I am trying to show that $d(a \wedge da)=0$ if $k$, the degree of k-form $a$ is even. I have said:
$=da \wedge da + (-1)^k a \wedge d^2a$
I believe the first term is zero due to repeated indices and the second is zero since $d^2=0$. However, a subsequent exercise relies on the fact that $d(a \wedge da)$ is not alway zero if k is odd. Can anybody point out the error that I have made?
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As levap writes in their answer, it is true that $da \wedge da = 0$ for forms $a$ of even degree $k$: The exterior derivative $da$ has degree $k + 1$, and so by the commutation rule $$\alpha \wedge \beta = (-1)^{|\alpha| |\beta|} \beta \wedge \alpha,$$ (where $|\gamma|$ denotes the degree of $\gamma$) we have $$da \wedge da = (-1)^{(\textrm{odd})(\textrm{odd})} = - da \wedge da ,$$ and so $da \wedge da = 0$.
On the other hand, if $a$ has odd degree, so that $da$ has even degree, $$da \wedge da = (-1)^{(\textrm{even})(\textrm{even})} = da \wedge da ,$$ which is a tautology.
Indeed, it is possible that for forms $a$ of odd degree that $da \wedge da \neq 0$. A simple example is $$a := x^1 \,dx^2 + x^3 \,dx^4,$$ on $\Bbb R^4$, for which $da$ is the standard symplectic form, $$da = dx^1 \wedge dx^2 + dx^3 \wedge dx^4 ,$$ and $da \wedge da$ is (a multiple) of the standard volume form, $$da \wedge da = 2 \,dx^1 \wedge dx^2 \wedge dx^3 \wedge dx^4 .$$
It is the second term that is always zero (because $d^2a = 0$), no matter what $k$ is. The first term is zero because $da$ has odd degree. In general
$$ x \wedge y = (-1)^{|x||y|} y \wedge x$$
where $|x|,|y|$ are the degrees of $x,y$. In particular, if $x$ has odd degree then $x \wedge x = 0$.