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Let $\mathfrak{g}$ be a semisimple Lie Algebra, $\mathfrak{t}$ a Cartan Subalgebra, $\Phi$ the roots with respect to $\mathfrak{t}$, and $E = \text{Span}_{\mathbb R}(\Phi)$ a Euclidean space with the Euclidean inner product $(\cdot, \cdot)_E$

For $\alpha \in E$ define $\check\alpha : E \rightarrow \mathbb R$ by: $\check\alpha(\lambda) = \frac{2(\alpha,\lambda)_E}{(\alpha,\alpha)_E}$.

Then $(\Phi, E)$ is a root system, i.e. it satisfies:

  1. $0 \not\in \Phi, \; E = \text{Span}_{\mathbb R}(\Phi)$

  2. If $\alpha, \beta \in \Phi$ then $\check\beta(\alpha) \in \mathbb Z$

  3. Define $\omega_\alpha : E \rightarrow E $ by $\omega_\alpha(\lambda) = \lambda - \check\alpha(\lambda)\alpha$.

    Then $\alpha \in \Phi \Rightarrow \omega_\alpha(\Phi) = \Phi$

  4. If $\alpha \in \Phi$ and $c\alpha \in \Phi$ then $c = \pm 1$

Consider now $\check\Phi = \{\check\alpha \mid \alpha \in \Phi\}$ and $E^*$ the dual space of $E$. I am asked to prove that $(\check\Phi, E^*)$, the dual root system, is indeed a root system.

I have shown that properties 1 and 4 hold for $(\check\Phi, E^*)$ but I am struggling with 2 and 3.

Specifically, I am unsure of what $(\cdot, \cdot)_{E^*}$ is. I suspect that I will want to define the functions:

$\check{\check\alpha} : E^* \rightarrow \mathbb R$ and $\omega_{\check\alpha} : E^* \rightarrow E^*$ for a given $\check\alpha$ such that for $\check\lambda \in E^*$ we have:

$\check{\check{\alpha}}(\check\lambda) = \frac{(\check\alpha, \check\lambda)_{E^*}}{(\check\alpha, \check\alpha)_{E^*}}$, and $\omega_{\check\alpha}(\check\lambda) = \check\lambda - \check{\check\alpha}(\check\lambda)\check\alpha$

But I don't really understand what this means. Specifically, I suspect there ought to be some sort of relationship between $(\cdot, \cdot)_E$ and $(\cdot, \cdot)_{E^*}$ but I'm not sure what it is.

Should I instead endow $E^*$ with a specific inner product that I construct so that everything works? If I do that, how should I go about constructing one?

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user366818
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    There is a canonical isomorphism $f\colon E \to E^$ given by $f(v)(w) := (v,w)_E$. If you equip $E^$ with the unique inner product that makes $f$ an isometry, can you work it out? – Claudius Nov 14 '18 at 15:40
  • @Claudius So then $f(\frac{2\alpha}{(\alpha, \alpha)E}) = \check\alpha$ so we want $(\check\alpha, \check\beta){E^*} = (\frac{2\alpha}{(\alpha,\alpha)_E}, \frac{2\beta}{(\beta,\beta)_E})_E = \frac{4}{(\alpha,\alpha)_E(\beta,\beta)_E}(\alpha,\beta)_E$ ? – user366818 Nov 14 '18 at 17:00
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    Since you were asked to prove this, the next question here might be about the base of the dual root system, see this question. – Dietrich Burde Nov 14 '18 at 22:29
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    Yes, now use this to compute $\check{\check\alpha} (\check \beta)$ and $\omega_{\check \alpha}$. You might have to use that $\omega_\alpha$ is an isometry. – Claudius Nov 15 '18 at 07:28

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