Looking into the infinitesimal view of rotations from Lie, I noticed that the vector cross product can be written in terms of the generators of the rotation group SO(3). For example:

$$\vec{\mathbf{A}} \times \vec{\mathbf{B}} = (A^T \cdot J_x \cdot B) \>\> \hat{i} + (A^T \cdot J_y \cdot B) \>\> \hat{j} + (A^T \cdot J_z \cdot B) \>\> \hat{k}$$

where,

$$J_x = \begin{pmatrix}0&0&0\\0&0&-1\\0&1&0\end{pmatrix} \>\>\>;\>\>\>;J_y = \begin{pmatrix}0&0&1\\0&0&0\\-1&0&0\end{pmatrix} \>\>\>;\>\>\>;J_z = \begin{pmatrix}0&-1&0\\1&0&0\\0&0&0\end{pmatrix} $$

I couldn't help but think I was missing some profound connection here. Why is the vector cross product seemingly intimately related to the generators of the SO(3) rotation group?

$$\vec{\mathbf{A}} \times \vec{\mathbf{B}} = (A^T \cdot J_x \cdot B) \>\> \hat{i} + (A^T \cdot J_y \cdot B) \>\> \hat{j} + (A^T \cdot J_z \cdot B) \>\> \hat{k}$$

where,

$$J_x = \begin{pmatrix}0&0&0\\0&0&-1\\0&1&0\end{pmatrix} \>\>\>;\>\>\>;J_y = \begin{pmatrix}0&0&1\\0&0&0\\-1&0&0\end{pmatrix} \>\>\>;\>\>\>;J_z = \begin{pmatrix}0&-1&0\\1&0&0\\0&0&0\end{pmatrix} $$

I couldn't help but think I was missing some profound connection here. Why is the vector cross product seemingly intimately related to the generators of the SO(3) rotation group?

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