Verification of the identity $\langle \nabla _{[X,Y]}Z,Z \rangle = \frac{1}{2}[X,Y]\langle Z,Z \rangle$

Question

In the book Riemannian Geometry, page 91, Do Carmo writes:

$$\langle \nabla _{[X,Y]}Z,Z \rangle = \frac{1}{2}[X,Y]\langle Z,Z \rangle$$

I could not understand how this happens. Can someone explain this to me please?

Answer 1

For vector fields $U, V, W$ on a ($C^1$) Riemannian manifold, we have the Leibniz Rule $$ W \langle U, V \rangle = \nabla_W \langle U, V \rangle = \langle \nabla_W U, V \rangle + \langle U, \nabla_W V \rangle \textrm{.} $$ Now, take $U := Z, V := Z, W := [X, Y]$. (In fact, this argument works for any connection $\nabla$ compatible with the metric.)

Answer 2

If $(M,g)$ is Riemannian manifold, the Levi-Cevita connection is the unique affine connection on $TM$ which is torsion-free and compatible w.r.t. $g$ :

• $\nabla_XY - \nabla_YX = [X,Y]$
• $X \cdot\langle Y,Z\rangle = \langle \nabla_XY,Z\rangle + \langle Y,\nabla_XZ\rangle$

So, in particular, $$[X,Y]\cdot\langle Z,Z\rangle = \langle \nabla_{[X,Y]}Z,Z\rangle +\langle Z,\nabla_{[X,Y]}Z\rangle $$ that is: $$\dfrac{1}{2} [X,Y]\cdot \langle Z,Z\rangle = \langle \nabla_{[X,Y]}Z,Z\rangle.$$