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Suppose we have a partition $\mu$ of $n$. There is an associated polynomial irreducible representation $\phi_{\mu}$ of $GL_n(\mathbb C)$.

How do I obtain a new representation of $GL_n(\mathbb C)$ from the dual representation $\phi_{\mu}^{*}$? What is the relation between $\mu$ and the partition associated to this new representation?

I tried to think about Young tableaux, but how can I find an isomorphism between a dual space representation and something to which I can associate a partition?

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    No, it is not the transpose. See below.2012-01-11

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Partitions index the irreducible POLYNOMIAL representations of $GL(V)$. But the dual of a polynomial representation is not going to be polynomial. Recall that, if a representation $W$ is given by the map $\rho: GL_n \to GL_N$, then $W^{\ast}$ is given by $(\rho^T)^{-1}$. The presence of that inverse means that it is unlikely to be a polynomial map.

We can broaden the class of polynomal representation to that of algebraic representations, meaning that the map $\rho$ must be made of rational functions whose denominators are powers of the determinant. Irreducible rational representations of $GL_n$ are indexed by $n$-tuples of integers $(\lambda_1, \lambda_2, \ldots, \lambda_n)$ with $\lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_n$; the polynomial representations correspond to the subset where $\lambda_1 \geq 0$. Then we can state that taking the dual replaces $(\lambda_1, \lambda_2, \ldots, \lambda_n)$ by $(-\lambda_n, -\lambda_{n-1}, \cdots, - \lambda_1)$.

Since I don't know how you are most comfortable thinking of the classification of representations of $GL_n$, I won't try to justify this, but it is pretty easy from most perspectives.

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    Thank you! So I see it is possible to take the product of the dual with some power of the determinant and get back a polynomial representation. What power would be enough to, in some sense, clear the denominator? To this polynomial representation, what partition can I associate?2012-01-11