1
$\begingroup$

Let $R$ a root system and $\Delta$ be a simple system of roots of a Lie algebra $\mathfrak g$, $\Delta'\subset \Delta$ and $R(\Delta')=R\cap \mathbb Z(\Delta')$. Define $$p(\Delta')=\mathfrak h \bigoplus_{\alpha \in R(\Delta')} \mathfrak g_{\alpha} \bigoplus_{\alpha \in R^+ \setminus R^+(\Delta')}\mathfrak g_{\alpha}$$ the parabolic subalgebra associated to $\Delta'$.

If $\alpha$ is a simple root in $R^+(\Delta)\setminus R^+ (\Delta')$, then $\beta(h_\alpha)=0$ for all $\beta$ in $R(\Delta')$???

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ If I understand your notation correctly, the answer is clearly no. Here the simple root $\alpha$ is any simple root not in $\Delta'$ and it need not be orthogonal to all roots in the subsystem. $\endgroup$
    – Jim Humphreys
    Dec 4, 2010 at 20:55
  • $\begingroup$ Can you give me an example where it fails? $\endgroup$
    – Binai
    Dec 4, 2010 at 21:11
  • 1
    $\begingroup$ Take $\mathfrak{g} =\mathfrak{sl}_3$ with $\Delta'$ containing just one simple root $\beta$ and $\alpha$ being the other simple root. $\endgroup$
    – Jim Humphreys
    Dec 4, 2010 at 22:13
  • $\begingroup$ This is a duplicate: math.stackexchange.com/questions/12996/parabolic-subalgebra $\endgroup$
    – Alex B.
    Dec 5, 2010 at 2:20
  • $\begingroup$ There isn't necessarily any problem, but it's good to give links between the two. $\endgroup$
    – j.c.
    Dec 5, 2010 at 22:59

1 Answer 1

Reset to default
1
$\begingroup$

Jim gave me the answer! It is false in general!

Thanks.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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