Well if $\Sigma$ is a submanifold of $R^{n+p}$ and $\{e_i,e_\alpha\}$ is orthonormal frame over $\Sigma$ where the $e_i$'s are tangent and the $e_\alpha$'s are normal to $\Sigma$.

Can anyone prove (with an adequate frame) that $\nabla_{e_i}^{\perp} e_\alpha=0$?

Obs: The result is pretty easy when we have only one normal direction, but in this case there are more.

  • $\begingroup$ Nobody can prove what is not true in general. Compare this with the Frenet-Serret formulas. $\endgroup$ – Yuri Vyatkin Nov 23 '12 at 20:54
  • $\begingroup$ The title of the question is quite misleading. The only Levi-Civita connection here is the standard connection in $\Bbb R^n$ whereas $\nabla^{\perp}$ is not the Levi-Civita connection, but a connection in the normal bundle (which is in this case, of course, a part of the ambient L-C connection) $\endgroup$ – Yuri Vyatkin Nov 23 '12 at 21:59

When we have only one normal direction the picture is somewhat simpler than in the general case.

Introduce $h_i{}^\beta{}_\alpha$ by $$ \nabla^{\perp}_{e_i}e_{\alpha} = h_i{}^\beta{}_\alpha e_\beta $$ (Einstein summation assumed), $i=1,\dots,n$, $\alpha, \beta = n+1,\dots,n+p$, as it is assumed in the question.

Claim. $h_i{}^\beta{}_\alpha = - h_i{}^\alpha{}_\beta$.

Proof. $0 = \nabla_{e_i}\langle e_\alpha , e_\beta \rangle = \langle \nabla^{\perp}_{e_i}e_{\alpha} , e_\beta \rangle + \langle e_\alpha , \nabla^{\perp}_{e_i}e_{\beta} \rangle = h_i{}_\beta{}_\alpha + h_i{}_\alpha{}_\beta$.

Corollary. When $p=1$ (i.e. in the case of hypersurfaces) $$ \nabla_{e_i}^{\perp} e_\alpha=0 $$

Proof. The only skew-symmetric $1\times1$-matrix is $(0)$.


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