In section 5.2 (p.95) of Representations of Semisimple Lie Algebras in the BGG Category $\mathcal{O}$.

Let $\mu\le \lambda$ if $\lambda-\mu\in \Gamma$, where $\Gamma$ is the set of $\mathbb{Z}^{\ge 0}$-linear combinations of simple roots.

Let $\lambda$ be a regular, integral, antidominant weight. It is easy to show $(s_\alpha w) \cdot\lambda < w \cdot \lambda \iff s_\alpha w < w$ in the Bruhat ordering. According to Humphreys, we can show $w'\cdot\lambda<w\cdot\lambda\iff w'<w$ by iteration. enter image description here

How to do iteration in order to show $w'\cdot\lambda<w\cdot\lambda\implies w'<w$?

Or is this a typo? Any conterexample if it is a typo?

  • $\begingroup$ If $w' < w$ in Bruhat order then there exists a sequence of reflections $s_{\alpha_1}, s_{\alpha_2}, \ldots, s_{\alpha_k}$ so that $\ell(s_{\alpha_i} \cdot s_{\alpha_{i-1}} \cdots s_{\alpha_1}\cdot w') = \ell( s_{\alpha_{i-1}} \cdots s_{\alpha_1}\cdot w') + 1$ for all $i$ and $w=s_{\alpha_k} \cdots s_{\alpha_1} w'$. This is a standard fact (see Theorem 2.2.6 of Björner-Brenti). With this fact the induction is clear, yes? $\endgroup$ – Sam Hopkins Aug 13 '19 at 1:25
  • 1
    $\begingroup$ Thank you. I understand this implication. I am asking about the other direction, which is obviously harder. $\endgroup$ – James Cheung Aug 13 '19 at 1:34
  • $\begingroup$ I see, let me think for a second about this... $\endgroup$ – Sam Hopkins Aug 13 '19 at 1:38
  • $\begingroup$ Here's the beginning of a proof by induction on the length of $w'$, but I can't complete it. If the length is $0$, then the result is obvious. Otherwise, if $s$ is a simple reflection such that $\ell(s w') < \ell(w')$, then $s w' < w' \implies s w'\lambda < w'\lambda < w\lambda \implies s w' < w$. Then either $w' \le w$ or $w' \le s w$ (Proposition (c), p. 5). Obviously we don't have $w' = w$; but I don't see how to rule out the possibility that $w' \le s w$. $\endgroup$ – LSpice Aug 13 '19 at 1:52
  • 1
    $\begingroup$ Actually we only need to rule out the possibility that $w' \le s w$ and $w < s w$ (although I'm still not sure how to do that). $\endgroup$ – LSpice Aug 13 '19 at 1:59

This question was answered correctly by Sam Hopkins. Let me just abuse this space for answers for pointing out that James Humphreys keeps up to date corrections on the AMS webpage of the book. Direct link to pdf of the errata.

(In this particular case he says that one should replace both occurences of $<$ symbol with $\uparrow$ and $\leq$ respectively.)

| cite | improve this answer | |
  • $\begingroup$ What does the upward arrow denote? $\endgroup$ – Sam Hopkins Aug 13 '19 at 20:29
  • 2
    $\begingroup$ Transitive reflexive closure of $s_\alpha \cdot \lambda < \lambda.$ $\endgroup$ – Vít Tuček Aug 13 '19 at 20:44

I think the implication you are asking about might not in fact hold.

If my computations are correct, we can see this already for $S_4$.

Let me ignore the dot action part of the question (I don't think that matters) and try to simplify it as follows: let $\lambda$ be any regular anti-dominant weight, and define a partial order $\preceq$ on $W$ by $u \preceq v$ if and only if $u\lambda \leq v\lambda$ (this is the usual partial order on weights with $\mu \leq \lambda$ if and only if $\lambda-\mu=\sum_{i}c_i\alpha_i$ with $c_i\in\mathbb{Z}_{\geq 0}$). The question becomes: is $\preceq$ the same partial order as $\leq$, the usual (strong) Bruhat order?

My answer is that, no, they are not the same and this can be seen already for $S_4$. For instance, let us take $\lambda =(1,2,3,4)$ as our regular anti-dominant weight so that $w\lambda$ is just the one-line notation of the permutation $w \in S_4$. Then I claim that $(1,4,2,3) \preceq (2,3,4,1)$ but $(1,4,2,3)\not \leq (2,3,4,1)$. That $(2,3,4,1)-(1,4,2,3)= 1*(1,-1,0,0) + 2*(0,0,1,-1)$ shows $(1,4,2,3) \preceq (2,3,4,1)$. And my computer tells me that $(1,4,2,3)\not \leq (2,3,4,1)$ (actually I think that this is also easy to see since we cannot ever move the $4$ rightward in $(1,4,2,3)$ when going up in Bruhat order).


Regarding the dot action aspect of the question, here is a larger portion of the text in question which shows the notions of regular and antidominant are meant relative to the dot action:

enter image description here

| cite | improve this answer | |
  • 1
    $\begingroup$ As I mentioned in an earlier version of this answer, with regards to the context of this question, the criterion for the existence of a nonzero homomorphism between Verma modules is stated always in terms of Bruhat order and not the (simpler) partial order determined by the positive roots; see e.g. Wikipedia: en.wikipedia.org/wiki/… $\endgroup$ – Sam Hopkins Aug 13 '19 at 4:15
  • $\begingroup$ Why shouldn't the dot-action part mattter? Doesn't ignoring it amount to replacing $\lambda$ by $\lambda + \rho$, which might destroy regularity or anti-dominance? $\endgroup$ – LSpice Aug 13 '19 at 11:31
  • $\begingroup$ We can choose a very anti dominant lambda, right? In the text surrounding the passage in question, Humphreys notes that at least the “regular” condition is meant relative to the dot action. $\endgroup$ – Sam Hopkins Aug 13 '19 at 13:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.