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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, since $\sum_\mu m_\mu$ equals $\dim(R)$, we have that $R$ agrees on the centre with $z I_N \mapsto z^{\ell\dim(R)}I_{\dim(R)}$, so $\det \circ R$ agrees on the centre with $z I_N \mapsto z^{N\ell\dim(R)} = \det(z I_N)^{\ell\dim(R)}$, so $A(R)$ equals $\ell\dim(R)$. As you point out, we can use the Weyl dimension formula to compute $\dim(R)$ in terms of $\lambda$ if desired.

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.

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. 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, since $\sum_\mu m_\mu$ equals $\dim(R)$, we have that $R$ agrees on the centre with $z I_N \mapsto z^{\ell\dim(R)}I_{\dim(R)}$, so $\det \circ R$ agrees on the centre with $z I_N \mapsto z^{N\ell\dim(R)} = \det(z I_N)^{\ell\dim(R)}$, so $A(R)$ equals $\ell\dim(R)$. As you point out, we can use the Weyl dimension formula to compute $\dim(R)$ in terms of $\lambda$ if desired.

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, since $\sum_\mu m_\mu$ equals $\dim(R)$, we have that $R$ agrees on the centre with $z I_N \mapsto z^{\ell\dim(R)}I_{\dim(R)}$, so $\det \circ R$ agrees on the centre with $z I_N \mapsto z^{N\ell\dim(R)} = \det(z I_N)^{\ell\dim(R)}$, so $A(R)$ equals $\ell\dim(R)$. As you point out, we can use the Weyl dimension formula to compute $\dim(R)$ in terms of $\lambda$ if desired.

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 $\sum m_\mu\mu$ (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 $\det \circ R = \det(\cdot)^N$.

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. 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, since $\sum_\mu m_\mu$ equals $\dim(R)$, we have that $R$ agrees on the centre with $z I_N \mapsto z^{\ell\dim(R)}I_{\dim(R)}$, so $\det \circ R$ agrees on the centre with $z I_N \mapsto z^{N\ell\dim(R)} = \det(z I_N)^{\ell\dim(R)}$, so $A(R)$ equals $\ell\dim(R)$. As you point out, we can use the Weyl dimension formula to compute $\dim(R)$ in terms of $\lambda$ if desired.