This question was asked in my assignment on tensors and I am stuck on this question.
Question: Let $V$ be a vector space. An element $ w\in A^k (V)$ is called decomposable if $w = \phi_1 \wedge \phi_2 \wedge \ldots\wedge \phi_k$ for some $\phi_i \in A^{1} (V) $ for $1 \leq i \leq k$. Show that if $\operatorname{dim}(V) =3$ then every $w \in \Omega^2 (V)$ is decomposable.
$ w\in A^{k} (V)$ is given by $\alpha(v_1,...,v_n) = det[v_1,...,v_n]$ and $\Omega^{k}(M) $ denote the set of k-forms on a manifold M.
Attempt: I have been following the textbook: Introduction to Smooth Manifolds by John Lee along with my class notes. But I am not sure which result I should use. Can you please give a couple of hints?
Thanks!
The claim is not true as written.
Counterexample Suppose $V$ is real, let $E$ denote the Euler vector field (i.e., the infinitesimal generator of scalar multiplication), and let $\epsilon \in \bigwedge^3 V^*$ be a volume form on $V$. We claim that the $2$-form field $$\iota_E \epsilon = \epsilon(E,\,\cdot\,,\,\cdot\,)$$ is not decomposable. Suppose it were; since pullbacks commute with the wedge product, so would the pullback $\omega$ of $\iota_E \epsilon$ to a sphere centered at the origin, say, $\omega = \alpha \wedge \beta$. But $\omega$ vanishes nowhere, so neither does, say, $\alpha$, which contradicts the Hairy Ball Theorem.
The same argument works for a vector space $V$ of any odd dimension $\geq 3$.
On the other hand, we do have the following result:
Proposition Suppose that $\phi$ is a $2$-form field on a vector space $V$, and that $\phi_v$ for a given $v \in V$. Then, there is a neighborhood $U$ of $v$ on which the restriction $\phi\vert_U$ decomposes as a wedge product, for essentially the same reason that the claim is true pointwise.