#### Question

Show that any Killing vector ##K^\mu## satisfies the relations mentioned in the text:\begin{align}

\nabla_\mu\nabla_\sigma K^\rho=R_{\ \ \ \sigma\mu\nu}^\rho K^\nu&\phantom {10000}(1)\nonumber\\

K^\lambda\nabla_\lambda R=0&\phantom {10000}(2)\nonumber

\end{align}

#### Answers

Wilhelm Karl Joseph Killing 1847-1923 |

\begin{align}

\nabla_{(\mu}K_{\nu)}\equiv\frac{1}{2}\left(\nabla_\mu K_\nu+\nabla_\nu K_\mu\right)=0&\phantom {10000}(3)\nonumber\\

\Rightarrow\nabla_\mu K_\nu=-\nabla_\nu K_\mu&\phantom {10000}(4)\nonumber

\end{align}##\nabla_\mu K_\nu## is antisymmetric.

The Killing vectors in (1) and (2) are given in contravariant form.

The Riemann tensor is antisymmetric* in its last two indices like (7) and has other interesting symmetries, not all independent, e.g. (6) comes form (7) and (8):

\begin{align}

&R_{\rho\sigma\mu\nu}=g_{\tau\rho}R_{\ \ \ \sigma\nu\mu}^\tau&\phantom {10000}(5)\nonumber\\

\text{Swap first two }~~~~~~~~~

&R_{\rho\sigma\mu\nu}=-R_{\sigma\rho\mu\nu}&\phantom {10000}(6)\nonumber\\

\text{Swap last two }~~~~~~~~~

&R_{\rho\sigma\mu\nu}=-R_{\rho\sigma\nu\mu}&\phantom {10000}(7)\nonumber\\

\text{Swap first and last pair }~~~~~~~~~

&R_{\rho\sigma\mu\nu}=R_{\mu\nu\rho\sigma}&\phantom {10000}(8)\nonumber\\

\text{Rotate last three }~~~~~~~~~

&R_{\rho\sigma\mu\nu}+R_{\rho\mu\nu\sigma}+R_{\rho\nu\sigma\mu}=0&\phantom {10000}(9)\nonumber\\

\text{Permute last three }~~~~~~~~~

&R_{\rho\left\lceil\sigma\mu\nu\right\rceil}=0&\phantom {10000}(10)\nonumber

\end{align}(6) implies

\begin{align}

g_{\tau\rho}R_{\ \ \ \sigma\mu\nu}^\tau=-g_{\tau\sigma}R_{\ \ \ \rho\mu\nu}^\tau&\phantom {10000}(11)\nonumber\\

\Rightarrow g^{\lambda\rho}g_{\tau\rho}R_{\ \ \ \sigma\mu\nu}^\tau=-g^{\lambda\rho}g_{\tau\sigma}R_{\ \ \ \rho\mu\nu}^\tau&\phantom {10000}(12)\nonumber\\

\Rightarrow R_{\ \ \ \sigma\mu\nu}^\lambda=-R_{\ \sigma\ \ \rho\mu\nu}^{\ \ \ \ \lambda}&\phantom {10000}(13)\nonumber

\end{align}We surmise that a similar manoeuvre on (7) would get us (14) which was the assertion * above and that (8) would get us (15)

\begin{align}

R_{\ \ \ \sigma\nu\mu}^\tau=-R_{\ \ \ \sigma\mu\nu}^\tau&\phantom {10000}(14)\nonumber\\

R_{\ \ \ \sigma\nu\mu}^\tau=R_{\nu\mu\ \ \ \sigma}^{\ \ \ \ \ \ \tau}&\phantom {10000}(15)\nonumber

\end{align}These might will not be useful, although it might will be better to work with fully covariant Riemann tensors.

As I found out after much effort, it is also important to remember the definition of the Riemann tensor when the connection is torsion free. It measures the difference between taking covariant derivatives of a vector going the two opposite ways round a path (Carroll 3.112)

\begin{align}

\left[\nabla_\rho,\nabla_\sigma\right]X^\mu=R_{\ \ \ \nu\sigma\rho}^\mu X^\nu&\phantom {10000}(16)\nonumber

\end{align}That equation is pretty similar to (1) and then there's (4).

The second part is very simple if you remember that if a Killing vector exists then it is always possible to find a coordinate system where it is one of the basis vectors and the metric will be independent of the coordinate for that basis vector. Carroll stated this after the Killing equation at his 3.174.

I was unable to answer the questions but was guided by a solution I stumbled across on Semantic Scholar by Professor Alan Guth.

More on my struggles and links to Guth solution at

Ex 3.12 Derivatives of Killing vectors.pdf (4 pages)

Your (16), as well the corresponding one in the PDF, has a sign error. Also, Guth's answer is actually quite clear, although there is a typo in the last term in (2.10). (Things like "a symmetric tensor contracting an antisymmetric tensor gives 0" are actually easy to prove and very useful.)

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