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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!

Travis Willse
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  • See https://math.stackexchange.com/q/1784312/472818 and https://math.stackexchange.com/q/1314067/472818 – mr_e_man May 05 '22 at 12:47
  • @mr_e_man Unfortunately, I am not able to solve the question using the answers you mentioned. Can you please write an answer for this question? I shall be really thankful. –  Jul 04 '22 at 09:43
  • "Attempt: [...] I am not sure which result I should use." - See https://math.meta.stackexchange.com/a/27933/472818 – mr_e_man Jul 04 '22 at 17:34
  • Ted Shifrin's answer here can be adapted to your situation immediately: https://math.stackexchange.com/questions/1314067/are-all-k-vectors-in-mathbbr3-simple – Travis Willse Jul 09 '22 at 13:59
  • @TravisWillse I am really sorry but I am not able to understand how his answer can be used in my situation. Can you please elaborate your comment prefarably by writing an answer? I am not so good in study of manifolds. –  Jul 09 '22 at 16:44
  • To be clear, are you using to mean $\Omega^2(V)$ to be the set $\bigwedge^2 V^*$ of $2$-forms on $V$? If so, just pick any basis $e_1, e_2, e_3$ of $V$ and apply Ted Shifrin's proof verbatim. – Travis Willse Jul 09 '22 at 16:54
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    I insist that you read the linked math.meta answer, and update your post accordingly. In particular: What is $A^k(V)$? What is $\Omega^k(V)$? Are the components of $w$ (with respect to a basis of $V$) just numbers, or are they smooth functions of position in $V$? – mr_e_man Jul 10 '22 at 23:42
  • @mr_e_man Thanks, I will edit it soon. –  Jul 11 '22 at 06:58
  • @mr_e_man Kindly have a look now. –  Jul 11 '22 at 15:15
  • @TravisWillse I am sorry , I didn't meant that. I have edited my question to add the definition to avoid any confusion. Can you please help me? –  Jul 11 '22 at 15:23
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    @Avenger Are you certain that the question didn't ask about the decomposability of elements of $\bigwedge^2 V^*$, rather than that of $2$-forms, i.e., elements of $\Omega^2(M)$? Offhand, I don't think the latter is even true globally (though it is locally, i.e., in some neighborhood of each point), but the former can be proved quickly: https://math.stackexchange.com/questions/1196387/why-are-2-covectors-on-mathbbr3-decomposable – Travis Willse Jul 11 '22 at 16:05
  • Your definition of $A^k(V)$ doesn't make sense. – mr_e_man Jul 11 '22 at 18:21
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    @TravisWillse - Indeed it's not true globally. For example, if $w=x,e_y\wedge e_z+y,e_z\wedge e_x+z,e_x\wedge e_y$ is the Hodge dual of the position vector (that is the tangent bivector to the sphere centred at the origin), then a decomposition $w=\phi_1\wedge\phi_2$ would give $\phi_1$ as a nowhere-vanishing vector field tangent to the sphere. (I'm using the standard metric to identify vectors with covectors.) This contradicts the hairy ball theorem. – mr_e_man Jul 11 '22 at 18:29

1 Answers1

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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.

Pick linear coordinates $(x, y, z)$ on $V$, so that $$\phi = a(x, y, z) \,dy \wedge dz + b(x, y, z) \,dz \wedge dx + c(x, y, z) \,dx \wedge dy .$$ By relabeling variables we may as well as assume $a(v) \neq 0$. Then, on $U := \{w : a(w) \neq 0\}$, we have $$\phi\vert_U = (c \,dx - a \,dz) \wedge \left(-\frac{b}{a} dx + dy\right) .$$

Travis Willse
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