Let $(M, g)$ be a Riemannian manifold with Levi-Civita-connection $\nabla$, and let $N \subseteq M$ be an embedded submanifold with a $g$-induced Riemannian metric $h$. I now want to show that the Levi-Civita-connection $\tilde \nabla$ of $(N, h)$ is given by

$$(\tilde \nabla_X Y)_p = \mathrm{pr}_{T_p M} (\nabla_{X_p} Z)_p\tag{$*$}$$


  • $\mathrm{pr}_{T_p M} : T_p M \to T_p N \subseteq T_p M$ is the $g$-orthogonal projection, and

  • $Z \in \mathfrak X(M)$ is a vector field on $M$, which is identical to $Y \in \mathfrak X(N)$ locally around $p \in N$.

Now I think this is mostly a matter of disentangling all the definitions and substituting them correctly wherever necessary, but I keep getting lost. Now my first problem is understanding why such a vector field $Z$ as desired even exists, and why the right-hand side of $(*)$ is independent of which $Z$ with this property I choose.

But even assuming that this is the case, where would I go from there? I thought about maybe using the Koszul formula or one of the basic properties of the Levi-Civita connection (because there's not much else that I know about it that might be helpful) but I'm not sure what exactly to do with them.


1 Answer 1


A possible way to prove this is to remember that the LC connection is the unique torsion free connexion for which the metric tensor is parallel.

The fact that your formula gives a connexion is obvious.

To check that it is torsion free, note that $g_N(\tilde \nabla _X Y, Z)= g(\nabla _X Y, Z)$ for every triple of tangent vector fields on $N$

To check that it preserves the induced tensor metric let $X,Y,Z$ three tangent vector fields on $N$. We can extend these fields on $M$ to compute :

$(\tilde \nabla _X g) (X,Y)= X. g(Y,Z)-g(\tilde \nabla _X Y, Z)- g(Y, \tilde \nabla _XZ)=X. g(Y,Z)-g( \nabla _X Y, Z)- g(Y, \nabla _XZ) $

  • $\begingroup$ Why $g_N(\tilde{\nabla}_XY,Z)=g(\nabla_X Y,Z)$ and how does that implie that it is torsion free? $\endgroup$
    – Gomes93
    May 14, 2019 at 4:45
  • 1
    $\begingroup$ The identity is just linear algebra, as ${\tilde \nabla} _X Y-\nabla _X Y$ is orthogonal to $Z$ by definition. For torsion free, consider two vector fields tangent to $N$ and compute $T(X,Y)=\ \nabla _X Y- \nabla _Y X-[X,Y]$. Then project orthogonaly on $N$ and you get the result. $\endgroup$
    – Thomas
    May 15, 2019 at 5:30
  • $\begingroup$ @Thomas For the connection to make sense, the vector fields on the submanifold need to be extended locally to vector fields $\widetilde{X}, \widetilde{Y}$ in a neighbourhood of the submanifold. This is easy enough: just pass to local coordinates such that we can identify the submanifold with some Euclidean space. Then you need to show that the definition is independent of the choice of extension -- Again, this is not too bad, just write out $\nabla^N$ in terms of Christoffel symbols. $\endgroup$ Apr 12, 2021 at 3:48

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