5
$\begingroup$

Let the standard symplectic structure on $B^4$ (viewed in $\mathbb{R}^4$ or $\mathbb{C}^2$) be given by $\omega=(1/2) d \eta$, for \begin{align*} \eta &:= x_1 \, dy_1 - y_1 \, dx_1 + x_2 \, dy_2 - y_2 \, dx_2 \\ &=\tfrac{i}{2}(z \, d\bar{z}-\bar{z} \, dz +w \, d\bar{w} - \bar{w}\, dw). \end{align*} The restriction of $\eta$ to the 3-sphere of radius $r$ defines a contact structure $\alpha_r:= \eta\mid_{S^3_r}$.

Definition. Let $F$ be a smooth oriented surface properly embedded in the unit 4-ball $B^4 \subset \mathbb{C}^2$, assumed to miss the origin and be transverse to $\partial B^4 = S^3 \subset \mathbb{C}^2$. We say that $F$ is ascending if

  1. the distance function $\rho(z,w)=\sqrt{|z|^2+|w|^2}$ is Morse when restricted to $F$, and
  2. at all regular points of $\rho\mid_F$, we have $(d\rho \wedge \eta)\mid_F>0$.

The definition comes from this paper of Boileau and Orevkov. They make the following assertion, which they claim follows immediately from the definition of an ascending surface: Let $F\subset B^4$ be an ascending surface. If $p \in F$ is a critical point of $\rho\mid_F$, then $T_p F = \ker \alpha_r$, where $r=\rho(p)$; that is to say that $T_p F$ is a complex plane.

Question. Why does $T_p F = \ker \alpha_r$ at critical points $p$? Isn't the following a counterexample?

Example. Let $V$ be the affine plane $\{(\tfrac{1}{2},t,0,s) \in \mathbb{R}^4\}$ and $F=V \cap B^4$. Then $\rho\mid_F$ is Morse with a unique critical point at $(\tfrac{1}{2},0,0,0)$. For condition (2), we note that $F$ intersects $S_{1/2}$ in the above critical point and $S_{1/2+\epsilon}$ in a circle parametrized by $\gamma(t)=(\tfrac{1}{2},t,0,\sqrt{\epsilon^2+\epsilon -t^2})$. We can evaluate $\alpha_{1/2+\epsilon}$ on $F \cap S_{1/2+\epsilon}$ for $\epsilon>0$: \begin{equation*} (\alpha_{1/2+\epsilon})_{\gamma(t)} \gamma'(t) = \big(\tfrac{1}{2}\, dy_1-t \,dx_1+0 \, dy_2 - \sqrt{\epsilon^2+\epsilon-t^2} \, dx_2\big)\begin{bmatrix}0 \\1 \\ 0 \\ \frac{-t}{\sqrt{\epsilon^2+\epsilon-t^2}}\end{bmatrix}= \tfrac{1}{2}>0. \end{equation*} Therefore $F$ is an ascending surface. But at the critical point $p=(\tfrac{1}{2},0,0,0)$, we see that $T_p F$ contains $(0,1,0,0)$ and $\big(\alpha_{1/2}\big)_{(1/2,0,0,0)}(0,1,0,0)= 1/2$. Therefore $T_p F \not \subset \ker \alpha_{1/2}$.

| cite | improve this question | | | | |
$\endgroup$
4
$\begingroup$

To see why $T_pF=\ker\alpha_r$ at critical points $p$ of $\rho|_F$, note that Morse-ness of $\rho|_F$ means that $d(\rho|_F)$ vanishes to precisely first order at $p$. On the other hand, we have $$ (d\rho\wedge\eta)|_F=fd\mathrm{vol}_F $$ for some $f\colon F\to\mathbb R_{\ge0}$. Because $f$ is nowhere negative, it can only vanish to even order, which means that $\eta|_F$ must also vanish to at least first order at $p$. But this says precisely that $$ T_pF\subset(\ker d\rho\cap\ker\eta)=\ker\alpha_r, $$ and equality follows from equality of the dimensions.

The reason your example doesn't contradict this is that you have only parametrized half the circle. To parametrize the other half with the same orientation, you would take $$ \gamma_2(t)=\left(\frac12,-t,0,-\sqrt{\epsilon^2+\epsilon-t^2}\right), $$ and there the same calculation gives $\eta_{\gamma_2(t)}(\gamma_2'(t))=-\frac12$.

| cite | improve this answer | | | | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.