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Let $(M,g,\nabla)$ be a Riemannian manifold of dimension $n$ and let $\gamma:\mathbb{R}\to M$ be a geodesic, namely, $\nabla_{\dot{\gamma}}\dot{\gamma}=0$.

It is well know that, given a basis $\{e_{j}(p)\}_{j=1}^{n}$ on the tangent space $T_{p}M$, where $p=\gamma(0)$, one can build a parallel transported frame $\{E_{0}(\gamma(t)):=\xi(\gamma(t)),E_{1}(\gamma(t)),\ldots,E_{n}(\gamma(t))\}$ on any tangent space $T_{\gamma(t)}M$ along the geodesic. In particular, this means that, for any $l$, $\nabla_{\xi}E_{l}=0$ and $E_{l}(\gamma(0))\equiv e_{l}(p)$.

Now, setting $\xi:=\dot{\gamma}$, let us introduce the symmetric endomorphism: $$ Ricc_{\xi}(p):=R(\xi(p),J(p))\xi(p),\qquad \forall J(p)\in T_{p}M $$ where $R$ is the Riemann curvature tensor, i.e, $R(X,Y)Z|_{p}=R^{i}_{jkl}(p)X^{j}(p)X^{k}(p)Z^{l}(p)e_{i}(p)$ with $\{e_{i}(p)\}_{i=1}^{n}$ is a orthonormal basis on the tangent space $T_{p}M$.

I know that $Ricc_{\xi}$ is symmetric because of the properties of the Riemann tensor, therefore, I can always diagonalized $Ricc_{\xi}$ on (in principle any tangent space) the initial point of the geodesic, that is, $p=\gamma(0)$ and I will get an eigensystem whose eigenvectors can be used as a basis on $T_{\gamma(0)}M$. Now, supposing that I know the isomorphism (parallel transport map) $\Gamma_{t}:T_{p}M\to T_{\gamma(t)}M$ since I solved the geodesic equation, my question is:

Can a parallel transported frame be obtained evolving the eigenvectors of the operator $Ricc_{\xi}$ through the map $\Gamma_{t}$?

RTS
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  • I am not sure about your question. Can it be rephrased as "can we define a parallel frame along $\gamma$, say ${\gamma'(t),E_1(t),\ldots,E_{n-1}(t)}$ such that $E_i(t)$ is an eigenvector of $\mathrm{Ricc}{\xi}(t)$?", or do you just want to know if the parallel transport of an eigenbasis of $\mathrm{Ricc}{\xi}(0)$ is a parallel frame? – Didier Aug 04 '21 at 13:29
  • Dear Didier, I am interesting in both the questions but the first your question is probably that which better matches my question. – RTS Aug 04 '21 at 16:58
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    I think the first one is in general not true (and more than that: almost never true) for the following heuristic reason. Writing $R(\xi,E_i)\xi = \lambda_i E_i$ with $E_i$ parallel along $\gamma$ and applying the covariant derivative along $\gamma$ brings the equality $\left(\nabla_{\xi}R\right)(\xi,E_i)\xi = \lambda_i' E_i$, and $R$ and $\nabla_{\xi} R$ would have too much symmetries in common. The second question is trivially true: the parallel transport of any orthonormal basis is an orthonormal frame along the geodesic, so this works with an eigenbasis of $\mathrm{Ricc}_{\xi}$. – Didier Aug 04 '21 at 17:04
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    Note that if the eigenvalues are not all distinct, there might not even be a smooth frame of eigenvectors. – Deane Aug 04 '21 at 17:11
  • @Didier, I followed your reasoning until "they would have too much symmetries in common". Did you mean that they should have too much symmetries in order for the $R_{\xi}$-eigenvectors to be a parallel transported frame? I do not understand why $(\nabla_{\xi}R)(\xi,E_{I})\xi=\dot{\lambda}{I}E{I}$ should prove that this is false. – RTS Aug 04 '21 at 17:11
  • One way to analyze this is to look at whether what you're asking necessarily holds at a single point up to second order. – Deane Aug 04 '21 at 17:16
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    @LLOR I meant than in general, the tensors $R$ and $\nabla_{\xi} R$ do not share any eigendirection in common. – Didier Aug 04 '21 at 17:16

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