# Smooth $r$-form as the exterior product of $r$ $1$-forms

Let $$M$$ be a $$n$$-dimensional manifold and $$\alpha$$ a smooth $$r$$-form in $$M$$. It is clear that $$\alpha$$ can be written locally as $$\alpha=\alpha_1\wedge \ldots \wedge \alpha_r$$ for some $$1$$-forms $$\alpha_1,\ldots,\alpha_r$$ just by considering its expression in a local chart of $$M$$.

In general, can we have this descomposition $$\alpha=\alpha_1\wedge \ldots \wedge \alpha_r$$ in the whole manifold? Maybe with the use of bump functions?

• In Euclidean $\mathbb{R}^3$, how do we write $dx_1 \wedge dx_2 + dx_2 \wedge dx_3 + dx_3 \wedge dx_1$ as $\alpha_1 \wedge \alpha_2$? Aug 16, 2022 at 14:27
• @aschepler $(dx_1 - dx_3)\wedge(dx_2 - dx_1)$ Aug 16, 2022 at 15:17

In looking for the easiest counter example, my initial answer was incorrect. However, it is still not true, even locally. For example, if we take $$dx\wedge dy+dz\wedge dw$$ on $$\mathbb{R}^4$$, you cannot write this as a wedge of $$1$$-forms, but trying to is probably a good exercise.