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  "body": "\n## Definition\n\nLet $M$ be a [[smooth manifold|F45VQ47GPV]], and let $C^\\infty(M)$ be the algebra of [[smooth functions|0V33BNKZ9S]] on $M$.\nA smooth vector field $X$ on $M$ is a smooth [[section|WWX7H5W1RP]] of the [[tangent bundle|WY6K3P4YCG]]; it acts on $f \\in C^\\infty(M)$ by $(Xf)(p) = X_p(f)$.\nFor smooth vector fields $X$ and $Y$ on $M$, the *Lie bracket* $[X, Y]$ is the vector field given by\n$$[X, Y] f = X(Y f) - Y(X f), \\qquad f \\in C^\\infty(M).$$\n\n## Remarks\n\nIn local coordinates, with $X = \\sum_i X^i \\partial_i$ and $Y = \\sum_j Y^j \\partial_j$,\n$$[X, Y] = \\sum_{j} \\sum_{i} \\left( X^i \\frac{\\partial Y^j}{\\partial x^i} - Y^i \\frac{\\partial X^j}{\\partial x^i} \\right) \\partial_j.$$\nExample on $\\R^2$: for $X = \\partial_x$ and $Y = x \\, \\partial_y$, one gets $[X, Y] f = \\partial_x(x f_y) - x f_{yx} = f_y$, so $[X, Y] = \\partial_y$.\n\nThe product $f \\mapsto X(Y f)$ alone is not a vector field.\nFor $X = Y = \\partial_x$ on $\\R$, it is $f \\mapsto f''$, and $(fg)'' = f'' g + 2 f' g' + f g''$, so the product rule fails.\nIn the bracket the second-order terms cancel, and the product rule holds.\n\nThe bracket is bilinear and antisymmetric, and it satisfies the Jacobi identity $[X, [Y, Z]] + [Y, [Z, X]] + [Z, [X, Y]] = 0$.\nSo the smooth vector fields on $M$ form a Lie algebra.\nThe bracket $[X, Y]$ equals the [[Lie derivative|PJWK0PKN6Q]] $\\mathcal{L}_X Y$.\n\n## Explanations\n\n### L1 human\n\nThe Lie bracket of two vector fields is the commutator $XY - YX$ of the two derivations they define, which is again a vector field.\n",
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