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Let $M$ be a manifold and let $g$ be a tensor on it, say for example a metric $g\in\Gamma(T^{\ast}M\otimes T^{\ast}M)$. I know how to perform any computation on $g$. For instance, taking its derivative respect to a connection $\nabla$, evaluating it at a point, taking its Lie derivative, obtaining the curvature of the Levi-Civita connection etc.

However, there is a dual formulation on the frame bundle $F(M)$ of $M$, but I never knew how to do the same calculations on the frame bundle, and as I understand it is sometimes simpler to work on the frame bundle. I would like to know how a tensor on $M$ is represented from the point of view of the frame bundle, and how are the typical operations (curvature, Lie derivative etc) implemented. A tensor in $M$ is a section of the corresponding tensor vector bundle. How is this mapped to the frame bundle?

For example, given an open set $U$ of the atlas of $M$ I can write $g$ in coordinates as follows

$g = g_{ab}\,dx^{a}\otimes dx^{b}$

What would be the analog local expression from the point of view of the frame bundle?

Finally, I would like to know a reference where these things are explained in detail.

Thanks.

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    $\begingroup$ The tangent bundle is an associated bundle to the frame bundle, so any tensor on the base can be 'lifted' to a function on the frame bundle with values in a representation that is equivariant under the action of the group. I'm pretty sure you can find this point of view in Sternberg's book Lectures on Differential Geometry, amongst others. Probably in Kobayashi-Nomizu I too. $\endgroup$ – Paul Reynolds Mar 1 '15 at 13:22
  • $\begingroup$ If you have some time, could you give an explicit example of the correspondence between tensors and equivariant functions on the frame bundle? For example for a vector field. I am checking the references that you mentioned but there is nothing too explicit. Thanks Paul. $\endgroup$ – Bilateral Mar 1 '15 at 15:33
  • $\begingroup$ I guess there is now no need, after Peter's post. $\endgroup$ – Paul Reynolds Mar 1 '15 at 19:09
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There are some computations in Appendix F of MSM 144 (AMS), following Hamilton's Harnack estimate paper. Under the fair use rule, we've put a link to this at:

www.math.ucsd.edu/~benchow/TensorCalculusFrameBundle.pdf

The material is geared toward an application in Ricci flow.

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See 18.12 of here, for example.

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I interpret the question differently from the other answers and comments. It appears to me that the question refers to the method of moving frames, where one uses differential forms to study Riemannian geometry. Most expositions start with a choice of an orthonormal frame of tangent vectors and work with the corresponding basis of dual 1-forms. However, all of this simplifies, if one works on the frame bundle instead. Instead of having to choose a moving frame, you work with a canonical set of 1-forms defined on the whole bundle that effectively do the calculations in all possible moving frames in one shot. One advantage of this, for example, is that the frames and calculations involving them become global rather than local, when the manifold is orientable.

Unfortunately, I don't know of a reference for this. I learned it from Robert Bryant, and we can hope that he'll answer this question. Otherwise, I'll try to add more details later.

ADDED: I should add that the approach I'm referring to is mathematically the same as but dual to the approach Peter and Paul describe. It's often easier to work with, largely because the exterior derivative is easier to work with than the Lie bracket.

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  • $\begingroup$ That is indeed interesting Deane. I hope Robert can comment on this. In fact, I think he uses this formalism in his popular paper "Remarks on $G_{2}$-structures", if I am not mistaken. $\endgroup$ – Bilateral Mar 2 '15 at 17:03
  • $\begingroup$ Could this be related to using instead of the principal bundle associated to the tangent bundle, the principal bundle associated to the cotangent bundle? $\endgroup$ – Bilateral Mar 2 '15 at 17:14
  • $\begingroup$ There is a natural identification between a tangent frame and a cotangent frame, so it's moot where you want to do it. Geometrically, it's more natural (at least for me) to work with differential forms on frames of tangent vectors. $\endgroup$ – Deane Yang Mar 2 '15 at 17:45
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Take a vector space $V$. Take a manifold $M$ of the same dimension as $V$, not necessarily equipped with a Riemannian metric, and let $FM$ be the set of all pairs $(m,u)$ where $m \in M$ and $u \colon T_m M \to V$ a linear isomorphism. Given a tensor $\tau$ on $M$, each element $u$ identifies $\tau(m)$ with a tensor on $V$, say $u_* \tau(m)$. So we make a map $t \colon FM \to \bigotimes V$, by $t(m,u)=u_* \tau(m)$. This map $t$ is $GL(V)$-equivariant. Conversely, given any map $t \colon FM \to \bigotimes V$ with image in a finite dimensional subspace of $\bigotimes V$, we can try to define a suitable tensor, by $u_* \tau(m)=t(m,u)$, but easily check that this is well defined just precisely when $t$ is $GL(V)$-equivariant. To work with orthonormal frames of a Riemannian metric $g$, just replace $FM$ by the orthonormal frame bundle $F^g M \subset FM$. A connection, from this point of view, is a $GL(V)$-equivariant 1-form $\gamma$ on $FM$ (not on $M$) valued in $\mathfrak{gl}(V)$, which agrees with the Maurer--Cartan 1-form on the fibers of $FM \to M$. Define covariant derivatives by $dt+\rho(\gamma)t = (\nabla t) \omega$ where $\rho$ is the representation of $GL(V)$ on $\bigotimes V$. This $\nabla t$ is the function on $FM$ which represents the tensor $\nabla \tau$. So you can do all of your work without touching the ground.

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