Let $(M,g)$ be a Riemannian manifold, $V \subset \mathbb{R}^2$ be an open subset and $\alpha: V \rightarrow M; (s,t) \mapsto \alpha(s,t)$ a smooth map.

for $(s,t) \in V$ one can define

$$ \frac{\partial \alpha}{\partial s}(s,t) := [\sigma \mapsto \alpha(s+ \sigma ,t) ]\in T_{\alpha(s,t)}M $$ $$\frac{\partial \alpha}{\partial t}(s,t) := [\tau \mapsto \alpha(s ,t + \tau) ]\in T_{\alpha(s,t)}M $$

Where the curve in brackets is a tangent vector. For a chart $(U,x)$ one has

$$ \frac{\partial \alpha}{\partial s}(s,t) = \sum \frac{\partial \tilde{\alpha}_i}{\partial s}(s,t) \cdot \frac{\partial}{\partial x_i}\vert_{\alpha(s,t)} $$ and $$ \frac{\partial \alpha}{\partial t}(s,t) = \sum \frac{\partial \tilde{\alpha}_i}{\partial t}(s,t) \cdot \frac{\partial}{\partial x_i}\vert_{\alpha(s,t)} $$

with $\tilde{\alpha}_i= x \circ \alpha \cdot e_i$.

I want to show that the Lee bracket disappears i.e $[ \frac{\partial \alpha}{\partial s}(s,t), \frac{\partial \alpha}{\partial t}(s,t)] = 0$ (needed to prove the jacobi equation).

If $$\frac{\partial \alpha}{\partial s}(s,t) = \sum a_i \cdot \frac{\partial}{\partial x_i}\vert_{\alpha(s,t)},$$

$$ \frac{\partial \alpha}{\partial s}(s,t) = \sum b_i \cdot \frac{\partial}{\partial x_i}\vert_{\alpha(s,t)},$$

$$[ \frac{\partial \alpha}{\partial s}(s,t), \frac{\partial \alpha}{\partial t}(s,t)] = \sum c_i\frac{\partial}{\partial x_i}\vert_{\alpha(s,t)} $$

then local computations for the Lie-Bracket yield

$$ c_k = \sum_i a_i \frac{\partial}{\partial x_i} b_k - b_i \frac{\partial}{\partial x_i} a_k$$

together with the above, one has

$$ c_k= \frac{\partial \tilde{\alpha}_i}{\partial s}(s,t) \frac{\partial}{\partial x_i} \frac{\partial \tilde{\alpha}_k}{\partial t}(s,t) - \frac{\partial \tilde{\alpha}_i}{\partial t}(s,t) \frac{\partial}{\partial x_i} \frac{\partial \tilde{\alpha}_k}{\partial s}(s,t)$$

But the partial derivatives with $x_i$ dont make much sense. Somewhere I made a mistake. I am gratefull for suggestions.


1 Answer 1


There is no mistake, but it is confusing. The reason is that $ \frac{\partial \alpha}{\partial s} $ defines a vector in $ T_{\alpha(s,t)}M $ for each $ s, t $, but this does not necessarily define a vector field. So it is not clear how to interpret the lie bracket.

If $ \alpha $ is locally a diffeomorphism, then there are inverse function $ s(x_1,x_2) $ and $ t(x_1,x_2) $, so that $ \alpha (s(x_1,x_2), t(x_1,x_2) ) = (x_1,x_2) $. So in the expressions like: $$ \frac{\partial \tilde{\alpha}_k}{\partial t}(s,t) = $$ the $ s $ and $ t $ really depend on $ x_1, x_2 $ so that the derivatives like: $$ \frac{\partial}{\partial x_i} \frac{\partial \tilde{\alpha}_k}{\partial t}(s,t) $$ make perfect sense.

  • $\begingroup$ If $\alpha$ is smooth the Lie bracket is defined at least locally by the implicit function theorem. The statement I want is something like Lemma 9.2 in Milnors Morse Theory (Page 52). $\endgroup$
    – asterisk
    Aug 28, 2015 at 5:53
  • $\begingroup$ In lemma 9.2, they are covariant derivatives. Covariant derivatives are different then lie brackets. In particular, in order for $ \nabla_u v $ to be well defined at $ p \in M $, $ v$ needs to be a vector field around $ p$, while $ u$ only need be a vector in $ T_p M$. Whereas, in $ [u,v] $ both need to be vector fields. $\endgroup$
    – user226970
    Aug 28, 2015 at 9:28
  • $\begingroup$ But there you also have the curvature Tensor in the equation, which makes no sense unless the vector fields are locally defined, for reasons which you explained. Therefore I believe that locally the Vector fields are defined and therefore one should be able to compute them. $\endgroup$
    – asterisk
    Aug 28, 2015 at 13:57
  • $\begingroup$ The curvature tensor makes perfect sense even if the vectors fields are not locally defined. Why can't I do $ R(X,Y) Z $ if $X, Y, Z $ are in the same tangent space? Also, what did you mean by saying that the lie bracket is defined by implicit function theorem? Finally, why do you need this fact for Jacobi theorem? $\endgroup$
    – user226970
    Aug 28, 2015 at 18:54
  • $\begingroup$ In the context of Variations of geodesics I have seen arguments like the vector fields commute in the sense that the lee Bracket disappears. (For example math.upenn.edu/~wziller/math660/… page 7 above equation (2.4)). Therefore I assumed that there must be an easy explanation why the vector fields are defined locally. This is not true in general, as you have rightfully wondered. For Example the constant Variation, where all paths are just the same path, does not admit a locally defined vector field. $\endgroup$
    – asterisk
    Aug 29, 2015 at 10:45

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