# Flows of vector fields and push-backs

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.

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 $$t = 0$$.
• 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? – User1 May 6 at 16:06
• "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? – User1 May 6 at 16:08
• "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). – hedgehog enthusiast May 6 at 18:47