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](http://www.math.univ-toulouse.fr/~orevkov/PXMA1945.pdf) 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 $T\_p F \not \subset \ker \alpha\_{1/2}$.



Ascending surfaces in the 4-ball
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 $\eta\_{\gamma\_2(t)}(\gamma\_2'(t))=-\frac12$.


