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Let $M$ be a riemannian manifold and $N$ be a submanifold of $M$. Let $v$ be a vector field in $N$. Then $v$ can be covariantly differentiated along $\gamma$ resulting in new field $u$. (Here consider Levi-Civita connection)

Because $N$ is embedded in $M$ then $\gamma$ and $v$ can be "carried" in $M$. Let us denote them in $M$ the same way as in $N$. Let's consider Levi-Civita connection in $M$ and differentiate $v$ along $\gamma$ resulting in $w$ field.

Is it true that the orthogonal projection of $w$ in $T_p M$ on $T_p N$ is equal to $u$?

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What is $\gamma$? –  Henning Makholm Jan 15 '12 at 22:16
1  
It is a smooth curve in N –  user1038085 Jan 15 '12 at 22:20

1 Answer 1

Yes, this is true. Let $p :TM\vert_N \to TN$ be the orthogonal projection. By uniqueness of the Levi-Civita connection, it is sufficient to show that the connection $\nabla^N$ on $N$ defined by $\nabla^N_X Y = p(\nabla^M_X Y)$ where $X$ and $Y$ are vector fields on $N$ (technically to take $\nabla^M_X Y$ we need $Y$ to be a vector field on $M$ but $\nabla^M_X Y$ depends only on $Y$ along $X$ so we can extend it to $M$ anyway we want and this will be well-defined) is compatible with the metric and torsionfree. For compatibility with the metric, let $X, Y, Z$ be vector fields on $N$ with $\tilde X, \tilde Y, \tilde Z$ denoting extensions to $M$. Then on $N$ we have $$ X \langle Y, Z\rangle = \tilde X \langle \tilde Y ,\tilde Z\rangle = \langle \nabla^M_{\tilde X} \tilde Y, \tilde Z\rangle + \langle \tilde Y, \nabla^M_{\tilde X} \tilde Z\rangle = \langle \nabla_{\tilde X}^M \tilde Y, Z\rangle + \langle Y, \nabla_{\tilde X}^M \tilde Z\rangle$$ $$ = \langle p(\nabla_{\tilde X}^M \tilde Y), Z\rangle + \langle Y, p(\nabla_{\tilde X}^M \tilde Z)\rangle = \langle \nabla^N_{X} Y, Z \rangle + \langle Y, \nabla_X^N Z\rangle.$$

Torsionfree comes from the fact that $p([\tilde X, \tilde Y]) = [X,Y]$ (which can be easily checked in local slice coordinates).

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