# Covariant derivative and the product rule

Let $(M,g)$ be a Riemannian manifold with a connection $\nabla$. It is required to satisfy the product rule: $$\nabla_X(fY) = f\nabla_X(Y) + \nabla_X(f)Y$$ where $\nabla_X(f)$ is the directional derivative of $f$ along $X$.

We can consider another product rule of the form: $$\nabla_X(Yf) = \nabla_X(Y)f + Y\nabla_X(f).$$ However, I think this fails. For instance, let me take $X = \partial/\partial u_j$, $Y = \partial/\partial u_k$ in local coordinates $(u_1,u_2,\dots)$ so that $\nabla_X(Y) = \sum_i\Gamma^i_{jk}\partial/\partial u_i$ in which case we have: $$\nabla_X(Yf) = \nabla_X(\partial f/\partial u_k) = \partial^2f/\partial u_j\partial u_k$$ while the right hand side of the expression is: $$\sum_i \Gamma^i_{jk}\partial f/\partial u_i + \partial^2f/\partial u_j\partial u_k$$ and the two expressions clearly don't match (unless $\Gamma^i_{jk} = 0$ identically).

Is there a mistake in my verification or does the second rule indeed fail? If so, what is special about the first rule vs the second and why do we require one to hold and not the other?

• The covariant derivative acts on functions as follows: $\nabla_X f = X(f)$. According to the rule you are using, I get $X(Y(f)) = (\nabla_X Y)f+Y(X(f))$, that is $\nabla_X Y = [X,Y]$. However this $\nabla$ is not $C^{\infty}$-linear in $X$. So I do not think that 'product rule' holds. Jun 8, 2018 at 19:15
• What is $Yf$ exactly?
– user99914
Jun 9, 2018 at 3:42
• Y applied to f, that is, the directional derivative of f along Y Jun 9, 2018 at 3:50

The second Leibniz rule

$$\nabla _X (Yf) = (\nabla _XY) f + Y(\nabla _Xf)$$

cannot be true. Indeed, re-write as

$$Y(\nabla_X f) = (\nabla _X Y) f - \nabla_X (Yf):= A(X, Y)$$

Note that the right hand side is $C^\infty$-linear in $X$: $A(gX, Y) = gA(X, Y)$. But the left hand side is not:

$$Y( \nabla_{gX} f) = Y(g\nabla_X f) = Yg (Xf) + g Y (\nabla_X f) \neq g Y (\nabla_X f).$$

Indeed,

$$\nabla_X (Yf) - \nabla _Y(Xf) = [X, Y] f$$

is the Lie bracket and is well known to be not tensorial.