e^1\wedge e^2$ since $e^1\wedge e^2(v_1,v_2)= v_1^1 v_2^2-v^1_2v^2_1$. But then $z\wedge z=0$ so I think your $z$ is not what you were looking for. – GFR Sep 22 '14 at 22:08
• is there a mistake in my expansion then? Because my sum isn't $0$... – Hawk Sep 22 '14 at 22:20
• Yes, in your sum the second term should come with a minus sign. – GFR Sep 22 '14 at 22:47
• My friend got $2z(v_1,v_2)z(v_3,v_4) - 4z(v_1,v_3)z(v_2,v_4)$, I think he is missing a term. Am I missing a term? – Hawk Sep 22 '14 at 22:51

Hint: If you're familiar with the terminology of differential forms, try computing $\omega \wedge \omega$ where: $$\omega = x \ dx \wedge dy + y \ dz \wedge dw$$

• I was going to do that too, but I wanted to come up with a more "fundamental" answer, free of differentials. – Hawk Sep 22 '14 at 22:04
• Those aren't differentials; they're just a different notation (and, quite common) notation for basis covectors. – Muphrid Sep 22 '14 at 22:43
• Well.. if $x_i: \Bbb R^4 \to \Bbb R$ are the coordinate functions for $\Bbb R^4$, then fixed $p \in \Bbb R^4$, $(dx_i)_p: T_p\Bbb R^4 \to T_{x_i(p)}\Bbb R$ is the differential of $x_i$ at $p$. The map $dx_i$, which maps $p$ to $(dx_i)_p$ is called a $1-$form. – Ivo Terek Sep 22 '14 at 22:50
• @IvoTerek, David Bachman. But this question, I found it online. I've done enough applications of the wedge product, so I am trying out theoretical problems. – Hawk Sep 22 '14 at 23:07
• No. The form $dx_i$ takes a point $p \in \Bbb R^4$ to a linear functional $(dx_i)_p$ in $(T_p\Bbb R^4)^*$. Ok? – Ivo Terek Sep 23 '14 at 1:37