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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 $R(\Delta')$???

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    $\begingroup$ 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. $\endgroup$ Commented Dec 4, 2010 at 20:55
  • $\begingroup$ Can you give me an example where it fails? $\endgroup$
    – Binai
    Commented Dec 4, 2010 at 21:11
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    $\begingroup$ Take $\mathfrak{g} =\mathfrak{sl}_3$ with $\Delta'$ containing just one simple root $\beta$ and $\alpha$ being the other simple root. $\endgroup$ Commented Dec 4, 2010 at 22:13
  • $\begingroup$ This is a duplicate: math.stackexchange.com/questions/12996/parabolic-subalgebra $\endgroup$
    – Alex B.
    Commented 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.
    Commented Dec 5, 2010 at 22:59

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

Thanks.

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