Suppose I have a smooth Riemannian manifold $(\tilde{M},g)$ and an embedded submanifold $M \subset \tilde{M}$. I fix a point $p \in M$ and $X_p \in T_pM$.

Suppose that $q \in M$ is joined to $p$ by two paths (which we can assume are smoothly homotopic): 1) A path $\gamma : [0,1] \to M$ with $\gamma(0) = p$ and $\gamma(1) = q$ and 2) A geodesic $\tilde{\gamma}$ in $\tilde{M}$.

I can of course parallel transport $X_p$ along $\gamma$ in $M$ to give me $X_q \in T_qM$.

Separately, I can parallel transport $X_p$ along the geodesic $\tilde{\gamma}$ in $\tilde{M}$ that joins $p$ to $q$. So at $q$ I now also have $\tilde{X}_q \in T_q\tilde{M}$.

I would like now to compare $X_q$ and $\tilde{X}_q$. Is there a sense in which the difference $|\tilde{X}_q - X_q|$ is somehow given by an integral of $A(\dot{\gamma},X)$ along $\gamma$? (where $A$ is the second fundamental form of the embedding $M \hookrightarrow \tilde{M}$)

  • $\begingroup$ If $\gamma$ and the geodesic $p \rightsquigarrow q$ are smoothly homotopic then there should be some expression in terms of a double integral (integrating over both the curve parameter and the homotopy parameter). $\endgroup$ – Anthony Carapetis Nov 20 '15 at 5:05
  • $\begingroup$ What is $A(\dot\gamma,X)$? $\endgroup$ – Ted Shifrin Nov 20 '15 at 5:15
  • $\begingroup$ I've edited now to ask the question I genuinely wanted to know rather than trying to simplify... $\endgroup$ – Thompson Nov 20 '15 at 16:48

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