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While there generally exist on differentiable manifolds $ M $ and for a choice of a torsion-free connection $ \nabla $ locally geodesic vector fields $ V \in \mathfrak{X}(M) $ extending a given (non zero) vector $ v \in TM $ in the sense that $ \nabla_V V = 0 $ on some neighbourhood of $ \pi(v) $ where $ \pi : TM \to M $ denotes the projection onto the base point, it is not generally true that some choice of $ V $ exists verifying locally $ \nabla_W V = 0 $ for all vector fields $ W \in \mathfrak{X}(M) $.

There may, however, exist an extension of $ v $ verifying $ \nabla_w V = 0 $ for all directions $ w \in T_{\pi(v)} M $ at that point. Trivial as it may be, I struggle to find sources commenting this question specifically. The case of concircular vector fields similarly deals with extensions verifying $ \nabla_W V = f W $ for some scalar field $ f \in C^\infty(M) $, either locally or globally, without recalling the constraints of that same condition at a single point, which signals that the answer should be plain.

By the inverse function theorem, the exponential map taking each $ w \in T_{\pi(v)} M $ to $ \exp(w) \in M $ a unit parameter away along the geodesic through $ \pi(v) $ parametrised with velocity $ w $ defines local coordinate functions $ x^i : t \mapsto \exp(te_i) $ for any basis $ e_i $ of $ T_{\pi(v)} $, such that: $$ \nabla_{e_i} \tfrac{\partial}{\partial x^j} = \Gamma^k_{i\!j}(\pi(v)) e_k = 0 $$ implicitly summing over $ k $. For $ v = e_i $ one of the basis vectors, this process yields the extension $ V = \frac{\partial}{\partial x^i} $ verifying the desired condition by linearity of $ \nabla V : w \mapsto \nabla_w V $. Alternatively, any $ V $ verifies: $$ \nabla V = \left( \tfrac{\partial V^k}{\partial x^i} + V^j \Gamma^k_{i\!j} \right) \tfrac{\partial}{\partial x^k} \otimes \mathrm{d} x^i $$ evaluating at $ w = w^i e_i \in T_{\pi(v)} M $ in this choice of coordinates as: $$ \nabla_w V = w^i \left( \left. \tfrac{\partial V^k}{\partial x^i} \right\vert_{\pi(v)} + v^j \Gamma^k_{i\!j}(\pi(v)) \right) e_k = v^i \left. \tfrac{\partial V^k}{\partial x^i} \right\vert_{\pi(v)} e_k $$ in which each coefficient function $ V^i $ can be set by its existence theorem to satisfy the homogeneous differential equation $ v^j \frac{\partial V^i}{\partial x^j} \vert_{\pi(v)} = d V^i \!\cdot\! v = 0 $. As far as I understand, resorting to the diffeomorphism of $ \exp $ is indispensable, and may at best manifest as the assumption of existence of normal coordinates. I have a feeling that any reference to choices of coordinates is beside the point, and only complicates the mental picture as hinted at by the completing of $ v $ into a basis of $ T_{\pi(v)} M $.

Is there some straighter path taking elementary considerations surrounding the exponential map to clear implications for extending $ v $ such that $ \nabla V = 0 $ at $ \pi(v) $? In particular, if we restrict the choice of $ V $ to a subset of $ \mathfrak{X}(M) $, such as the $ g $-horizontal fields of Riemannian submersions $ \sigma : (M,g) \to (N,h) $ where $ \nabla $ is the Levi-Civita connection associated to $ g $, how to get a feel for how that restriction transfers to $ \nabla V $? Can the $ g $-horizontal lift of an element of $ \mathfrak{X}(N) $ verify the condition for $ g $-horizontal vectors at one point? These questions should be as obvious to answer as the main existence of $ V $ verifying $ \nabla V = 0 $ at one point, nevertheless I struggle with them for the lack of a clear understanding.

Septillion
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    Since you are only concerned at one point $p=\pi(v)$, you may assume $\nabla$ is "constant" in a local trivialization $\Gamma^k_{ij}(q)=\Gamma^k_{ij}(p)$ for all $q$. Can you finish it off from here? – user10354138 Apr 21 '21 at 14:12
  • Indeed, this does answer the question. My main concern would be that, though trivialisation makes mechanical sense, I fail to understand essentially how free we are to set an extension such that $ \nabla V $ vanishes on some subset of $ T_{\pi(v)}M $ in one context or another. It is possible that my lack of understanding limits my ability to phrase my difficulty correctly. – Septillion Apr 22 '21 at 10:02

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Perhaps I'm misinterpreting your question, but it is straightforward enough to find an extension $V\in\mathfrak{X}M$ of $v\in T_pM$ such that $\nabla V|_p=0$ using normal coordinates:

Choose normal coordinates $x^1,\cdots,x^n$ centered at $p$, and note that in these coordiantes $\Gamma^i{}_{jk}(p)=0$. Thus, it suffices to find a vector field $V$ with $V(p)=v$ and whose components $V^i$ are locally constant around $p$ in these coordintes, since in that case we have $$ \nabla_iV^j(p)=\frac{\partial V^j}{\partial x_i}(p)+\Gamma^j{}_{ik}(p)V^k(p)=0 $$ Such a vector field exists, and can be constructed using smooth bump functions.

Kajelad
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  • That correctly answers the question, assuming adequate coordinates eliminates the problem of finding an extension such that $ \nabla V $ vanishes at one point. I also understand the matter of normal coordinates separately. In this case, one can resort to normal coordinates to construct the extension. In other cases, there may not exist an extension under certain constraints such that $ \nabla V = 0 $ on the whole of $ T_p M $. When dealing with submanifolds, simply consider it in its own right. When dealing with submersions, I start having doubts as to what I can and cannot assume about V. – Septillion Apr 22 '21 at 10:07
  • I believe I finally understand it: We could select a submanifold the $ \mathfrak{X} $ of which be the subset of $ \mathfrak{X}(M) $ to which I desire to restrict my choice of $ V $, making its tangent space at a point the subset of $ T_{\pi(v)} M $ where $ \nabla V $ may vanish? In particular, for Riemannian submersions, some $ g $-horizontal $ V $ verifies $ \nabla V = 0 $ at one point for all $ g $-horizontal vectors, correct? Does the constraint on the extension we want with maximally null $ \nabla V $ consistently match with the tangent plane of the submanifold integral to that constraint? – Septillion Apr 22 '21 at 17:27
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    If you have a Riemannian submersion $\pi:M\to N$ and a horizontal vector $v\in T_pM$, there is a horizontal vector field $V\in \mathfrak{X}M$ such that $V(p)=v$ and $(\nabla_XV)(p)=0$ for all horizontal $X$. This can be seen by finding an extension of $d_p\pi(v)$ as above and taking its horizontal lift. Is this what you're asking? – Kajelad Apr 23 '21 at 04:53
  • thank you for your comment, yes precisely this is the example I have presented. Indeed, for a horizontal $ v $ there exists an extension of $ \mathrm{d} \pi \cdot v $ verifying the condition, and its horizontal lift $ V $ will be an extension of $ v $ and verify $ \nabla_w V = 0 $ for horizontal $ w \in T_{\pi(v)} M $. If we then consider an anisotropic metric $ g_v $, and define $ g_v $-horizontality with respect to, say, a pseudo-Finsler submersion, we could still have $ (\nabla_w V)(v) $ for all $ g_v $-horizontal $ w \in T_{\pi(v)} M $, do I understand correctly? – Septillion Apr 26 '21 at 07:29