I would like to see how to derive the connection 1 form on the bundle of oriented orthonormal frames for a manifold embedded in Euclidean space.

It seems that the geometry resides in a collection of 1 forms on the frame bundle the arise naturally from the embedding. Given a frame (e1,...,en) over a point in the manifold, map it to e1. This is a smooth map of the frame bundle into Euclidean space. The 1 forms are then defined as

wi = de1.ei 1=2,...n

The action of the rotation group on the frame bundle leaves these forms invariant because it preserves the Euclidean inner product. That is

dL(e1).L(ei) = Ld(e1).L(ei) = de1.ei

It is tempting to think that these 1 forms are the components of a connection 1 form but I do not not see how to formalize it.


1 Answer 1


No, you're confusing the frame bundle and the sphere bundle. For the $\mathfrak{so}(n)$-valued $1$-form on the frame bundle, you need $$\omega_{ij} = de_i\cdot e_j\,, \quad 1\le i,j \le n\,.$$

  • $\begingroup$ So how do you obtain these forms from the embedding? s it just the matrix of all of the forms that you get from the all of the maps,ei, of the frame bundle into Euclidean space? $\endgroup$
    – lavinia
    May 12, 2013 at 1:10
  • $\begingroup$ Yes. And to get the extrinsic geometry — the second fundamental form — you might want an orthonormal basis for the normal space, too, $e_{n+1},\dots,e_N$ (with $M\subset\mathbb R^N$). $\endgroup$ May 12, 2013 at 2:42
  • $\begingroup$ Right I see this. Thank you. $\endgroup$
    – lavinia
    May 17, 2013 at 20:06
  • $\begingroup$ I see that now. Thanks. What still eludes me is how to obtain the connection as a Lie algebra valued 1 form and to show that is obeys the required equations. I see that the forms are skew symmetric when one differentiates along the tangent bundle but do not see how to map the tangent space to the fiber into the Lie algebra of SO(n) $\endgroup$
    – lavinia
    May 17, 2013 at 20:10
  • $\begingroup$ Well, the matrix $[\omega_{ij}]$ is skew-symmetric, so it's giving you an $\mathfrak{so}(n)$-valued $1$-form. When you change frame by $a\colon M\to SO(n)$, it transforms by $\omega \rightsquigarrow a\omega a^{-1} + da\cdot a^{-1}$, which is what it should. [You should come take my graduate Riemannian geometry course this fall! :)] $\endgroup$ May 19, 2013 at 17:58

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