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The Lie bracket of two elements of the Lie algebra of every matrix group is given by the matrix commutator, $[A_1, A_2] = A_1A_2 - A_2A_1$

## Skew-symmetric matrix

The vector cross product can be expressed as the product of a skew-symmetric matrix and a vector:

$\mathbf{a} \times \mathbf{b} = \overline{\mathbf{a}} \mathbf{b}$

where $\overline{\mathbf{a}}$ is the dual of $\mathbf{a}$:

$\overline{\mathbf{a}} \quad \stackrel{\rm def}{=} \quad \begin{bmatrix}\,\,0&\!-a_3&\,\,\,a_2\\\,\,\,a_3&0&\!-a_1\\\!-a_2&\,\,a_1&\,\,0\end{bmatrix}$

One actually has

$\overline{(\mathbf{a \times b})} = \overline{\mathbf{a}}\overline{\mathbf{b}} - \overline{\mathbf{b}}\overline{\mathbf{a}} = [\overline{\mathbf{a}},\overline{\mathbf{b}}]$

i.e., the commutator of skew-symmetric three-by-three matrices can be identified with the cross-product of three-vectors.

$[\overline{\mathbf{a}},\overline{\mathbf{b}}] = \overline{\mathbf{a}}\overline{\mathbf{b}} - \overline{\mathbf{b}}\overline{\mathbf{a}} = \left[ \begin{matrix} 0 & a_2b_1-b_2a_1 & a_3b_1-b_3a_1 \\ a_1b_2-b_1a_2 & 0 & a_3b_2-b_3a_2 \\ a_1b_3-b_1a_3 & a_2b_3-b_2a_3 & 0 \end{matrix}\right] = \mathbf{b} \wedge \mathbf{a}$

## Lie bracket of vector fields

In the mathematical field of differential topology, the Lie bracket of vector fields, also known as the Jacobi–Lie bracket or the commutator of vector fields, is an operator that assigns to any two vector fields X and Y on a smooth manifold M a third vector field denoted [X, Y].

Conceptually, the Lie bracket [X, Y] is the derivative of Y along the flow generated by X. A generalization of the Lie bracket is the Lie derivative, which allows differentiation of any tensor field along the flow generated by X. The Lie bracket [X, Y] equals the Lie derivative of the vector Y (which is a tensor field) along X, and is sometimes denoted $\mathcal{L}_X Y$ (read "the Lie derivative of Y along X").

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