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I´m working myself right now through this article and have trouble understanding a part of the proof of Proposition 6.4, case II) on page 345-346. I try to give a short outline of the situation:

Suppose $V$ is an irreducible $\mathfrak{g}$ module, with highest weight $\lambda$. And $\mathfrak{g}$ being a semisimple Lie-algebra, with a subalgebra $\mathfrak{s} \simeq \mathfrak{sl}(2,\mathbb{C})$. Also let $V$ be a complex vector space [Edit] that as an $\mathfrak{s}$-module is the direct sum $V=V(1)\oplus V_0$, where $V(1)$ is the usual 2-dimensional (irreducible) representation of $\mathfrak{s}$, and $V_0$ is a direct sum of trivial 1-dimensional representations of $\mathfrak{s}$ [/Edit, JL]. Then the statement i want to verify goes as follows:

Then there exists a dominant root $\alpha$ for $\mathfrak{g}$, s.th. $(\lambda, \alpha^{\lor})=1, (w_0 \lambda, \alpha^{\lor})=-1, (\mu, \alpha^{\lor})=0$ for all weights $\mu$ with $w_0 \lambda < \mu < \lambda$. Here $w_0$ is the longest element in the Weyl group $\mathfrak{W}$ , and $<$ the usual ordering.

My thoughts so far:

  • By the theorem of the highest weight there exists a highest weight $\lambda$ satisfying $$2\frac{(\lambda, \alpha)}{(\alpha,\alpha)} \in \mathbb{Z},$$ which coincides with $(\lambda, \alpha^{\lor})$, using the definition of the coroot of a semisimple Lie-algebra.
  • Using the Properties of the longest element $w_0$ in the Weyl-group of a semisimple Lie-algebra $\mathfrak{g}$ and concidering that $\mathfrak{sl}(n,\mathbb{C})$ has root system $A_{n-1}$, $w_0$ satisfies $w_0\lambda = - \lambda$. So in particular we have $$(w_0 \lambda, \alpha^{\lor}) = - (\lambda, \alpha^{\lor}).$$
  • The trivial representation of $\mathfrak{s}$ has a single weight, $0$.
  • Let $V(1)=\mathbb{C}^2$ be the standard representation. Write $e_1,e_2$ for the standard basis. Note that $$H=\begin{pmatrix} 1 & 0\\ 0 & -1\\ \end{pmatrix}.$$ Then $He_1 = e_1$ and $He_2 = -e_2$. Thus the set of weights of $V(1)$ is $\{\pm1\}$.
user12345
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  • You probably mean a positive* root. – Alvaro Martinez Jul 06 '22 at 14:53
  • Consider $\mathfrak{sl}_2$ and its irreducible representation with highest weight $\lambda=2$ (this means, if $v$ is the highest weight vector, then $H*v=2v$). The positive roots are $n\alpha$ for $n>0$, and you should be able to conclude from that. – Alvaro Martinez Jul 06 '22 at 14:57
  • The representation above is also known as the adjoint representation. – Alvaro Martinez Jul 06 '22 at 14:57
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    I'm confused by the second paragraph. The claim most certainly does not hold for all the irreducible $\mathfrak{g}$-module as the highest weight can be well inside the dominant chamber, far from all the walls. I suspect that an existential quantifier is missing or something. You make it sound like every dominant weight is minuscule?? – Jyrki Lahtonen Jul 07 '22 at 13:22
  • You´re right. I failed to mention, that we have a subalgebra of $\mathfrak{g}$ isomorphic to $\mathfrak{sl}(2)$ in the assumptions. I corrected this now. – user12345 Jul 07 '22 at 13:35
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    @user12345 That doesn't really change much. Every complex semisimple Lie algebra has a subalgebra isomorphic to $\mathfrak{sl}_2$. Among other things this is the basis for the theory of roots. – Callum Jul 07 '22 at 14:15
  • Then i may ask about the minuscule part in another question, as this one is about the above statement. – user12345 Jul 07 '22 at 14:45
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    Still looks very confusing. My point is that the claim cannot possibly hold for all dominant weights. For example with $\mathfrak{g}=\mathfrak{sl}_3$, and $\lambda=2\lambda_1+3\lambda_2$ we have $\langle\lambda,\alpha\rangle=2,3$ or $5$ depending on which of the three positive roots you try as $\alpha$. In this case all the roots have the same length, so I can normalize $\langle \alpha,\alpha\rangle=2$, so there is no difference between a root and a coroot. – Jyrki Lahtonen Jul 07 '22 at 21:31
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    Can you expand the post to contain the actual problem you want to solve? The answer to the current version, the way it is phrased, is trivially "No!" What is the origin of this question? Where did you find it? – Jyrki Lahtonen Jul 07 '22 at 21:33
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    I mean, surely you know this to be false for $\mathfrak{sl}_2$. The sole highest weight of an irreducible representation is unbounded! There is only a single positive root, and $\langle \lambda,\alpha\rangle$ can be as large as you want! – Jyrki Lahtonen Jul 07 '22 at 21:34
  • @JyrkiLahtonen Firstly, thank you for commenting. This problem is part of a proof, i´m trying to understand. I´m linking the article above. – user12345 Jul 07 '22 at 21:53
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    @user12345 You left out a key piece of information about the structure of $V$ as an $\mathfrak{s}$-module. I added it to an IMHO correct place. Do check my edit. – Jyrki Lahtonen Jul 08 '22 at 04:11

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