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I've been both trying to prove and looking for a proof in a couple of book and on the Internet, and I can't find it.

How can I prove that a $k$-$1$ differential form defined on a $k$ dimensional manifold is simple? That is, it can be written as a single wedge product of $k$-$1$ $1$-forms?

MyUserIsThis
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  • Dont they both have dimension $k$ ? – Rene Schipperus Jun 19 '14 at 05:16
  • @ReneSchipperus Yes, but is that enough? I don't see it. – MyUserIsThis Jun 19 '14 at 05:18
  • $k-1$ wedges are $k-1$ forms, so in the space of all $k-1$ forms you have a subspace of $k-1$ wedges and the spaces have the same dimension. – Rene Schipperus Jun 19 '14 at 05:19
  • @ReneSchipperus Hm? I see they must be a linear combination of the $k$ linearly independent $k$-$1$ forms, but I don't see it. I promise I've given it many thoughts. And the fact I've always seen it stated without proof makes me feel stupid xD – MyUserIsThis Jun 19 '14 at 05:22
  • @ReneSchipperus I'm sorry, in your first comment, with both you were talking about the space of $k$-$1$ forms and what else? – MyUserIsThis Jun 19 '14 at 05:24
  • Yeah you are right its not clear yet. – Rene Schipperus Jun 19 '14 at 05:24
  • If you fix an inner product and an orientation on $\mathbb R^k$ then you can use the Hodge star to interpret a $k-1$-form as a $1$-form. Then choose an oriented orthonormal basis so that this $1$-form is $C dx^1$, and the original $k-1$-form will then be $C dx^2 \wedge \cdots \wedge dx^k$. This works for a single vector space, I haven't thought about how much of an issue there is generalizing it to manifolds. – Anthony Carapetis Jun 19 '14 at 05:33
  • I see that particular case, thank you. I was trying to relate the fact that 1 forms are obviously simple and that the space they span has the same dimension as the space of k-1 forms. – MyUserIsThis Jun 19 '14 at 05:46

2 Answers2

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Just using vector notation, we show that every $k-1$ form on a $k$ dimensional space is a wedge of $1$ forms. The key point is that the sum of two wedge forms is again a wedge form. (think the term is decomposable).

But this is trivial for if you have a the sum of two $k-1$ wedges,

$$x_1\wedge \cdots \wedge \hat{x}_i \wedge \cdots \wedge x_k +x_1\wedge \cdots \wedge \hat{x}_j \wedge \cdots \wedge x_k =\pm(x_i+x_j)\wedge x_1\wedge \cdots \hat{x}_i \cdots \hat{x}_i \cdots \wedge x_k $$

Now you can just write the same thing in the cotangent space of a manifold with functions as coefficients.

Rene Schipperus
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For any vector space $\mathbb{V}$ there is a canonical vector space isomorphism $\Phi: \Lambda^{n - 1} \mathbb{V}^* \to \mathbb{V} \otimes \Lambda^n \mathbb{V}^*$ whose inverse is the contraction map. So, since $\dim \Lambda^n \mathbb{V}^* = 1$, given $\alpha \in \Lambda^{n - 1} \mathbb{V}^*$, we can write $\Phi(\alpha)$ as a simple element $v \otimes \nu$. If we extend $v$ to a basis $(v, w_2, \ldots, w_n)$ of $\mathbb{V}$ such that $\nu(v, w_2, \ldots, w_n) = 1$, say, with dual basis $(\eta_1, \ldots, \eta_n)$, then by construction $$\alpha = \nu(v, \, \cdot \, , \cdots , \, \cdot \,) = \eta_2 \wedge \cdots \wedge \eta_n,$$ and in particular $\alpha$ is decomposable.

Travis Willse
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