The standard contact structure on $\mathbb R^{2n+1}=(x_1,y_1,\dots,x_n,y_n,z)$ is given by $\ker\alpha$, where $\alpha=dz-\sum_{i=1}^ny_idx_i$. But is there a reason why this contact structure is called "standard"? Is it just convention, or because it's the "nicest" contact structure, or is this contact structure actually natural in some important way?
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9$\begingroup$ Given a function $f: \mathbb{R}^n \rightarrow \mathbb{R}$, consider the graph of f and its gradient, $G = \{ (x_1, \partial_1f, \dots, x_n,\partial_nf, f) \}$. The standard contact form $\alpha$ vanishes when restricted to this graph. Conversely, any $n$-dimensional submanifold on which $\alpha$ vanishes but the $n$-form $dx_1 \wedge\cdots\wedge dx_n \ne 0$ at every point is the graph of a function and its gradient. This is how the concept of a contact structure first arose. $\endgroup$– Deane YangCommented Dec 1, 2020 at 4:56
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13$\begingroup$ Like in symplectic geometry, there is a Darboux theorem in contact geometry, which tells you that on a manifold with contact structure there is always a chart such that the given contact structure looks like the standard contact structure on $\mathbb R^n$. $\endgroup$– Panagiotis KonstantisCommented Dec 1, 2020 at 5:42
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2$\begingroup$ Forgiving historical inaccuracies (and perhaps eliding constants), I would wager it has something to do with the restriction of the (canonical) Liouville one form on $T^{*}\mathbb{R}^{n}$ to the unit co-sphere bundle with respect to the Euclidean metric on $\mathbb{R}^{n}.$ $\endgroup$– Andy SandersCommented Dec 1, 2020 at 17:36
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2 Answers
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I agree that Darboux' theorem is a very good reason for calling this "standard".
Another reason is that it (one-point-)compactifies to the standard contact structure on the $(2n+1)$-sphere: this is the space of complex tangencies to $S^{2n+1}$, viewed as the unit sphere in $\mathbb{C}^{n+1}$.
answered Dec 1, 2020 at 9:13
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I think Panagiotis Konstantis comment is the correct answer. To close the question: Any contact manifold is locally isomorphic to the standard contact structure on $\mathbb R^{2n+1}$ by the contact version of Darboux' theorem.
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3$\begingroup$ That is a bad reason. Actually, Darboux theorem shows that all contact structures are locally isomorphic. It can't be a reason to call one structure "standard". $\endgroup$ Commented Dec 1, 2020 at 12:36
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2$\begingroup$ You can find an embedding of $\mathbb{R}^{2n+1}$ with the standard contact structure (not with the standard contact form) around each point of any contact $2n+1$-manifold. It's not just a local statement. $\endgroup$ Commented Dec 1, 2020 at 14:31