Show that $\alpha \wedge \mathrm d \alpha =0$ when $\alpha \in \Omega ^1(M)$ and $d(f\alpha) = 0$ for some nowhere zero function $f$

Let $\alpha \in \Omega ^1(M)=\text{Tens}_1(M)$ be a differential form of degree $1$ on the smooth manifold $M$. Suppose that there is $f\in \mathcal C^\infty (M)$ s.t. $f(x)\neq 0$ for all $x\in M$ and s.t. $\mathrm d (f\alpha )=0$. Show that $$\alpha \wedge \mathrm d \alpha =0.$$

My work

Don't we need the dimension of $M$ ? If yes, I'll suppose that it's dimension is $n$. Since $\dim \Omega ^1(M)=n$, if $(U,\varphi)$ is a chart in $x$, then

$$\alpha =g_1 \mathrm d x^1+...+g_n\mathrm d x^n$$

where $g_i\in \mathcal C^\infty (M)$. We have that

$$\mathrm d \alpha =\mathrm d g\wedge \mathrm d x=\sum_{i=1}^n\sum_{j=1,\ j\neq i}^ng_{ij}\mathrm d x^i\wedge \mathrm d x^j$$

where $g_{ij}=\frac{\partial g_i}{\partial x^j}$.

Finally,

$$\alpha \wedge \mathrm d \alpha =\sum_{k=1}^n\sum_{i=1,\ i\neq k}^n\sum_{j=1, \ j\neq i,k}^n g_{ij}\mathrm d x^k\wedge \mathrm d x^i\wedge \mathrm d x^j.$$

Now,

$$0=\mathrm d (f\alpha )=\alpha \mathrm d f +f\mathrm d \alpha$$

but I don't see how to conclude.

• Your expansion of $d(f\alpha)$ is not quite correct: $d(f\alpha) = df\wedge\alpha + fd\alpha$. Commented Jan 9, 2016 at 15:18

Hint: The conditions $d(f\alpha) = 0$ and $f$ nowhere zero allow you to obtain an expression for $d\alpha$. Once you have this expression, use it to simplify $\alpha\wedge d\alpha$.