Motivation: Feel free to skip. Let $\Sigma$ be a complex curve in $\mathbb{C}^2$. If $\Sigma$ is transverse to a 3-sphere $S^3 \subset \mathbb{C}^2$ of radius $r$, then $\Sigma \cap S^3$ is a link, sometimes called a transverse $\mathbb{C}$-link. These include links of singularities and can shed light on the topology of $\Sigma$. Results of Rudolph and of Boileau-Orevkov show that the set of transverse $\mathbb{C}$-links is equal to the set of quasipositive links, which are defined in purely braid-theoretic terms. (For those interested, quasipositive braids are products of conjugates $w \sigma_i w^{-1}$, where $w$ is any word in the braid group $B_n$ and $\sigma_i$ is a positive generator.) In particular, Boileau and Orevkov prove the following in this paper:
Theorem. If a complex curve $\Sigma \subset \mathbb{C}^2$ intersects $S^3$ transversally, then $\Sigma \cap S^3$ is the closure of a quasipositive braid.
To prove this, they show that a smooth piece $(\Sigma,\partial \Sigma)\subset (B^4,S^3)$ of a complex curve in $\mathbb{C}^2$ is an ascending surface (defined below), and then use properties of ascending surfaces to show that $\partial \Sigma$ is a quasipositive link.
We can define a 1-form $\eta$ in $\mathbb{C}^2$ using either of the two equivalent definitions: \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 $\alpha_r:= \eta\mid_{S^3_r}$.)
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 $(d\rho \wedge \eta)\mid_F>0$.
Boileau and Orevkov make the following assertion, which they claim follows immediately from the definition of an ascending surface: Let $F\subset B_1$ 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 $\{(\tfrac{1}{2},t,0,s) \in \mathbb{C}^2\}$ and $F=V \cap B_1$. 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 $\gamma(t)=(\tfrac{1}{2},t,0,\sqrt{\epsilon^2+\epsilon -t^2})$. To evaluate $\alpha_{1/2+\epsilon}$ on $F \cap S_{1/2+\epsilon}$, we use the definitions $z=x_1+iy_1$ and $w=x_2+iy_2$ to write $\eta$ as $x_1 \, dy_1 - y_1 \, dx_1 + x_2 \, dy_2 - y_2 \, dx_2$. Then 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 $T_p F \not \subset \ker \alpha_{1/2}$.