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Consider $\Delta$ a root system. Let $\alpha, \beta \in \Delta$. Denote by $s_\alpha$ and $s_\beta$ the reflections in the Weyl group associated with $\alpha$ and $\beta$, respectively. Show that $(s_\alpha s_\beta)^i =1$ with $i \in \{2,3,4,6 \}$.

Suppose that $\alpha$ and $\beta$ are not connected in the Dynkin diagram. In that case, I will resolve it by brute force. We know that $$ s_\alpha (\lambda) = \lambda - 2 \frac{\langle \lambda, \alpha \rangle}{\|\alpha \|^2} \alpha \ \ \ \ \text{and} \ \ \ \ s_\beta (\lambda) = \lambda - 2 \frac{\langle \lambda, \beta \rangle}{\|\beta \|^2} \beta .$$ Then \begin{eqnarray} s_\alpha s_{\beta} (\lambda) & = & \lambda - 2 \frac{\langle \lambda, \beta \rangle}{\|\beta \|^2}\beta - 2 \frac{\langle \lambda, \alpha \rangle}{\|\alpha \|^2} \alpha + 4 \frac{\langle \lambda, \beta \rangle}{\|\beta \|^2} \frac{\langle \beta, \alpha \rangle}{\|\alpha \|^2} \alpha \\ & = & \lambda - 2 \frac{\langle \lambda, \beta \rangle}{\|\beta \|^2}\beta - 2 \frac{\langle \lambda, \alpha \rangle}{\|\alpha \|^2} \alpha \end{eqnarray} and \begin{eqnarray*} (s_\alpha s_\beta)^2 (\lambda) & = & \lambda - 2 \frac{\langle \lambda, \beta \rangle}{\|\beta \|^2} \beta - 2 \frac{\langle \lambda, \alpha \rangle}{\|\alpha \|^2} \alpha \\ & & - \frac{2}{\|\alpha \|^2} \Big\langle \lambda - 2 \frac{\langle \lambda, \beta \rangle}{\|\beta \|^2} \beta - 2 \frac{\langle \lambda, \alpha \rangle}{\|\alpha \|^2} \alpha , \beta \Big\rangle \beta \\ & & - \frac{2}{\|\alpha \|^2} \Big\langle \lambda - 2 \frac{\langle \lambda, \beta \rangle}{\|\beta \|^2} \beta - 2 \frac{\langle \lambda, \alpha \rangle}{\|\alpha \|^2} \alpha , \alpha \Big\rangle \alpha \\ & = & \lambda - 2\frac{\langle \lambda, \beta \rangle}{\|\beta \|^2}\beta - 2 \frac{\langle \lambda, \alpha \rangle}{\|\alpha \|^2} \alpha - 2 \frac{\langle \lambda, \beta \rangle}{\|\beta \|^2} \beta \\ & & + 4 \frac{\langle \lambda, \beta \rangle}{\|\beta \|^2} \beta - 2 \frac{\langle \lambda, \alpha \rangle}{\|\alpha \|^2} \alpha + 4 \frac{\langle \lambda, \alpha \rangle}{\|\alpha \|^2} \alpha \\ & = & \lambda \end{eqnarray*} for all $\lambda \in \Delta$. The other cases will be more complicated. Does anyone have a simple way to prove this?

Thanks!

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    $\begingroup$ Work geometrically in $\text{span}(\alpha, \beta)$. $s_{\alpha}$ and $s_{\beta}$ are two reflections in this plane, whose product is a rotation. Work out the angle of this rotation; you will see it depends only on the angle between $\alpha$ and $\beta$. $\endgroup$ Commented Jan 30 at 17:18

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To expand Qiaochu Yuan's comment to a hint-answer:

Arguably, the whole point of the root system machinery being set in $\mathbb R^n$ is so that we can use our intuition (plus basic linear-algebra / geometric knowledge) of Euclidean space.

You are given two roots $\alpha, \beta$ with corresponding reflections (through the hyperplanes orthogonal to them) $s_\alpha, s_\beta$. Want to show something about the order of the composition $s_\alpha s_\beta$.

Well, first of all, both of them operate trivially on the orthogonal complement of $span(\alpha, \beta)$, so w.l.o.g. we can restrict to this two-dimensional (the case $\alpha =\pm \beta$ is trivial) space.

Now in a two-dimensional space, the product of two reflections is a rotation, cf. e.g. Are rotations the result of composing two reflections?(Linear Algebra), Show that the composition of two reflection is a rotation, Composition of two reflections (non-parallel lines) is a rotation, Product of reflections is a rotation, by elementary vector methods, Composition of two reflections is a rotation, Understanding a Theorem on Rotations as the Product of Reflections.

Every rotation is rotation by some angle. I leave it to you to figure out ...

  1. ... what the angle is. It depends on the roots $\alpha$ and $\beta$ but careful, it is not exactly the angle between them. (Visualize what you showed with all that algebra: If the angle between $\alpha$ and $\beta$ is $90$ degrees, $s_\alpha s_\beta$ has order 2 i.e. is a rotation by 180 degrees.) If it helps, remember that $\cos(\angle(\alpha, \beta)) = \dfrac{\langle \alpha, \beta \rangle}{\| \alpha \| \|\beta \|}$ and get out some basic trigonometric identities.

  2. ... how that answers your question. Remember that the integrality condition $2\dfrac{\langle \alpha, \beta \rangle}{\| \alpha \|^2 } \in \mathbb Z$ implies there are only very few angles which can occur between two roots, and via 1) their list is closely related to your list of orders.

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    $\begingroup$ This. The point really is to get away with 2-dimensional pictures! I once prepared an animation showing how the composition of two reflections is a rotation. Not sure it's the best of way of visualizing this fact, but anyway (a picture proof with angles in-between clearly marked might serve the end better)! $\endgroup$ Commented Jan 31 at 5:32
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    $\begingroup$ @JyrkiLahtonen You know what's funny, I was not (consciously) aware of this fact until way after I got my Ph.D. One day I played around a bit with 2D reflections and rotations as example of groups, for a possible bonus class for gifted high schoolers, and kind of stumbled upon this result myself. I still wonder how I could write a thesis dealing with root systems without such basics. -- Like your animation. The multitude of links is due to that I'm still not too happy with each individual proof. $\endgroup$ Commented Feb 1 at 18:14
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    $\begingroup$ I think the basic visual proof should be: a reflection keeps all angles rigid but reverses orientation / chirality / direction of angles. So after two reflections you end up with something with all rigid angles, now again in the right direction. But the intersection point of the reflection axes remains fixed. So whatever picture one starts with, it can only have rotated around this point now. $\endgroup$ Commented Feb 1 at 18:16
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    $\begingroup$ I guess I was lucky in the sense that my dissertation work required playing with the reflections a lot (with the affine Weyl groups because I was doing algebraic groups rep theory in characteristic $p$). And again lucky, because it was obvious how to apply what I had learned to the use of Weyl group orbits in the design of radio signal constellations. $\endgroup$ Commented Feb 1 at 18:51

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