- #1
Markus Kahn
- 112
- 14
- Homework Statement
- Calculate ##(\nabla_X Y)^i## and ##(\nabla_X \omega)_i##, where ##X## is a vector field and ##\omega## is a covector field.
- Relevant Equations
- Definition of the covariant derivative:
A covariant differentiation on a manifold ##\mathcal{M}## is a mapping ##\nabla## which assigns to every pair ##X, Y## of ##C^\infty## vector fields on ##\mathcal{M}## another ##C^\infty## vector field ##\nabla_X Y## with the following properties:
1) ##\nabla_{X} Y \text { is bilinear in } X \text { and } Y##,
2) ##\text { if } f \in \mathcal{F}(\mathcal{M}), \text { then }##
$$\begin{array}{l}{\nabla_{f X} Y=f \nabla_{X} Y} \\ {\nabla_{X}(f Y)=f \nabla_{X} Y+X(f) Y}\end{array}.$$
My attempt so far:
$$\begin{align*}
(\nabla_X Y)^i &= (\nabla_{X^l \partial_l}(Y^k\partial_k))^i=(X^l \nabla_{\partial_l}(Y^k\partial_k))^i\\
&\overset{2)}{=} (X^l (Y^k\nabla_{\partial_l}(\partial_k) + (\partial_l Y^k)\partial_k))^i = (X^lY^k\Gamma^n_{lk}\partial_n + X^lY^k{}_{,l}\partial_k)^i\\
&= (X^lY^k\Gamma^n_{lk}\partial_n + X^lY^k{}_{,l}\partial_k) x^i\\
&= X^lY^k\Gamma^i_{lk}+ X^lY^i{}_{,l}
\end{align*}$$
This seems reasonable to me, at least all the dummy indices disappear. But when I try to do the same for the covector field I quickly run into problems...
$$\begin{align*}
(\nabla_X \omega)_i &= (\nabla_{X^l \partial_l}(\omega_kdx^k))_i = (X^l\nabla_{\partial_l}(\omega_kdx^k))_i \\
&= (X^l(\omega_k \nabla_{\partial_l}(dx^k) +( \partial_l \omega_k) dx^k))_i \overset{(*)}{=}(X^l\omega_k \nabla_{\partial_l}(dx^k) +X^l \omega_{k,l} dx^k)\partial_i \\
&= X^l\omega_k \nabla_{\partial_l}(dx^k) (\partial_i) +X^l \omega_{k,l} dx^k(\partial_i) = X^l\omega_k \nabla_{\partial_l}(dx^k) (\partial_i) +X^l \omega_{i,l}
\end{align*}$$
After ##(*)## I started guessing, but now I have to evaluate ##\nabla_{\partial_l}(dx^k) ## and I'm not really sure what this is... If I had to guess I'd probably say
$$\nabla_{\partial_l}(dx^k) = U_{ln}^{k}dx^n,$$
where ##U_{ln}^{k}## is somehow related to ##\Gamma^n_{lk}##. Can someone maybe help me here? How exactly is ##\nabla_{\partial_l}(dx^k) ## defined?
$$\begin{align*}
(\nabla_X Y)^i &= (\nabla_{X^l \partial_l}(Y^k\partial_k))^i=(X^l \nabla_{\partial_l}(Y^k\partial_k))^i\\
&\overset{2)}{=} (X^l (Y^k\nabla_{\partial_l}(\partial_k) + (\partial_l Y^k)\partial_k))^i = (X^lY^k\Gamma^n_{lk}\partial_n + X^lY^k{}_{,l}\partial_k)^i\\
&= (X^lY^k\Gamma^n_{lk}\partial_n + X^lY^k{}_{,l}\partial_k) x^i\\
&= X^lY^k\Gamma^i_{lk}+ X^lY^i{}_{,l}
\end{align*}$$
This seems reasonable to me, at least all the dummy indices disappear. But when I try to do the same for the covector field I quickly run into problems...
$$\begin{align*}
(\nabla_X \omega)_i &= (\nabla_{X^l \partial_l}(\omega_kdx^k))_i = (X^l\nabla_{\partial_l}(\omega_kdx^k))_i \\
&= (X^l(\omega_k \nabla_{\partial_l}(dx^k) +( \partial_l \omega_k) dx^k))_i \overset{(*)}{=}(X^l\omega_k \nabla_{\partial_l}(dx^k) +X^l \omega_{k,l} dx^k)\partial_i \\
&= X^l\omega_k \nabla_{\partial_l}(dx^k) (\partial_i) +X^l \omega_{k,l} dx^k(\partial_i) = X^l\omega_k \nabla_{\partial_l}(dx^k) (\partial_i) +X^l \omega_{i,l}
\end{align*}$$
After ##(*)## I started guessing, but now I have to evaluate ##\nabla_{\partial_l}(dx^k) ## and I'm not really sure what this is... If I had to guess I'd probably say
$$\nabla_{\partial_l}(dx^k) = U_{ln}^{k}dx^n,$$
where ##U_{ln}^{k}## is somehow related to ##\Gamma^n_{lk}##. Can someone maybe help me here? How exactly is ##\nabla_{\partial_l}(dx^k) ## defined?