Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
The dual representation is already a representation; what do you mean by obtaining a new representation from it? The corresponding partition is the transpose of $\mu$. –  Qiaochu Yuan Jan 10 '12 at 21:39
1  
No, it is not the transpose. See below. –  David Speyer Jan 11 '12 at 14:47

1 Answer 1

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.

share|improve this answer
    
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? –  Goran_ Jan 11 '12 at 19:56

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.