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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?

Also, what is this strange letter Spivak uses to denote this ideal, and does this ideal have a name (maybe annihilator of the distribution or something)?

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  • $\begingroup$ 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 $\endgroup$
    – Didier
    Nov 20, 2022 at 17:04
  • $\begingroup$ The letter — hardly strange — is a script I. $\endgroup$ Nov 20, 2022 at 17:58
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    $\begingroup$ @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? $\endgroup$ Nov 20, 2022 at 20:36
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    $\begingroup$ @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. $\endgroup$
    – Didier
    Nov 20, 2022 at 20:40

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