For the plane in polar coordinates $(r,\theta)$ with metric $$ds^2=dr^2+r^2d\theta^2,$$
the curl on a vector field $v^a\partial_a$ is given by the rank-2 antisymmetric tensor $\nabla_av_b-\nabla_bv_a$. This tensor has only one independent & non-zero component:
$$\nabla_rv_\theta-\nabla_\theta v_r=\partial_rv_\theta – \Gamma_{r\theta}^av_a -\partial_\theta v_r +\Gamma^a_{\theta r}v^a=\partial_rv_\theta-\partial_\theta v_r$$
since $\Gamma^a_{\theta r} = \Gamma^a_{r \theta }$. Expressed in contravariant vector components, $$\nabla_rv_\theta-\nabla_\theta v_r=\partial_r (g_{\theta a}v^a)-\partial_\theta (g_{rb}v^b)=\partial_r(r^2v^\theta)-\partial_\theta(v^r)=2rv^\theta+r^2\partial_rv^\theta-\partial_\theta v^r$$
I am trying to compare this to the formula for curl in vector calculus:
$$\nabla\times \textbf{v} = \frac{1}{r}\left[\frac{\partial}{\partial r}(rv^\theta) – \frac{\partial}{\partial\theta}v^r\right] \vec{\hat{z}}=\left[\frac{1}{r}v^\theta+\partial_rv^\theta-\frac{1}{r}\partial_\theta v^r\right]\vec{\hat{z}}$$, where $\vec{\hat{z}}$ is the unit vector pointing out of the plane.
The difference in the two formulas is because in manifolds we used the coordinate basis $\partial_\theta$ while in vector calculus we used the normalized coordinate basis $\frac{1}{r}\partial_\theta$. In the coordinate basis, $\textbf{v}=v^r\partial_r+v^\theta\partial_\theta$ while in the normalized coordinate basis, $\textbf{v}=v^r\partial_r+v^\theta\frac{1}{r}\partial_\theta$.
Thus to convert between the 'manifold' formula and the 'vector calculus' formula, we just have to rescale the $v^\theta$ component of the vector.
From the 'manifold' formula, replace $v^\theta\rightarrow \frac{v^\theta}{r}$ to get the 'vector calculus' formula:
$$\nabla_rv_\theta-\nabla_\theta v_r=2rv^\theta+r^2\partial_rv^\theta-\partial_\theta v^r \rightarrow2r\frac{v^\theta}{r}+r^2\partial_r\frac{v^\theta}{r}-\partial_\theta v^r =v^\theta+r\partial_rv^\theta -\partial_\theta v^r$$
However, this answer larger than the correct formula by a factor of $r$. Why is that so?
Best Answer
Say $(M,g)$ is a $2$-dimensional oriented Riemannian manifold, and let $\star$ be the Hodge star operator. Given a vector field $F$ on $M$, we define \begin{align} \text{curl}(F)&:=\star d(g^{\flat}F) \end{align} i.e we take our vector field $F$, convert it to a covector field $g^{\flat}(F)$, take its exterior derivative (so we now have a $2$-form) and finally take its Hodge dual to get a smooth function (since we're on a $2$-dimensional manifold). The fact that the curl of a vector field in $2$-dimensions yields a smooth function corresponds to your observation that there's only one non-vanishing term. The thing you're missing is the final Hodge star (the extra $r$ you have is the same $r$ in $dx\wedge dy=r\,dr\wedge d\theta$).
Explicitly, suppose we're in the plane and using polar coordinates. Suppose $F=f^r\mathbf{e}_r+ f^{\theta}\mathbf{e}_{\theta}$ is the expansion in terms of unit vector fields (as in vector calculus). Then, $g^{\flat}(F)=f^r\,dr +f^{\theta}\,rd\theta$ (here this factor of $r$ is the norm of $\frac{\partial}{\partial \theta}$, i.e $\sqrt{g_{\theta\theta}}=\sqrt{r^2}=r$). So, \begin{align} d(g^{\flat}(F)) &= \left(\frac{\partial f^r}{\partial r}\,dr+\frac{\partial f^r}{\partial \theta}\,d\theta\right)\wedge dr + \left(\frac{\partial (rf^{\theta})}{\partial r}\,dr+\frac{\partial (r f^{\theta})}{\partial \theta}\,d\theta\right) \wedge d\theta\\ &=\left(\frac{\partial (r f^{\theta})}{\partial r}-\frac{\partial f^r}{\partial \theta}\right)dr\wedge d\theta\tag{$*$}\\ &=\frac{1}{r}\left(\frac{\partial (r f^{\theta})}{\partial r}-\frac{\partial f^{r}}{\partial \theta}\right)\, rdr\wedge d\theta \end{align} In the last line, I simply divided and multiplied by $r$. The reason for this is that the area-form in the plane is $dx\wedge dy=r\,dr\wedge d\theta$, and the nice thing about the Hodge star is that for any function $\phi$ and the area form $\omega$, we have $\star(\phi \, \omega)=\phi (\star\omega)=\phi\cdot 1$, i.e $\star(r\,dr\wedge d\theta)=1$. Therefore, \begin{align} \text{curl}(F)&=\star d(g^{\flat}(F))=\frac{1}{r}\left(\frac{\partial (r f^{\theta})}{\partial r}-\frac{\partial f^{r}}{\partial \theta}\right) \end{align}
Where you went 'wrong'
Note that in your calculation, you stopped at $(*)$, i.e the expression you found is the component of a $2$-form (the coefficient of $dr\wedge d\theta$). Inherently, there's nothing wrong with that, but the classical definition of curl requires an extra Hodge star application. So really, it boils down to the fact that your initial expression $\nabla_av_b-\nabla_bv_a$ is the component of a $2$-form, but it is NOT the usual definition of curl.
Extra Ramblings.
For a vector field $F$ in $\Bbb{R}^3$, the definition of curl must be modified slightly: \begin{align} \text{curl}(F)&:=g^{\sharp}(\star d(g^{\flat}F)) \end{align} i.e we convert the vector field $F$ to a $1$-form $g^{\flat}(F)$, take its exterior derivative (so we have a $2$-form now), then take the Hodge star (so we get a $3-2=1$-form) and finally use $g^{\sharp}$ to convert it back to a vector field (as a side remark: from this one can already see that the curl of a vector field is an EXTREMELY unnatural operation, and it's unfortunate we use it so often in elementary vector calculus).
BTW, if you want to think of a vector field in $\Bbb{R}^2$ as a vector field in $\Bbb{R}^3$, you can do so by thinking of it as having no $z$ component, in which case applying this 3-D definition of curl will yield a vector field having only a $\mathbf{e}_z$ component, which matches with what we found above (so overall there's no inconsistency in jumping between $2$ and $3$ dimensions).