I'm sure this is something silly but I am trying to understand the following paper <http://www.sciencedirect.com/science/article/pii/S0001870807001636#> and something is not clear to me. The paper starts with "irreducible polynomial representations of $GL(r)$ are indexed by sequences $\lambda=(\lambda\_1\geq\ldots\geq\lambda\_r\geq 0)\in\mathbb{Z}^r$". I don't understand why we can't have negative entries for the highest weight? I mean, if $V$ is the natural representation of $GL(V)$ isn't $V^\*$ also an irreducible polynomial representation? Am I missing something here, perhaps some correspondence between $r$-tuples where $r = \dim V$?



Some question about polynomial representations of $GL(V)$
If you remove the assumption that $\lambda\_r$ is nonnegative, then you are indexing all rational representations of $GL(V)$, so the main point is that the author is focusing on *polynomial* representations, i.e., those whose matrix entries can be defined in terms of polynomials.


$V^\*$ is not polynomial because $GL(V)$ acts via the inverse. You need to use rational functions to get a formula for the inverse of a general matrix in terms of its coefficients.


EDIT. I'm going to add some more details about this example in response to Sam Hopkins's comment. In this partition convention, the highest weight for $V^\*$ is $(0,0,\dots, 0,-1)$. If you take $\det (V) \otimes V^\*$ (here $\det(V)$ is the 1-dimensional representation of $GL(V)$ given by determinant), then this becomes a polynomial representation with highest weight $(1,\dots,1,0)$ (in fact this is just the $(r-1)$st exterior power representation. Another way of thinking about that is that I only need to divide by one power of the determinant to compute the inverse of a matrix.


