A question about irreducible representation of symmetric group (permutation group) in tensor space and tensor contraction 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.
 A: 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 $T^{\tau(a)}{T'}_{\lambda(a)}=T^{a}{T'}_{\tau(\lambda(a))}=0$.
A: 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
$$T^{\tau(a)}T'_{\lambda(a)}=0.$$
