Eric Forgy Poisson Bracket

Classical Concepts

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

d f(X) = X(f)

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

d^2 = 0

d(A B) = (d A)B + (-1)^{|A|} A(d B)

[A,B] = A B - (-1)^{|A||B|} B A

\{A,B\} = A B + (-1)^{|A||B|} B A

d[A,B] = [d A,B] + (-1)^{|A|} [A,d B]

d\{A,B\} = \{d A,B\} + (-1)^{|A|} \{A,d B\}

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

[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\quad\implies\quad \stackrel{\leftarrow}{C^{\mu\nu}_\lambda} = -\stackrel{\leftarrow}{C^{\nu\mu}_\lambda}

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

d f = \stackrel{\leftarrow}{\partial_\mu f} d x^\mu

\begin{aligned} d(f g) &= (d f)g + f(d g) \\ &= [d f,g] + g(d f) + f(d g) \end{aligned}

Revised on April 27, 2010 at 07:12:36 by Eric Forgy