Let $ (M,g) $ be an $ n $-dimensional Riemannian manifold. Let $ S\subset M $ be an embedded oriented hypersurface of $ M $. Let $ \hat N $ be the unit normal vector field on $ S $.
If we pick a chart $ (U,\phi) $ adapted to $ S $ (meaning that $ \phi(U\cap S) = \{x\in \phi(U) : x^n = 0\} $) and apply the Gram-Schmidt process to the frame field $ \{\partial_1,\dots,\partial_n\} $ induced by $ (U,\phi) $, we obtain an orthonormal frame field $ \{e_1,\dots,e_n\} $ such that $ \{e_1,\dots,e_n\} $ is an orthonormal frame field for $ S $ and thus such that $ e_n = \hat N $.
Let $ \omega_g $ be the Riemannian volume form on $ M $. On $ U $ we can write $$ \omega_g = e^1\wedge\dots\wedge e^n $$ since $ \{e_1,\dots,e_n\} $ is orthonormal.
Now, I know that the Riemannian volume form $ \eta_g $ o $ S $ induced by the pullback metric is given by the receipt $$ \eta_g = \iota^*(\hat N\mathop{\lrcorner} \omega_g) $$ where $ \iota\colon S\hookrightarrow M $ is the canonical inclusion, but...
If we compute $ \iota^*(\hat N\mathop{\lrcorner} \omega_g) $ we discover that $$ \hat N\mathop{\lrcorner} \omega_g = \color{red}{(-1)^{n - 1}} e^1\wedge\dots\wedge e^{n - 1}\qquad \leadsto\qquad \eta_g = \color{red}{(-1)^{n - 1}}e^1\wedge\dots\wedge e^{n - 1} $$
if instead we compute directly $ \eta_g $ employing the fact that $ \{e_1,\dots,e_{n - 1}\} $ is orthonormal for $ S $, we obtain $$ \eta_g = e^1\wedge\dots\wedge e^{n - 1} $$
Something smells wrong to me. Why does that happens?
