0
$\begingroup$

Let $R$ be a finite-dimensional irreducible representation of $U(N)$, with the set of weights $W_R$. Each element of $W_R$ is a vector of length $N$ with integer entries.

Firstly, I would like to know if it is true that $$\sum_{\mu\in W_R}m_\mu\mu=A(R)(1,\ldots,1),$$ where $m_\mu$ is the multiplicity of the weight $\mu$ and $A(R)$ is some integer that varies with the representation. Note that we are considering $U(N)$ rather $SU(N)$, because I think the sum of weights of an irrep vanishes for $SU(N)$.

Secondly, I would like to know if there is a simple formula for $A(R)$, or if there is an existing name for it?

Improve this question
$\endgroup$
10
  • $\begingroup$ You are just taking a bare sum of weights, without any multiplicity (according to the dimension of the weight space)? $\endgroup$
    – LSpice
    Commented Jan 25, 2023 at 22:37
  • 2
    $\begingroup$ @LSpice Sorry I meant to include multiplicities; the question has been edited. $\endgroup$
    – Blind Miner
    Commented Jan 25, 2023 at 22:49
  • $\begingroup$ Since $m_\mu$ is constant on Weyl orbits (because we're dealing with a representation of $G$), $\sum m_\mu\mu$ is Weyl-fixed. The Weyl-fixed sublattice of the character lattice of $\operatorname{SU}(N)$ is trivial, which is why you always get $0$ there; and is spanned by the character $\operatorname{det} = (1, \dotsc, 1)$ you indicate for $\operatorname U(N)$, so, indeed, you always get your desired equality. Possibly $A(R)$ can be computed in terms of the highest weight by the Kostant multiplicity formula. $\endgroup$
    – LSpice
    Commented Jan 25, 2023 at 23:15
  • $\begingroup$ @LSpice Thanks for the reply! Do you know a simple proof (or reference) for why the weyl-fixed character lattice of $SU(N)$ is trivial? $\endgroup$
    – Blind Miner
    Commented Jan 26, 2023 at 0:08
  • 1
    $\begingroup$ @LSpice Thanks! now it's obvious... $\endgroup$
    – Blind Miner
    Commented Jan 26, 2023 at 0:32

1 Answer 1

Reset to default
2
$\begingroup$

As discussed in the comments, your sum is a Weyl-fixed character, so trivial for $G = \operatorname{SU}(N)$ and a multiple of $\det = (1, \dotsc, 1)$ for $\operatorname U(N)$.

To be concrete, as I guessed in the comments, one sees that your sum $\sum_\mu m_\mu\mu$ (the sum taking place in the character lattice $X^*(T)$ of the (implicitly chosen) maximal torus $T$, not in $\mathbb C[X^*(T)]$ as in the Weyl character formula) is precisely the character of $\det \circ R$. Again, since $\det \circ R$ is Weyl invariant and $\det$ spans the Weyl-invariant part of the character lattice, your integer $A(R)$ is precisely the integer $n$ such that we have $\det \circ R = \det(\cdot)^n$.

Inspired by your comment, I realise we can be a little more explicit. (On having written this, I realise that it was actually exactly what you were saying; I mistook your $\lambda_i$ for an indexing of weights, rather than a component of $\lambda$. Oops, sorry!)

Let $\lambda$ be the highest weight of $R$. Then all weights $\mu$ of $R$ agree on the centre of $\operatorname U(N)$ with $\lambda$. Specifically, they all act as $z I_N \mapsto z^\ell$ for some integer $\ell$. (If we think of $\lambda$ as an element of $\mathbb Z^N$, then $\ell$ is the sum of the components.) Thus, we have that $R$ agrees on the centre with $z I_N \mapsto z^\ell I_{\dim(R)}$, so $\det \circ R$ agrees on the centre with $z I_N \mapsto \det(z^\ell I_{\dim(R)}) = z^{\ell\dim(R)} = \det(z I_N)^{\ell\dim(R)/N}$, so $A(R)$ equals $\ell\dim(R)/N$. As you point out, we can use the Weyl dimension formula to compute $\dim(R)$ in terms of $\lambda$ if desired.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .