Bracket of Lie algebra-valued differential form

In this wikipedia article:
https://en.wikipedia.org/wiki/Lie_algebra-valued_differential_form
the bracket of Lie algebra-valued forms is defined. At one point it mentions that it is the bilinear product $[\cdot , \cdot]$ on $\Omega^* ( \mathfrak{g})$ such that,

$$\begin{equation} [(g_1 \otimes \alpha) , (g_2 \otimes \beta)] = [g_1, g_2] \otimes (\alpha \wedge \beta) \end{equation}$$
for all $g_1, g_2 \in \mathfrak{g}$ and all $\alpha, \beta \in \Omega^*(M)$.
From this expression it looks like for an odd form $\alpha$, $[\alpha , \alpha] = 0$ since the second term is $0$. But this is false according to the definition
$$\begin{equation} [\alpha, \beta](X_1,\dots,X_{p+q}) := \sum_{\sigma \in S_{p+q}} \text{sgn}(\sigma) \left[\alpha(X_{\sigma(1)},\dots,X_{\sigma(p)}),\beta(X_{\sigma(p+1)},\dots,X_{\sigma(p+q)})\right] \end{equation}$$
What am I missing?

Answer

It's true that the Lie bracket of things of the form $g \otimes \alpha$ with themselves are always zero. (This is really no more than the statement that $[g,g]=0$, as you note.)

But most Lie algebra-valued forms do not look like this. They're sums of such terms. If $g_i$ is a basis of the Lie algebra, then a Lie algebra valued form actually looks like $\sum g_i \otimes \alpha_i$. You can see how the brackets of such terms can fail to be zero. (Pick some nice 2-dinensional Lie algebra to play with.)

So things are only interesting when you're working with non-pure tensors.