The Unapologetic Mathematician

Mathematics for the interested outsider

Dual Root Systems

Given a root system \Phi, there’s a very interesting related root system \Phi^\vee, called the “dual” or “inverse” root system. It’s made up of the “duals” \alpha^\vee, defined by

\displaystyle\alpha^\vee=\frac{2}{\langle\alpha,\alpha\rangle}\alpha

This is the vector that represents the linear functional \underline{\hphantom{X}}\rtimes\alpha. That is, \beta\rtimes\alpha=\langle\beta,\alpha^\vee\rangle.

The dual root \alpha^\vee is proportional to \alpha, and so \sigma_{\alpha^\vee}=\sigma_\alpha. The dual reflections are the same as the original reflections, and so they generate the same subgroup of \mathrm{O}(V). That is, the Weyl group of \Phi^\vee is the same as the Weyl group of \Phi.

As we should hope, dualizing twice gives back the original root system. That is, \left(\Phi^\vee\right)^\vee=\Phi. We can even show that \left(\alpha^\vee\right)^\vee=\alpha. Indeed, we calculate

\displaystyle\begin{aligned}\left(\alpha^\vee\right)^\vee&=\frac{2}{\langle\alpha^\vee,\alpha^\vee\rangle}\alpha^\vee\\&=\frac{2}{\langle\frac{2}{\langle\alpha,\alpha\rangle}\alpha,\frac{2}{\langle\alpha,\alpha\rangle}\alpha\rangle}\frac{2}{\langle\alpha,\alpha\rangle}\alpha\\&=\frac{4}{\frac{4}{\langle\alpha,\alpha\rangle\langle\alpha,\alpha\rangle}\langle\alpha,\alpha\rangle\langle\alpha,\alpha\rangle}\alpha\\&=\alpha\end{aligned}

It turns out that passing to duals reverses the roles of roots, in a way, just as we might expect from a dualization. Specifically, \alpha^\vee\rtimes\beta^\vee=\beta\rtimes\alpha. Indeed, we calculate

\displaystyle\begin{aligned}\alpha^\vee\rtimes\beta^\vee&=\frac{2\langle\alpha^\vee,\beta^\vee\rangle}{\langle\beta^\vee,\beta^\vee\rangle}\\&=\frac{2\langle\frac{2}{\langle\alpha,\alpha\rangle}\alpha,\frac{2}{\langle\beta,\beta\rangle}\beta\rangle}{\langle\frac{2}{\langle\beta,\beta\rangle}\beta,\frac{2}{\langle\beta,\beta\rangle}\beta\rangle}\\&=\frac{\frac{8}{\langle\alpha,\alpha\rangle\langle\beta,\beta\rangle}\langle\alpha,\beta\rangle}{\frac{4}{\langle\beta,\beta\rangle\langle\beta,\beta\rangle}\langle\beta,\beta\rangle}\\&=\frac{2\langle\beta,\alpha\rangle}{\langle\alpha,\alpha\rangle}\\&=\beta\rtimes\alpha\end{aligned}

January 26, 2010 - Posted by | Geometry, Root Systems

4 Comments »

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