# If there exists a nontrivial vector field $V$ such that $\nabla_{X}V=0$ for any vector field $X$, the manifold must be flat?

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?

• Does that condition ever hold? – Mariano Suárez-Álvarez Apr 20 '14 at 4:57
• @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? – 346699 Apr 20 '14 at 5:08
• Have you tried $V=0$? – Ben McKay Apr 20 '14 at 8:29
• @BenMcKay Yes, it's my fault. $V$ must be nontrivial. – 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.
• 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 $i$. – Eric O. Korman Apr 20 '14 at 5:27
• I should say that above $n = \dim M$. – Eric O. Korman Apr 20 '14 at 14:26
• 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. – Renato G. Bettiol Apr 30 '14 at 1:05