# Poisson Bivectors

Differential Geometry

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

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Published

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

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

## 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.