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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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I'd like to give you some advice about the site: To get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people are much more willing to help you if you show that you've tried the problem yourself. If this is homework, please add the [homework] tag; people will still help, so don't worry. Also, many would consider your post rude because it is a command ("Show that..."), not a request for help, so please consider rewriting it. – Zev Chonoles Jul 19 '12 at 12:48
+1 interesting, where does it come from? – draks ... Jul 19 '12 at 12:57
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. – Henning Makholm Jul 19 '12 at 13:00
Hint: What happens if you use sums of basis vectors for $\lambda$ and $\mu$? – celtschk Jul 19 '12 at 14:30

1 Answer 1

up vote 4 down vote accepted

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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