# Thursday, January 27 :::{.remark} Goal: classify Legendrian knots up to (Legendrian) isotopy. Recall a knot $\gamma: S^1 \injects Y$ satisfies $\gamma^*(\alpha) = 0$, and a Legendrian isotopy is a 1-parameter family $\gamma_t$ which are Legendrian for all $t$. ::: :::{.example title="?"} $\gamma(s) = \tv{x(s), y(s), z(s)}$ and $\xi = \ker \alpha, \alpha = \dz - y\dx$. Then $\gamma^*(\alpha) = z' \ds - yx^1\ds = (z' - yx')\ds$, which is Legendrian iff $y=z'/x'$. ::: :::{.example title="?"} Let $f:\RR \to \RR$ and take the 1-jet $\gamma(s) = \tv{s, f'(s), f(s)}$ of the graph of $f$ -- this is like the graph of the 1st order Taylor expansion. This is Legendrian since $s'=1$ implies $z'/x' = f'/s' = f'$. ::: :::{.remark} There are two projections: - $\tv{x,y,z} \to \tv{x,z}$, a wave front projection, plotted with $y$ into the board, - $\tv{x,y,z} \to \tv{x,y}$, Lagrangian projection. ::: :::{.example title="?"} Let $\gamma(s) = \tv{s^2, {3\over 2}s, s^3}$, then the two projections are as follows: ![](figures/2022-01-27_11-25-37.png) ::: :::{.remark} The front projection uniquely determines $L$, since the $y$ coordinate can be recovered as $y=z'/x'$. So for example, there is no ambiguity about crossing order: the more negatively sloped line in a diagram is the over-crossing: ![](figures/2022-01-27_11-30-01.png) ::: :::{.example title="?"} A front diagram of the unknot: ![](figures/2022-01-27_11-34-42.png) ::: :::{.theorem title="?"} Every knot $K \injects \RR^3$ can be $C^0$ approximation by a Legendrian knot $L$. Idea: zigzags in an $\eps$ tube in the knot diagram, which will be Legendrian. How to measure: $\sup_{s\in I} \abs{\gamma_1(s) - \gamma_2(s)} \leq \eps$? ::: :::{.remark} Note that $\Lie(\SO_3) \da \T_e(\SO_3) = \liesu_2$, spanned by roll, pitch, and yaw generators: ![](figures/2022-01-27_11-48-58.png) So measuring the number of rotations along each generator after traversing $L$ in a full loop yields integer invariants. ::: :::{.definition title="The Thurston–Bennequin number"} A **framing** of a knot $K$ is a trivialization of its normal bundle, so an identification of $\nu(K) \cong S^1\times \DD^2$. The potential framings are in $\pi_1(\SO_2) \cong \pi_1(S^1) \cong \ZZ$, since a single vector field normal (?) to the knot determines the framing by completing to an orthonormal basis. The Reeb vector field is never tangent to a Legendrian knot, so this determines a framing called the **contact framing**. The **Thurston–Bennequin number** is the different between the 0-framing and the contact framing. The 0-framing comes from a Seifert surface. This is an invariant of Legendrian knots, since Legendrian isotopy transports frames. Note that adding zigzags adds cusps, and thus decreases this number. ::: :::{.remark} How to compute: take a pushoff and compute the linking number: ![](figures/2022-01-27_12-07-19.png) ::: :::{.proposition title="?"} \[ \mathrm{tb}(L) = w(L) - {1\over 2}C(L) ,\] where $w(L)$ is the writhe and $C(L)$ is the number of cusps. ::: :::{.proof title="?"} The linking number is ${1\over 2}(c_+(L) - c_-(L))$, half of the signed number of crossings. Here all 4 crossing have the same sign: ![](figures/2022-01-27_12-12-35.png) ::: :::{.example title="?"} TB for the knots from before: - The 3 unknots: - 2 cusps, so $-1$ - 4 cusps, so $-2$ - 4 cusps, so $-2$ - The 2 trefoils: - $3-{1\over 2}4 = 1$ - $3 - {1\over 2}6 = -6$. ::: :::{.remark} Since adding zigzags decreases $\mathrm{tb}$, define $\mathrm{TB}$ to be the max over all Legendrian representatives of $K$. This distinguishes mirror knots. In fact $\mathrm{tb}(L) \leq 2g_3(L) - 1$ (the Bennequin bound), involving the 3-genus. ::: :::{.definition title="Rotation number"} The **rotation number** of $L$ is the *turning number* $\rot(L)$ in the Lagrangian projection, i.e. how many times a tangent vector spins after traversing the knot. ::: :::{.example title="?"} ![](figures/2022-01-27_12-26-39.png) It turns out that \[ \rot(L) = {1\over 2}\qty{ \size\text{down cusps} - \size\text{up cusps}} .\] :::