I don't get the following theorem from my lecture: $M$ smooth manifold

Let $T \in \mathcal{T}^{r,s}(M)$ a tensor field, $\Phi$ the global flow of a complete vector field $X$ and $\varphi_t(p) := \Phi(t,p)$.

Then $\dfrac{\partial}{\partial t}|_{t=0} \varphi^{*}_t T = \mathcal{L}_X$ where $\mathcal{L}_X$ is the Lie-derivative.

What I don't get is, how do you take the partial derivative towards $t$ from this function?

I mean even if we take the case that $T=f \in C^{\infty}(M)$, then $t \mapsto \varphi^{*}_t f=f \circ \varphi_t$ is a function-valued function.


There are two ways I think about this. One is to work in coordinates and simply take derivatives of all component functions. Another is to apply the tensor to an arbitrary element of its domain. (Of course these are equivalent.)

I'm not sure I understand the objection to the fact that $t \mapsto \phi_t^{\ast}f$ is a function-valued function. For instance, you can take the derivative of $f(x,t) = xt$ with respect to $t$, at a specified value $t = t_0$. The result is a function of $x$. Maybe the issue is realizing that the derivative is taken at $t = 0$.

  • $\begingroup$ Hi, actually I realised that saying "function valued function" didn't make much sense because this is not my problem, my problem is if I look at $\varphi^{*}_t T$ for example for $T$ a vector field then $\varphi^{*}_t T$ is a vector field right? How do I take the partial derivative of a vector field? $\endgroup$ – User1 May 6 at 16:06
  • $\begingroup$ "Another is to apply the tensor to an arbitrary element of its domain." Do you mean applying my vector field to a function (then I get a $C^{\infty}$ function? But then the derivative would not be well defined, would it? $\endgroup$ – User1 May 6 at 16:08
  • $\begingroup$ "How do I take the partial derivatives of a vector field"--this was what I was trying to explain in the first part of my comment. Take coordinates and view it as a function $U \times (-\epsilon,\epsilon) \rightarrow \mathbb{R}^n$ which is differentiated in $t$ component-wise. You can convince yourself that this is coordinate-independent by relating it to the derivative of (the vector field applied to certain covectors). $\endgroup$ – hedgehog enthusiast May 6 at 18:47
  • $\begingroup$ Ok, I think I got it now, thanks $\endgroup$ – User1 May 8 at 12:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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