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Most often a contact manifold is given as a pair $(M,\xi)$ where $\xi$ is a completely non-integrable hyperplane field on $M$---that is, $\xi$ is given (locally) as the kernel of a 1-form $\alpha$ such that $\alpha\wedge(d\alpha)^n\neq 0$.

However, there is an alternative definition that I have never quite understood, that $(d\alpha)^n\vert_\xi\neq 0$. I don't quite know why this is equivalent to $\alpha\wedge(d\alpha)^n\neq 0$.

Any hints would be appreciated!

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Computational approach. If you don't mind I will do the computations for $n=1$.

Let $X_1,X_2,X_3$ be vector fields on $M$, one has the following equality: $$2(\alpha\wedge\mathrm{d}\alpha)(X_1,X_2,X_3)=\sum_{\sigma\in\mathfrak{S}_3}\varepsilon(\sigma)\alpha(X_{\sigma(1)})\mathrm{d}\alpha(X_{\sigma(2)},X_{\sigma(3)})\tag{1}.$$

Going back to the result we aim for.

Let $x\in M$, there exists $U$ a neighborhood of $x$ and $X,Y,Z\in\Gamma(TU)$ such that: $$\ker(\alpha)_{\vert U}=\textrm{Span}(X,Y),TU=\ker(\alpha)_{\vert U}\oplus\mathbb{R}Z.$$

  • Assume $\alpha\wedge\mathrm{d}\alpha=0$, then using $(1)$, one gets the following equality: $$\alpha(Z)\mathrm{d}\alpha(X,Y)=0.$$ Therefore, $\mathrm{d}\alpha(X,Y)=0$ and by linearity $\mathrm{d}\alpha_{\vert U}$ vanishes identically on $\ker(\alpha)_{\vert U}$.

  • Assume that $\mathrm{d}\alpha$ vanishes identically on $\ker(\alpha)$, then using $(1)$, one gets the following equality: $$\alpha\wedge\mathrm{d}\alpha(X,Y,Z)=0.$$ Therefore, by linearity $\alpha\wedge\mathrm{d}\alpha_{\vert U}$ vanishes identically on $TU$.

Whence the result taking the contrapositive.

In summary, we have established that $\alpha$ is a contact form if and only if $\mathrm{d}\alpha$ is a symplectic form on $\ker(\alpha)$.

Linear algebra approach. Let us prove the following general result:

Proposition. Let $E$ be a finite-dimensional real vector space, $\omega\in\Lambda^kE^*$ and $\alpha\in E^*\setminus\{0\}$, then one: $$\alpha\wedge\omega=0\iff\omega_{\vert\ker(\alpha)}=0\iff\exists\beta\in\Lambda^{k-1}E^*\textrm{ s.t. }\omega=\alpha\wedge\beta.$$

Proof. First, assume that $\omega=\alpha\wedge\beta$, then $\alpha\wedge\omega=0$.

Assume that $\alpha\wedge\omega=0$, let $\alpha_1:=\alpha\neq 0$ and let us complete $(\alpha_1)$ as a basis $(\alpha_1,\ldots,\alpha_n)$ of $E^*$. Therefore, one can write $\omega$ in a unique way as: $$\omega=\sum_{1\leqslant i_1<\ldots<i_k\leqslant n}\omega_{i_1,\ldots,i_k}\alpha_1\wedge\cdots\wedge\alpha_k,$$ from our assumption, one gets: $$\sum_{1<i_1<i_2<\ldots<i_k\leqslant n}w_{i_1,\ldots,i_k}\alpha\wedge\alpha_{i_1}\wedge\cdots\wedge\alpha_{i_k}=0,$$ but $\{\alpha\wedge\alpha_{i_1}\wedge\cdots\wedge\alpha_{i_k}\}_{1<i_1<i_2<\ldots<i_k\leqslant n}$ is linearly independent, so that if $1\not\in\{i_1,\ldots,i_k\}$, $\omega_{i_1,\ldots,i_k}=0$. Finally, one gets: $$\omega=\sum_{1<i_2<\ldots<i_k\leqslant n}\omega_{1,i_2,\ldots,i_k}\alpha\wedge\alpha_{i_2}\wedge\cdots\wedge\alpha_{i_k}=\alpha\wedge\beta.$$

Assume that $\omega=\alpha\wedge\beta$, then $\omega_{\ker(\alpha)}=0$ using that: $$\alpha\wedge\beta(x_1,\ldots,x_k)=\frac{1}{(k-1)!}\sum_{\sigma\in\mathfrak{S}_k}\varepsilon(\sigma)\alpha(x_{\sigma(1)})\beta(x_{\sigma(2)},\ldots,x_{\sigma(k)}).$$

Assume that $\omega_{\vert\ker(\alpha)}=0$, then $\alpha\wedge\omega=0$, using that: $$\alpha\wedge\omega(x_1,\ldots,x_{k+1})=\frac{1}{k!}\sum_{\sigma\in\mathfrak{S}_{k+1}}\varepsilon(\sigma)\alpha(x_{\sigma(1)})\omega(x_{\sigma(2)},\ldots,x_{\sigma(k+1)}),$$ along with the splitting $E=\ker(\alpha)\oplus\langle R_{\alpha}\rangle$.

Whence the result. $\Box$

Whence the result, working pointwise in the tangent bundle and taking the contrapositive of the proposition.

Addendum. If you are also wondering why $\ker(\alpha)$ is integrable if and only if $\alpha\wedge\mathrm{d}\alpha=0$, this comes from: $$\mathrm{d}\alpha(X,Y)=X(\alpha(Y))-Y(\alpha(X))-\alpha([X,Y]),$$ which holds for all vector fields on $M$ along with Frobenius theorem.

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  • $\begingroup$ Thanks, this is helpful! And also thank you for the explanation for the integrability condition. I had seen it attributed to Frobenius in numerous places but it was treated as some well-known thing which I could not find in any other reference. Thanks again! $\endgroup$
    – anak
    Dec 28, 2017 at 23:52
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    $\begingroup$ It is my pleasure! When reading on symplectic and contact topology, it is usually required to be familiar with various classical identities of differential topology and sometimes it is indeed difficult to fill the gaps of quick arguments like "from Frobenius theorem, one has". Good luck! :) $\endgroup$
    – C. Falcon
    Dec 28, 2017 at 23:59
  • $\begingroup$ Besides the usual Guillemin & Pollack or Milnor suggestion, do you have any additional suggestions for differential topology? En français, même. $\endgroup$
    – anak
    Dec 29, 2017 at 14:46
  • $\begingroup$ For basic constructions, I recommend Introduction aux variétés différentielles by J. Lafontaine (I bet an English version also exists). For a wide survey on differential topology, I like these lectures notes, a lot of geometric insight is provided! For more advanced stuff, I'd say Differential topology by M. Hirsch. $\endgroup$
    – C. Falcon
    Dec 29, 2017 at 15:07
  • $\begingroup$ Thanks again! I will for sure check them out. $\endgroup$
    – anak
    Dec 29, 2017 at 22:13

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