Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is remark 1.24 on p. 13 of the book Hamilton's Ricci Flow by Bennett Chow, but how to prove this conclusion?

If $\varphi (t): M^n \to M^n$ is the $1$-parameter family of diffeomorphism and $\alpha$ is a tensor, then $$ \frac{\partial}{\partial t} (\varphi(t)^\ast \alpha) = L_{X(t)} \varphi(t)^\ast \alpha ,$$ where $$ X(t_0) = \left. \frac{\partial}{\partial t} \right|_{t = t_0} (\varphi (t_0)^{-1} \circ \varphi(t)) . $$

share|cite|improve this question
attention:$\varphi (t)$ may not be local 1-parameter groups of diffeomorphism. – deng ya Nov 12 '11 at 1:55

One way to prove this for $\alpha$ a differential form is as follows:

  1. Show that both operators $\frac{\partial}{\partial t}\circ \phi(t)^*$ and $L_{X(t)} \phi(t)^*$ agree when $\alpha$ is a function.
  2. Show that they are both derivations with respect to the wedge product.
  3. Show that they commute with exterior differentiation.

Since wedge and $d$ generate all differential forms, this proves that the operators are the same. I imagine a similar inductive argument would work general tensors $\alpha$ (but I haven't checked the details).

share|cite|improve this answer

Denoting by $\mathcal{T}(M)$ the graded $\mathbb{R}$-algebra of the tensorial fields on $M$, a differential operator on $M$ is defined to be $D:\mathcal{T}(M)\to\mathcal{T}(M)$, a graded $\mathbb{R}$-linear map of degree $0$, which has the following properties:

  1. $D$ is a tensorial derivation, i.e. it commutes with contraction,
  2. $D$ is local, i.e. natural with respect to restrictions.

A theorem of Willmore (cf. the original paper here) states that the differential operators on $M$ are completely determined by their action on smooth functions and vector fields.

So, in order to prove the identity, you have just to verify that LHS and RHS have the same behaviour on smooth functions and vector fields.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.