# Clarifying the definition of Lie derivative for tensors

Let $$(M,g)$$ be a Riemannian manifold and $$T$$ be a smooth tensor field. The official definition of the Lie derivative of $$T$$ with respect to a field $$X\in\Gamma(TM)$$ is: $$\mathcal{L}_X(T)_p:=\left.\frac{d}{dt}\right|_{t=0}(\varphi_{-t})_*T_{\varphi_t(p)}$$

(where $$\varphi$$ is the flow of $$X$$)

What does this notation $$(\varphi_{-t})_*T_{\varphi_t(p)}$$ mean?

Taking for example $$T=g$$, I believe it would make sense to define $$\mathcal{L}_X$$ as follows: $$\mathcal{L}_X(g)_p(Y_p,Z_p):=\left.\frac{d}{dt}\right|_{t=0}g_{\varphi_{t}(p)}((d\varphi_t)_p(Y_p),(d\varphi_t)_p(Z_p))$$

But I don't know how to reconciliate this with the $$-t$$ in the official definition.

Let $$T$$ be a tensor field and $$X$$ a vector field. If you wish to look at the Lie derivative at a point $$p$$, $$(\mathcal{L}_XT)_p$$, you must look at the tensor at the point $$\phi(t,p)$$, where $$\phi(t,p)$$ is the flow of $$X$$ generated from $$p$$. You now need a way to push this tensor $$T_{\phi_t(p)}$$ from the point $$\phi_t(p)$$ back to the point $$p$$. Since $$\phi(0,p)=p$$, we have that $$\phi(-t,\phi(t,p))=\phi(-t+t,p)=\phi(0,p)=p.$$ So we use the pushforward of the function $$\phi_{-t}$$ at the point $$\phi_t(p)$$, that is, the map $$d(\phi_{-t})_{\phi_t(p)}$$ (I'm using this notation instead of the $$*$$ notation for subscript clarity). We then see that $$d(\phi_{-t})_{\phi_t(p)}T_{\phi_t(p)}$$ is now a tensor at $$p$$ for any defined $$t$$. We can now take the usual time derivative as normal since we're in a vector space, resulting in $$(\mathcal{L}_XT)_p$$.
• I knew that $g$ was superfulous, I just wanted to look at it as an example. I understand the motivation of pushing the tensor back to $p$, but I'm still confused when I try to do it explicitly for the tensor $g$. What would $(d\phi_{-t})_{\phi_t(p)}g_{\varphi_t(p)}(Y_{\varphi_t(p)},Z_{\varphi_t(p)})$ look like given the vector fields $Y,Z$? Oct 3 '18 at 20:49