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The context for this question is Schur-Weyl duality. Let $V$ be a vector space. For $r>0$ consider $\otimes^rV$. This has commuting actions of the symmetric group $S(r)$ and $G=GL(V)$.

My question is prompted by reading $\S 51$ in "Compact Lie groups and their representations" by D. P. Zelobenko.

Choose a Borel subgroup $B\subset G$ and let $N\subset B$ be the unipotent radical. Then by the algebra of highest weight tensors I mean the algebra of $N$-invariant tensors in the tensor algebra of $V$. Is there a description of this algebra? say, generators, relations and a basis indexed by standard tableaux?

Since I expect this to be known (perhaps implicitly) a supplementary question is whether the $q$-analogue of this algebra is a deformation of the plactic monoid algebra? More precisely it should be possible to put $q=0$ to get the plactic monoid algebra.

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Hmm, wait a moment. You live over an algebraically closed field of characteristic zero? Then all Borel subgroups of $G=\mathrm{GL}\left(V\right)$ are conjugate to $U_n$, the group of upper triangular unipotent matrices (where $n=\dim V$ and I identify $V$ with $k^n$), right? Now there is a theorem that for every irreducible representation $P$ of $G$, the space $P^{U_n}$ is $1$-dimensional, i. e. there is (up to scalars) one and only one $U_n$-invariant vector in $P$. (This is, for example, in math.unibas.ch/~kraft/Papers/KP-Primer.pdf 5.7 Corollary 1.) –  darij grinberg Nov 6 '10 at 23:14
    
The decomposition of $\otimes V$ into the Schur functors of $V$ is well-known - at least, nonconstructively. It now probably remains to actually locate these Schur functors inside $\otimes V$, and find their $U_n$-invariants. –  darij grinberg Nov 6 '10 at 23:16
    
Darij- I think you missed the point of his question. He wants to know the ring, not just the vector space. –  Ben Webster Nov 8 '10 at 4:13
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1 Answer

The highest weight vectors in the tensor algebra are obtained by applying certain Young symmetrizers to tensor products of standard basis vectors of $V$. For example, a highest weight vector is obtained from $$ e_1 \otimes e_1 \otimes e_2 \otimes e_2 $$ by applying the Young symmetrizer of the tableau of shape $(2,2)$ filled with $1,2,3,4$ from left-to-right top-to-bottom. There is a basis for the space of highest weight vectors indexed by pairs of tableaux $(T,S)$ where $T$ and $S$ have the same shape, row $i$ of $T$ consists of the number $i$ and $S$ is standard. I think this is written up in Fulton's book on Young Tableaux.

The product of two highest weight vectors must be "straightened" to be expressed as a integer linear combination of highest weight vectors of this form. This straightening algorithm is well know (in complete generality by Grosshans--Rota--Stein, but this case might be due to De Concini and Procesi, perhaps {way} eariler. You might look in Brian Taylor's MIT thesis to see that such relations form a Groebner basis).

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