Difference between revisions of "Corollary of Leibniz rule for Lie bracket"

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Let <math>X,Y</math> be smooth [[vector field]]s on a [[differential manifold]] <math>M</math> and <math>f</math> be in <math>C^\infty(M)</math>. We then have:
 
Let <math>X,Y</math> be smooth [[vector field]]s on a [[differential manifold]] <math>M</math> and <math>f</math> be in <math>C^\infty(M)</math>. We then have:
  
<math>f[X,Y] = [fX,Y] + (Yf)X</math>
+
<math>\! f[X,Y] = [fX,Y] + (Yf)X</math>
  
<math>f[X,Y] = [X,fY] - (Xf)Y</math>
+
<math>\! f[X,Y] = [X,fY] - (Xf)Y</math>
  
 
==Applications==
 
==Applications==

Latest revision as of 18:42, 24 July 2009

Statement

This is an identity that uses the Leibniz rule to measure the failure of the Lie bracket operation from being C^\infty-linear.

Let X,Y be smooth vector fields on a differential manifold M and f be in C^\infty(M). We then have:

\! f[X,Y] = [fX,Y] + (Yf)X

\! f[X,Y] = [X,fY] - (Xf)Y

Applications

Proof

First identity

We prove this by showing that for any g \in C^\infty(M), both sides evaluate to the same thing. Let's do this. Simplifying the right side yields:

f(X(Yg) - Y(Xg)) - f(X(Yg)) + Y((fX)g) = Y(f(Xg)) - f(Y(Xg))

Applying the Leibniz rule for Y on the product of functions f and Xg, this simplifies to:

(Yf)(Xg) + Y(Xg)(f) - f(Y(Xg)) = (Yf)(Xg)

which is precisely equal to the left side.

Second identity

We prove this as well by taking any test function g \in C^\infty(M). Simplifying the right side yields:

[X,fY]g - (Xf)(Yg) = X(f(Yg)) -f(YXg) - (Xf)(Yg) = (Xf)(Yg) + f(XYg) - f(YXg) - (Xf)(Yg) = f[X,Y]g

which is the same as the left side.