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 $F$ isascendingif

the distance function $\rho(z,w)=\sqrt{|z|^2+|w|^2}$ is Morse when restricted to $F$, andat 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 $T_p F \not \subset \ker \alpha_{1/2}$.