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

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

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

4 Comments »

[...] what about ? This is just the dual root system to ! The roots of squared-length are left unchanged, but the roots of squared-length are doubled. [...]

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[...] other observation we want to make is that duality of root systems is easily expressed in terms of Dynkin diagrams: just reverse all the oriented [...]

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[...] we can see that this root system is isomorphic to its own dual. Indeed, if is a short root, then the dual root is [...]

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[...] like the case, the root system is isomorphic to its own dual. The long roots stay the same length when dualized, while the short roots double in length and [...]

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This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).

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