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Nov 8, 2010 at 4:13 comment added Ben Webster Darij- I think you missed the point of his question. He wants to know the ring, not just the vector space.
Nov 8, 2010 at 3:17 answer added Andy B timeline score: 2
Nov 6, 2010 at 23:16 comment added darij grinberg 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.
Nov 6, 2010 at 23:14 comment added darij grinberg 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.)
Nov 6, 2010 at 16:49 history edited Bruce Westbury
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Nov 5, 2010 at 14:36 history asked Bruce Westbury CC BY-SA 2.5