# Smooth unitary irreducible finite-dimensional representations of U(n)

I want to make sure I completely understand the isomorphism classes of smooth unitary irreducible finite-dimensional representations of $U(n)$. We have the irreducible defining representation $R$, and we can apply any Young diagram to this, hitting it with a Schur functor, to get a load more irreps. We can also take the complex conjugate $R^*$ of the defining representation, and hit this with all the Young diagrams, to get another series of irreps. Some of these irreps will actually be the same, if the number of rows in our Young diagrams gets as large as $n$, but I'm not worried about that.

I understand how to tensor together representations that arise from applying Young diagrams to $R$, and decompose this into a direct sum of irreducible representations. But what if I tensor together an irrep coming from a Young diagram applied to $R$, and an irrep coming from a Young diagram applied to $R^\star$? How does this product decompose into a sum of irreps? I've convinced myself it just isn't possible when you're restricted to using the irreps that arise from applying Young diagrams to $R$ and $R^\star$ --- there must be some more of them. For example, consider $U(2)$ and the representation $R \otimes R^\star$, where $R$ is the defining representation. Since $R$ and $R^\star$ are dual, this must decompose as $R \otimes R^* \simeq 1 \oplus X$, where $X$ is a 3-dimensional self-dual possibly-reducible representation. But I don't think it's possible to build such an $X$ by taking direct sums of the irreps described in the first paragraph.

So, apart from applying Young diagrams and conjugate Young diagrams to the defining representation, how to you get the rest of the of the irreps? In particular, I don't think I know how to build any self-dual irreps, apart from the trivial irrep, but some have to exist by the argument in the previous paragraph.

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The irreps of $U(n)$ are indexed by decreasing $n$-tuples of integers $\lambda\_1 \geq \lambda\_2 \geq \cdots \geq \lambda\_n$. Note that I do NOT impose that $\lambda_n \geq 0$.

One way to see this is to note that the conjugacy classes of $U(n)$ are $(S^1)^n/S\_n$, where $S\_n$ acts by the obvious permutations. In other words, the conjugacy class of a unitary matrix is determined by its eigenvalues up to permutation. For $\theta \in \mathbb{R}^n$, write $[e^{i \theta}]$ for the conjugacy class with eigenvalues $e^{i \theta\_1}$, ..., $e^{i \theta\_n}$. Using the Fourier transform, a class function will look like $$f([e^{i \theta}]) = \sum_{\omega} a(\omega) e^{i \langle \omega, \theta \rangle}$$ function $a(\omega)$ on $\mathbb{Z}^n$ which unchanged by permuting the input. In the case of a character, this will be a finite sum, because in any representation of $U(n)$ the diagonal matrices will act diagonalizably. I'll write $x\_j$ for the character $x\_j(e^{i \theta}) = e^{i \theta\_j}$. So the character ring of $U(n)$ is the symmetric Laurent polynomials in the $x$'s, and you should believe that there is one irrep for each point in $\mathbb{Z}^n/S\_n$.

Now, how does this relate to the examples you raised? The defining representation has character $x\_1 + x\_2 + \cdots x\_n$. Applying any Schur functor to this will get you another polynomial in the $x$'s. So, just using Schur functors, you are only seeing the representations that correspond to polynomials; in other words, to the case that $\lambda_n \geq 0$. In other (more algebraic) contexts, these are called the polynomial representations, and I'm going to use that term here.

For a unitary matrix, the complex conjugate is the same as the inverse transpose; so taking the complex conjugate moves you to the dual representation. So, the complex conjugate of the dual representation has character $x\_1^{-1}+x\_2^{-1}+\cdots+x\_n^{-1}$. Taking Schur functors of this will give you polynomials in the variables $x\_i^{-1}$. On the $\lambda$'s, taking the dual sends $(\lambda\_1, \ldots, \lambda\_n)$ to $(-\lambda\_n, \ldots, -\lambda\_1)$.

You are right that there are representations you are not seeing, because there are symmetric Laurent polynomials which are neither polynomials in the original variables nor in the inverse variables. In terms of $\lambda$'s, it is possible to have an $n$-tuple of numbers, some of which are positive and some nonnegative. The simplest example is adjoint rep, which is $V\_{1,0,\ldots,0,-1} \oplus V\_{0,0,\ldots, 0}$, and has character $$(x\_1 + x\_2 + \cdots x\_n) (x\_1^{-1} + x\_2^{-1} + \cdots x\_n^{-1}).$$

Note that none of these representations are much different from the polynomial representations, though. By tensoring with a high enough power of the determinant, you can always get back into the polynomial representations. In other words, if you multiply a Laurent polynomial by a high enough power of $x\_1 x\_2 \cdots x\_n$, it becomes a polynomial.

This sounds complicated, but if you know the $SU(n)$ case well then you'll find this easier. In $SU(n)$, the torus is $T / S_n$ where $T$ is the subtorus of $(S^1)^n$ where the $n$-components add up to $0$. So you have to remember that the Schur polynomials $1$ and $x\_1 x\_2 \cdots x\_n$ represent the same representation. All of those little details go away in the $U(n)$ case!

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Thanks, I can tell that the theory is very elegant --- I should learn it! –  Jamie Vicary Nov 12 '09 at 16:20

I'd prefer to separate the problem into classifying representations of SU(n) and classifying representations of U(1), and then thinking about how U(n) is built from those two groups. This may not be "synthetic" enough for you, but it'll get the job done, and then you can figure out how to state the classification of U(n) reps in the manner you desire.

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Thanks, that is a very useful perspective! –  Jamie Vicary Nov 12 '09 at 16:19