Poisson Bivectors

Differential Geometry

Equivalence between Poisson brackets and Poisson bivectors will be studied in this post.


August 3, 2020

1 Derivations and Vector Fields

The Hadamard lemma is surprisingly useful in study of relations between derivations and vector fields.

Lemma 1 (Hadamard Lemma) For every smooth function \(f\in C^\infty(\mathbb{R}^n)\), there are smooth functions \(g_i\in C^\infty(\mathbb{R}^n)\) such that \[ f(x)-f(a)=\sum(x_i-a_i)\int_0^1\frac{\partial}{\partial x_i}f(tx+(1-t)a)\mathrm{d}x. \]

Proof. \[ \begin{aligned} f(x)-f(a) =&\int_0^1\frac{\mathrm{d}}{\mathrm{d}t}\left(f(tx+(1-t)a)\right)\mathrm{d}t\\ =&\int_0^1\left\{\sum\left[\left(\frac{\partial}{\partial x_i}f(tx+(1-t)a)\right)\cdot\frac{\mathrm{d}}{\mathrm{d}t}(tx_i+(1-t)a_i)\right]\right\}\mathrm{d}t\\ =&\int_0^1\left\{\sum\left[\left(\frac{\partial}{\partial x_i}f(tx+(1-t)a)\right)\cdot(x_i-a_i)\right]\right\}\mathrm{d}t\\ =&\sum\left[(x_i-a_i)\int_0^1\left(\frac{\partial}{\partial x_i}f(tx+(1-t)a)\right)\mathrm{d}t\right]\\ \end{aligned} \] Let \[ g_i=\int_0^1\left(\frac{\partial}{\partial x_i}f(tx+(1-t)a)\right)\mathrm{d}t. \]

Since \(f\) is smooth, so are \(g_i\).

Let \(U\) be an open submanifold of a smooth manifold \(M\) with the local coordinate chart \((x_1, \dots, x_n)\). Then a \(C^\infty(M)\)-derivation \(D\) on \(U\) is one-to-one corresponding to a vector field \(\sum\varphi_i\frac{\partial}{\partial x_i}\), where \(\varphi_i\) are smooth functions.

Proof. Let \(f\) be a smooth function. For any point \(a\in U\), by Hadamard Lemma, we can write \(f(x)=f(a)+\sum(x_i-a_i)g_i\) from some smooth functions \(g_i\). Therefore,
\[ \begin{aligned} &\left(D-\sum D(x_i)\frac{\partial}{\partial x_i}\right)(f)\\ =&\sum\left(g_iD(x_i)+(x_i-a_i)D(g_i)\right)-\sum\left(D(x_i) + (x_i-a_i)\frac{\partial g_i}{\partial x_i}\right) \end{aligned} \] It follows that \(D(f)(a)= \sum D(x_i)(a)\frac{\partial f}{\partial x_i}(a)\) for any \(a\in U\). Therefore, \(D=\sum D(x_i)\frac{\partial f}{\partial x_i}\) on \(U\).

Conversely, a vector field \(\sum\varphi_i\frac{\partial}{\partial x_i}\) defines a linear map \(C^\infty(U)\to C^\infty(U)\) that stratifies the Leibniz rule.

This ends the proof of the proposition.

The correspondence between derivatives and vector fields can be generalized to higher degrees.

Proposition 1 Let \(\{\cdot, \cdot\}:C^\infty(M)\times C^\infty(M)\to C^\infty(M)\) be a skew-symmetric bilinear map satisfying the Leibniz rule: \[ \{fg, h\}=f\{g, h\}+g\{f, h\}. \] Then on a open submanifold \(U\subset M\) with local coordinates \((x_1, \dots, x_n)\), we have \[ \{f, g\}=\sum_{1\le i<j\le n}\{x_i, x_j\}\frac{\partial f}{\partial x_i}\wedge\frac{\partial g}{\partial x_j}. \]

Proof. By the definition of \(\{\cdot,\cdot\}\), we note that \(\{c, g\}=\{c\cdot 1, g\}=c\{1,g\}+1\{c, g\}\). Therefore, \(\{c, g\}=0\) for any real number \(c\).

By the Hadamard lemma, at any point \(a\in U\), we have \(f(x)=f(a)+\sum(x_i-a_i)f_i(x)\) and \(g(x)=g(a)+\sum(x_i-a_i)g_i(x)\), where \(f_i=\int_0^1\frac{\partial}{\partial x_i}f(tx+(1-t)a)\mathrm{d} t\) and \(g_i=\int_0^1\frac{\partial}{\partial x_i}g(tx+(1-t)a)\mathrm{d} t\). Then \[ \begin{aligned} \{f, g\} =&\left\{\sum_{i=1}^n(x_i-a_i)f_i,\sum_{j=1}^n(x_j-a_i)g_j\right\}\\ =&\sum_{i=1}^n f_i\left\{x_i, \sum_{j=1}^n(x_j-a_i)g_j\right\}+\sum_{i=1}^n(x_i-a_i)\left\{f_i, \sum_{j=1}^n(x_j-a_i)g_j\right\}\\ =&\sum_{i=1}^n\sum_{j=1}^n f_ig_j\{x_i, x_j\}+\sum_{j=1}^n f_i\sum_j^n(x_j-a_i)\{x_i, g_j\} +\sum_{i=1}^n (x_i-a_i)\left\{f_i, \sum_{j=1}^n(x_j-a_i)g_j\right\} \end{aligned} \] Note that \(f_i(a)=\frac{\partial f}{\partial x_i}(a)\) and \(g_i(a)=\frac{\partial g}{\partial x_i}(a)\). Then \[ \{f, g\}(a)=\sum_i^n\sum_j^n \{x_i, x_j\}(a)\frac{\partial f}{\partial x_i}(a)\frac{\partial g}{\partial x_i}(a) \] for any \(a\in U\). Therefore, \[ \{f, g\}=\sum_i^n\sum_j^n \{x_i, x_j\}\frac{\partial f}{\partial x_i}\frac{\partial g}{\partial x_i} \] on \(U\). Because \(\{f, g\}=-\{f,g\}\), we know that \[ \{f, g\}=\sum_{1\le i <j\le n} \{x_i, x_j\}\frac{\partial f}{\partial x_i}\wedge\frac{\partial g}{\partial x_i}. \]

2 Poisson Bivector

Proposition 2 Let \(\pi\in\Gamma(M, \wedge^2TM)\) be a bivector field on a smooth manifold. Then the map \(\{\cdot, \cdot\}:C^\infty(M)\times C^\infty(M)\to C^\infty M\) defined by \(\{f, g\}=\pi(\mathrm{d}f\wedge\mathrm{d}g)\) is skew-symmetric bilinear and satisfies the Leibniz rule.

Proof. The skew-symmetry and bilinearity are from the fact that the wedge product is skew-symmetric and bilinear. The Leibniz rule follows from the fact that the differential operator \(\mathrm{d}\) satisfies the Leibniz rule.

Definition 1 (Poisson Bracket) Let \(M\) be a smooth manifold and \(\{\cdot,\cdot\}: C^\infty(M)\times C^\infty(M)\to C^\infty(M)\) a map. If

  1. \(\{\cdot, \cdot\}\) is a Lie bracket, i.e., it is bilinear, skew-symmetric and satisfies the Jacobi identity: \[ \{f, \{g, h\}\}+\{g, \{h, f\}\}+\{h, \{f, g\}\}=0 \] for any \(f\), \(g\), \(h\) in \(C^\infty(M)\), and

  2. \(\{\cdot, \cdot\}\) satisfies the Leibniz rule: \[ \{fg, h\}=f\{g,h\}+g\{f,h\} \] for any \(f\), \(g\), \(h\) in \(C^\infty(M)\),

then we call \(M\) a Poisson manifold with Poisson structure given by the Poisson bracket \(\{\cdot,\cdot\}\).

In the following, \(\Gamma(M, \wedge^pTM)\) is the space of degree \(p\) multivector fields and \(\bigwedge TM=\bigoplus_{p\ge0}\wedge^pTM\). As a \(C^\infty(M)\)-algebra, \(\Gamma(M, \bigwedge TM)=\bigoplus\wedge^p\Gamma(M, TM)\). We adapt the following definition from (Michor 1987)

3 Schouten–Nijenhuis bracket

Let \(M\) be a smooth manifold. The Schouten–Nijenhuis bracket associated to \([\cdot,\cdot]\) is the skew symmetric bilinear map \([\cdot, \cdot]:\Gamma(M ,\wedge^pTM)\times\Gamma(M,\wedge^qTM)\to \Gamma(M, \wedge^{p+q-1}TM)\) such that

  1. \([f, g]=0\) for any smooth functions \(f\) and \(g\),

  2. \([X, Y]\) is the Lie bracket for vector fields \(X\) and \(Y\),

  3. \[ [u_{1}\wedge \cdots \wedge u_{p},v_{1}\wedge\cdots \wedge v_{q}]=\sum _{i,j}(-1)^{i+j}[u_{i},v_{j}]\wedge(u_{1}\wedge\cdots \widehat{u_{i}} \cdots \wedge u_{p})\wedge (v_{1}\wedge \cdots \widehat{v_{j}} \cdots\wedge v_{q}) \] for vector fields \(u_i\), \(v_j\),

  4. \[ [f,v_{1}\wedge\cdots\wedge v_{q}]=-\iota _{\mathrm{d}f}(v_{1}\wedge\cdots \wedge v_{q}) \] for vector fields \(v_{j}\) and a smooth function \(f\), where \(\iota _{\mathrm{d}f}\) is the interior product, the unique antiderivation of degree −1 on the exterior algebra \(\Gamma(M, \bigwedge TM)\) such that \(\iota_{\mathrm{d}f}(v)=<\mathrm{d}f, v>\) for any vector field \(v\), and \[ \iota_{\mathrm{d}f}(u\wedge v)=\iota_{\mathrm{d}f}(u)\wedge v+(-1)^{\deg u}u\wedge\iota_{\mathrm{d}f}(v) \] for any multivector fields \(u\) and \(v\), where \(<\cdot, \cdot>\) is the dual pairing between vector fields and differential \(1\)–forms.

The Schouten–Nijenhuis bracket has the following two important properties which can be used to characterize the bracket (see for example, (Marle 1997) Proposition 3.1).

  1. For any multivector fields, \(u\) and \(v\), we have the antisymmetry \[ [u, v]=-(-1)^{(\deg u-1)(\deg v-1)}[v, u]. \]

  2. For any multivector fields, \(u\), \(v\) and \(w\), we have the Poisson identity \[ [u, v\wedge w]=[u,v]\wedge w+(-1)^{(\deg u-1)\deg v}v\wedge[u,w] \]

  3. The graded Jacobi identity \[ (-1)^{(\deg u-1)(\deg w-1)}[u, [v, w]]+(-1)^{(\deg v-1)(\deg u-1)}[v,[w,u]]+(-1)^{(\deg w-1)(\deg v-1)}[w, [u,v]] = 0. \]

  4. For any vector filed \(u\) and a function, vector field or multivector field \(v\), the Schouten–Nijenhuis bracket \([u, v]\) is the Lie derivative \(\mathcal{L}_u(v)\) of \(v\) with respect to \(u\).

Proposition 3 Let \(\pi\) be a bivector field. The skew-symmetric bilinear form \(\{\cdot, \cdot\}:C^\infty(M)\times C^\infty(M)\to C^\infty(M)\) defined by \(\{f, g\}=\pi(\mathrm{d}f\wedge\mathrm{d}g)\) is a Poisson bracket if the Schouten-Nijenhuis bracket \([\pi,\pi]=0\).

Proof. For any smooth functions \(f, g, h\), we have \[ \begin{aligned} \{fg, h\} =&\pi(\mathrm{d}(fg)\wedge\mathrm{d}h)=\pi(\mathrm{d}(fg)\wedge\mathrm{d}h)\\ =&\pi((f\mathrm{d}g+g\mathrm{d}f)\wedge \mathrm{d}h)\\ =&f\pi(\mathrm{d}g, \mathrm{d}h)+g\pi(\mathrm{d}f,\mathrm{h})\\ =&f\{g,h\}+g\{f,h\}. \end{aligned} \] This proves that \(\{\cdot,\cdot\}\) satisfies the Leibniz rule.

To show that \(\{\cdot,\cdot\}\) is a Lie bracket, it suffices to show that \[ \{f,\{g,h\}\}+\{g,\{h,f\}\}+\{h,\{f,g\}\}=0. \]

Using the notation \(\iota_{\mathrm{d}f}\) and the fact that \(\pi\) is a bivector, we can express the bracket \(\{f,g\}\) using the Schouten-Nijenhuis bracket \[ \{f, g\}=\pi(\mathrm{d}f\wedge \mathrm{d}g)=[g,[f,\pi]]. \] Therefore, \[ \{f,\{g,h\}\}=\{f, [h,[g,\pi]]\} = [[h,[g,\pi]],[f, \pi]]. \]

By the antisymmetry identity, we know \[ [f, \pi]=-(-1)^{(0-1)(2-1)}[\pi, f]=[\pi, f], \] \[ [f, [g,\pi]]=-(-1)^{(0-1)(1-1)}[[g, \pi], f]=-[[g, \pi], f], \] and \[ [[f,\pi], \pi]=-(-1)^{(1-1)(2-1)}[\pi, [f, \pi]]=-[\pi, [f, \pi]], \]

By the graded Jacobi identity, we have \[ \begin{aligned} 0=&-((-1)^{(2-1)(2-1)}[\pi, [f,\pi]]+(-1)^{(0-1)(2-1)}[f,[\pi,\pi]]+(-1)^{(2-1)(0-1)}[\pi,[\pi, f]])\\ =& [\pi, [f,\pi]]+[f,[\pi,\pi]]+[\pi,-(-1)^{(2-1)(0-1)}[f,\pi]]\\ =& [\pi, [f,\pi]]+[f,[\pi,\pi]]+[\pi,[f,\pi]], \end{aligned} \] and \[ \begin{aligned} 0=&-((-1)^{(0-1)(2-1)}[f, [g,\pi]]+(-1)^{(0-1)(0-1)}[g,[\pi,f]]+(-1)^{(2-1)(0-1)}[\pi,[f, g]])\\ =& [f, [g,\pi]]+[g,[\pi,f]]. \end{aligned} \]

Therefore, \[ -2[[f,\pi], \pi]=[f,[\pi,\pi]]\quad\quad \text{and}\quad\quad [f,[g,\pi]]=[g,[f,\pi]]. \]

By the Jacobi identity, we also have \[ \begin{aligned} &[\pi,[g,[h,\pi]]]\\ =&-(-1)^{(2-1)(1-1)}((-1)^{(0-1)(2-1)}[g,[[h,\pi],\pi]]+(-1)^{(1-1)(0-1)}[[h,\pi],[\pi, g]])\\ =&[g, [[h,\pi],\pi]]-[[h,\pi],[g,\pi]]\\ =&[g, [[h,\pi],\pi]]-(-1)(-1)^{(1-1)(1-1)}[[g,\pi], [h,\pi]]\\ =&[g, [[h,\pi],\pi]]+[[g,\pi], [h,\pi]] \end{aligned} \]

Because \([\cdot, \cdot]\) is an operator of degree −1 and \([f, g]=0\) for any smooth functions \(f\) and \(g\). Apply the Jacobi identity again, we have \[ \begin{aligned} &[[h,[g,\pi]], [f, \pi]]\\ =&-(-1)^{(0-1)(2-1)}((-1)^{(0-1)(0-1)}[f,[\pi, [h, [g,\pi]]]]+(-1)^{(2-1)(0-1)}[\pi, [[h,[g,\pi]],f]])\\ =&-[f,[\pi, [h, [g,\pi]]]]-[\pi, [[h,[g,\pi]],f]]\\ =&-[f,[\pi, [h, [g,\pi]]]]\\ %=&(-1)^{(2-1)(1-1)}([f, -(-1)^{(0-1)(2-1)}[h, [[g,\pi], \pi]]]+[f, -(-1)^{(1-1)(0-1)}[[g,\pi], [\pi, h]]])\\ =&-[f, [h, [[g,\pi], \pi]]]-[f, [[h,\pi], [g,\pi]]]\\ =&\frac12 [f, [h, [g,[\pi, \pi]]]]-[f, [[g,\pi], [h,\pi]]]. \end{aligned} \]

Further more, \[ \begin{aligned} &[h, [g,[\pi, \pi]]]\\ =&-(-1)^{(0-1)(3-1)}((-1)^{(0-1)(0-1)}[g, [[\pi,\pi], h]]+(-1)^{(3-1)(0-1)}[[\pi,\pi],[h, g]])\\ =&[g, [[\pi,\pi], h]]\\ =&-(-1)^{(0-1)(3-1)}[g, [h,[\pi,\pi]]]\\ =&-[g, [h,[\pi,\pi]]] \end{aligned} \]

\[ \begin{aligned} &[f, [g, [h,[\pi, \pi]]]]\\ &=-(-1)^{(0-1)(2-1)}((-1)^{(0-1)(0-1)}[g,[[h,[\pi,\pi]],f]]+(-1)^{(2-1)(0-1)}[[h,[\pi,\pi]], [f,g]])\\ &=-[g,[[h,[\pi,\pi]],f]]\\ &=-(-(-1)^{(0-1)(2-1)}[g, [f, [h, [\pi,\pi]]]])\\ &=-[g, [f, [h, [\pi,\pi]]]]. \end{aligned} \]

Therefore, \[ \begin{aligned} &-2(\{f,\{g,h\}\}+\{g,\{h,f\}\}+\{h,\{f,g\}\})\\ =&[f, [g, [h,[\pi, \pi]]]]+[g, [h, [f,[\pi, \pi]]]]+[h, [f, [g,[\pi, \pi]]]]\\ &-2[f, [[g,\pi], [h, \pi]]]-2[g, [[h,\pi], [f, \pi]]]-2[h, [[f,\pi], [g, \pi]]]\\ =&[f, [g, [h,[\pi, \pi]]]]+[f, [g, [h,[\pi, \pi]]]]+[f, [g, [h,[\pi, \pi]]]]\\ &-2[f, [[g,\pi], [h, \pi]]]-2[g, [[h,\pi], [f, \pi]]]-2[h, [[f,\pi], [g, \pi]]]\\ =&[f, [g, [h,[\pi, \pi]]]]+2([f, [g, [h,[\pi, \pi]]]]-[f, [[g,\pi], [h, \pi]]]-[g, [[h,\pi], [f, \pi]]]-[h, [[f,\pi], [g, \pi]]])\\ =&[f, [g, [h,[\pi, \pi]]]]+2(-2[f, [g, [[h,\pi], \pi]]]-[f, [[g,\pi], [h, \pi]]]-[g, [[h,\pi], [f, \pi]]]-[h, [[f,\pi], [g, \pi]]]). \end{aligned} \]

Apply the Jacobin identity and antisymmetry, we get \[ \begin{aligned} &2[f, [g, [[h,\pi], \pi]]]+[f, [[g,\pi], [h, \pi]]]+[g, [[h,\pi], [f, \pi]]]+[h, [[f,\pi], [g, \pi]]]\\ =&[f, [g, [[h,\pi], \pi]]]+[f, [[g,\pi], [h, \pi]]]+[g, [h, [[f,\pi], \pi]]]+[g, [[h,\pi], [f, \pi]]]+[h, [[f,\pi], [g, \pi]]])\\ =&[f, [\pi,[g,[h,\pi]]]]+[g, [\pi,[h,[f,\pi]]]]+[h, [[f,\pi], [g, \pi]]]. \end{aligned} \] Note that \[ \begin{aligned} &[f, [\pi,[g,[h,\pi]]]]\\ =&-(-1)^{(0-1)(0-1)}((-1)^{(2-1)(0-1)}[\pi,[f, [g,[h,\pi]]]]+(-1)^{(0-1)(2-1)}[[g,[h,\pi]],[f,\pi]])\\ =&-[\pi,[f,[g,[h,\pi]]]]-[[g,[h,\pi]],[f,\pi]] \end{aligned} \]

Then, \[ \begin{aligned} &[f, [\pi,[g,[h,\pi]]]]+[g, [\pi,[h,[f,\pi]]]]+[h, [[f,\pi], [g, \pi]]]\\ =&-[[g,[h,\pi]], [f,\pi]]-[[h,[f,\pi]], [g,\pi]]+[h, [[f,\pi], [g, \pi]]]\\ =&-[[g,[h,\pi]], [f,\pi]]-[[h,[f,\pi]], [g,\pi]]-[[f,\pi],[[g,\pi],h]]-[[g,\pi],[h,[f,\pi]]]\\ =&-[[g,[h,\pi]], [f,\pi]]-[[h,[f,\pi]], [g,\pi]]+[[f,\pi],[h,[g,\pi]]]+[[h,[f,\pi]],[g,\pi]]\\ =&-[[g,[h,\pi]], [f,\pi]]+[[f,\pi],[h,[g,\pi]]]\\ =&[[f,\pi],[g,[h,\pi]]]+[[f,\pi],[h,[g,\pi]]]\\ =&[[f,\pi],[g,[h,\pi]]]-[[f,\pi],[g,[h,\pi]]]\\ =&0. \end{aligned} \]

Consequently, \[ -2(\{f,\{g,h\}\}+\{g,\{h,f\}\}+\{h,\{f,g\}\})=[f, [g, [h,[\pi, \pi]]]]. \] Therefore, the Jacobi identity holds for the Poisson bracket if and only if the Schouten-Nijenhuis bracket \([\pi,\pi]=0\) for the bivector \(\pi\).

A bivector \(\pi\) is called a Poisson bivector if \([\pi,\pi]=0\).


Marle, Charles-Michel. 1997. “The Schouten-Nijenhuis Bracket and Interior Products.” J. Geom. Phys. 23 (3-4): 350–59. https://doi.org/10.1016/S0393-0440(97)80009-5.
Michor, Peter W. 1987. “Remarks on the Schouten-Nijenhuis Bracket.” In Proceedings of the Winter School on Geometry and Physics (Srnı́, 1987), 207–15. 16.