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I have a particular problem, the following.

$T^{a_1 \dots a_p;b_1 \dots b_p}$ is a tensor with the following symmetries.

1) $a_i$'s and $b_i$'s are completely antisymmetric, ie restricted to either$a_i's $ or $ b_i's$

2) Total interchange of $a_i$'s with $b_i$'s are symmetric, that is $T^{a_1 \dots a_p;b_1 \dots b_p} = T^{b_1 \dots b_p;a_1 \dots a_p}$

Is this an irreducible representation of the full $S_{2p}$ group? if not how to decompose it? And what are/is the associated young tableau.

In general I would be grateful if anyone can provide me with a reference on how to go about finding the associated tableau given arbitrary symmetries, such as cyclic symmetry etc..

I am guessing the usual procedure of the Littlewood-Richardson rule can be constrained further when additional symmetries are present?

Thanks for you help.

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$\textbf {Question 1}$

An arbitrary Tensor $T$ is decomposed into irrep's, ie, it is projected by a young operator. I call one of the projectors, $P_1$. So the tensor corresponding to this particular irrep is given by $T_{irrep_1} = P_1.T$. Now I have some additional information about T, that is a symmetrisation of certain indices vanish. I take this symmetrisation operator to be $\mathbf S$. So the additional information I have is

$\mathbf S.T =0$.

Q1) Does this in general imply $\mathbf S.T_{irrep_1} =0$. For this to be true the two operators should commute, $[P_1, S]=0$. Is this true in general? If this is true, $S.T = 0 \implies S.T_{irrep_1} =0$.

Let me make my question clear by taking an example. HERE EVERYTHING IS DONE IN SYMMETRIC BASIS. Consider $T^{ab,cd}$ symmetric in the comma separated indices. My question is that How do I go about decomposing this one with the additional constraint that $(abc)=0$. So I think of T as the product of two tensors with the symmetry {(ab)X(cd)} and decompose this. That would give me i) {ab,cd} ii){abc,d} iii){abd,c} iv) {abcd} (HERE COMMA SEPARATES THE ROWS). At this point I want to apply the additional constraint $(abc) =0$. and this where my confusion lies. I DO understand that these components do not mix and I can treat them sepearately.

Q) If I assume the additional symmetry is inherited by the each component I will just apply this to i,ii,iii, iv and I indeed see only the component {ab,cd} satisfies. So it somehow works? Does that mean the two projectors always commute? Or It works by someother reason?

Another way to see {ab,cd} is the only one is to decompose using two different ways (symmetric basis)

1) Decomposing ${(ab) X (cd)}$

2) Decomposing ${ab,c X d}$ the right component/s should be whatever that is the intersection of these two operations and I see only the {ab,cd} one is shared by 1) 2). This method- I see why it works. but the previous method is puzzling for me.

Question 2

Given $T^{a[p],b[p]} $computing $1^p \otimes 1^p$ gives ${a[p]b[i],b[p-i]}$ with the young condition this tensor corresponds to one of the irreducible representation. I two questions regarding this,

2.1) For the same reason I was confused in question 1, Is saying the tensor T, satisfies condition two, ie, total pairwise interchange of a's and b's is symmetric, imply, the pairwise interchange of an irreducible component symmetric.

2.2) You said that anti-symmetrising over all b's in ${a[p]b[i],b[p-i]}$ yields a tensor satisfying condition one and lies in the same invariant sub-space. I AGREE. but is this the only tensor in this invariant sub-space that satisfies condition one? If I take for example $a[p-1]b[i+1],ab[p-i-1]$ this tensor lies in the invariant subspace. If I antisymmetrise over all b's do I get another independent tensor that satisfied condition 1?

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    $\begingroup$ I do not know the recipe for all possible cases, but for your particular one... First compute $1^p\otimes 1^p$ to find Young diagrams made of two columns with the first of height $(p+i)$ and the second $p-i$, these all obey 1). Now we try to impose the additional constraint 2) of being symmetric under the exchange of all a's with b's. You can use the Young condition that the anti-symmetrization of all indices corresponding to the first column to one index from the second column vanishes to exchange the two groups. The exchange produces a sign $(-)^i$, so only for even $i$ your 2) is met. $\endgroup$ – John Mar 20 '14 at 19:07
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    $\begingroup$ As an example you can try the totally symmetric rank-$2p$ tensor and see that it obeys 2) when p is even. Another check is for the tensor with the symmetry of $[p,p]$ diagram, i.e. $i=0$, it obeys 2), as it is easy to see. Often such subproblems appear in the study and can be solved just from Young conditions and knowledge of tensor products. $\endgroup$ – John Mar 20 '14 at 19:08
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    $\begingroup$ I used the identity $T^{a[k]b[k-i],b[i]}=(-)^iT^{b[k]a[k-i],a[i]}$ with the obvious meaning of condensed notation $a[k]=a_1...a_k$. In principle one can use the symmetric base, where everything is split into pairs of symmetric indices, but this is just less convenient. There is no inconsistency, any computation can be done in any base. $\endgroup$ – John Mar 21 '14 at 17:47
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    $\begingroup$ yes, $a[k]$ is a group of $k$ anti-symmetric indices. I derived the identity step by step. First, consider the Young condition: $T^{[a[k]b[i],b] b[k-i-1]}\equiv0$ where the enclosing brackets mean antisymmetrization over $k+i+1$ indices. Let us write this as $T^{a[k][b[i],b]b[k-i-1]}+(-)^{(i+1)k}T^{b[i+1][a[k-1],a]b[k-i-1]}\equiv0$. From this we can move one $b$ to the first group of indices and $a$ to the second. Repeating this and being careful about signs, you can derive the identity above. You may look at Appendix A,E of arxiv.org/pdf/1401.2975.pdf but this is mostly about symmetric $\endgroup$ – John Mar 22 '14 at 20:08
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    $\begingroup$ 2) can be seen to be satisfied for $i$ even from $T^{a[k]b[i],b[k-i]}=(-)^iT^{b[k]a[i],a[k-i]}$, which can be derived step by step from Young condition. Young condition for a tensor of $[k+i,k-i]$ states that anti-symmetrization of first $(k+i)$ indices with at least one more is zero. This was I showed how to use - one can exchange the two groups of indices eventually and see what is the sign cost, if any. Comma indeed separates the two groups of indices $\endgroup$ – John Mar 23 '14 at 7:37

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