# A question about Killing vector and Riemann curvature tensor

In Sean Carroll's Spacetime and Geometry, a formula is given as $${\nabla _\mu }{\nabla _\sigma }{K^\rho } = {R^\rho }_{\sigma \mu \nu }{K^\nu },$$

where $K^\mu$ is a Killing vector satisfying Killing's equation ${\nabla _\mu }{K_\nu } +{\nabla _\nu }{K_\mu }=0$ and the convention of Riemann curvature tensor is

$$\left[\nabla_{\mu},\nabla_{\nu}\right]V^{\rho}={R^\rho}_{\sigma\mu\nu}V^{\sigma}.$$

So how to prove the this formula (the connection is Levi-Civita)?

Permit me the use of Latin indices instead of Greek indices and the convention $$\nabla_a K_b=K_{b;a}$$. So we wish to prove $$\newcommand{\Tud}[3]{{#1}^{#2}_{\phantom{#2}{#3}}}$$ $$\Tud{K}{a}{;b c} = \Tud{R}{a}{b c d} K^d$$ where $$\Tud{V}{a}{;b c} - \Tud{V}{a}{;c b} = \Tud{R}{a}{d c b} V^d$$ and $$K_{a ; b} + K_{b ; a} = 0$$

Differentiating the last equation, we get $$K_{a ; b c} + K_{b ; a c} = 0$$ so, relabelling and summing, $$K_{a ; b c} + K_{b ; a c} - K_{b ; c a} - K_{c ; b a} + K_{c ; a b} + K_{a ; c b} = 0$$ hence, $$K_{a; b c} + K_{a ; c b} = R_{b d a c} K^d + R_{c d a b} K^d$$ By the interchange symmetry $$R_{a b c d} = R_{c d a b}$$, and raising indices, we get $$\Tud{K}{a}{;b c} - \Tud{R}{a}{b c d} K^d = -(\Tud{K}{a}{;c b} - \Tud{R}{a}{c b d} K^d)$$

On the other hand, by the first Bianchi identity and antisymmetry, we have $$\Tud{R}{a}{d c b} = \Tud{R}{a}{b c d} + \Tud{R}{a}{c d b}$$ Hence we get $$\Tud{K}{a}{;b c} = \Tud{K}{a}{; c b} + \Tud{R}{a}{d c b} K^d = \Tud{K}{a}{;c b} + \Tud{R}{a}{b c d} K^d + \Tud{R}{a}{c d b} K^d$$ and therefore $$\Tud{K}{a}{;b c} - \Tud{R}{a}{b c d} K^d = \Tud{K}{a}{;c b} - \Tud{R}{a}{c b d} K^d$$ The conclusion follows.

• Is there any simpler proof? This doesn't seem to be the kind that I can prove myself. Dec 31, 2011 at 11:35
• @Karsus: It's just algebraic manipulations of symbols using well-known identities. If you're asking for a high-level conceptual proof, then I'm afraid I don't have an answer for you. Dec 31, 2011 at 11:43
• And I don't understand how ${K^a}_{;bc}={R^a}_{bcd}K^d$ follows from your last line. Dec 31, 2011 at 11:45
• It follows from the last equation and the equation fourth from the bottom together. Dec 31, 2011 at 14:51
• Where does the penultimate line come from? In the fourth from the bottom, ${K^a}_{;bc}$ and ${K^a}_{;cb}$ have differing sign on each side of the equation, but in the penultimate line they have the same sign.
– T.H
Feb 16, 2021 at 15:27

@C.R. This is my 'simpler proof'; I'm pretty sure it's correct, and simpler than Zhen's one as well.

From the first Bianchi identity [Carroll, (3.132)] $$R_{\mu\nu\rho\sigma}+R_{\mu\rho\sigma\nu}+R_{\mu\sigma\nu\rho}=0$$ we have that, for every vector $V^\rho$, $$\nabla_{[\mu}\nabla_\nu V_{\rho]}=\tfrac{1}{6}\big(R_{\rho\alpha\mu\nu}+R_{\mu\alpha\nu\rho}+R_{\rho\mu\nu\alpha}\big)V^\alpha=0,$$ where in the last equation I used the symmetry properties of the indices in the Riemann tensor to reduce it to the Bianchi identity. This is a very useful formula! I'm quite sure about the index placement thanks to the metric compatibility, $\nabla_\mu g_{\nu\rho}=0$ [Carroll, (1.32)], which implies that the metric $g_{\nu\rho}$ commutes with the covariant derivative $\nabla_\mu$. Now, we take the Killing vector $K^\mu$ and we expand the antisymmetrization in the previous equation: $$0=6\nabla_{[\mu}\nabla_\nu K_{\rho]}=\nabla_\mu\nabla_\nu K_\rho + \nabla_\nu \nabla_\rho K_\mu +\nabla_\rho\nabla_\mu K_\nu- \nabla_\nu\nabla_\mu K_\rho-\nabla_\mu\nabla_\rho K_\nu-\nabla_\rho\nabla_\mu K_\nu.$$ Now we use the Killing's equation [Carroll, (3.174)], $\nabla_{(\nu}K_{\rho)}=0$ or $\nabla_\nu K_\rho =-\nabla_\rho K_\nu$, to simplify the previous equation: $$\nabla_\mu\,\nabla_\nu K_\rho=-\nabla_\mu\,\nabla_\rho K_\nu\quad \Rightarrow\quad 0=2\nabla_\mu\nabla_\nu K_\rho+(\nabla_\nu\nabla_\rho-\nabla_\rho\nabla_\nu)K_\mu+\nabla_\rho\nabla_\mu K_\nu-\nabla_\nu\nabla_\mu K_\rho.$$ Again, using Killing's equations $\nabla_\rho\nabla_\mu K_\nu=-\nabla_\rho\nabla_\nu K_\mu$ and $\nabla_\nu\nabla_\mu K_\rho=-\nabla_\nu\nabla_\rho K_\mu$ we get: $$0=2\big(\nabla_\mu\nabla_\nu K_\rho+(\nabla_\nu\nabla_\rho-\nabla_\rho\nabla_\nu)K_\mu\big)=2\big(\nabla_\mu\nabla_\nu K_\rho+R_{\mu\alpha\nu\rho}K^\alpha\big)$$ $$\nabla_\mu\nabla_\nu K_\rho=-R_{\mu\alpha\nu\rho}K^\alpha=R_{\rho\nu\mu\alpha}K^\alpha.$$ Then, we can rise the index $\rho$ multiplying with the metric $g^{\rho\sigma}$ (and using the metric compatibility): $$\nabla_\mu\nabla_\nu K^\sigma=R^\sigma_{\phantom{\sigma}\nu\mu\alpha}K^\alpha.$$ That's all folks! I hope it would be helpful (and correct)!

• I've left the field of math and physics and am no longer able to decipher these symbols. Sorry about your first answer. But someone else who has the same problem may one day find you answer helpful. Sep 10, 2013 at 2:18
• anyone have any tips on how the second equation comes into being? Where does $\nabla_{[\mu}\nabla_\nu K_{\rho]}$ come from? May 21, 2021 at 20:28

A solution that is simpler still, requiring no differentiation or fancy identities other than one of the Bianchi identities, is as follows.

We know that the Riemann tensor can measure how much the covariant derivatives commute with each other, e.g $$$$[ \nabla _{\mu } ,\nabla _{\nu }] k_{\alpha } =-R^{\beta }{}_{\alpha \mu \nu } k_{\beta }$$$$ We can shuffle the indices in (4) and use the antisymmetry of the Lie bracket to get the three equations $$\begin{gather} [ \nabla _{\nu } ,\nabla _{\mu }] k_{\alpha } =R^{\beta }{}_{\alpha \mu \nu } k_{\beta } \tag{5A}\\ [ \nabla _{\mu } ,\nabla _{\alpha }] k_{\nu } =R^{\beta }{}_{\nu \alpha \mu } k_{\beta } \tag{5B}\\ [ \nabla _{\alpha } ,\nabla _{\nu }] k_{\mu } =R^{\beta }{}_{\mu \nu \alpha } k_{\beta } \end{gather}$$ We now look at $$\displaystyle ( 5\mathrm{C}) -( 5\mathrm{B}) -( 5\mathrm{A})$$: $$\begin{equation*} [ \nabla _{\alpha } ,\nabla _{\nu }] k_{\mu } -[ \nabla _{\mu } ,\nabla _{\alpha }] k_{\nu } -[ \nabla _{\nu } ,\nabla _{\mu }] k_{\alpha } =\left( R^{\beta }{}_{\mu \nu \alpha } -R^{\beta }{}_{\nu \alpha \mu } -R^{\beta }{}_{\alpha \mu \nu }\right) k_{\beta } \end{equation*}$$ Applying the first Bianchi identity, $$\begin{equation*} R^{\beta }{}_{\alpha \mu \nu } +R^{\beta }{}_{\mu \nu \alpha } +R^{\beta }{}_{\nu \alpha \mu } =0 \end{equation*}$$ This simplifies to $$\begin{equation*} [ \nabla _{\alpha } ,\nabla _{\nu }] k_{\mu } -[ \nabla _{\mu } ,\nabla _{\alpha }] k_{\nu } -[ \nabla _{\nu } ,\nabla _{\mu }] k_{\alpha } =2R^{\beta }{}_{\mu \nu \alpha } k_{\beta } \end{equation*}$$ Expanding the Lie brackets, $$\begin{gather*} [ \nabla _{\alpha } ,\nabla _{\nu }] k_{\mu } -[ \nabla _{\mu } ,\nabla _{\alpha }] k_{\nu } -[ \nabla _{\nu } ,\nabla _{\mu }] k_{\alpha }\\ =( \nabla _{\alpha } \nabla _{\nu } -\nabla _{\nu } \nabla _{\alpha }) k_{\mu } -( \nabla _{\mu } \nabla _{\alpha } -\nabla _{\alpha } \nabla _{\mu }) k_{\nu } -( \nabla _{\nu } \nabla _{\mu } -\nabla _{\mu } \nabla _{\nu }) k_{\alpha }\\ =\nabla _{\alpha } \nabla _{\nu } k_{\mu } -\nabla _{\nu } \nabla _{\alpha } k_{\mu } +\nabla _{\alpha } \nabla _{\mu } k_{\nu } -\nabla _{\mu } \nabla _{\alpha } k_{\nu } +\nabla _{\mu } \nabla _{\nu } k_{\alpha } -\nabla _{\nu } \nabla _{\mu } k_{\alpha } \end{gather*}$$ Using the linearity of $$\displaystyle \nabla$$, $$\begin{gather*} [ \nabla _{\alpha } ,\nabla _{\nu }] k_{\mu } -[ \nabla _{\mu } ,\nabla _{\alpha }] k_{\nu } -[ \nabla _{\nu } ,\nabla _{\mu }] k_{\alpha }\\ =\nabla _{\alpha }( \nabla _{\nu } k_{\mu } +\nabla _{\mu } k_{\nu }) +\nabla _{\mu }( \nabla _{\nu } k_{\alpha } -\nabla _{\alpha } k_{\nu }) -\nabla _{\nu }( \nabla _{\alpha } k_{\mu } +\nabla _{\mu } k_{\alpha }) \end{gather*}$$ Using Killing's equation $$\displaystyle \nabla _{( \rho } k_{\sigma )} =0$$ the first and third terms vanish, leaving $$\begin{equation*} [ \nabla _{\alpha } ,\nabla _{\nu }] k_{\mu } -[ \nabla _{\mu } ,\nabla _{\alpha }] k_{\nu } -[ \nabla _{\nu } ,\nabla _{\mu }] k_{\alpha } =\nabla _{\mu }( \nabla _{\nu } k_{\alpha } -\nabla _{\alpha } k_{\nu }) =2\nabla _{\mu } \nabla _{\nu } k_{\alpha } \end{equation*}$$ Hence, $$$$\boxed{\nabla _{\mu } \nabla _{\nu } k_{\alpha } =R^{\beta }{}_{\mu \nu \alpha } k_{\beta }}$$$$

We could try to solve it the other way that if a vector obeys the first condition then it must be a Killing vector.

We assume, $$\nabla_{\mu} \nabla_{\sigma} K^{\rho} = R_{\sigma \mu \nu}^{\rho} k^{\nu}-\text { (i) }$$ $$where [\nabla_{\mu}, \nabla_{\sigma}] V^{\rho} = R_{\sigma,\mu,\nu}^{\rho} V^{\sigma}$$ Subtracting $$\nabla_{\sigma} \nabla_{\mu} K^{\rho}$$ from (i) $$\begin{array}{l} {\left[\nabla_{\mu}, \nabla_{\sigma}\right] K^{\rho} = R_{\sigma, \mu \nu}^{\rho} K^{\nu} - \nabla_{\sigma} \nabla_{\mu} K^{\nu}} \\ \nabla_{\sigma} \nabla_{\mu} K^{\rho}=\left(R_{\sigma, \mu \nu}^{\rho}-R_{\nu, \mu \sigma}^{\rho}\right) K^{\nu} \end{array}$$ $$\begin{array}{r} \nabla_{\sigma} \nabla_{\mu} K_{\rho}=\left(R_{\rho} _{\sigma}_{\mu \nu}- R_{\mu \sigma \rho \nu}\right) K^{\nu} \\ \left(\because R_{a b c d}=R_{c d a b}\right) \end{array}$$ using (1), $$\nabla_{\sigma} \nabla_{\mu} K_{\rho}=\nabla_{\mu} \nabla_{\sigma} K_{\rho}-\nabla_{\rho} \nabla_{\sigma} K_{\mu}$$ Add $$\nabla_{\sigma} \nabla_{\rho} K_{\mu}$$

$$\nabla_{\sigma}\left(\nabla_{\mu} K_{\rho}+\nabla_{\rho} K_{\mu}\right)=\nabla_{\mu} \nabla_{\sigma} K_{\rho}+\left[\nabla_{\sigma}, \nabla_{\rho}\right] K_{\mu}$$

$$=R_{\rho \sigma \mu \alpha} K^{\alpha}+ R_{\mu \alpha \sigma \rho} V^{\alpha}$$ $$=R_{\rho \sigma \mu \alpha} V^{\alpha}+R_{\sigma \rho \mu \alpha} V^{\alpha}$$ $$=R_{\rho \sigma \mu \alpha} V^{\alpha}-R_{\rho \sigma \mu \alpha} V^{\alpha}=0$$ $$\because \left( R_{a b c d}=R_{c d a b}\right) and$$ $$\left( R_{a b c d}=-R_{b a c d}\right)$$

$$\therefore \nabla_{\sigma}\left(\nabla_{\mu} K_{\rho}+\nabla_{\rho} K_{\mu}\right)=0$$ $$\because x^{\sigma}$$ is an arbitrary direction, $$\nabla_{\mu} K_{\rho}+\nabla_{\rho} K_{\mu}=0$$

$$\therefore K$$ is a Killing vector.