Currently reading Mathematical Gauge theory by Hamilton and in the section on the lie group $G_2$ he defines this map:
$$b:\mathbb{R}^7 \times \mathbb{R}^7\rightarrow \Lambda^7\mathbb{R}^{7*}$$ $$b:(x,y)\rightarrow \frac{1}{6}\phi_{x}\wedge \phi_y\wedge \phi$$ Where $\phi$ is a three form on $\mathbb{R}^7$ given by:
$$\phi=\omega^{123}+\omega^1(\omega^{45}+\omega^{67})+\omega^2(\omega^{46}-\omega^{57})-\omega^3(\omega^{47}+\omega^{56})$$ And: $$\omega^{1}\omega^{23}=\omega^1\wedge\omega^2\wedge\omega^3$$ And finally $\phi_x$ is the contraction of $\phi$ with $x\in\mathbb{R}^7$ given by: $$\phi_x(y,z)=\phi(x,y,z)$$ So, now that all the notation and jargon is out of the way, Hamilton off the bat calls the map $b$ obviously symmetric. Now, I've only been studying alternating tensors for a couple months now, and from what I can see this doesn't look symmetric at all.
Let $a,b,c,d,e,f,g\in\mathbb{R}^7$, then I have that: $$b(x,y)=\phi_x(a,b)\wedge \phi_y(c,d)\wedge\phi(e,f,g)$$ Then by the definition of the contraction I have: $$b(x,y)=\phi(x,a,b)\wedge\phi(y,c,d)\wedge\phi(e,f,g)$$ And similarly: $$b(y,x)=\phi(y,a,b)\wedge\phi(x,c,d)\wedge\phi(e,f,g)$$ Which to me seems like it would pick up a minus sign once I switch $x$ and $y$. I even tried just swapping $x$ with and $y$ with their neighbors position by position and I still got an odd number of permutations. I am clearly missing something here but I don't know what it is, so any clarification here would be great.
Note that $\phi_x$ and $\phi_y$ are two-forms, so $\phi_y\wedge\phi_x = \phi_x\wedge\phi_y$. Therefore
$$b(y, x) = \frac{1}{6}\phi_y\wedge\phi_x\wedge\phi = \frac{1}{6}\phi_x\wedge\phi_y\wedge\phi = b(x, y)$$
so $b$ is symmetric.