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Given a linear function $f$ on euclidean space $\mathbb{R}^n$ we know that $f$ is of form $f(x) = b^{T}x$ and its gradient is constant.

My question of similar thing on riemannian manifolds. specifically, is there function $f: \mathcal{M} \rightarrow \mathbb{R}$ s.t riemannian gradient $\text{grad} f(x) = v$. Now this can't be possible, a tangent vector $v$ is not tangent vector at other points.

consider simple relaxation $\text{grad} f(x) = \text{PT}^{x_{0} \rightarrow x}(v) $ where $\text{PT}^{x_{0} \rightarrow x}$ is parallel transport from $x_{0}$ to $x$ along the minimizing geodesic (and assume such minimizing geodesic is unique). I am putting question for clarity below clarity

Is there are function $f: \mathcal{M} \rightarrow \mathbb{R}$ s.t $\text{grad} f(x) = \text{PT}^{x_{0} \rightarrow x}(v) $ where $v \in T_{x_{0}} \mathcal{M}$

focusing
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    Nice question but it is ill posed. The parallel transport $x_0\to x$ is not well-defined. This is not Euclidean space, you need to specify a path from $x_0$ to $x$. The parallel transport will in general depend on this path. If the parallel transport is path-independent then the space is flat: https://math.stackexchange.com/q/852574/8157 (this is for surfaces, I guess the same holds for higher-dimensional manifolds). – Giuseppe Negro Sep 05 '22 at 09:46
  • @GiuseppeNegro consider it along minimizing geodesic along x_{0} to x and assume it is unique – focusing Sep 05 '22 at 09:47
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    @focusing You should then state it in your question – Didier Sep 05 '22 at 09:47
  • @Didier I have added it – focusing Sep 05 '22 at 09:48
  • @focusing: OK, but are you sure there are nontrivial examples of manifolds with a unique geodesic joining any pair of points? I am genuinely asking because I do not know. For example, the sphere $\mathbb S^2$ does not satisfy this, because if you take two antipodal points then they are joined by uncountable geodesics. – Giuseppe Negro Sep 05 '22 at 09:59
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    @GiuseppeNegro I think these manifolds are diffeomorphic to $\Bbb R^n$: maybe some adaptation of the Cartan-Hadamard Theorem will work. – Didier Sep 05 '22 at 15:16
  • @Didier: thanks, I kind of suspected that. Would that diffeomorphism be a conformal mapping? If that is the case, this completely answers this question: the maps with constant gradients are precisely the affine linear ones, up to conformal mapping. – Giuseppe Negro Sep 05 '22 at 15:52
  • @GiuseppeNegro Conformal to what? The euclidean metric? I almost sure that the answer would be negative: I believe that there are plenty of nonpositively curved metrics on $\Bbb R^n$ that are not conformal to a flat metric (I might be wrong). – Didier Sep 05 '22 at 16:38

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