= Solution
Choose the outward <normal vector> $n=(\cos\varphi,\sin\varphi,0)$ and the positive convention
$$
K_{ij}=e_i^ae_j^b\nabla_a n_b=e_j\cdot\partial_i n.
$$
The ambient connection vanishes in Cartesian coordinates. Since $\partial_\varphi n=e_\varphi/\rho$ and $\partial_{\widehat z}n=0$, the <extrinsic curvature> components are $K_{\varphi\varphi}=\rho$, $K_{\varphi\widehat z}=K_{\widehat z\widehat z}=0$. Taking the trace using the <induced metric>,
$$
\boxed{K=\gamma^{ij}K_{ij}=\frac1\rho.}
$$
\b[Reversing the normal or using the negative extrinsic-curvature convention gives $-1/\rho$]. The two <principal curvatures> are $1/\rho$ and zero in the chosen convention; the averaged <mean curvature> would be $K/2$, so it must not be confused with the requested trace.
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