A question about Killing vector and Riemann curvature tensor

Question

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)?

Answer 1

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.

Answer 2

@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)!

Answer 3

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

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We know that the Riemann tensor can measure how much the covariant derivatives commute with each other, e.g \begin{equation} [ \nabla _{\mu } ,\nabla _{\nu }] k_{\alpha } =-R^{\beta }{}_{\alpha \mu \nu } k_{\beta } \end{equation} 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, \begin{equation} \boxed{\nabla _{\mu } \nabla _{\nu } k_{\alpha } =R^{\beta }{}_{\mu \nu \alpha } k_{\beta }} \end{equation}

Answer 4

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.