# Spivak, Volume I, page 292

Define the ring $$\Omega(M)$$ to be the direct sum of the rings of $$l$$-forms on $$M$$, for all $$l$$. If $$\Delta$$ is a $$k$$-dimensional distribution on $$M$$, then $$\mathscr{I}(\Delta)\subset\Omega(M)$$ will denote the subring generated by the set of all forms $$\omega$$ with the property that (if $$\omega$$ has degree $$l$$) $$\omega\left( X_1,\ldots,X_l \right) =0\quad \text{whenever }X_1,\ldots,X_l\text{ belong to }\Delta .$$ It is clear that $$\omega_1+\omega_2\in\mathscr{I}(\Delta)$$ if $$\omega_1,\omega_2\in\mathscr{I}(\Delta)$$, and that $$\epsilon\wedge\omega\in\mathscr{I}(\Delta)$$ if $$\omega\in\mathscr{I}(\Delta)$$ [thus, $$\mathscr{I}(\Delta)$$ is an ideal in the ring $$\Omega(M)$$]. Locally, the ideal $$\mathscr{I}(\Delta)$$ is generated by $$n-k$$ independent $$1$$-forms $$\omega ^{k+1},\ldots,\omega ^{n}$$. In fact, around any point $$p\in M$$ we can choose a coordinate system $$(x,U)$$ so that $$\left.\frac{\partial }{\partial x^{1}} \right|_p,\ldots,\left.\frac{\partial }{\partial x^{k}} \right|_p\ \ \text{span }\Delta _p .$$ Then $$dx^{1}(p)\wedge\cdots\wedge dx^{k}(p)\ \ \text{is non-zero on }\Delta _p .$$ By continuity, the same is true for $$q$$ sufficiently close to $$p$$, which by Corollary 4 implies that $$dx^{1}(q),\ldots,dx^{k}(q)$$ are linearly independent in $$\Delta _q$$. Therefore, there are $$C^{\infty}$$ functions $$f_{\beta}^{\alpha}$$ such that $$dx^{\alpha}(q)=\sum_{\beta=1}^{k} f_{\beta}^{\alpha}(q)dx^{\beta}(q)\ \ \textit{restricted to }\Delta _q\quad \alpha =k+1,\ldots,n .$$ We can therefore let $$\omega ^{\alpha}=dx^{\alpha}-\sum_{\beta=1}^{k} f_{\beta}^{\alpha}dx^{\beta} .$$
My question is why are $$f_\alpha^\beta$$ not $$0$$? Since $$\{dx^1, ..., dx^n\}$$ is a dual basis to $$\{\frac{\partial}{\partial x^1}, ..., \frac{\partial}{\partial x^n}\}$$, wouldn't we simply get $$dx^\alpha(\frac{\partial}{\partial x^\beta})=f_\alpha^\beta=0,\; 1\leq \beta\leq k$$? In other words, why can't we just take $$\omega^\alpha=dx^\alpha,\; (k+1)\leq\alpha\leq n$$ as the generators of the ideal in question?
• There is no reason for $\partial/\partial x^{\beta}$ to be tangent to $\Delta$ at any point $q$ distinct from $p$, even as close as you want. Anyway, the idea is this: if $\Delta = \ker (dx - dy)$ in $\Bbb R^2$, then on $\Delta$, $dx|_{\Delta} = dy|_{\Delta}$ by construction even though in $\Bbb R^2$, $dx$ and $dy$ are independent. So you can't really tell things on differential form destricted to $\Delta$ just with the data of the ambient space Nov 20, 2022 at 17:04
• @Didier, I see. I assumed that Spivak chose the coordinates $(x, U)$ so that $\{\frac{\partial}{\partial x^\beta}\}$ span $\Delta$ at every point of $U$. I realize now this is only possible for integrable $\Delta$, and we do not assume $\Delta$ to be integrable. You are saying, he only chose the coordinates $(x, U)$ so that $\{\frac{\partial}{\partial x^\beta}\}|_p$ span $\Delta|_p$, correct? But $\{dx^\beta\}|_q,\; 1\leq\beta\leq k$ still span the dual of $\Delta|_q$, for any $q\in U$, so that we still have $dx^\alpha=\sum f_\beta^\alpha dx^\beta,\; k<\alpha$ on $\Delta|_q$. Is that correct? Nov 20, 2022 at 20:36
• @user1104937 Your comment is perfectly correct. Restricting a vector field to a distribution does not always give a tangent vector field, while restricting a differential form always gives a differential form: this is the main difference between vector fields and $1$-forms. Nov 20, 2022 at 20:40