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Sorry for the initial mistake. $\tau\lambda^a\mu^b\lambda^c\mu^d=0$ should read $\tau_{abcd}\lambda^a\mu^b\lambda^c\mu^d=0$. However, my approach to this problem is to introduce vectors, $\alpha$ and $\beta$ such that, I could use them for expansion. i.e. $\tau_{abcd}\lambda^a(\alpha^b+\beta^b)\lambda^c(\alpha^d+\beta^d)$. How to get out from there is my problem.\

A type $(0,4)$ tensor $\tau_{abcd}$ satisfies $\tau \lambda^a\mu^b\lambda^c\mu^d=0$ for all contravariant vectors $\lambda^a$ and $\mu^b$. Show that its components satisfy $$\tau_{abcd}+\tau_{cbad}+\tau_{adcb}+\tau_{cdab}=0$$

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  • $\begingroup$ +1 interesting, where does it come from? $\endgroup$ – draks ... Jul 19 '12 at 12:57
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    $\begingroup$ Notationwise, it should be $\tau_{abcd}\lambda^a\mu^b\lambda^c\mu^d$, not $\tau\lambda^a\mu^b\lambda^c\mu^d$. And I can never remember which is which, but either the tensor or the vectors are contravariant. $\endgroup$ – hmakholm left over Monica Jul 19 '12 at 13:00
  • $\begingroup$ Hint: What happens if you use sums of basis vectors for $\lambda$ and $\mu$? $\endgroup$ – celtschk Jul 19 '12 at 14:30
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We have for every $\lambda$, $\xi$, $\mu$: \begin{align*} 0 &= \tau_{abcd}(\lambda^a + \xi^a)\mu^b(\lambda^c + \xi^c)\mu^d\\ &= \tau_{abcd}\lambda^a\mu^b\lambda^c\mu^d + \tau_{abcd}\xi^a\mu^b\lambda^c\mu^d + \tau_{abcd}\lambda^a\mu^b\xi^c\mu^d + \tau_{abcd}\xi^a\mu^b\xi^c\mu^d\\ &= \tau_{abcd}\xi^a\mu^b\lambda^c\mu^d + \tau_{abcd}\lambda^a\mu^b\xi^c\mu^d\\ &= (\tau_{abcd} + \tau_{cbad})\xi^a\mu^b\lambda^c\mu^d \end{align*} and hence, for every $\lambda$, $\xi$, $\mu$, $\nu$: \begin{align*} 0 &= (\tau_{abcd} + \tau_{cbad})\xi^a(\mu+\nu)^b\lambda^c(\mu+\nu)^d\\ &= (\tau_{abcd} + \tau_{cbad})\xi^a\mu^b\lambda^c\mu^d+ (\tau_{abcd} + \tau_{cbad})\xi^a\nu^b\lambda^c\mu^d + (\tau_{abcd} + \tau_{cbad})\xi^a\mu^b\lambda^c\nu^d + (\tau_{abcd} + \tau_{cbad})\xi^a\nu^b\lambda^c\nu^d\\ &= (\tau_{abcd} + \tau_{cbad})\xi^a\nu^b\lambda^c\mu^d + (\tau_{abcd} + \tau_{cbad})\xi^a\mu^b\lambda^c\nu^d\\ &= (\tau_{abcd} + \tau_{cbad} + \tau_{adcb} + \tau_{cdab})\xi^a\mu^b\lambda^c\nu^d \end{align*} As $\xi$, $\mu$, $\nu$ and $\lambda$ were arbitrary, is follows \[ 0 = \tau_{abcd} + \tau_{cbad} + \tau_{adcb} + \tau_{cdab}. \]

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