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Here $\delta=\frac{1}{2}\sum_{\alpha>0}\alpha$, where the sum is over all positive simple roots for some root system.

Lemma: Suppose $\mu\in\Lambda^+$ (dominant weight), and $\nu=\sigma^{-1}\mu$ for some $\sigma\in W$ (Weyl group). Then $(\nu+\delta,\nu+\delta)\leq(\mu+\delta,\mu+\delta)$, with equality holding iff $\sigma=1$ or $\mu=0$.

Proof: We have

\begin{aligned} (\nu+\delta,\nu+\delta) &= (\mu+\sigma\delta,\mu+\sigma\delta)\\ &= (\mu,\mu)+(\sigma\delta,\mu)+(\mu,\sigma\delta)+(\sigma\delta,\sigma\delta)\\ &= (\mu,\mu)+2(\mu,\sigma\delta)+(\delta,\delta)+(\delta,\mu)+(\mu,\delta)-2(\mu,\delta)\\ &= (\mu+\delta,\mu+\delta)-2(\mu,\delta-\sigma\delta). \end{aligned}

The claim follows if $(\mu,\delta-\sigma\delta)\geq 0$. Since $\delta$ is strongly dominant, $\sigma\delta\preceq\delta$. Thus $\delta-\sigma\delta\geq 0$, thus if $\delta-\sigma\delta=\sum_i c_i\alpha_i$, we have $c_i\geq 0$ for all $i$. Then $$ (\mu,\delta-\sigma\delta)=\sum_i c_i(\mu,\alpha_i)\geq 0 $$ since $(\mu,\alpha_i)\geq 0$ as $\mu$ is dominant. Then equality holds iff $(\mu,\delta-\sigma\delta)=0$. If $\mu\neq 0$, then $\delta-\sigma\delta=0$, or $\delta=\sigma\delta$. Since $\delta$ is strongly dominant, necessarily $\sigma=1$.

At the end, I don't follow why if $\mu\neq 0$, necessarily $\delta-\sigma\delta=0$. Couldn't $\mu$ be orthogonal to the $\alpha_i$ for which $c_i\neq 0$, but possibly not orthogonal to simple roots $\alpha_j$ where $c_j=0$? If feel like you could only make the conclusion if $\mu$ is assumed to be strongly dominant, not just dominant.

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In my copy the proof continues as follows

... with equality only if $(\mu,\delta-\sigma\delta)=0$, i.e. $(\mu,\delta)=(\mu,\sigma\delta)=(\nu,\delta),$ or $(\mu-\nu,\delta)=0$. But $\mu-\nu$ is a sum of positive roots, and $\delta$ is strongly dominant, so $\mu=\nu$.

The point is that as $\mu\preceq\nu$ the inner product $(\nu-\mu,\delta)$ can vanish only if $\nu=\mu$. The inner product of $\delta$ and any positive root is, as you seem to know, strictly positive.


May be this is a version conflict? Mine is the 3rd printing, revised in 1980.

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    $\begingroup$ Because the key step is verbatim from Humphreys I am switching to CW. $\endgroup$ – Jyrki Lahtonen Mar 10 '17 at 8:11
  • $\begingroup$ Thanks, that's not present in mine, I'm using an old library copy from 1972. Perhaps I should get a newer version... $\endgroup$ – Nastassja Mar 10 '17 at 8:14

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