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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 $\left(\mathbf{A},\xi,\eta\right).$

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    $\begingroup$ For three 3-vectors, there is the triple product $a\cdot(b \times c)$. $\endgroup$ – achille hui Jul 15 '19 at 15:03
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In the standard $(x,y,z)$ coordinates, write $\mathbf A = (a,b,c)$ and let $\omega^2_{\mathbf A} = a\,dy\wedge dz + b\,dz\wedge dx + c\,dx\wedge dy$. Now check, expanding in cofactors along the first column, that $$\omega^2_{\mathbf A}(\xi,\eta) = \det(\mathbf A,\xi,\eta).$$

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  • $\begingroup$ It works! I get $\omega_{\mathbf{A}}^{2}(\xi,\eta)=a\left(\xi^{2}\eta^{3}-\xi^{3}\eta^{2}\right)+b\left(\xi^{3}\eta^{1}-\xi^{1}\eta^{3}\right)+c\left(\xi^{1}\eta^{2}-\xi^{2}\eta^{1}\right),$ which is equal to $\det(\mathbf{A},\xi,\eta)$. Thanks. $\endgroup$ – Peter4075 Jul 16 '19 at 7:20
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Arnold's product of $3$ vectors is simply the determinant of the respective $3\times3$-matrix.

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