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I am trying to understand how a root system of a semisimple Lie algebra $\mathfrak{g}$ is independent (up to isomorphism) of the choice of Cartan subalgebra $\mathfrak{h}$. Given two such Cartan subalgebras $\mathfrak{h_1}, \mathfrak{h_2}$, I understand that there is always an automorphism $\psi: \mathfrak{g} \rightarrow \mathfrak{g}$ such that $\psi(\mathfrak{h_1}) = \mathfrak{h}_2$.

What I don't understand is how this induces an isomorphism of the respective root systems. Suppose $R_1$ is a root system in $\mathfrak{h}_1^*$, and $R_2$ is a root system in $\mathfrak{h}_2^*$. There are two things which I don't see at the moment:

  1. Why for $\alpha \in R_1$, $\psi^*(\alpha) \in R_2$,
  2. Why for $\alpha,\beta \in R_1, n(\alpha,\beta) = n(\psi^*(\alpha),\psi^*(\beta))$ (i.e. why Cartan integers are preserved),

where $\psi^*$ is the isomorphism induced by $\psi$. I don't really understand what $\psi^*$ ought to be. Any help with that, and the other two points would be appreciated.

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1 Answer 1

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Let $\psi : \mathfrak g \to \mathfrak g$ the automorphism with $\psi(\mathfrak h_1) = \mathfrak h_2$. Recall the construction of a root sytem : we fix a basis of $\mathfrak g$ so that $\mathfrak g = \bigoplus_{\alpha \in R} \mathfrak g_{\alpha}$, where $R \subset \mathfrak h^*$, and for any $x \in \mathfrak g_{\alpha}$ we have $[h,x] = \alpha(h)x$.

Now let $\psi : \mathfrak g \to \mathfrak g$ your automorphism. We notice that $$[\psi(h), \psi(x)] = \psi([h,x])= \psi(\alpha(h)(x)) = \alpha(h) \psi(x)$$ So $\psi(h)$ acts with eigenvalues $\alpha : \mathfrak h_1 \to \Bbb C$. Let $h' = \psi(h)$, then $[\psi(h), \psi(x)] = \alpha(h) \psi(x)$ can be written $[h', y] = \alpha (\psi^{-1}(h'))y$. So $\psi^* \alpha := \alpha \circ \psi^{-1}$ is the root associated to $\psi(h)$, i.e the root system of $\mathfrak h_2$ is indeed $\psi^* R_1$.

2) An unsatisfactory explanation is that $n_{\alpha, \beta}$ can be recovered from the Killing form, preserved by $\psi$, so indeed Cartan integers are preserved.

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