In his book, "Gravity and Cosmology", Weinberg talks about relations between homogeneous metric spaces and Klling vectors. First he says about infinitesimal isometrics $$ x^{\alpha}{'} = x^{\alpha} + \varepsilon k_{\lambda}{}^\alpha. \qquad (1) $$ Then he gets Killing equations and a fact that each of Killing vectors $k_{\lambda}$ can be represented near point X as Taylor series $$ k_{\lambda}{}^\mu (X, x) = A_{\lambda}{}^\nu (X, x) k_{\nu}{}^\mu + B_{\lambda}{}^{\rho \delta}D_{\rho}(X, x) k_{\delta}{}^\mu, \quad A, B \neq f(k_{\lambda}). $$ Then he gives the definition of homogeneous spaces: they are the spaces which refer to the existence of isometries $(1)$ which make parallel shifts from point X to any point of its surroundings. By the other words, he says, it means that this property of space requiers that the metric of corresponding space must have Killing vectors which can take all possible values ​​at any given point X. So it leads to the statement that in N-dimensional space we can choose N Killing vectors which have the property $$ k_{\lambda}{}^\mu(X, X) = \delta_{\lambda}^{\mu}. $$ I don't understand words in italics. Can you explain them to me?

  • 4
    $\begingroup$ Instead of using \quad, try breaking apart indices with {} instead. Thus, $\delta_\lambda{}^\mu$ produces $\delta_\lambda{}^\mu$. $\endgroup$
    – Stan Liou
    Dec 13, 2013 at 3:08

1 Answer 1


For reference, Weinberg p. 378:

A metric space is said to be homogeneous if there exist infinitesimal isometries (13.1.3) that carry any given point $X$ into any other point in its immediate neighborhood.

Equation (13.1.3) defines an infinitesimal transformation and (13.1.5) concludes the Killing equation $\xi_{\sigma;\rho} + \xi_{\rho;\sigma} = 0$ from it. Now, pick a particular point $X$. By the quoted definition, every direction from $X$ is fair game, so there must exist some Killing vector fields $\xi^{\mu}$ whose values at $X$ form a basis for the $X$'s tangent space.

Thus, we can write any vector at $X$ as a linear combination of $\xi^{(\mu)}(X)$, any take the coefficients to define a new vector field $K$. By linearity of the Lie derivative, $K$ is a Killing vector field, $$\mathcal{L}_K(g) = \sum_\mu a_{(\mu)}\mathcal{L}_{\xi^{(\mu)}}(g) = \sum_\mu a_{(\mu)}(0) = 0\text{.}$$

That is, the metric must admit Killing vectors that at any given point take all possible values.

By construction, the value of $K$ at $X$ is whatever vector we started with. Since it was arbitrary, we can have a Killing vector field that takes any value we want at $X$.

In particular, in an $N$-dimensional space we can choose a set of $N$ Killing vectors $\xi_\lambda {}^{(\mu)}(x;X)$ with $\xi_\lambda {}^{(\mu)}(X;X) = \delta_\lambda{}^\mu$.

Okay, $\xi_\rho{}^{(\mu)}(x;X)$ is the $\rho$-component of $\xi^{(\mu)}$ at $x$ written in terms of the value of the Killing vector field (and its derivatives) at $X$. So all it means is that we can have Killing vector fields whose values at $X$ are the coordinate vectors, $$\xi^{(\mu)}|_X = \partial_\mu|_X = \delta^{\lambda\mu}\partial_\lambda|_X\text{.}$$ That follows from the $K$-construction above, one for each of them individually.


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