Eric Forgy
Poisson Bracket

Classical Concepts

Hamiltonian Vector Field

$$dH(Y)=\omega (X_H,Y)$$d H(Y) = \omega(X_H,Y)

Vector Field

$$df(X)=X(f)$$d f(X) = X(f)

Gradient Vector Field

$$dH(Y)=g(\nabla H,Y)$$d H(Y) = g(\nabla H,Y)

Differential Graded Algebra

$$d^2=0$$d^2 = 0

$$d(AB)=(dA)B+(-1{)}^{\mid A\mid }A(dB)$$d(A B) = (d A)B + (-1)^{|A|} A(d B)

Graded Brackets

$$[A,B]=AB-(-1{)}^{\mid A\mid \mid B\mid }BA$$[A,B] = A B - (-1)^{|A||B|} B A

$$\{A,B\}=AB+(-1{)}^{\mid A\mid \mid B\mid }BA$$\{A,B\} = A B + (-1)^{|A||B|} B A

$$d[A,B]=[dA,B]+(-1{)}^{\mid A\mid }[A,dB]$$d[A,B] = [d A,B] + (-1)^{|A|} [A,d B]

$$d\{A,B\}=\{dA,B\}+(-1{)}^{\mid A\mid }\{A,dB\}$$d\{A,B\} = \{d A,B\} + (-1)^{|A|} \{A,d B\}

Noncommutative Concepts

Left Structure Constants

$$[d{x}^{\mu },x^{\nu }]=\overleftarrow{C_{\lambda }^{\mu \nu }}d{x}^{\lambda }$$[d x^\mu,x^\nu] = \stackrel{\leftarrow}{C^{\mu\nu}_\lambda} d x^\lambda

$$[x^{\mu },x^{\nu }]=0\Rightarrow \overleftarrow{C_{\lambda }^{\mu \nu }}=\overleftarrow{C_{\lambda }^{\nu \mu }}$$[x^\mu,x^\nu] = 0\quad\implies\quad \stackrel{\leftarrow}{C^{\mu\nu}_\lambda} = \stackrel{\leftarrow}{C^{\nu\mu}_\lambda}

$$\{x^{\mu },x^{\nu }\}=0\Rightarrow \overleftarrow{C_{\lambda }^{\mu \nu }}=-\overleftarrow{C_{\lambda }^{\nu \mu }}$$\{x^\mu,x^\nu\} = 0\quad\implies\quad \stackrel{\leftarrow}{C^{\mu\nu}_\lambda} = -\stackrel{\leftarrow}{C^{\nu\mu}_\lambda}

Right Structure Constants

$$[d{x}^{\mu },x^{\nu }]=d{x}^{\lambda }\overrightarrow{C_{\lambda }^{\mu \nu }}$$[d x^\mu,x^\nu] = d x^\lambda \stackrel{\rightarrow}{C^{\mu\nu}_\lambda}

Left Components

$$df=\overleftarrow{{\partial }_{\mu }f}d{x}^{\mu }$$d f = \stackrel{\leftarrow}{\partial_\mu f} d x^\mu

Product Rule

$$\begin{array}{rl}d(fg)& =(df)g+f(dg)\\ & =[df,g]+g(df)+f(dg)\end{array}$$\begin{aligned} d(f g) &= (d f)g + f(d g) \\ &= [d f,g] + g(d f) + f(d g) \end{aligned}