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.