# When can a k-form be expressed as a wedge product of one forms? [duplicate]

Suppose I am handed a k-form $$\omega$$. What can I do to check if $$\omega$$ can be expressed in the form

$$\omega = \alpha_1 \wedge \dots \wedge \alpha_k?$$

The context for this question is an application in differential geometry. A codimension-k surface is characterized by a k-form field of the above form; I am trying to ascertain when a specified k-form field defines a codimension-k surface

EDIT: An equivalent question is, "when is a given k-form the unique top-level form of some k-dimensional subspace?"

• @cmk I'm not sure what you mean by this. I think the statement I'm making should be basis-independent. – user_35 Jun 17 '19 at 23:44
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• A wedge of 1 forms without superpositions? If this is true that's good news, but I'm surprised. In R4, how can $dx \wedge dy + dz \wedge dw$ be expressed as a wedge product of 1 forms? – user_35 Jun 17 '19 at 23:48
• Briefly, you need to have $\omega \wedge \iota_v\omega = 0$ for every element of a basis set of vectors $v$. – Ted Shifrin Jun 18 '19 at 0:25
• What if $\omega$ is a $d/2+1$ or higher form in a d-dimensional space? Then just by counting your condition is trivially satisfied. Does this suggest that any sufficiently high-rank form can be written in the form I suggested in my question? – user_35 Jun 18 '19 at 0:36