Computation of Liebracket for Vectorfields assosiated with a Variation of Geodesics

Question

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.

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.