If there exists a nontrivial vector field $V\not=0$ in Riemannian manifold $M$ and an open set $U\subset M$ such that $\nabla_{X}V=0$ in $U$ for any vector field $X$ in $M$, then dose $U$ have to be flat?

That is, if a Riemannian maniflod exists a vector field $V$ parallel transport along any vector field, then is this maniflod flat?

  • $\begingroup$ Does that condition ever hold? $\endgroup$ – Mariano Suárez-Álvarez Apr 20 '14 at 4:57
  • $\begingroup$ @Mariano Suárez-Alvarez If it is a flat manifold such as $\mathbb{R}^3$, then any constant vector field parallel transport along any vector field. So does the converse hold? $\endgroup$ – 346699 Apr 20 '14 at 5:08
  • $\begingroup$ Have you tried $V=0$? $\endgroup$ – Ben McKay Apr 20 '14 at 8:29
  • $\begingroup$ @BenMcKay Yes, it's my fault. $V$ must be nontrivial. $\endgroup$ – 346699 Apr 20 '14 at 8:38

If you take a manifold like $M = \mathbb R \times M'$ with the usual metric on $\mathbb R$ and where $M'$ is some Riemannian manifold, the vector field induced from $\mathbb R$ will satisfy this condition but $M$ will not be flat in general.

However, if you impose instead that you have $\dim M$ many independent parallel vector fields then by computing in this frame you immediately see that the curvature tensor vanishes.

  • $\begingroup$ Thanks, this is exactly what I want. Can you give me the proof of your second claim ? Thanks. $\endgroup$ – 346699 Apr 20 '14 at 5:19
  • 2
    $\begingroup$ Say $V_1,\ldots,V_n$ is a collection of linearly independent (everywhere on $U$) vector fields that are parallel, i.e. $\nabla_X V_j = 0$ for all vector fields $X$. For any vector fields $X,Y$, $R(X,Y)$ is an endomorphism of the tangent bundle, where $R$ is the Riemann curvature tensor. Since it is an endomorphism (as opposed to a differential operator), to show it is zero it suffices to show it vanishes on a frame of the tangent bundle. But $R(X,Y) V_i = \nabla_X \nabla_Y V_i -\nabla_Y\nabla_X V_i - \nabla_{[X,Y]} V_i = 0$ for all $i$. $\endgroup$ – Eric O. Korman Apr 20 '14 at 5:27
  • $\begingroup$ I should say that above $n = \dim M$. $\endgroup$ – Eric O. Korman Apr 20 '14 at 14:26
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    $\begingroup$ Actually, by de Rham's decomposition theorem, a complete simply-connected manifold that admits a nonvanishing parallel vector field $\nabla V=0$ must split as a product $\mathbb R\times M$. The more linearly independent parallel vector fields you have, the further you can decompose it as a product. $\endgroup$ – Renato G. Bettiol Apr 30 '14 at 1:05

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