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If $f:M\to \bar{M}$ is a immersion, $\bar{M}$ riemannian manifold with Levi-Civita connection $\bar{\nabla}$ then, if we pull-back the metric of $\bar{M}$ to $M$, and let ${\bar{\nabla}}^\perp$ the projection on $T_pM$ of the connection $\bar{\nabla}$ then $\bar{\nabla}^{\perp}$ is the Levi-Civita connection on $M$.

I already showed that $\bar{\nabla}^{\perp}$ is a connection on $M$, how do I show that it is metric compatible?

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  • $\begingroup$ To compute covariant derivatives you only need to compute vector fields along curves. But every ($TM$-)vector field along a curve in $M$ can be considered a ($T\bar M$-)vector field along a curve in $\bar M$, so... $\endgroup$ – Llohann Aug 30 '16 at 16:37
  • $\begingroup$ so the claim follows immediately. $\endgroup$ – L.F. Cavenaghi Aug 30 '16 at 16:51
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If $X,\ Y,\ Z$ are vector fields on $(M,g)$ where $\overline{M}$ has a metric $\overline{g}$, then let $\overline{X}=f_\ast X$

Then $$ \overline{X}\overline{g}(\overline{Y},\overline{Z}) =\overline{g}(\overline{\nabla}_{\overline{X}}\overline{Y}, \overline{Z}) + \overline{g}(\overline{\nabla}_{\overline{X}} \overline{Z},\overline{Y}) =\overline{g}(\overline{\nabla}_{\overline{X}}^\top\overline{Y}, \overline{Z}) + \overline{g}(\overline{\nabla}_{\overline{X}}^\top \overline{Z},\overline{Y}) $$ where $ \overline{\nabla}^\top$ is a projection

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