# Lie Derivative of a Metric in Coordinate Expression.

I apologize in advance if this question is too vague, but I have no idea where to start.

I am trying to find the Killing vector fields of a certain manifold with respect to a metric which is written as $ds^2 = ...$ where the right hand side of the equation is a combination of certain one-forms.

I have two questions:

1. I have only ever seen "metrics" written as tensors. I know how to the compute the Lie derivative of tensors, but this time the metric is written as "ds" which is like an infinitessimal line segment so I'm confused how to go about finding the Killing vector fields since I need to compute the Lie derivative of the metric.

2. Is there a book or website that has some good examples of problems like this? I am reading Lee's smooth manifolds and nothing really looks quite like this. The closest thing I have seen is his section on Riemannian metrics.

• where exactly did you find this problem? – magma Nov 29 '17 at 7:34
• @magma I didn't really "get" it from anywhere. This metric is from a paper my adviser is making me read. The paper itself is even harder to understand. – pictorexcrucia Nov 29 '17 at 13:38
• Yes, and the paper is publicly available? If yes, please quote the paper. – magma Nov 30 '17 at 5:40

In physics text the metric tensor $g$ usually written in a specific coordinate chart, or even if we don't mention the particular metric tensor $g$ it is also written in coodinate expression $ds^2 =g= g_{ij}dx^i dx^j$. In my opinion, the symbol $ds^2$ here is just physicist notation for metric tensor, probably motivated by previous experince in Euclidean space $\Bbb{R}^n$. If my memories serves me right, at first, Riemann also write metric tensor in his paper as $ds^2$.

The coordinate representation of any Lie derivative of metric $ds^2 = g_{\mu\nu} dx^{\mu}dx^{\nu}$ with respect to a vector field $V=V^{\mu } \partial_{\mu}$ is $$\mathscr{L}_V g_{\mu\nu} = V^{\alpha}\partial_{\alpha}g_{\mu\nu} + (\partial_{\mu}V^{\alpha})g_{\alpha\nu} + (\partial_{\nu}V^{\alpha})g_{\mu\alpha}.$$ This can be derived directly from coordinate invariant definition of Lie derivative of a tensor field, that is through a comparison of a tensor field along a a flow of a vector field using its pullback map. You can read the derivation in Hawking & Ellis's Large Scale Structure of Space-time or Frankel's Geometry of Physics text.

$\textbf{Update :}$

Actually there is a result in Lee's $\textit{Introduction to Smooth Manifold, 2nd ed}$ about how to compute Lie derivative of a tensor field (look $\textit{Corollary 12.33}$ ), but this result presented in coordinate invariant way. Is says that : For any smooth covariant $k$-tensor field $A \in \mathscr{T}(M) = \Gamma(T^kT^*M)$ and smooth vector fields $V,X_1,\dots X_k \in \Gamma(TM)$ we have $$(\mathscr{L}_VA)(X_1,\dots,X_k) = V(A(X_1,\dots,X_k)) - A([V,X_1],X_2,\dots,X_k) - \cdots -A(X_1,\dots,X_{k-1},[V,X_k]).$$ In your case, we have local expression $ds^2 = g = g_{ij}dx^i \otimes dx^j$ and Killing vector field $V = V^i\partial_i$ in a smooth chart $(U,(x^i))$. By applying the formula locally we obtained the $ij$-th component of the covariant smooth $2$-tensor field $\mathscr{L}_Vg = (\mathscr{L}_Vg)_{ij} dx^i \otimes dx^j$ as \begin{align*} (\mathscr{L}_Vg)_{ij} = (\mathscr{L}_Vg)(\partial_i,\partial_j) &= V^k\partial_k (g_{ij}) - g([V,\partial_i],\partial_j) - g(\partial_i,[V,\partial_j]) \\ &= V^k \partial_kg_{ij} + (\partial_iV^k)g_{kj} + (\partial_j V^k)g_{ik} . \end{align*}

• Ok, but what if my metric isn't written in that form? in fact, it's written like this: $$ds^2 = \alpha^2 dr^2 + \gamma^2(\omega^a)^2 + \beta^2(\omega^a - \frac{1}{2} \omega'^a) ^2$$ Where $\omega$ and $\omega'$ are left-invariant 1-forms – pictorexcrucia Nov 28 '17 at 16:50
• You can expand $\omega^a$ and $\omega'^a$ in terms of the basis $dx^a$. – Sou Nov 28 '17 at 16:52
• well, specifically we are looking at the manifold $R^4$ x $S^3$ and $\omega^a$ and $\omega'^a$ are "the left invariant one forms on $S^3$ and $S'^3$ respectively" – pictorexcrucia Nov 28 '17 at 16:55
• For example if $\omega^a = A_i dx^i$, so $(\omega^a)^2 = A_iA_j dx^idx^j$. And then you can apply the rule above. – Sou Nov 28 '17 at 16:59
• @pictorexcrucia I know what $S^3$ is, but what is $S'^3$ ? – magma Nov 30 '17 at 5:55