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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?

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  • $\begingroup$ I noticed a mistake in the computations in my initial post. I've fixed it, but unfortunately the same issue remains. $\endgroup$ – Mathstudent1996 Jan 12 at 16:39
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    $\begingroup$ 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. $\endgroup$ – Alfred Yerger Jan 12 at 17:16
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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 $$\frac{\partial}{\partial_x} = \left(\frac{d\phi}{dy}(y)\right)^{-1}\frac{\partial}{\partial_y}.$$

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    $\begingroup$ Thanks very much! Such a silly error. $\endgroup$ – Mathstudent1996 Jan 12 at 21:04

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