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How to understand that boosts, rotations and translations are Killing vector fields.

+ 2 like - 0 dislike
139 views

If we consider the usual Minkowski space-time, then the Killing equation reduces to:

$$ \partial_\mu X_\nu + \partial_\nu X_\mu = 0  $$

where $X$ is the Killing vector field. Apparently the $X^\mu$ corresponds to the generators of the Poincare group, but I have difficulties understanding this.

asked Apr 23, 2014 in Theoretical Physics by Hunter (510 points) [ revision history ]
edited Apr 23, 2014 by Hunter

1 Answer

+ 3 like - 0 dislike

If you do the infinitesimal space-time transformation $x^\mu \rightarrow x+ \epsilon X^\mu$, the first-order change in the metric is equal to the Killing expression. Poincare group consists of all the transformations that don't change the metric, so the solutions to the equations are infinitesimal Poincare transformations. It's true by definition.

answered Apr 23, 2014 by Ron Maimon (7,535 points) [ revision history ]
edited Apr 23, 2014 by Ron Maimon

@RonMaimon thanks for your answer. I understand your first sentence, because that it how the Killing equation is derived (at least in the book I'm reading). For the second part of your message: let us consider for defineteness a rotation about the $z$-axis:

$ L_z = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & i & 0 \\ 0 & -i & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} $

which is a diffeomorphism $L_z : M \to M$. Let $x$ denote the original coordinates and $x'$ the infinitesimally transformed coordinates. Then, is the following correct:

$ \frac{\partial x'^\alpha}{\partial x^\mu} \frac{\partial x'^\beta}{\partial x^\nu} g_{\alpha \beta} = g_{\mu \nu} $?

And this is true for all the generators of the Poincare group and so you can conclude that the generators are Killing vector fields?

It's true, but when you state it your way, the infinitesimal displacement vector is $V^\mu(x) = \pm iL^\mu_\nu x^\nu$, so in this case, the vector field at the point (t,x,y,z) is (0,-y,x,0) (or the opposite sign, depending on your generator convention) and you can check the Killing equation is trivially satisfied. The sign is by whatever convention. I personally never use the i, I just use real generators for this stuff, the i is useless if you aren't doing QM, mathematicians usually don't put it in.

In the i-free convention, for linear transformations, where the coordinates are transformed by a matrix, the Killing equation just says that the matrix L is antisymmetric as a rank-two tensor, that is if you define $L_{\mu\nu} = L^\alpha_\mu g_{\alpha\nu}$, then $L_{\mu\nu}=-L_{\nu\mu}$. That's the definition of the generators of the rotation or Lorentz groups, the sign of the metric is the only thing that breaks the antisymmetry when you raise an index to act on a coordinate.

These conditions are too trivial in flat space-time, at least for constant metrics. the Killing vectors are for isometries of manifolds with nontrivial metrics.

@RonMaimon great thanks! I think I have understood it.

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