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(the notation here is compatible with J.E. Humphrey's "Introduction to Lie Algebras and Representation Theory")

Let $\Phi \subset E$ be a root system. Let $\Delta \subset \Phi$ be a base. I already know that $\Phi^\vee \subset E$ is a root system. I'd like to show that $\Delta^\vee \subset \Phi^\vee$ is a base.
(the notation $\alpha^\vee$ means $\frac{2\alpha}{(\alpha,\alpha)}$)

I tried 2 approaches and was stuck with about the same problem.

Here's one attempt: I'll denote $\Delta=\Delta(\gamma)$ for some regular $\gamma \in E$. I'd like to show that $\Delta(\gamma)^\vee=\Delta(\gamma^\vee)$. I can easily show that $\Phi^+(\gamma)^\vee=(\Phi^\vee)^+(\gamma^\vee)$. But then, how do I show that the indecomposable elements in $(\Phi^\vee)^+(\gamma^\vee)$ are exactly the duals of the indecomposable elements in $\Phi^+(\gamma)$? (If this is true, than it goes the other way around too, so I'll try to show that the dual of a decomposable element is decomposable). Denote $\alpha=\beta_1+\beta_2$ for some $\alpha,\beta_1,\beta_2 \in \Phi^+(\gamma)$. I'd like to show that $\alpha^\vee$ is decomposable too (in $(\Phi^+(\gamma))^\vee$). It is not true that $\alpha^\vee=\beta_1^\vee+\beta_2^\vee$ as I've seen by checking 2-dimensional examples. That's where I'm stuck.

(the other attempt was to go straight from the definition of a base, and there I had a problem with proving the integrality of the coefficients)

Another idea was to use the correspondence between Weyl chambers and bases. Both $\Phi$ and $\Phi^\vee$ define the same Weyl chambers. This gives a correspondence between the bases of $\Phi$ and the bases of $\Phi^\vee$. The correspondence is probably given by taking the dual of a base, but that's what I'm stuck at showing.

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    My question is a duplicate of this, but it has been answered. So perhaps this should be merged or closed as a duplicate.2014-05-11
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    @spin How can this be a duplicate? It was asked almost *two years before your own question*. Your question is the duplicate!2014-05-11
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    @spin You're kidding me, right?!2014-05-11
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    @FlybyNight: You don't understand my point, or maybe you didn't read my other comment. Yes, exactly like I said, my question is a duplicate. But this question has no answers, while my question already has two answers that solve the problem. Which is why this should be closed as a duplicate or merged..2014-05-12
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    @FlybyNight [According to the network meta](http://meta.stackexchange.com/a/147651/249378), age is just about the least important factor for determining the direction of duplication.2014-05-12
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    @epimorphic So because 28 people agree, it should be? If you say so.2014-05-12
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    @FlybyNight There's no need for you to get as dismissive as you have been in this thread. There are numerous additional meta-questions that were marked as duplicates of the one I linked to, in the sidebar of that page. Feel free to check them out.2014-05-12

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