Effect of a Lutz twist on Euler number

Question

I already asked this question on the Math Stack Exchange but did not get an answer. I am currently working through Geiges proof of the Martinet-Lutz theorem, which can be found here, and am trying to figure out the effect of a half Lutz twist on the Euler class of the contact structure. During the proof of Propositioin 3.15, he basically claims that given a contact structure $\xi$ on $M$ and a transverse link $L$, then performing a half Lutz twist along the components of $L$ yields a contact structure $\eta$ such that $$e(\eta) = e(\xi) - 2PD[L]$$ where $PD: H_1(M)\to H^2(M)$ is Poincaré-Duality. I am however fairly certain, that I have proven the same result with $2PD[L]$ replaced by $PD[L]$, here is my proof:

• Choose a tubular neighbourhood $V$ of $L$, such that on each component the contact structure is given by $\ker d\theta + r^2d\phi$.
• Start with a vector field $X$ tangent to $\xi$, which is in general position with the zero section, does not vanish over $V$ and is given by $\partial_r$ on the boundary of $V$, then $e(\xi) = PD[X\cap M]$.
• After the Lutz twist the $X$ I have in mind is not tangent to $\eta$, however we can replace $X$ on $V$ by $\partial_r$, such that $X$ now vanishes along $L$. The new Euler class is thus given by $$e(\eta) = PD[X\cap M] = e(\xi) + PD[L].$$

Can anyone spot a mistake in my argument, or is the statement in Geiges wrong?

Answer 1

I have actually solved this by now, his statement is indeed correct but the proof is a bit misleading. I will only discuss the case where $K$ is a knot.

Instead of choosing a field which does not vanish in a tubular neighbourhood of $K$, one simply chooses a generic vector field which is given by the radial vector field $\partial_r$ in a tubular neighbourhood of $K$, which vanishes along $K$.

After the Lutz twist the same vector field is also tangent to $\eta$ and still vanishes along $K$, however, the Lutz twist has made $K$ a negatively transverse knot such that the difference $e(\eta) - e(\xi)$ is given by $-2PD[K]$.

Answer 2

I think the proof as written in the book is more along the following lines. Let $K$ be the positive transverse knot along which the Lutz twist on $\xi_0$ is performed. Take a neighborhood $S^1\times D^2$ of $K$ [with $K$ being identified orientation-preserving with $S^1\times \{0\}$] on which $\xi_0$ is given by the kernel of the 1-form $$d\theta + r^2d\varphi = d\theta + u\,dv-v\,du.$$ Here $\theta$ is the coordinate on $S^1$, and $(u,v)$, $(r,\varphi)$ are Cartesian/polar coordinates on $D^2$. If $\xi_1$ is the result of the Lutz twist, then in the same neighborhood, $\xi_1$ is given as the kernel of the 1-form $$h_1(r)\,d\theta + h_2(r)\,d\varphi.$$ Here $h_1,h_2\colon [0,\infty)\to\mathbb{R}$ satisfy

• $h_1(r) = -1$ and $h_2(r) = -r^2$ near $r=0$,
• $h_1(r) = 1$ and $h_2(r) = r^2$ for $r\geq \delta$,
• $(h_1(r),h_2(r))$ is never parallel to $(h_1'(r),h_2'(r))$ for $r\neq 0$.

Let $X_0$ be a generic section of $\xi_0$ which on $S^1\times D^2$ is equal to $$\frac{1}{\sqrt{1+v^2}}\Big(\frac{\partial}{\partial u}+ v\frac{\partial}{\partial \theta}\Big)$$ [i.e., on $S^1\times D^2$, $X_0$ is the projection of $\partial_u$ on $\xi_0$ scaled to unit length]. On $S^1\times (D^2-\{0\})$ we can express $X_0$ in cylindrical coordinates as $$\frac{1}{\sqrt{1+r^2\sin^2\varphi}}\bigg[\cos \varphi\frac{\partial}{\partial r}-\sin\varphi\Big(\frac{1}{r}\frac{\partial}{\partial \varphi} - r\frac{\partial}{\partial \theta}\Big)\bigg].$$ Let $\rho\colon [0,\infty)\to[0,1]$ be a smooth non-decreasing function with $\rho(r)=r^2$ near $r=0$ and $\rho(r)=1$ for $r\geq\delta$. We define the section $X_1$ of $\xi_1$ to vanish on $K$, to equal \begin{multline*}\frac{\rho(r)}{\sqrt{1+r^2\sin^2\varphi}}\bigg[\cos \varphi\frac{\partial}{\partial r}\\-\sin\varphi\sqrt{\frac{1+r^2}{r^2h_1(r)^2+h_2(r)^2}}\Big(h_1(r)\frac{\partial}{\partial \varphi} - h_2(r)\frac{\partial}{\partial \theta}\Big)\bigg]\end{multline*} on $S^1\times (D^2-\{0\})$, and to equal $X_0$ on the rest of $M$. With respect to the metric $d\theta^2+du^2+dv^2$ we have $$\lim_{r\to 0^+}\frac{X_1(\theta,r,\varphi)}{\lVert X_1(\theta,r,\varphi)\rVert}= \cos 2\varphi \frac{\partial}{\partial u}+ \sin 2\varphi \frac{\partial}{\partial v}.$$ Thus $X_1$ has index $-2$ zeros along $K$ with respect to the orientation on $\xi_1$. [Note $K$ is a negative transverse knot for $\xi_1$.] Therefore $$e(\xi_1) = e(\xi_0 ) -2PD[K].$$