# How much does the Lie algebra of Killing fields constrain the metric?

MOTIVATION

Suppose $$(M ^n, g)$$ is a (pseudo-)Riemannian manifold with a connected isometry group. This means we can fully determine $$\text{Iso}(M, g)$$ from the Killing vector fields on $$M$$. Moreover, once we find a Killing vector field $$X$$, we can simplify parts of the metric by using the Killing equation $$\nabla_a X_b + \nabla_b X_a = 0$$. Now repeat this procedure to extract all the information available from the structure constants. In other words, the symmetries constrain the metric, and the structure constants partially constrain the symmetries, so information on the Lie algebra should partially constrain the metric.

A good place to start is to invoke the Frobenius theorem to find a coordinate system $$(x ^k)$$ where the Killing fields $$X_i$$ look like $$X_i = \sum_{j = 1} ^m \alpha_i ^j(x_1, \dots, x_n) \partial_j,$$ where $$m$$ is the dimension of the integral manifold generated by the $$X_i$$'s (here we already have a problem since the same Lie algebra of Killing fields can have different $$m$$'s). A further reduction can be achieved by looking for the largest commutative sub-algebra of $$\{X_i\}$$. We can then coordinate the integral manifold and renumber the $$X_i$$'s in such a way that $$X_i = \delta ^j_i$$ for $$j = 1, \dots, k$$. Finally, we can use our knowledge of the structure constants to write equations for the components of the $$X_i$$'s $$[X_i, X_j] = C_{ij} ^k X_k,$$ which we can hopefully solve for the coefficients of the Killing fields. These fields can then be used to find a form of the metric. Note I am not looking to determine the metric exactly, I only want all the information I can glean from the Killing fields. I can show what I mean with the following example.

EXAMPLE

To illustrate, let us set up the proof of Birkhoff's theorem. In this case, $$M$$ is a spherically symmetric Lorentzian manifold, which means there exist Killing vector fields $$X, Y$$, and $$Z$$ s.t. $$[X, Y] = Z, \quad [X, Z] = - Y, \quad [Y, Z] = X,$$ where the orbit manifolds are of dimension $$m = 2$$. Following the steps outlined before, we can set $$X = \partial_1, \ \ Y = a \partial_1 + b \partial_2, \ \ Z = c \partial_1 + d \partial_2,$$ where $$a, b, c,$$ and $$d$$ are functions of $$(x_1,\dots, x_4)$$. The structure equations $$[X, Y] = Z$$ and $$[X, Z] = -Y$$ force \begin{align*} \partial_1 a &= c, \quad -\partial_1 c = a \Rightarrow \partial_1 ^2 a + a = 0, \\ \partial_1 b &= d, \quad - \partial_1 d = b \Rightarrow \partial_1 ^2 b + b = 0, \end{align*} from which we deduce that $$a = a_1(x_2, x_3, x_4) \sin(x) + a_2(x_2, x_3, x_4) \cos(x)$$ and $$b = b_1(x_2, x_3, x_4) \sin(x) + b_2(x_2, x_3, x_4) \cos(x)$$. Plugging this into the third structure equation with $$y = (x_2, x_3, x_4)$$ gives: \begin{align*} b_2(y) \partial_2a_1(y)-b_1(y) \partial_2a_2(y)-a_1(y){}^2-a_2(y){}^2&=1, \\ b_2(y) \left(\partial_2b_1(y)-a_2(y)\right)-b_1(y) \left(a_1(y)+\partial_2b_2(y)\right) &= 0. \end{align*} These equations are far from having a unique solution, but if I put $$b_1 = 1 = 1 - b_2$$, then the second line forces $$-a_1(y) = 0$$, which in turn simplifies the first line to $$-\partial_2 a_2(y) = 1 + a_2(y) ^2 \Rightarrow a_2(y) = \cot(x_2)$$. In other words, \begin{align*} Y &= \cos(x_1) \cot(x_2) \partial_1 + \sin(x_1)\partial_2, \\ Z &= -\sin(x_1) \cot(x_2) \partial_1 + \cos(x_1) \partial_2. \end{align*} The Killing equation $$\alpha_i ^{\sigma} \partial_{\sigma} g_{\gamma \beta} + g_{\gamma \sigma} \partial_{\beta} \alpha_i ^{\sigma} + g_{\beta \sigma} \partial_{\gamma} \alpha_i ^{\sigma} = 0$$ shows $$\partial_1 g_{\gamma \beta} = 0$$ when $$i = 1$$, so the metric is independent of $$x_1$$. Thus, for $$i = 2$$ and $$i =3$$ we get \begin{align} \sin(x_1) \partial_2 g_{\gamma \beta} &+ g_{\gamma 1} \partial_{\beta} \cos(x_1) \cot(x_2) + g_{\gamma 2} \partial_{\beta} \sin(x_1) \nonumber \\ &+ g_{\beta 1} \partial_{\gamma} \cos(x_1)\cot(x_2) + g_{\beta 2} \tag{1} \partial_{\gamma} \sin(x_1) = 0, \\ \cos(x_1) \partial_2 g_{\gamma \beta} &- g_{\gamma 1} \partial_{\beta} \sin(x_1) \cot(x_2) + g_{\gamma 2} \partial_{\beta} \cos(x_1) \nonumber \\ &- g_{\beta 1} \partial_{\gamma} \sin(x_1)\cot(x_2) + g_{\beta 2} \tag{2} \partial_{\gamma} \cos(x_1) = 0. \end{align} Multiply the first equation by $$\cos(x_1)$$ and the second by $$\sin(x_1)$$ and take the difference to get \begin{align*} 0 &= g_{\gamma 1}\cot(x_2)(\cos(x_1)\partial_{\beta} \cos(x_1) + \sin(x_1)\partial_{\beta} \sin(x_1)) + g_{\gamma 1} \partial_{\beta} \cot(x_2) \\ &+g_{\gamma 2}(\cos(x_1) \partial_{\beta} \sin(x_1) - \sin(x_1) \partial_{\beta} \cos(x_1)) \\ &+ g_{\beta 1}\cot(x_2)(\cos(x_1)\partial_{\gamma} \cos(x_1) + \sin(x_1)\partial_{\gamma} \sin(x_1)) + g_{\beta 1} \partial_{\gamma} \cot(x_2) \\ &+g_{\beta 2}(\cos(x_1) \partial_{\gamma} \sin(x_1) - \sin(x_1) \partial_{\gamma} \cos(x_1)). \end{align*} By using the identities \begin{align} \cos(x_1)\partial_{\beta} \cos(x_1) + \sin(x_1)\partial_{\beta} \sin(x_1) &= 0, \tag{3} \\ \cos(x_1) \partial_{\gamma} \sin(x_1) - \sin(x_1) \partial_{\gamma} \cos(x_1) &= \partial_{\gamma} x_1, \nonumber \end{align} we obtain \begin{align*} g_{\gamma 1} \partial_{\beta} \cot(x_2) +g_{\gamma 2}\partial_{\beta} x_1 + g_{\beta 1} \partial_{\gamma} \cot(x_2) +g_{\beta 2}\partial_{\gamma} x_1 = 0. \end{align*} Plugging values of $$\gamma$$ and $$\beta$$ gives \begin{align*} (\gamma, \beta) = (1, 1) &: 2g_{12} = 0, \\ (\gamma, \beta) = (1, 2) &: -g_{11}\csc(x_2) ^2 + g_{22} = 0 \Rightarrow g_{11} = g_{22} \sin(x_2) ^2, \\ (\gamma, \beta) = (1, 3) &: g_{32} = 0, \\ (\gamma, \beta) = (1, 4) &: g_{42} = 0, \\ (\gamma, \beta) = (2, 2) &: -2g_{21} \csc(x_2) ^2 = 0, \\ (\gamma, \beta) = (2, 3) &: -g_{31}\csc(x_2) ^2 = 0, \\ (\gamma, \beta) = (2, 4) &: -g_{41} \csc(x_2) ^2 = 0. \end{align*} Multiplying $$(1)$$ by $$\sin(x_1)$$ and $$(2)$$ by $$\cos(x_1)$$ and simplifying the sum of the results with the identities $$(3)$$ gives: \begin{align*} \partial_2 g_{\gamma \beta} - g_{\gamma 1} \cot(x_2) \partial_{\beta} x_1 - g_{\beta 1} \cot(x_2) \partial_{\gamma} x_1 = 0. \end{align*} For $$\beta, \gamma \in \{3, 4\}$$ we have $$\partial_2 \gamma_{\gamma \beta} = 0$$, and similarly for $$\gamma = \beta = 2$$, whence, recalling that $$\partial_1 g_{\gamma \beta} = 0$$ we have: $$g = \begin{bmatrix} g_{22}(x_3, x_4)\sin(x_2) ^2 & 0 & 0 & 0 \\ 0 & g_{22}(x_3, x_4) & 0 & 0 \\ 0 & 0 & g_{33}(x_3, x_4) & g_{34}(x_3, x_4) \\ 0 & 0 & g_{43}(x_3, x_4) & g_{44}(x_3, x_4) \end{bmatrix}.$$ From this representation we can obtain the usual spherically symmetric metric by constraining the behavior of $$g_{22}$$ and making a change of coordinates, and I am perfectly satisfied with this answer. This is as far as the specification of the Lie algebra could take us.

QUESTION

My question is: given the structure constants and the dimension of the integral manifolds, how much can we constrain the form of the metric? Put another way, can we determine the Killing fields once we are told the Lie algebra structure and integral submanifold dimension?

The choice $$b_1 = 1 = 1 - b_2$$ was pretty arbitrary, and I could have chosen any other values and obtained different solutions for the $$a_i's$$. Is this a case of fixing a gauge or are these solutions fundamentally different in the sense that different choices lead to different metrics?