(I've been self-learning differential geometry for some time now, and math.stackexchange has been of sooo much help, that I thought I might get my answer here.)
From Lee's "Introduction to Riemannian manifolds", the shape operator $s$ of a hypersurface is the Weingarten map, which is defined as $(\tilde{\nabla}_X N)^{\top} = -W_N (X)$, where $\tilde{\nabla}$ is the embedding manifold's connection, $X$ is a tangent vector and $N$ is the normal.
Also, for a tangent vector $Y$, $(\tilde{\nabla}_X Y)^{\top} = \nabla_X Y$, where $\nabla$ is the surface's connection. Then the total derivative on the surface can be obtained by $\nabla Y = (\tilde{\nabla}Y)^{\top} = \mathsf{P}\,(\tilde{\nabla} Y)\,\mathsf{P}$, where $\mathsf{P} = 1 - N\otimes N$.
This means that I can get the shape operator like \begin{equation} s = (\tilde{\nabla} N)^{\top} = \mathsf{P}\,(\tilde{\nabla} N)\,\mathsf{P}. \end{equation} When the embedding manifold is, for example, $\mathbb{R}^3$, this gives a $3 \times 3$ matrix . However, the shape operator using the surface connection would be only a $2 \times 2$ matrix.
My question: what is the relation between the shape operator $s_{ij} = - \nabla N$ with indices $i,j = 1,2$ and $s_{ij} = - \mathsf{P}\,(\tilde{\nabla} N)\,\mathsf{P} $ with indices $i,j = 1,2,3$?
In addition, I know that the Gauss curvature is given by $K = det(s)$. So, under my logic, I get \begin{align} K &= det(s) \\ &= det(\mathsf{P}\,(\tilde{\nabla} N)\,\mathsf{P}) \\ &= det(\mathsf{P})det(\tilde{\nabla} N)det(\mathsf{P})\\ &= 0, \end{align} as $det(\mathsf{P}) = 0$.
My question: How can this be? I know it is not zero for all surfaces. Also I should get the product of two (not three) principal curvatures. Where's my error?