1

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.

asterisk
  • 376

1 Answers1

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.

user226970
  • 638
  • 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). – asterisk Aug 28 '15 at 05:53
  • 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. – user226970 Aug 28 '15 at 09:28
  • 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. – asterisk Aug 28 '15 at 13:57
  • 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? – user226970 Aug 28 '15 at 18:54
  • 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 https://www.math.upenn.edu/~wziller/math660/ComparisonTheorems-Eschenburg.pdf 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. – asterisk Aug 29 '15 at 10:45
  • I therefore assumed that in order to show $D_XD_Y Z = D_XD_YZ + R(X,Y)Z$ in the context of Variations I first needed to show $D_{[X,Y]}Z=0$ which follows from $[X,Y]=0$.

    However this approach turns out to be wrong, because the vector fields need not to be defined but instead one can do local calculations which go through.

     – asterisk Aug 29 '15 at 10:49