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I found this on an old qualifying exam. I started the problem, but I'm not sure what my next step should be:

Let $T^2$ be the standard 2-dimensional torus with $\mathbb{Z}$-periodic coordinates $(x,y)$. Consider the vector field $$ V = \sin(2\pi y) \dfrac{\partial}{\partial x}$$

Prove that any vector field $Y$ on $T^2$ which commutes with $V$, i.e., $[Y,V] = 0$, has zero coefficient in front of $\frac{\partial}{\partial y}$.

I started by letting $Y = f(x,y) \frac{\partial}{\partial x} + g(x,y) \frac{\partial}{\partial y}$ be an arbitrary vector field that commutes with $V$. I then tried calculating the bracket:

$$0 = [Y,V] = 2\pi\cdot g(x,y) \cos(2\pi y) \frac{\partial}{\partial x} - \sin(2\pi y) \frac{\partial f}{\partial x} \cdot \frac{\partial}{\partial x} - \sin(2\pi y) \frac{\partial g}{\partial x} \cdot \frac{\partial}{\partial y}$$

I then deduced that since $\sin(2\pi y) \frac{\partial g}{\partial x} = 0$, and $\sin(2\pi y)$ is not always zero, so $ \frac{\partial g}{\partial x} = 0$. This means that $g$ is a function of only $y$.

However, at this point I am stuck. Any suggestions?

phenomenalwoman4
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1 Answers1

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Then you need to have $$\frac{\partial f}{\partial x} = 2\pi g(y)\cot(2\pi y)\,,$$ and there is no $1$-periodic solution $f$ unless $g=0$.

Ted Shifrin
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