Timeline for Sum of weights of an irreducible representation of $U(N)$
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 26, 2023 at 18:12 | comment | added | LSpice | Correction: the character lattice of $\operatorname{SU}(N)$ is more naturally identified with the quotient of $\mathbb Z^N$ by $\mathbb Z(1, \dotsc, 1)$. (What I wrote is naturally identified with the cocharacter lattice.) Nonetheless, one still computes that $(\mathbb Z^N/\mathbb Z(1, \dotsc, 1))^{\mathfrak S_N}$ is trivial. (I also, there and elsewhere, switched $N$s and $n$s.) | |
| Jan 26, 2023 at 16:06 | comment | added | LSpice | I thought I used an idea from your comment to make my answer more explicit, but now I realise that I just wasn't understanding your comment properly, and what I added to my answer is exactly what you said. Sorry about that! | |
| Jan 26, 2023 at 14:56 | vote | accept | Blind Miner | ||
| Jan 26, 2023 at 14:27 | answer | added | LSpice | timeline score: 2 | |
| Jan 26, 2023 at 10:09 | comment | added | Blind Miner | @LSpice If we sum over the components of the formula, we get $N A(R)=\sum_{\mu\in W_R}m_\mu\sum_{i=1}^N\mu_i$, but now $\sum_i\mu_i=\sum_i\lambda_i$ where $\lambda$ is the highest weight, because all weights differ from each other by a linear combination of roots $\alpha$, and $\sum_i\alpha_i=0$. Therefore $\sum_{\mu\in W_R}m_\mu\sum_{i}\mu_i=(\sum_i\lambda_i)\sum_{\mu\in W_R}m_\mu=\text{dim}(R)\sum_i\lambda_i$, and we get $A(R)=\frac{\text{dim}(R)}{N}\sum_i\lambda_i$. I'm not sure if this formula can be further simplified. One could use the Weyl dimension formula to compute $\text{dim}(R)$. | |
| Jan 26, 2023 at 1:03 | comment | added | LSpice | Hmm, I wonder whether $A(R)$ is the integer $N$ such that $\det \circ R = \det(\cdot)^N$ …. | |
| Jan 26, 2023 at 0:32 | comment | added | Blind Miner | @LSpice Thanks! now it's obvious... | |
| Jan 26, 2023 at 0:24 | comment | added | LSpice | Re, it's a straightforward computation: the character lattice-with-$W$-action is naturally identified with the subspace of $\mathbb Z^n$-with-$\mathfrak S_n$ action on which the sum of coördinates is $0$. The $\mathfrak S_n$-fixed lattice in $\mathbb Z^n$ is spanned by $(1, \dotsc, 1)$, and only the trivial vector in that span has the sum of its coördinates $0$. | |
| Jan 26, 2023 at 0:08 | comment | added | Blind Miner | @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? | |
| Jan 25, 2023 at 23:15 | comment | added | LSpice | 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. | |
| Jan 25, 2023 at 22:49 | comment | added | Blind Miner | @LSpice Sorry I meant to include multiplicities; the question has been edited. | |
| Jan 25, 2023 at 22:48 | history | edited | Blind Miner | CC BY-SA 4.0 |
added 57 characters in body
|
| Jan 25, 2023 at 22:37 | comment | added | LSpice | You are just taking a bare sum of weights, without any multiplicity (according to the dimension of the weight space)? | |
| S Jan 25, 2023 at 17:45 | review | First questions | |||
| Jan 26, 2023 at 7:17 | |||||
| S Jan 25, 2023 at 17:45 | history | asked | Blind Miner | CC BY-SA 4.0 |