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?



Sum of weights of an irreducible representation of $U(N)$
As discussed in the [comments](https://mathoverflow.net/questions/439350/sum-of-weights-of-an-irreducible-representation-of-un#comment1133244_439350), 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](https://mathoverflow.net/questions/439350/sum-of-weights-of-an-irreducible-representation-of-un#comment1133257_439350), 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](https://mathoverflow.net/questions/439350/sum-of-weights-of-an-irreducible-representation-of-un#comment1133302_439350), 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](https://mathoverflow.net/questions/439350/sum-of-weights-of-an-irreducible-representation-of-un#comment1133302_439350), we can use the [Weyl dimension formula](https://en.wikipedia.org/wiki/Weyl_character_formula#Weyl_dimension_formula) to compute $\dim(R)$ in terms of $\lambda$ if desired.


