The surface gradient operator $\nabla_\Gamma$ is defined as $\nabla_\Gamma = \nabla - \mathbf{n(n} \cdot \nabla)$, where $\Gamma$ is a surface.
The surface curl operator on a vector tangent to $\Gamma$ is defined as $ \nabla_\Gamma \times [] = \{ \nabla_\Gamma \cdot (\mathbf{n} \times []) \} \mathbf{n}$.
Let $\mathbf{P}$ be a tangent vector to a surface $\Gamma$. The surface curl of $\mathbf{P}$ is as follows:
$ \nabla_\Gamma \times \mathbf{P} \\ = [\nabla_\Gamma \cdot (\mathbf{n} \times \mathbf{P})]\mathbf{n} \\ = [(\nabla_\Gamma \times \mathbf{n}) \cdot \mathbf{P} - (\nabla_\Gamma \times \mathbf{P} )\cdot \mathbf{n}] \mathbf{n} \quad [1] $
In the previous expression, I applied the identity $\nabla \cdot (\mathbf{A \times B) = (\nabla \times A)\cdot B - (\nabla \times B)\cdot A}$.
Moreover, for any surface $\nabla_\Gamma \times \mathbf{n} = 0$. Thus, we obtain the following:
$ \nabla_\Gamma \times \mathbf{P} = - [ (\nabla_\Gamma \times \mathbf{P} )\cdot \mathbf{n}] \mathbf{n} \quad [2]$
Now, we can expand the term on the right hand side of the previous equation by using the first line of Equation [1], which is $ \nabla_\Gamma \times \mathbf{P} = [\nabla_\Gamma \cdot (\mathbf{n} \times \mathbf{P})]\mathbf{n}$ :
$ [ (\nabla_\Gamma \times \mathbf{P} )\cdot \mathbf{n}] \mathbf{n} = \Big ( [\nabla_\Gamma \cdot (\mathbf{n} \times \mathbf{P})]\mathbf{n} \cdot \mathbf{n} \Big ) \mathbf{n} = \Big ( [\nabla_\Gamma \cdot (\mathbf{n} \times \mathbf{P})] \Big ) \mathbf{n} = \nabla_\Gamma \times \mathbf{P} $
Then, evaluating in Equation [2], we get that $\nabla_\Gamma \times \mathbf{P} = 0$.
I am pretty sure this is wrong since I could define a vector $v$ on the $x-y$ plane as $v = < f(x,y), g(x,y)>$, where $f$ and $g$ are arbitrary functions.
Then $\nabla_\Gamma \times \mathbf{P} = (\partial_x g - \partial_y f)\hat{z}$, which is not always zero.
Would you please let me find my error in the derivation I did?