# Lie derivative of Riemannian metric

Let $$X$$ be a vector field in a $$(M,g)$$ riemannian manifold. Let $$F^t$$ de local flows of $$X$$. Why is true the next equality?

$$L_X g(u,v) = \frac{d}{dt}|_{t = 0}g(DF^t(u),DF^t(v))$$

• The answer for why this equality is true depends on what definition you start with for the Lie derivative. What's your definition of the Lie derivative? – jgon May 15 at 23:34
• Lie derivative of a (0,2) tensor, the definition that Im using can be found in Peter Petersen and in this case $X(g) - g([X,U],V) - g(U,[X,V])$ – hal97 May 15 at 23:38

I'm still a bit new to differential geometry, so this might not be the simplest answer, but I'm pretty sure it's at least correct. Any feedback is appreciated.

Since you're using the definition that $$\newcommand\Ell{\mathcal{L}}\Ell_Xg(U,V)=X(g(U,V))-g([X,U],V)-g(U,[X,V]),$$ and the identity you want to prove involves a local flow of $$X$$, it would be good to recall identities for computing $$Xf$$ and $$[X,U]$$ where $$f$$ is a smooth function and $$U$$ is a vector field in terms of a local flow for $$X$$.

Let $$F_t$$ be a local flow for $$X$$ at some point. Then as $$t$$ varies, $$F_t(p)$$ is an integral curve for $$X$$ through $$p$$. In particular $$\frac{d}{dt}|_{t=0} F_t(p) = X_p$$. Thus we can use this curve to compute $$Xf_p$$. $$Xf_p = \frac{d}{dt}|_{t=0} f(F_t(p))$$.

In our particular case, this yields $$X(g(U,V))_p = \frac{d}{dt}|_{t=0} g(U_{F_t(p)},V_{F_t(p)}) = g(\frac{d}{dt}|_{t=0}U_{F_t(p)},V_p) + g(U_p,\frac{d}{dt}|_{t=0}V_{F_{t(p)}}).$$

Now $$[X,U]f_p = X(Uf)_p - U(Xf)_p = \frac{d}{dt}|_{t=0} (Uf)(F_t(p)) - U\left(\frac{d}{dt}|_{t=0} f(F_t(q))\right)(p)$$ $$=\frac{d}{dt}|_{t=0} (Uf)(F_t(p)) - \frac{d}{dt}|_{t=0} U(f\circ F_t)(p).$$ We have to be careful when pulling out the $$\frac{d}{dt}$$ from inside of $$U$$, since generally limits and derivatives don't commute. However, in this case, $$U$$ should be continuous in the germ, so we can pull the limit with respect to $$t$$ out, and $$1/t$$ is then just some real number, and $$U$$ is $$\Bbb{R}$$ linear.

Note that the first term is $$U_{F_t(p)}f$$ and the second is $$(DF_t\,U)f$$.

Thus we conclude that $$[X,U]_p = \frac{d}{dt}|_{t=0} U_{F_t(p)} - \frac{d}{dt}|_{t=0} DF_t\,U$$.

Substituting everything into the right hand side of the definition of $$\Ell_Xg(U,V)$$, we end up with $$X(g(U,V)) - g([X,U],V)-g(U,[X,V]) = g(\frac{d}{dt}|_{t=0} DF_t\, U,V) + g(U,\frac{d}{dt}|_{t=0} DF_t\, V),$$ which by product rule simplifies to $$\frac{d}{dt}|_{t=0} g(DF_t\, U, DF_t\, V),$$ as desired.

• Sorry but if you have $d/dt g(U,V) = g(DU,V) + g(U,DV)$ i dont know why you use the last expresion – hal97 May 16 at 7:13
• @hal97, I can't parse your comment, the equation you're giving doesn't seem to be any of the equations I've written, and is similar to two different equations I wrote, could you clarify a bit what you meant? – jgon May 16 at 11:48