Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $$\tau_{abcd}+\tau_{cbad}+\tau_{adcb}+\tau_{cdab}=0$$

share|cite|improve this question
+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
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 \[ 0 = \tau_{abcd} + \tau_{cbad} + \tau_{adcb} + \tau_{cdab}. \]

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.