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In chapter 13 of the textbook of Group Theory in Physics by Wu-Ki Tung, Lemma 2 discusses the equivalence of two irreducible representations of GL(m) on ${T^i}_j$. In its proof, it simply mentioned (without deeper arguments as if it is quite obvious) that the contraction two tensors is zero if their indices belong to different symmetry types. I have thought it over for some time but really cannot figure out a proof of this statement.

Let me try to state the question more clearly below. Consider a tensor $T^a$ where $a$ refers to a list of contravariant indices, like $T^{1234}$. Now one may symmetrize this tensor according to a given Young Tableaux, namely $\tau$, and the resulting tensor is denoted by

$$T^{\tau(a)}.$$

It is noted that, according to theorems on representations of symmetric group $S_n$, $\tau$ corresponds to a irreducible representation of $S_n$. Now one considers a covariant tensor ${T'_b}$ and a different Young Tableaux $\lambda$, where $b$ contains the same number of indices, $\tau$ and $\lambda$ are not equivalent (one may assume $\tau$ > $\lambda$ without loss of generality). It is stated that the contraction

$$T^{\tau(a)}T'_{\lambda(a)}=0.$$

It is obviously true when $\tau$ corresponds to the symmetrizer and $\lambda$ is the anti-symmetrizer. But I cannot think of the proof for a general case.

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  • $\begingroup$ What I was trying was to build a proof following similar arguments one might use for the simple case of symmetrizer and anti-symmetrizer, in this case exchange any two index in the contraction will do the job. I found out (according to Lemma IV.3 of the same book) when $\tau$ and $\lambda$ are inequivalent, there must be two numbers in the same row of Young Tableaux $\tau$ also appear in a same column of $\lambda$. And I thought that these two numbers can be considered as the two indices to be exchanged to obtain the proof except I could not. $\endgroup$
    – gamebm
    May 15, 2015 at 21:48
  • $\begingroup$ One need to multiply a transposition (of these two numbers) before $\tau$, while multiply the same transposition after $\lambda$ to get the extra -1 from $\lambda$. But exchanging two indices in the contraction, one need to multiply the transposition before both tensors. So I got nowhere... $\endgroup$
    – gamebm
    May 15, 2015 at 21:50
  • $\begingroup$ Following this argument, the best I can get is when at least one of the Young Tableaux (between $\tau$ and $\lambda$) is a symmetrizer or anti-symmetrizer, the contraction gives zero. $\endgroup$
    – gamebm
    May 16, 2015 at 19:00

2 Answers 2

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If $\sigma$ is a permutation, then $T^{\sigma(a)}{T'}_{a}=T^{a}{T'}_{\sigma^{-1}(a)}$. If a permutation is part of a symmetrizer for a Young tableaux, then so is its inverse, so $T^{\tau(a)}{T'}_{a}=T^{a}{T'}_{\tau(a)}$. So $T^{\tau(a)}{T'}_{\lambda(a)}=T^{a}{T'}_{\tau(\lambda(a))}=0$.

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  • $\begingroup$ Thanks a lot for the proof! If $\sigma$ and $\sigma^{-1}$ are both part of symmetrizer for a Young tableaux $\tau$, $\sigma\tau=\sigma^{-1}\tau=\tau$, then why $T^{\tau(a)}T'_a=T^aT'_{\tau(a)}$? Just by looking at the expression, all I can think is that symmetrizer does not have inverse in general(?), why can one exchange the summed indices in $T$ and $T'$? $\endgroup$
    – gamebm
    May 16, 2015 at 16:08
  • $\begingroup$ I got it. You mean $(T^{e(a)}+T^{\sigma_1(a)}+T^{\sigma_2(a)}+\cdots)T'_a=T^a(T'_{e(a)}+T'_{{\sigma_1}^{-1}(a)}+T'_{{\sigma_2}^{-1}(a)}+\cdots)$ and $e+\sigma_1+\sigma_2+\cdots=\lambda=e+\sigma_1^{-1}+\sigma_2^{-1}+\cdots$, where $\lambda \equiv e_\lambda$. $\endgroup$
    – gamebm
    May 16, 2015 at 18:22
  • $\begingroup$ Since for an idempotent corresponding to a general Young Tableaux, the inverse of a permutation is not necessarily a part of the sum. One may get the following $\lambda \equiv e_\lambda=s_\lambda a_\lambda=(e+\sigma_1+\sigma_2+\cdots)(1+(-1)^{\nu_{\omega_1}}\omega_1+(-1)^{\nu_{\omega_2}}\omega_2+\cdots)=(e+\sigma_1^{-1}+\sigma_2^{-1}+\cdots)(1+(-1)^{\nu_{\omega_1}}\omega_1^{-1}+(-1)^{\nu_{\omega_2}}\omega_2^{-1}+\cdots)$. $\endgroup$
    – gamebm
    May 16, 2015 at 20:08
  • $\begingroup$ But it seems that we do not have $T^{\sigma\omega(a)}T'_a=T^aT'_{\sigma^{-1}\omega^{-1}(a)}$ (?) $\endgroup$
    – gamebm
    May 16, 2015 at 20:09
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I figured out the following proof. Considering the following primitive idempotent

$$\tilde{e}_\lambda \equiv a_\lambda s_\lambda,$$

one can show that it is equivalent to $e_\lambda$, since $\tilde{e}_\lambda e_\lambda = a_\lambda s_\lambda s_\lambda a_\lambda = \eta a_\lambda e_\lambda \ne 0$. Therefore the left ideal generated by $e_\tau$ can be rewritten in terms of $\tilde{e}_\tau$ as $e_\tau(a)=\sum_p \zeta_p \tilde{e}_\tau^p(a)$. Now, one has assumes that $\tau > \lambda$, so there are two numbers, namely $i$ and $j$, locate on the same line of the Young Tableaxu $\Theta_\tau$ and simultaneously on the same column of $\Theta_\lambda$. If not, following the argument used in the Appendix IV of the textbook, the two Young Tableaux must be the same, which leads to a contradiction. Therefore, if one multiplies the transportation $(ij)$ from the left of $e_\tau$ gives no effect: $(ij)e_\tau =e_\tau$, while an extra $-1$ is obtained in the case of $\tilde{e}_\lambda$: $(ij)\tilde{e}_\lambda^p= - \tilde{e}_\lambda^p$.

So one can apply $(ij)$ to both the contravariant indices and covariant indices of the contraction

$$T^{\tau(a)}T'_{\lambda(a)}=T^{(ij)\tau(a)}T'_{(ij)\lambda(a)},$$

and the above arguments give $$T^{(ij)\tau(a)}T'_{(ij)\lambda(a)}=-T^{(ij)e_\tau (a)}T'_{(ij)\sum_p \zeta_p \tilde{e}_\lambda^p (a)}=-T^{\tau(a)}T'_{\lambda(a)},$$

and therefore

$$T^{\tau(a)}T'_{\lambda(a)}=0.$$

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