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  "body": "\n## Definition\n\nLet $M$ be a [[smooth manifold|F45VQ47GPV]].\nFor $p \\ge 0$, let $\\mathfrak{X}^p(M)$ be the space of $p$-vector fields, the smooth sections of the [[exterior power|0MDMZ965VE]] $\\Lambda^p TM$.\nSo $\\mathfrak{X}^0(M) = C^\\infty(M)$, and $\\mathfrak{X}^1(M)$ is the space of [[vector fields|0299VCW1TH]].\nThe *Schouten–Nijenhuis bracket* is the unique family of $\\R$-bilinear maps $[\\cdot, \\cdot] \\colon \\mathfrak{X}^p(M) \\times \\mathfrak{X}^q(M) \\to \\mathfrak{X}^{p+q-1}(M)$ with these properties, for $P \\in \\mathfrak{X}^p(M)$, $Q \\in \\mathfrak{X}^q(M)$ and $R \\in \\mathfrak{X}^r(M)$.\n\n1. For functions $f, g$: $[f, g] = 0$.\n2. For a vector field $X$: $[X, P] = \\mathcal{L}_X P$, the [[Lie derivative|PJWK0PKN6Q]]; in particular $[X, f] = X(f)$ and $[X, Y]$ is the [[Lie bracket|68FGQEBD6J]].\n3. Graded antisymmetry: $[P, Q] = -(-1)^{(p-1)(q-1)} [Q, P]$.\n4. Graded Leibniz rule: $[P, Q \\wedge R] = [P, Q] \\wedge R + (-1)^{(p-1)q}\\, Q \\wedge [P, R]$.\n\n## Remarks\n\nFor vector fields $X_1, \\dots, X_p$ and $Y_1, \\dots, Y_q$ with $p, q \\ge 1$, the bracket is\n$$[X_1 \\wedge \\dots \\wedge X_p,\\ Y_1 \\wedge \\dots \\wedge Y_q] = \\sum_{i=1}^{p} \\sum_{j=1}^{q} (-1)^{i+j} [X_i, Y_j] \\wedge X_1 \\wedge \\dots \\widehat{X_i} \\dots \\wedge X_p \\wedge Y_1 \\wedge \\dots \\widehat{Y_j} \\dots \\wedge Y_q,$$\nwhere a hat marks an omitted factor.\nThe bracket also satisfies a graded Jacobi identity, so $\\mathfrak{X}^{\\bullet + 1}(M)$, with degrees shifted by one, is a graded Lie algebra.\n\nA bivector field $\\pi \\in \\mathfrak{X}^2(M)$ defines the bracket $\\{f, g\\} = \\pi(df, dg)$ on functions.\nThis bracket satisfies the Jacobi identity, so that $\\pi$ is a [[Poisson structure|A2KQ49Y4NF]], exactly when $[\\pi, \\pi] = 0$.\n\nExample: on $\\R^2$, let $\\pi = \\partial_x \\wedge \\partial_y$.\nAll brackets of the coordinate vector fields vanish, so the formula gives $[\\pi, \\pi] = 0$, and $\\{f, g\\} = f_x g_y - f_y g_x$ is the standard Poisson bracket.\n\nNon-example: the wedge product $(P, Q) \\mapsto P \\wedge Q$ is bilinear but is not this bracket.\nIt fails property 2: for a vector field $X$ and a function $f$, it gives $X \\wedge f = fX$, while property 2 requires the function $X(f)$.\n",
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