# Some question about polynomial representations of $GL(V)$

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 $r = \dim V$?

• Proposition 1 in sciencedirect.com/science/article/pii/S0021869307004619 maybe helps you. – emiliocba May 14 '14 at 16:25
• This is just about the difference between "rational" and "polynomial". Let's consider $\text{GL}(1)$. The irreducible rational representations of $\text{GL}(1)=\mathbb{C}^\times$ (or $\mathbb{G}_m$) are indexed by integers $k\in \mathbb{Z}$, with $k\in \mathbb{Z}$ corresponding to the 1-dimensional representation on which $z\in \mathbb{C}^\times$ acts by $z^k$. This is a rational function of $z$, matching the terminology. But $z^k$ is only a polynomial when $k\geq 0$, so the irreducible polynomial representations are indexed by $k\geq 0\in \mathbb{Z}$. – Tom Church May 14 '14 at 16:46

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.
• But I seem to remember some statement along the lines that any rational representation of $GL(n,\mathbb{C})$ is a product of the determinant to some nonpositive power and a polynomial representation, so there is only a slight distinction between these two classes of representations. – Sam Hopkins May 14 '14 at 16:15
• Yes, but when you tensor with determinant, you have to add 1 to all of the entries of $\lambda$. – Steven Sam May 14 '14 at 16:15
• While this is a justification, it doesn't explain why people would do it. The answer is that every polynomial irrep of $GL(r)$ extends to a functor ${\bf Vec}\to {\bf Vec}$ (where $GL(r)$ is considered as a one-object subcategory of $\bf Vec$). – Allen Knutson May 14 '14 at 22:32