# Lie brackets of vector fields along a geodesic to obtain the Jacobi equation

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?

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.

• 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?
– cely
Apr 5, 2021 at 6:16
• Yup, that's right :) (sorry for the late reply) Apr 6, 2021 at 15:26
• Thanks again a lot!
– cely
Apr 6, 2021 at 15:50