I know that the lie derivative of a metric $g_{ab}$ along a vector field $X$ can be written as $L_X g_{ab} = \nabla_a X_b + ∇_b X_a$. Now my question is, how would you use this to actually compute the lie derivative?
Say for example, if $X = (X^1, X^2) = (-y, x)$
Apologies for being unable to insert the correct notation.
On
Assume that $n=2$. First, you have to find the metric components $g_{ab}$ and the connection components $\Gamma^c_{ab}$. Then, you may use the following lowering index to find $X_a$ $$X_a=g_{ab}X^b$$For example $$X_1=g_{11}X^1+g_{12}X^2$$ Then, $$\nabla_aX_b=\frac{\partial X_b}{\partial x^a}-X_c\Gamma^c_{ab}$$ For example, $$\nabla_1X_1=\frac{\partial X_1}{\partial x^1}-X_1\Gamma^1_{11}-X_2\Gamma^2_{11}$$
Recall that the Lie derivative is the tensorial derivation characterized $\renewcommand\vec[1]{{\mathbf #1}}$by $L_{\vec{X}}f = \vec{X}(f)$ and $L_{\vec{X}}\vec{Y} = [\vec{X},\vec{Y}]$. Then, if $T$ is an $(r,s)$ tensor field along your manifold, we have $$\begin{align}(L_{\vec{X}}T)(\alpha^1,\ldots,\alpha^r, \vec{X}_1,\ldots,\vec{X}_s) &= L_{\vec{X}}(T(\alpha^1,\ldots,\alpha^r, \vec{X}_1,\ldots,\vec{X}_s) ) - \\ &\qquad - \sum_{i=1}^r T(\alpha^1,\ldots,L_{\vec{X}}\alpha^i,\ldots,\alpha^r,\vec{X}_1,\ldots,\vec{X}_s) - \\ & \qquad - \sum_{i=1}^sT(\alpha^1,\ldots, \alpha^r,\vec{X}_1,\ldots,L_{\vec{X}}\vec{X}_i,\ldots,\vec{X}_s).\end{align}$$In the particular case of the metric $g$, which is a $(0,2)$-tensor, we obtain a $(0,2)$-tensor $L_{\vec{X}}g$ given by $$(L_{\vec{X}}g)(\vec{Y},\vec{Z}) = \vec{X}(g(\vec{Y},\vec{Z})) - g([\vec{X},\vec{Y}],\vec{Z}) - g(\vec{Y},[\vec{X},\vec{Z}]).$$You can go further if you have more information on $g$. You can express that condition in a neater way if you use $g$'s Levi-Civita connection, for example.