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Let me set up the notation I am using. $(abc,de)$ denotes the standard Young tableau where the first row is $abc$ and the second row is $de$. Each young tableau corresponds to the young symmetriser, and I use the convention that given the Young tableau $\lambda$ young symmetriser is given by,

$P_{\lambda} = Na.b$ where $a = \sum_{\sigma \in Rowgroup} \sigma$, and $b = \sum_{\sigma \in Column-group} (sgn \,\sigma) \sigma$. N stands for normaliser such that $P_{\lambda}$ is idempotent ($P_{\lambda}^2 = P_{\lambda}$).

The action of the symmetric group on an abstract tensor is by means of shuffling the indices. I know that the following identity is true.

$\mathbb{1} = \sum_iP_{\lambda_i}$ Here $\lambda_i$ are all the Standard young tableau of a given number of boxes. Applying this identity to any abstract tensor should give the same tensor back. And if this abstract tensor is completely arbitrary, i.e does not have any symmetries in its indices this decomposition yields components living in each irreps. OK. My question is when we have tensor products, then we use Littlewood-Richardson rule to find a one of the basis tensors representing each irreducible space. However I am interested in not just treating the tensor product as a vector space and finding the disjoint union of irreducible vector spaces, that each young tableau corresponds to and acting with the Young symmetriser yields one of the basis tensors.

I would like to find out the given tensor in each irreducible space. Let me give an example. Suppose I have a rank 3 tensor that is antisymmetric in the first 2 indices. I can treat this as a tensor product of $(a,b) \otimes c$(Here ab is a column as per my notation.) Applying LRH I get, $(ac,b),(a,b,c)$. Using the young symmetriser I defined above, I act with this normalised young symmetriser on the tensor $T^{ab|c}$ (here the bar is used to group anti-symmetric indices). I get,

$T_1^{abc}= P_{ac,b} T^{ab|c} = \frac{2}{3}(T^{ab|c} + T^{cb|a})$ (1)

$T_2^{abc}=P_{a,b,c} T^{ab|c} = \frac{1}{3}(T^{ab|c} + T^{bc|a} + T^{ca|b})$ (2)

Now $T_1 + T_2 \neq T$ as is clear. However when I project each component using the young symmetrisers associated with the factors of the tensor product, in this case it is $P_{a,b}$ I get T back. That is

$T = P_{a,b}(T_1 + T_2) $ Explicitly, $P_{a,b} T_1^{abc} = \frac{2}{3} T^{ab|c} + \frac{1}{3}(T^{cb|a} - T^{ca|b})$,

Therefore, $P_{a,b}(T_1^{abc} + T_2^{abc}) = \frac{2}{3} T^{ab|c} + \frac{1}{3}(T^{cb|a} - T^{ca|b})+ \frac{1}{3}(T^{ab|c} + T^{bc|a} + T^{ca|b}) = T^{ab|c}$.

Is this relationship true in general i,e suppose I have a tensor product which I write as $T^{\lambda_1|\lambda_2}$ where $\lambda_1,\lambda_2$ denote the factors belonging to irreps corresponding to the Standard tableaux. Then is the following relationship TRUE, and how do I prove this?

$T^{\lambda_1|\lambda_2} = P_{\lambda_1}P_{\lambda_2}\times$(irreducible components we get by using young symmetrisers generated by LRR and acting on $T^{\lambda_1|\lambda_2}$). In short I want to know if the component of this tensor product living in the irreducible rep characterised by the tableau $\nu$ generated by LRR is,

$\mathbf{P_{\lambda_1}P_{\lambda_2} T_{\nu}}?$

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    $\begingroup$ Crossposted to math.stackexchange.com/q/874778/11127 Dear user40469. In general, it is frown upon to cross-post simultaneously, because it may waste potential answerer's time. As a minimum OP should mention the cross-post (on both sites!). The preferred procedure is to not cross-post, and if the post hasn't received an acceptable answer after, say, a couple of days, then OP could flag for migration. $\endgroup$ – Qmechanic Jul 22 '14 at 13:52
  • $\begingroup$ I am sorry I was not aware of this practice. This question straddles between pure math and physics.. $\endgroup$ – user40469 Jul 22 '14 at 14:49
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    $\begingroup$ First comment, the phenomenon you observed exists even for a trivial tensor product when you have tensor product of an irreducible tensor $T^A$ given in anti-symmetric basis, as in your example, and a trivial scalar tensor. Then if you apply Young symmetrizer in the form (SA) you will change the basis from antisymmetric to symmetric. Applying in another order (AS) does nothing. $\endgroup$ – John Jul 22 '14 at 19:00
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    $\begingroup$ So, the relatioship is true in general, but of course (SA)T will never be equal (AS)T for some tensor T, which is not a scalar, so please compare, expand or do anything else always in the same basis, symmetric or anti-symmetric, or some else basis. $\endgroup$ – John Jul 22 '14 at 20:09
  • $\begingroup$ Thanks @John I am afraid you may have misunderstood what I was asking. Say I have a tensor $1^p \otimes 1^p $ in your notation $a[p]|b[p]$ one of the basis tensors corresponding to the irreps is $a[p]b[k]|b[p-k]$ another one would be $a[p-1]b[p+1]|ab[p-k-1]$ If i antisymmetrise both of these tensors in a and b I see that they are proportional. But how do I show this in general? Would you some idea as to how I can prove this result? $\endgroup$ – user40469 Jul 22 '14 at 21:26

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