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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.

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First of all, Lie derivatives are defined on any smooth manifold, so the Riemannian structure is superfluous.

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 $(\mathcal{L}_XT)_p$.

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  • $\begingroup$ 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 $Y,Z$? $\endgroup$ – rmdmc89 Oct 3 '18 at 20:49

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