I started reading the book "Foundations of Differential Geometry" from Kobayashi & Nomizu and have a question regarding the possibility to define a metric on the frame bundle in a certain way.

Let $(M,g)$ be a Pseudo-Riemannian manifold, i.e. $g$ is a Pseudo-Riemannian metric on $M$. Let $LM$ be the frame bundle and $OM$ the bundle of $g$-orthonormal frames over $M$. An isometry $f$ of $M$ can be "lifted" to a diffeomorphism $\tilde{f}$ of the frame bundle, where $\tilde{f}(p):=(Df(p_1),...,Df(p_n))$ for a frame $p\in LM$. Since $f$ is an isometry, $\tilde{f}$ leaves $OM$ invariant. So we can restrict $\tilde{f}$ to a diffeomorphism $\tilde{f}$ of $OM$.

I wonder if it's possible to define a Pseudo-Riemannian metric $\tilde{g}$ on $OM$ such that $\tilde{f}$ is an isometry of $OM$ with respect to that metric. If so, is this metric uniquely determined by the metric $g$ on $M$? I know that there is a $1:1$ correspondence between metric compatible connections on $(M,g)$ and connection forms on $OM$. Can one use the connection form of the Levi-Civita-connection of $(M,g)$ to define such a metric $\tilde{g}$ on $OM$?


It is possible to define such a metric, but it is not unique and indeed weaker than the structure you get from the Levi-Civita connection. Indeed, let $\theta$ be the usual soldering form on $OM$ (so this has values in $\mathbb R^n$, where $n$ is the dimension of $M$) and let $\gamma$ be the connection form of the Levi-Civita connection. Then $\theta\oplus\gamma$ can be viewed as a one-form on $OM$ with values in $\mathbb R^n\oplus\mathfrak o(p,q)$ (where $(p,q)$ is the signature of the initial pseudo-Riemannian metric). It turns out that this one-form trivializes the tangent bundle of $OM$ and each of the maps $\tilde f$ you describe has the property that $f^*(\theta\oplus\gamma)=\theta\oplus\gamma$. (This is the basis of the description of pseudo-Riemannian metrics as a Cartan geometry.)

This shows that choosing any inner product on $\mathbb R^n\oplus\mathfrak o(p,q)$ you get a pseudo-Riemannain metric on $OM$ (via the trivialization) for which $f$ is an isometry. Of course there are natrual choices, like the standard inner product of signature $(p,q)$ plus the negative of the Killing form or something like that. However, the property of preserving a trivialization of the tangent bundle is much stronger than the property of being an isometry for a pseudo-Riemannian metric. For example, it is relatively easy to show that the diffeomorphisms preserving a trivialization form a Lie group.

  • $\begingroup$ I have two questions: 1. Can we take an arbitrary metric compatible connection for this construction? Or is the torsion-freenes needed anywhere? 2. Can we also take a positive definite inner product on $\mathbb{R}^n\oplus \mathfrak{o}(k,l)$, which then gives a Riemannian metric on $OM$ for which $\tilde{f}$ is an isometry? $\endgroup$ – user247741 Oct 15 '16 at 19:31
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    $\begingroup$ Torsion freeness is crucial, since it makes sure that $\tilde f$ preserves $\theta\oplus\gamma$. (An isometry is not compatible with a general metric connection, say with torsion $T$. The pullback is again metric, but has torsion $f^*T$, which in general is different from $T$.) On the other hand, you certainly can take a positive definite inner product, but this is not naturally related to anything that is around. (And as I said, preserving a trivialization anyway is stronger than preserving a metric.) $\endgroup$ – Andreas Cap Oct 17 '16 at 6:57
  • $\begingroup$ Your answers were really helpful to me. I have one more question. Reading further along in Kobayashi's book he actually uses such a metric in a proof. He mentions that it is clear that the group of lifted isometries is a closed subgroup of the group of isometries of $(OM, \tilde{g})$. Why is this obvious? $\endgroup$ – user247741 Oct 19 '16 at 15:35
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    $\begingroup$ It is clear that if $\tilde f$ and $\tilde g$ preserve $\theta\oplus\gamma$, then so do $\tilde g\circ\tilde f$ and $\tilde f^{-1}$, so you have a subgroup. Also, if you have a sequence of diffeomorphisms preserving $\theta\oplus\gamma$ which converges in a reasonable topology, then the limit will preserve $\theta\oplus\gamma$, too. Hence the subgroup is closed. $\endgroup$ – Andreas Cap Oct 20 '16 at 8:15
  • $\begingroup$ The topology is the co-topology, right? I don't see the last argument. Why does the limit also preserve the $\theta\oplus\gamma$? $\endgroup$ – user247741 Oct 20 '16 at 16:13

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