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.


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

  • $\begingroup$ 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

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.