Take $n^an_a=-1$. The <first fundamental form>, or <induced metric>, is the <spatial projection tensor>
$$
\boxed{h_{ab}=g_{ab}+n_an_b},\qquad h_a{}^b=\delta_a{}^b+n_an^b.
$$
With the convention used in the question, define the <second fundamental form> by
$$
\boxed{K_{ab}=-h_a{}^ch_b{}^d\nabla_cn_d}.
$$
Locally $n_a$ is proportional to the gradient of a defining function for the <hypersurface>, so $h_a{}^ch_b{}^d\nabla_{[c}n_{d]}=0$ and $K_{ab}=K_{ba}$. Both derivative indices are projected tangentially. If two extensions of $n$ agree on $\Sigma$, their difference vanishes there, as does its tangential derivative, so $K_{ab}$ is independent of the extension.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 311 / 2 / i / Solution

Under the <Euclidean isometry> $\widetilde{\mathbf r}=\mathbf a+R\mathbf r$, where $R^TR=I$,
$$
\partial_u\widetilde{\mathbf r}=R\partial_u\mathbf r.
$$
Therefore
$$
\widetilde g
=(R\mathbf r_u)\mathbin{\cdot}(R\mathbf r_u)
=\mathbf r_u\mathbin{\cdot}\mathbf r_u=g.
$$
Since the integration parameter is unchanged,
$$
\widetilde S(u)=\int_0^u\sqrt{\widetilde g(u')}\,du'
=S(u).
$$
Both the <induced metric> and <arc length> are invariant.

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 344 / 3 / a / i / Solution

{wiki}

An immersion into a Riemannian manifold inherits a metric by taking ambient inner products of tangent vectors. For a parametrized curve $\mathbf r(u)$ in Euclidean space, the induced metric coefficient is $g=\partial_u\mathbf r\cdot\partial_u\mathbf r$.


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