Suppose I define the action of the symmetric group on abstract tensors as shuffling indices. I know this is very naive. I apologise, I am a physicist and working on a problem that involves tensors with symmetries. I guess mathematicians use an equivalent notion of Schur functions to define rep's?.

So I have a tensor product of two irreducible factors, which I denote $T^{\lambda_1} \otimes T^{\lambda_2}$ where $\lambda_1, \lambda_2$ are standard Young tableaux and $T^{\lambda_i} \equiv C_{\lambda_i} X $. Here $X$ is some arbitrary abstract tensor with no symmetries and $C_{\lambda_i}$ are Young symmetrisers. Now under the action of permutation group by means of shuffling the indices $T^{\lambda_i}$ generates an irreducible subspace which can be defined as,

$V_{\lambda_i} = span [\sigma T^{\lambda_i}], \sigma \in S_n $ where $\sigma$ acts on $T^{\lambda_i}$ by means of shuffling indices. By using Littlewood-Richardson I can find irreducible spaces for the tensor product. My question is the following.

Let $\nu$ be a standard tableau generated by the Littlewood-richardson. I define $T^{\nu} = \ C_{\nu} (T^{\lambda_1} \otimes T^{\lambda_2})$. Where $C_\nu$ is the associated Young symmetriser and of course $T_{\nu}$ is one of the basis tensors of this irreducible sub space. We can use this to define the corresponding subspace to be,

$V_{\nu} = span[\sigma T^{\nu}], \sigma \in Symmetriec group$

Now I notice in my calculations the following, Considering the product of young symmetrisers of the individual tableax $\lambda_1, \lambda_2$, $C_{\lambda_1} . C_{\lambda_2}$ and their action on $V_{\nu}$,

$C_{\lambda_1} .C_{\lambda_2} :V_{\nu} \to V_{\nu}$. THE DIMENSION OF THE IMAGE OF THIS MAP IS ONE ??. Is this a result that is true in general. Can anyone please guide me to a reference where this is proven or help me with the proof? Sorry if this is too trivial, and thanks for your help.

  • $\begingroup$ With "generated by the Littlewood-richardson" do you mean that the LR coefficient $c_{\lambda_1\lambda_2}^\nu$ is nonzero, or else what do you mean? I am having trouble reading your question. Since the actions of Young symmetrizers are isotypical projectors, we should have $C_{\lambda_1}C_{\lambda_2}$ equal to $0$ if $\lambda_1\ne\lambda_2$ (so its image will always be $0$) and equal to a rational multiple of $C_{\lambda_1}$ if $\lambda_1=\lambda_2$, in which case it either acts by zero on $V_\nu$ hence image has dim=0 (if $\nu\ne\lambda_1$) else it acts by a scalar, so is an isomorphism. No? $\endgroup$ – blue Jul 23 '14 at 22:45
  • $\begingroup$ By generated I mean LR coefficient is non-zero. $C_{\lambda_1},C_{\lambda_2}$ are young projectors corresponding to tableaus with different labels. for example (ab,c) (de,f) first one acts on labels a,b,c and the next on labels d,e,f. Therefore they commute. $\endgroup$ – user40469 Jul 23 '14 at 23:05

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