Using the commutator to show unit vectors in polar coordinates are a noncoordinate basis? I'm having trouble getting the algebra right here and don't know where I'm going wrong:
Show that the unit basis vector fields for polar coordinates in the Euclidean plane,
$$\hat{\mathbf{r}} = \cos\theta \hat{\mathbf{x}} + \sin\theta \hat{\mathbf{y}} \\
\hat{\mathbf{\theta}} = -\sin\theta \hat{\mathbf{x}} + \cos\theta \hat{\mathbf{y}}
$$
where $\hat{\mathbf{x}}  = \partial/\partial x $, $\hat{\mathbf{y}}  = \partial/\partial y $, are a noncoordinate basis.
I started by noting that $x=r\cos\theta$, $y=r\sin\theta$, therefore $\dfrac{\partial}{\partial x}\cos\theta = \dfrac{1}{r}$ and $\dfrac{\partial}{\partial x}\sin\theta= \dfrac{1}{r}$.
Then I wrote out the commutator components:
$$ [\hat{\mathbf{r}}, \hat{\mathbf{\theta}}] = \left[r^i {\partial\over\partial x^i}, \theta^j {\partial\over\partial x^j}\right] \\
 = \left(r^i {\partial\theta^j\over\partial x^i}- \theta^i {\partial r^j\over\partial x^i}\right){\partial\over\partial x^j} \\
=-cos\theta{\partial \sin\theta \over \partial x}{\partial  \over \partial x} + \cos\theta{\partial \cos\theta \over \partial x}{\partial  \over \partial y} 
-\sin\theta{\partial \sin\theta \over \partial y}{\partial  \over \partial x}
+ \sin\theta{\partial \cos\theta \over \partial y}{\partial  \over \partial y} \\
+ \sin\theta{\partial cos\theta \over \partial x}{\partial  \over \partial x}
+ \sin\theta{\partial sin\theta \over \partial x}{\partial  \over \partial y}
- \cos\theta{\partial cos\theta \over \partial y}{\partial  \over \partial x}
- \cos\theta{\partial sin\theta \over \partial y}{\partial  \over \partial y} \\
= 0 + \cos\theta {1\over r}{\partial \over \partial y} - \sin\theta {1\over r}{\partial \over \partial x} + 0 + \sin\theta {1\over r}{\partial \over \partial x}  +0+0-\cos\theta {1\over r}{\partial \over \partial y} = 0
$$
but the answer is supposed to be non-zero, so I've included terms that I shouldn't have somewhere.
 A: I'm note sure what you mean by "I've started noting that $x = r \cos \theta, y = r \sin \theta$". If you perform the change of coordinates $x = r \cos \theta, y = r \sin \theta$, the resulting vector fields $\frac{\partial}{\partial r}$ and $\frac{\partial}{\partial \theta}$ are by definition a coordinate basis. What is the relation between $\frac{\partial}{\partial r}, \frac{\partial}{\partial \theta}$ and $\frac{\partial}{\partial x} = \hat{\mathbf{x}}, \frac{\partial}{\partial y} = \hat{\mathbf{y}}$? We have
$$ \frac{\partial}{\partial x} = \cos \theta \frac{\partial}{\partial r} - r \sin \theta \frac{\partial}{\partial \theta}, \\
\frac{\partial}{\partial y} = \sin \theta \frac{\partial}{\partial r} + r\cos \theta \frac{\partial}{\partial \theta}. $$
Solving for $\frac{\partial}{\partial r}, \frac{\partial}{\partial \theta}$ we get
$$ \frac{\partial}{\partial r} = \cos \theta \frac{\partial}{\partial x} + \sin \theta \frac{\partial}{\partial y} = \hat{
\mathbf{r}}, \\
\frac{\partial}{\partial \theta} = \frac{1}{r} \left( -\sin \theta \frac{\partial}{\partial x} + \cos \theta \frac{\partial}{\partial y} \right) = \frac{1}{r} \mathbf{\hat{\theta}}. $$
Thus, the difference between $\frac{\partial}{\partial r}, \frac{\partial}{\partial \theta}$ and $\mathbf{\hat{r}}, \mathbf{\hat{\theta}}$ is that $\frac{\partial}{\partial \theta}$ is not normalized and $\hat{\theta}$ points in the same direction as $\frac{\partial}{\partial \theta}$, only it is of unit length.
Finally, since $[\frac{\partial}{\partial r}, \frac{\partial}{\partial \theta}] = 0$, we have
$$ [\hat{\mathbf{r}}, \hat{\mathbf{\theta}}] = \left[ \frac{\partial}{\partial r}, r\frac{\partial}{\partial \theta} \right] = \frac{\partial r}{\partial r} \frac{\partial}{\partial \theta} + r \left[ \frac{\partial}{\partial r}, \frac{\partial}{\partial \theta} \right] = \frac{\partial}{\partial \theta} = \frac{1}{r} \mathbf{\hat{\theta}} $$
where we used the fact that
$$ [X, fY] = (Xf)Y + f[X,Y]. $$
A: I think what Ievap said is wrong, and the right answer would be $-\frac{1}{r}\hat{\theta}$. 
$\frac{\partial}{\partial\theta} = r\hat\theta$, and then $[ \hat{\mathbf{r}},\hat{\mathbf{\theta}}] = [ \frac{\partial}{\partial r},\frac{1}{r}\frac{\partial}{\partial \theta}] = -\frac{1}{r^2}\frac{\partial}{\partial \theta} = -\frac{\hat{\mathbf{\theta}}}{r}$
