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Poisson Bivectors

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

Author

Affiliation

Fei Ye

QCC-CUNY

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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 $f\in C^{\mathrm{\infty }}({\mathbb{R}}^n)$, there are smooth functions $g_i\in C^{\mathrm{\infty }}({\mathbb{R}}^n)$ 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. $$\begin{array}{rl}f(x)-f(a)=& {\int }_0^1\frac{\mathrm{d}}{\mathrm{d}t}(f(tx+(1-t)a))\mathrm{d}t\\ =& {\int }_0^1\{\sum [(\frac{\partial }{\partial x_i}f(tx+(1-t)a))\cdot \frac{\mathrm{d}}{\mathrm{d}t}(t{x}_i+(1-t)a_i)]\}\mathrm{d}t\\ =& {\int }_0^1\{\sum [(\frac{\partial }{\partial x_i}f(tx+(1-t)a))\cdot (x_i-a_i)]\}\mathrm{d}t\\ =& \sum [(x_i-a_i){\int }_0^1(\frac{\partial }{\partial x_i}f(tx+(1-t)a))\mathrm{d}t]\end{array}$$ Let $$g_i={\int }_0^1(\frac{\partial }{\partial x_i}f(tx+(1-t)a))\mathrm{d}t.$$

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 $C^{\mathrm{\infty }}(M)$-derivation $D$ on $U$ is one-to-one corresponding to a vector field $\sum {\phi }_i\frac{\partial }{\partial x_i}$, where ${\phi }_i$ 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,
$$\begin{array}{rl}& (D-\sum D(x_i)\frac{\partial }{\partial x_i})(f)\\ =& \sum (g_{i}D(x_i)+(x_i-a_i)D(g_i))-\sum (D(x_i)+(x_i-a_i)\frac{\partial g_i}{\partial x_i})\end{array}$$ 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 $\sum {\phi }_i\frac{\partial }{\partial x_i}$ defines a linear map $C^{\mathrm{\infty }}(U)\to C^{\mathrm{\infty }}(U)$ 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 $\{\cdot ,\cdot \}:C^{\mathrm{\infty }}(M)\times C^{\mathrm{\infty }}(M)\to C^{\mathrm{\infty }}(M)$ 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 $$\begin{array}{rl}\{f,g\}=& \{\sum _{i=1}^n(x_i-a_i)f_i,\sum _{j=1}^n(x_j-a_i)g_j\}\\ =& \sum _{i=1}^n{f}_i\{x_i,\sum _{j=1}^n(x_j-a_i)g_j\}+\sum _{i=1}^n(x_i-a_i)\{f_i,\sum _{j=1}^n(x_j-a_i)g_j\}\\ =& \sum _{i=1}^n\sum _{j=1}^n{f}_i{g}_j\{x_i,x_j\}+\sum _{j=1}^n{f}_i\sum _j^n(x_j-a_i)\{x_i,g_j\}+\sum _{i=1}^n(x_i-a_i)\{f_i,\sum _{j=1}^n(x_j-a_i)g_j\}\end{array}$$ 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 $\pi \in \mathrm{\Gamma }(M,{\wedge }^{2}TM)$ be a bivector field on a smooth manifold. Then the map $\{\cdot ,\cdot \}:C^{\mathrm{\infty }}(M)\times C^{\mathrm{\infty }}(M)\to C^{\mathrm{\infty }}M$ 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 $\{\cdot ,\cdot \}:C^{\mathrm{\infty }}(M)\times C^{\mathrm{\infty }}(M)\to C^{\mathrm{\infty }}(M)$ 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 $C^{\mathrm{\infty }}(M)$, and

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

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

In the following, $\mathrm{\Gamma }(M,{\wedge }^{p}TM)$ is the space of degree $p$ multivector fields and $\bigwedge TM=\underset{p\ge 0}{⨁}{\wedge }^{p}TM$. As a $C^{\mathrm{\infty }}(M)$-algebra, $\mathrm{\Gamma }(M,\bigwedge TM)=⨁{\wedge }^p\mathrm{\Gamma }(M,TM)$. 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 $[\cdot ,\cdot ]:\mathrm{\Gamma }(M,{\wedge }^{p}TM)\times \mathrm{\Gamma }(M,{\wedge }^{q}TM)\to \mathrm{\Gamma }(M,{\wedge }^{p+q-1}TM)$ 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 $\mathrm{\Gamma }(M,\bigwedge TM)$ such that ${\iota }_{\mathrm{d}f}(v)=<\mathrm{d}f,v>$ for any vector field $v$, and $${\iota }_{\mathrm{d}f}(u\wedge v)={\iota }_{\mathrm{d}f}(u)\wedge v+(-1{)}^{\mathrm{deg}u}u\wedge {\iota }_{\mathrm{d}f}(v)$$ 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 $$[u,v]=-(-1{)}^{(\mathrm{deg}u-1)(\mathrm{deg}v-1)}[v,u].$$

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

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

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 $\{\cdot ,\cdot \}:C^{\mathrm{\infty }}(M)\times C^{\mathrm{\infty }}(M)\to C^{\mathrm{\infty }}(M)$ 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 $$\begin{array}{rl}\{fg,h\}=& \pi (\mathrm{d}(fg)\wedge \mathrm{d}h)=\pi (\mathrm{d}(fg)\wedge \mathrm{d}h)\\ =& \pi ((f\mathrm{d}g+g\mathrm{d}f)\wedge \mathrm{d}h)\\ =& f\pi (\mathrm{d}g,\mathrm{d}h)+g\pi (\mathrm{d}f,\mathrm{h})\\ =& f\{g,h\}+g\{f,h\}.\end{array}$$ 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 $$\begin{array}{rl}0=& -((-1{)}^{(2-1)(2-1)}[\pi ,[f,\pi ]]+(-1{)}^{(0-1)(2-1)}[f,[\pi ,\pi ]]+(-1{)}^{(2-1)(0-1)}[\pi ,[\pi ,f]])\\ =& [\pi ,[f,\pi ]]+[f,[\pi ,\pi ]]+[\pi ,-(-1{)}^{(2-1)(0-1)}[f,\pi ]]\\ =& [\pi ,[f,\pi ]]+[f,[\pi ,\pi ]]+[\pi ,[f,\pi ]],\end{array}$$ and $$\begin{array}{rl}0=& -((-1{)}^{(0-1)(2-1)}[f,[g,\pi ]]+(-1{)}^{(0-1)(0-1)}[g,[\pi ,f]]+(-1{)}^{(2-1)(0-1)}[\pi ,[f,g]])\\ =& [f,[g,\pi ]]+[g,[\pi ,f]].\end{array}$$

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

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

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

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

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

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

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

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

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 $[\pi ,\pi ]=0$.

References

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.