In Arnold's Mathematical Methods of Classical Mechanics, page 179, he says:
“In an oriented euclidean three-space every vector $\mathbf{A}$ corresponds to a 1-form $\omega_{\mathbf{A}}^{1}$ and a 2-form $\omega_{\mathbf{A}}^{2}$ defined by the conditions$$\omega_{\mathbf{A}}^{1}\left(\mathbf{\xi}\right)=\left(\mathbf{A},\xi\right)$$ $$\omega_{\mathbf{A}}^{2}\left(\mathbf{\xi,\eta}\right)=\left(\mathbf{A},\xi,\eta\right)$$
$$\xi,\eta\in\mathbb{R}^{3}.”$$
I take $\left(\mathbf{A},\xi\right)$ to be a scalar product. So he's saying the 1-form $\omega_{\mathbf{A}}^{1}$ acts on the vector $\xi$ to give the same number as the scalar product $\left(\mathbf{A},\xi\right)$. But what does $\left(\mathbf{A},\xi,\eta\right)$ mean? Is it some sort of scalar product of three vectors (multiplying out the respective components)? Thanks.
EDIT
I understand that the exterior product of two 1-forms is given by$$\left(\omega_{1}\wedge\omega_{2}\right)\left(\xi,\eta\right)=\left|\begin{array}{cc}
\omega_{1}\left(\xi\right) & \omega_{1}\left(\eta\right)\\
\omega_{2}\left(\xi\right) & \omega_{2}\left(\eta\right)
\end{array}\right|.$$ But I can't see why this is equal to the determinant of $\left(\mathbf{A},\xi,\eta\right).$
Best Answer
Arnold's product of $3$ vectors is simply the determinant of the respective $3\times3$-matrix.