Smooth vector field over $S^1$

Solution 1:

First, you need to figure out how to change between the two coordinate charts $x_1$ and $x_2$. Let $(y,z)$ be a point on the unit circle $y^2+z^2=1$. Using triangle ratios, you can find $x_1 = \dfrac y{1-z}$ and $x_2 = \dfrac y{1+z}$. Then $x_1x_2 = \dfrac{y^2}{1-z^2} = 1$, since $y^2+z^2=1$. Thus, whenever both $x_1,x_2$ are both defined, $x_1x_2 = 1$.

Now we want to get $\dfrac\partial{\partial x_1}$ in terms of $x_2$. By the chain rule, this is $\dfrac{\partial x_1}{\partial x_2}\dfrac\partial{\partial x_2}$. Then using $x_1 = \dfrac1{x_2}$, $\dfrac{\partial x_1}{\partial x_2} = \dfrac{-1}{x_2^2}$. Thus $\dfrac\partial{\partial x_1} = \dfrac{-1}{x_2^2}\dfrac\partial{\partial x_2}$. The point $N$ corresponds to $x_2 = 0$, so we see that this quantity is diverging. So no, the vector field is not smooth.