does_Lie_derivative_commute_with_pullback@20250813_173140

user: does Lie derivative commute with pullback? [Timestamp: 2025/08/13 17:31:40]

ai: Yes, the Lie derivative does commute with pullback, under suitable conditions.

Suppose $f:M\to N$ is a diffeomorphism, $v$ is a vector field on $N$, and $\omega$ is a differential form on $N$. Then, the following commutation relation holds:

$$f^{\ast }(L_v\omega )=L_{f^{\ast }v}(f^{\ast }\omega )$$

In words:
The pullback of the Lie derivative of $\omega$ along $v$ equals the Lie derivative (with respect to the pullback vector field $f^{\ast }v$) of the pullback form $f^{\ast }\omega$.

This result can be seen as a consequence of the definitions and of the more fundamental property that the pullback commutes with the interior product (see the proposition in the note: $f^{\ast }(i_v\omega )=i_{f^{\ast }v}{f}^{\ast }\omega$) and with the exterior derivative (for smooth maps: $f^{\ast }(d\omega )=d(f^{\ast }\omega )$). Since the Lie derivative can be written using Cartan’s formula:

$$L_v\omega =d(i_v\omega )+i_v(d\omega )$$

pullbacks commute with all operations in the formula, so the commutation relation follows.

Summary:
Yes, for diffeomorphisms $f$, vector fields $v$ on $N$, and forms $\omega$,
$f^{\ast }(L_v\omega )=L_{f^{\ast }v}(f^{\ast }\omega )$

Sources:

• Differential Forms [Timestamp: 2025/08/13 17:32:14]