I am reading John M. Lee's Riemannian Manifolds: An Introduction to Curvature. In Lemma $5.2$, it is said that the following conditions are equivalent for a linear connection $\nabla$ on a Riemannian manifold:

(a) $\nabla$ is compatible with $g$. i.e., for any vector fields $X,Y,Z$, $$ \nabla_X g(Y,Z) = g(\nabla_X Y,Z) + g(Y,\nabla_X Z) $$ (b) $\nabla g\equiv 0.$

How do we go from (a) to (b) (and (b) to (a))?


1 Answer 1


Hint: Use Lemma $4.6$ (ii) (i.e. the formula displayed here) to deduce that

$$(\nabla_X g)(Y, Z) = \nabla_Xg(Y, Z) - g(\nabla_XY, Z) - g(Y, \nabla_XZ).$$

As $g(Y, Z)$ is a smooth function, $\nabla_Xg(Y, Z) = Xg(Y, Z)$ so the metric compatibility condition can also be written as

$$Xg(Y, Z) = g(\nabla_XY, Z) + g(Y, \nabla_XZ).$$

  • $\begingroup$ @MichaelAlbenese, please, I am a little confused. By definition, the connection is a map $\nabla \mathcal X(M) \times \mathcal X(M) \to X(M)$, where $\mathcal X(M)$ is the set of smooth vector fields on the manifold $M$. Can a smooth function be seen as a smooth vector field? $\endgroup$ Commented Jul 24, 2020 at 18:24
  • 1
    $\begingroup$ @DaniloGregorin: A connection $\nabla$ on $TM$ can be viewed as such a map, but it induces a connection on all tensor bundles, in particular the trivial line bundle $M\times\mathbb{R}$. A smooth section of the trivial line bundle is nothing but a smooth function and for the induced connection we have $\nabla_Xf = Xf$. See Lemma 4.15 of Lee's Introduction to Riemannian Manifolds (second edition). $\endgroup$ Commented Jul 24, 2020 at 19:37
  • $\begingroup$ Thank you very much. I am having trouble with do Carmo's, he hides too many things. $\endgroup$ Commented Jul 24, 2020 at 19:52

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