Is the Lie bracket always invariant under coordinate transformations?

This is my first question on StackExchange. I think it's probably quite easy, but it's been baffling me for a while.

I'm doing computations to determine invariant properties of the quantity $$X\circ Y= \nabla_Y X$$ where $$X$$ and $$Y$$ are vector fields, when we change coordinate choices (when $$\nabla$$ is a flat torsion-free connection).

So naturally, we start with the simplest case, $$\mathbb{R}$$ with its usual connection. However we encounter the following problem.

Start with vector fields $$X=f(x)dx$$ and $$Y=g(x)dx.$$ We have $$X\circ Y = g(x)\frac{df}{dx}dx.$$ We then make a general change of coordinates $$\phi(y)=x.$$ So $$X=f(\phi(y))\frac{d\phi}{dy}dy$$ and $$Y=g(\phi (y))\frac{d\phi}{dy}dy.$$ Then $$X \circ' Y = g(\phi(y))\frac{d\phi}{dy}(\frac{df(\phi(y))}{d \phi}\frac{d \phi}{dy}+f(\phi(y))\frac{d^2\phi}{d y^2})dy$$ in the new coordinates. Attempting to return back to $$x$$-coordinates we get $$X \circ' Y =(\frac{d \phi}{dy})^2(g(x)\frac{df}{dx}dx)+ g(x)f(x)\frac{d^2 \phi}{dy^2}dx$$. The rightmost term is symmetric in $$X$$ and $$Y$$ and for the purposes of this question can be neglected.

In this case $$X\circ Y-Y\circ X = [Y,X]$$ as $$\nabla$$ is flat torsion free. But in the expression above we pick up a factor of $$(\frac{d\phi}{d y})^2$$ when we compute the bracket. But the bracket is coordinate independent! What's gone wrong?

• I noticed a mistake in the computations in my initial post. I've fixed it, but unfortunately the same issue remains. – Mathstudent1996 Jan 12 at 16:39
• The bracket of two vector fields is a vector field by commutativity of mixed partials, and so should be invariant, yes. This doesn't answer the question in the body of the text, as I haven't yet found the error, but it answers the question in the title. – Alfred Yerger Jan 12 at 17:16

The mistake is in the change of coordinates. If $$x=\phi(y)$$, then $$dx = d\phi = \frac{d\phi}{dy}(y)\,dy$$ is the change of coordinates for one-forms, not vectors. For vectors you have the dual formula $$\frac{\partial}{\partial_x} = \left(\frac{d\phi}{dy}(y)\right)^{-1}\frac{\partial}{\partial_y}.$$