# Contact Forms and Structures (Thursday, January 13) :::{.definition title="Contact form"} A **contact form** on $Y^3$ is a 1-form $\alpha$ with $\alpha \wedge d\alpha > 0$. A **contact structure** is a 2-plane field $\xi = \ker \alpha$ for some contact form. ::: :::{.remark} Forms are more rigid than structures: if $f>0$ and $\alpha$ is contact, then $f\cdot \alpha$ is also contact with $\ker( \alpha ) = \ker(f \alpha)$. ::: ## Examples of Contact Structures :::{.example title="Standard contact structure"} On $\RR^3$, a local model is $\alpha\da \dz - y\dx$. :::{.exercise title="?"} Show $\alpha \wedge d\alpha = dz\wedge dx \wedge dy$. ::: Write $\xi = \spanof_\RR(\del y, y \del z + \del x)$, which yields planes with a corkscrew twisting. Verify this by writing \( \alpha = 0 \implies \dd{z}{x} = y \), so the slope depends on the $y\dash$coordinate. ::: :::{.example title="Rotation of the standard structure"} On $\RR^3$, take \( \alpha_2 \da dz + x\dy \) and check \( \alpha_2 \wedge d \alpha_2 = dz \wedge dx \wedge dy \). This is a rigid rotation by $\pi/2$ of the previous \( \alpha \), so doesn't change the essential geometry. ::: :::{.example title="Radially symmetric contact structure"} Again on $\RR^3$, take \( \alpha_3 = dz + {1\over 2}r^2 d\theta \). Check that \( d \alpha_3 = r\dr \wedge \dtheta \) and \( \alpha_3 \wedge d \alpha_3 = r\dz \wedge \dr \wedge \dtheta \). Then $\xi = \spanof_\RR(\del r, {1\over 2} r^2 \del z + \del \theta )$. Note that as $r\to \infty$, the slope of these planes goes to infinity, but doesn't depend on $z$ or $\theta$. ::: :::{.example title="Rectangular version of radially symmetric structure"} Set \( \alpha_4 = \dz + {1\over 2}(x\dy - y\dx) \), then this is equal to \( \alpha_3 \) in rectangular coordinates. ::: :::{.example title="Overtwisted"} Set \[ \alpha_4 = \cos(r^2)\dz + \sin(r^2)\dtheta .\] :::{.exercise title="?"} Compute the exterior derivative and check that this yields a contact structure. ::: Now note that \[ \alpha = 0 \implies \dd{z}{\theta} = -{\sin(r^2)\over \cos(r^2)} = -\tan(r^2) ,\] which is periodic in $r$. So a fixed plane does infinitely many barrel rolls along a ray at a constant angle $\theta_0$. This is far too twisty -- to see the twisting, consider the graph of $(r, \tan(r^2))$ and note that it flips over completely at odd multiples of $\pi/2$. In the previous examples, the total twist for $r\in (-\infty, \infty)$ was less than $\pi$. ::: :::{.definition title="Contactomorphisms"} A **contactomorphism** is a diffeomorphism \[ \psi: (Y_1^3, \xi_1) \to (Y_2^3,\xi_2) \] such that $\phi_*(\xi_1) = \xi_2$ (tangent vectors push forward). A strict contactomorphism is a diffeomorphism \[ \phi: (Y_1^3, \ker \alpha_1) \to (Y_2^3, \ker \alpha_2) .\] such that $\phi^*( \alpha_2) = \alpha_1$ (forms pull back). ::: :::{.remark} Strict contactomorphisms are more important for dynamics or geometric applications. ::: :::{.exercise title="?"} Prove that \( \alpha_1,\cdots, \alpha_4 \) are all contactomorphic. ::: :::{.remark} Recall that $X$ has a cotangent bundle $\T\dual X \mapsvia{\pi} X$ of dimension $2 \dim X$. There is a canonical 1-form $\lambda \in \Omega^1(\T\dual X)$, i.e. a section of $T\dual(T\dual X)$. Given any smooth section $\beta\in \Gamma(\T\dual X\slice X)$ there is a unique 1-form $\lambda$ on $\T\dual X$ such that $\beta^*( \lambda) = \beta$, regarding $\beta$ as a smooth map on the left and a 1-form on the right. In local coordinates $(x_1,\cdots, x_n)$ on $X$, write $y_i = dx_i$ on the fiber of $\T\dual X$. Why this works: the fibers are collections of covectors, so if $x_i$ are horizontal coordinates there is a dual vertical coordinate in the fiber: ![](figures/2022-01-13_11-50-52.png) So we can write \[ \lambda = \sum y_i \dx_i \in \Omega^1(\T\dual X) ,\] regarding the $y_i$ as functions on $\T\dual X$ and $\dx_i$ as 1-forms on $\T\dual X$. :::{.exercise title="?"} Find out what $\beta = \sum a_i \dx_i$ is equal to as a section of $\T\dual X$. ::: ::: :::{.remark} To get a contact manifold of dimension $2n+1$, consider the 1-jet space $J^1(X) \da T\dual X \times \RR$. Write the coordinates as $(x,y)\in \T\dual X$ and $z\in \RR$ and define $\alpha = \dz - \lambda$, the claim is that this is contact. For dimension $2n-1$, choose a cometric on $X$ and take $\SS\T\dual X$ the unit cotangent bundle of unit-length covectors. Then $\alpha \da -\ro{\lambda}{\SS \T\dual X}$ is contact. ::: :::{.exercise title="?"} Check that $\RR^3 = J^1(\RR)$ and $\SS \T\dual(\RR^2) = \RR^2 \cross S^1$. ::: :::{.remark} A neat theorem: the contact geometry of $\SS\T\dual \RR^3$ is a perfect knot invariant. This involves assigning to knots unique Legendrian submanifolds. ::: ## Perturbing Foliation :::{.example title="?"} Define \[ \alpha_t = \dz - ty\dx \qquad t\in \RR \] to get a 1-parameter family of 1-forms. Check that \( \alpha_t \wedge d \alpha_t = t(\dz \wedge \dx \wedge \dy) \). Consider $t\in (-\eps, \eps)$: - $t>0 \implies \alpha = \dz - y\dx$ yields a positive contact structure, - $t>0 \implies \alpha = \dz$ is a foliation, - $t<0 \implies \alpha = \dz + y\dx$ is a negative contact structure. ::: :::{.remark} What is a (codimension $r$) foliation on an $n\dash$manifold? A local diffeomorphism $U\cong \RR^n \times \RR^{n-r}$ with *leaves* $\pt\cross \RR^{n-r}$. For example, $\RR^3\cong \RR \cross \RR^2$ with coordinates $t$ and $(x, y)$. We're leaving out a lot about how many derivatives one needs here! > For a fiber bundle or vector bundle to admit an interesting foliation, one needs a flat connection. ::: :::{.definition title="Integrability"} Any $\xi \da \ker \alpha$ is **integrable** iff for all vector fields $X, Y \subseteq \xi$, their Lie bracket $[X, Y] \subseteq \xi$. ::: :::{.theorem title="Frobenius Integrability"} For $\alpha$ nonvanishing on $Y^3$, $\ker \alpha$ is tangent to a foliation by surfaces iff $\alpha \wedge d\alpha = 0$. ::: :::{.example title="?"} Consider $\alpha = \dz - y\dx$, so $\ker \alpha = \spanof_\RR\ts{\del y, y\del z + \del x}$ which bracket to $\del z \not \in \ker \alpha$. This yields a non-integrable contact structure. On the other hand, for $\alpha = \dz$, $\ker \alpha = \spanof_\RR\ts{\del x, \del y}$ which bracket to zero. So this yields a foliation. ::: :::{.remark} A theorem of Eliashberg and Thurston: taut foliations can be perturbed to a (tight) positive contact structure. :::