I have done this question in mathstack but someone has suggested me it is more appropriate to ask this here.
With reference in https://archive.org/details/GeneralRelativity/page/n82/mode/1up, where it is said $[u,v]=0$ that in this case are the vector fields below.

Given two vector fields along a geodesic $\gamma$ in a manifold given by: $$\partial_t \Gamma=\frac{\partial x^\alpha}{\partial t}\frac{\partial}{\partial x^\alpha}\quad \quad \partial_s \Gamma=\frac{\partial x^\alpha}{\partial s}\frac{\partial}{\partial x^\alpha}$$

With $\gamma:\mathbb{I}\rightarrow \cal{M}$ and the geodesic variation of $\gamma$ given by $\Gamma:[-\epsilon,\epsilon]\times\mathbb{I}\rightarrow \cal{M}$ s.t. $\Gamma(0,t)=\gamma(t)$ and $\forall s\in [-\epsilon,\epsilon]$ we have that $t\rightarrow \Gamma(s,t)$ is a geodesic.

Now let $d\Gamma_p:\mathbb{R}\times\mathbb{R}\rightarrow T_{\Gamma(p)}\cal{M}$ s.t. $\partial_s\Gamma=(d\Gamma)_p(\frac{d}{ds},0)$ and $\partial_t\Gamma=(d\Gamma)_p(0,\frac{d}{dt})$.
In particular when we consider $p\in\gamma$ then $\partial_t\Gamma(0,t)=\dot \gamma(t)$ and $\partial_s\Gamma(0,t):=J(t)$.
These two last vector fields along $\gamma$ are the vector fields of my interest that I have written in coordinates considering $\Gamma(t,s)=x^{\alpha}(s,t)$. In addition I know that $\nabla_s\partial_t\Gamma=\nabla_t\partial_s\Gamma$ and obviously $\nabla_t\partial_t\Gamma=0$, since $\gamma$ is a geodesic.

$\textbf{I want to prove that $[\partial_t \Gamma,\partial_s \Gamma]=0$}$, where $\textbf{[}\quad\textbf{]}$ represent the Lie brackets.

To do this I have developed the following idea: $$[\partial_t \Gamma,\partial_s \Gamma]=\Big[\frac{\partial x^\alpha}{\partial t}\frac{\partial}{\partial x^\alpha}, \frac{\partial x^\alpha}{\partial s}\frac{\partial}{\partial x^\alpha}\Big]=\Big(\frac{\partial x^\alpha}{\partial t}\frac{\partial}{\partial x^\alpha}\frac{\partial x^\beta}{\partial s}-\frac{\partial x^\alpha}{\partial s}\frac{\partial}{\partial x^\alpha}\frac{\partial x^\beta}{\partial t}\Big)\frac{\partial}{\partial x^\beta}=\Big(\frac{\partial^2 x^\beta}{\partial t\partial s}-\frac{\partial^2 x^\beta}{\partial s\partial t}\Big)\frac{\partial}{\partial x^\beta}=0$$ $\textbf{My doubt: }$I am not sure of the second and above all of the third equality, where I have thought to write $\displaystyle\frac{\partial x^\alpha}{\partial s}\frac{\partial}{\partial x^\alpha}\frac{\partial x^\beta}{\partial t}=\frac{\partial^2 x^\beta}{\partial s\partial t}$.

Do you think what I have done it is correct? If not can you tell me where there are the mistakes and how can I prove that the lie brackets give me a null vector in this case?


1 Answer 1


Your reasoning is correct, but I think you might be missing two cancelling terms in the Lie bracket, so I will rewrite it again.

It is convenient to set the bracket to act on something, say a function $v$. (It is good practice not to use the same letter twice for dummy indices) \begin{align} [\partial_t \Gamma,\partial_s \Gamma]=&\Big[\frac{\partial x^\alpha}{\partial t}\frac{\partial}{\partial x^\alpha}, \frac{\partial x^\beta}{\partial s}\frac{\partial}{\partial x^\beta}\Big] v\\ =& \frac{\partial x^\alpha}{\partial t}\left(\frac{\partial}{\partial x^\alpha} \frac{\partial x^\beta}{\partial s}\right)\frac{\partial}{\partial x^\beta}v + \frac{\partial x^\alpha}{\partial t} \frac{\partial x^\beta}{\partial s}\left(\frac{\partial^2}{\partial x^\alpha\partial x^\beta}v\right)\\ &- \frac{\partial x^\beta}{\partial s}\left(\frac{\partial}{\partial x^\beta} \frac{\partial x^\alpha}{\partial t}\right)\frac{\partial}{\partial x^\alpha}v -\frac{\partial x^\beta}{\partial s} \frac{\partial x^\alpha}{\partial t}\left(\frac{\partial^2}{\partial x^\beta\partial x^\alpha}v\right) \end{align} Now cancel the second and third term as they are the same (partial derivatives commute), and rename $\alpha\to\beta, \beta\to\alpha$ in the third term to group it all as $$ = \left(\frac{\partial x^\alpha}{\partial t}\frac{\partial}{\partial x^\alpha} \frac{\partial x^\beta}{\partial s} -\frac{\partial x^\alpha}{\partial s}\frac{\partial}{\partial x^\alpha} \frac{\partial x^\beta}{\partial t}\right) \frac{\partial}{\partial x^\beta}v $$ Now, use the chain rule, $\frac{\partial x^\alpha}{\partial t}\frac{\partial}{\partial x^\alpha}\equiv \frac{\partial}{\partial t}$ $$ = \left(\frac{\partial^2 x^\beta}{\partial t \partial s} - \frac{\partial^2 x^\beta}{\partial s \partial t}\right) \frac{\partial}{\partial x^\beta}v = 0$$ which vanishes again due to the fact that partial derivatives commute.

P.S. The way I quickly remind myself of the fact $[\frac{\partial}{\partial t},\frac{\partial}{\partial s} ]=0$ is by viewing the variation geodesic as a map between manifolds, as you say $$\Gamma:[-\epsilon,\epsilon]\times\mathbb{I}\rightarrow \cal{M}$$ The domain is the manifold $[-\epsilon,\epsilon]\times\mathbb{I}$ completely covered by coordinates $s, t$, so $\frac{\partial}{\partial t},\frac{\partial}{\partial s}$ need to be coordinate vector fields, i.e., they must commute.

  • $\begingroup$ Thanks a lot! So my intuition was correct apart from the fact that I have not used the product rule of derivation in writing the lie brackets (so I will correct this fact), right? $\endgroup$
    – cely
    Apr 5, 2021 at 6:16
  • $\begingroup$ Yup, that's right :) (sorry for the late reply) $\endgroup$
    – Mateo
    Apr 6, 2021 at 15:26
  • $\begingroup$ Thanks again a lot! $\endgroup$
    – cely
    Apr 6, 2021 at 15:50

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