Suppose $M$ is a manifold, and $E$ a vector bundle over $M$ equipped with a connection $\nabla $. If $F$ is the frame bundle of $E$, is there an explicit construnction of a connection on $F$ associated with $\nabla$ such that in this way connections on $E$ and $F$ are $1$-$1$ correspondent?

Edit for the bounty:

I really need an answer to this question, and as it was already posted I think that putting a bounty on it is the most sensible way to go.

To rephrase the question in my own terms: Let $M$ be a smooth $n$-manifold. We can associate the following principal $GL(n)$-bundle to it: $$F = \{(m,\theta)|m\in M, \theta:\mathbb{R}^n\to T_mM\mathrm{\ lin.\ isom.}\}$$ with right action given by $(m,\theta)g = (m,\theta g)$. Its tangent space is defined (as for any other manifold) as a quotient of the space of paths on $F$. In order to get a more concrete representation, we need a way to differentiate "paths of frames," but as such paths can be seen as tuples of paths of vectors on $M$, it is enough to specify a connection $\nabla$ on $M$ to obtain the identification $$T_{(m,\theta)}F \cong \{(\hat{m},\hat{\theta})|\hat{m}\in T_mM,\hat{\theta}:\mathbb{R}^n\to T_mM\}$$ where we identify the equivalence class of paths $[\gamma(t),\theta(t)]$ with $(\dot{\gamma}(0),(\nabla_{\dot{\gamma}}\theta)(0))$. This gives us a map $$\{\mathrm{connections\ on\ }M\}\longrightarrow\{\mathrm{principal\ connections\ on\ F}\}$$ mapping $\nabla$ to $A([\gamma,\theta]) = \theta^{-1}\nabla_{\dot{\gamma}}\theta\in\mathfrak{gl}(n)$.

I believe there should be a way to invert this map (maybe only on a subset of the principal connections, though) but I cannot see how. Does anyone have an idea or a solution?

Remark 1: My question is in fact a special case of the original question on vector bundles, namely if we take $E=TM$.

Remark 2: I took a look at Taubes' book, as suggested in the answers, but I didn't find what I need (or maybe I found it, but wasn't smart enough to realize it).

  • 1
    $\begingroup$ There will be more connections on $F$ in general... but I think that if you restrict yourself to principal connections on $F$ then they are in 1-1 correspondence with connections in $E$. Hopefully someone who knows this better than I do can give an answer. $\endgroup$ – Anthony Carapetis Sep 13 '13 at 2:51
  • $\begingroup$ Did you look in Kobayashi-Nomizu? $\endgroup$ – Gunnar Þór Magnússon May 20 '15 at 12:31
  • $\begingroup$ @GunnarÞórMagnússon I took a fast look, there seems to be some stuff relating the second fundamental form of an immersed submanifold with connections on the normal frame bundle, but I've not found a direct answer to the question above. $\endgroup$ – Daniel Robert-Nicoud May 20 '15 at 12:46
  • $\begingroup$ I'm going from memory, but I think they talk about the two in the chapter on connections. I could be delirious though. $\endgroup$ – Gunnar Þór Magnússon May 20 '15 at 13:50
  • $\begingroup$ @GunnarÞórMagnússon It's probably just me not looking with enough attention. I think I'm getting it on my own anyway (see my answer below). If I really cannot prove everything I will take a deeper look. $\endgroup$ – Daniel Robert-Nicoud May 20 '15 at 15:33

Ok, I had an inspiration, and found the following answer.

Identify $TM$ as the associated bundle $E = F\times_{GL(n)}\mathbb{R}^n$, where the action of $GL(n)$ on $\mathbb{R}^n$ is given simply by left multiplication. The identification between the two is given by $$[(m,\theta),v]\in E\longmapsto\theta v\in T_mM.$$ Then we have a bijective correspondence between vector fields on $M$ and sections of $E$ where to $X:M\to TM$ we associate $$\overline{X}(m) = [m,\theta,\theta^{-1}X(m)]$$ for any choice $\theta$ of frame at $m$. Now a connection $A\in\Omega^1(F;\mathfrak{gl}(n))$ gives us a way to differentiate $\overline{X}$, namely $d_A\overline{X}\in\Omega^1(M;E)$ ($1$-forms on $M$ with values in $E$).

Claim: The assignment $$d_A\overline{X}\longmapsto\nabla X$$ is the required bijection, where from $\nabla$ we can recover $A$ by noticing that $d_A = d + \rho(A)$.

Proof: We have the further identification of $T^*M$ with $E^*=F\times_{GL(n)}(\mathbb{R}^n)^*$, where the action of $GL(n)$ on $(\mathbb{R}^n)^*$ is given by $g\cdot v^* = v^*\circ g$. This gives a correspondence between $\Omega^1(M)$ with sections of $E^*$ by $$\overline{\beta}(m) = [m,\theta,\beta(m)\circ\theta]$$ in a way similar to the above. We have a natural isomorphism $$\Omega^1(M;E)\cong\Omega^0(M;E^*),$$ thus the only thing we are left to show is that $$d_A\overline{X} = \overline{\nabla X}.$$ This goes: $$\begin{align}\overline{\nabla X}(\hat{m}) = & \langle[m,\theta,\nabla_{\theta-}X],[m,\theta,\theta^{-1}\hat{m}]\rangle\\=&\langle[m,\theta,d_A\overline{X}(\theta-)],[m,\theta,\theta^{-1}\hat{m}]\rangle\\=&d_A\overline{X}(\hat{m})\end{align}$$ for $\hat{m}\in T_mM$, where $\langle\cdot,\cdot\rangle$ is the natural pairing.

  • $\begingroup$ It's a pity I can't assign the bounty to myself. $\endgroup$ – Daniel Robert-Nicoud May 27 '15 at 9:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.