The question asks: "If $v_a$ are the components of a vector, show that in an arbitrary coordinate system that $A_{ab}$ are components of a rank-2 tensor, where:"
$$A_{ab}= \partial_bv_a-\partial_av_b $$
I've tried using the transformation law. However, when I expand the differentials using the chain rule and then change the $v_a$ into a new frame, I get a two second derivative terms that don't transform as a tensor. I cannot get rid of these and hence any help would be appreciated.
Do I need to use the connection somewhere in this transformation? So far I have not.
Thanks!
The 2nd derivative terms I get are:
$$ \frac{\partial x^c}{\partial x'^b}\frac{\partial^2x^d}{\partial x^c\partial x'^a} - \frac{\partial x^c}{\partial x'^a}\frac{\partial^2x^d}{\partial x^c\partial x'^b} $$
where c and d are dummy indices and a and b are free indices. Why do these two terms cancel?
By the chain rule
$$\frac{\partial x^c}{\partial x'^b}\frac{\partial^2x^d}{\partial x^c\partial x'^a}=\frac{\partial^2x^d}{\partial x'^b\partial x'^a}$$
and similarly the other term becomes
$$\frac{\partial^2x^d}{\partial x'^a\partial x'^b}.$$
Then these terms cancel because, as ziggurism says in the comments, of "equality of mixed partial derivatives" i.e. the fact that
$$\frac{\partial^2w}{\partial u\;\partial v}=\frac{\partial^2w}{\partial v\;\partial u}$$ for arbitary $w(u,v)$.