The Unapologetic Mathematician

Mathematics for the interested outsider

Pairs of Roots

When we look at a root system, the integrality condition \beta\rtimes\alpha\in\mathbb{Z} puts strong restrictions on the relationship between any two vectors in the root system. Today we’ll look at this in two ways: one more formal and the other more visual.

First off, let’s look a bit deeper at the condition

\displaystyle\beta\rtimes\alpha=\frac{2\langle\beta,\alpha\rangle}{\langle\alpha,\alpha\rangle}=k\in\mathbb{Z}

We know a lot about the relation between the inner product and the lengths of vectors and the angle between them. Specifically, we can write

\displaystyle\frac{2\langle\beta,\alpha\rangle}{\langle\alpha,\alpha\rangle}=\frac{2\lVert\beta\rVert\lVert\alpha\rVert\cos(\theta)}{\lVert\alpha\rVert^2}=2\frac{\lVert\beta\rVert}{\lVert\alpha\rVert}\cos(\theta)

Then we can consider the product

\displaystyle(\beta\rtimes\alpha)(\alpha\rtimes\beta)=\left(2\frac{\lVert\beta\rVert}{\lVert\alpha\rVert}\cos(\theta)\right)\left(2\frac{\lVert\alpha\rVert}{\lVert\beta\rVert}\cos(\theta)\right)=4\cos(\theta)^2

This has to be an integer between {0} and 4. And it can only hit 4 if \theta is a multiple of \pi, in which case we already know that \beta must be \pm\alpha. So some simple counting shows the possibilities. Without loss of generality we’ll assume that \lVert\beta\rVert\geq\lVert\alpha\rVert.

\begin{tabular}{rrcc}\hline\(\alpha\rtimes\beta\)&\(\beta\rtimes\alpha\)&\(\theta\)&\(\frac{\lVert\beta\rVert^2}{\lVert\alpha\rVert^2}\)\\\hline0&0&\(\frac{\pi}{2}\)&arbitrary\\1&1&\(\frac{\pi}{3}\)&1\\-1&-1&\(\frac{2\pi}{3}\)&1\\1&2&\(\frac{\pi}{4}\)&2\\-1&-2&\(\frac{3\pi}{4}\)&2\\1&3&\(\frac{\pi}{6}\)&3\\-1&-3&\(\frac{5\pi}{6}\)&3\\\hline\end{tabular}

We see that if \alpha and \beta are orthogonal then we have no information about their relative lengths. This is to be expected, since when we form the coproduct of two root systems the roots from each side are orthogonal to each other, and we should have no restriction on their relative lengths. On the other hand, if they aren’t orthogonal, then there are only a handful of possibilities.

Let’s try to look at this more visually. The vectors \alpha and \beta span some two-dimensional space. We can choose an orthogonal basis (and thus coordinates) that makes \alpha=(1,0), and then \beta=(x,y). We calculate

\displaystyle\begin{aligned}\beta\rtimes\alpha=\frac{2\langle(x,y),(1,0)\rangle}{\langle(1,0),(1,0)\rangle}=2x&=k\in\mathbb{Z}\\\alpha\rtimes\beta=\frac{2\langle(1,0),(x,y)\rangle}{\langle(x,y),(x,y)\rangle}=\frac{2x}{x^2+y^2}&=l\in\mathbb{Z}\end{aligned}

The first of these conditions says that x=\frac{k}{2} for some integer k. We can manipulate the second to tell us that \left(x-\frac{1}{l}\right)^2+y^2=\left(\frac{1}{l}\right)^2 for some integer l. Let’s plot some of these!

The vector pointing to (1,0) is \alpha. The blue vertical lines correspond to the first condition for k=\pm1,\pm2,\pm3,\pm4, while the red circles correspond to the second condition for l=\pm1,\pm2,\pm3,\pm4. Any larger values of k will put lines further out than any of the circles can touch, while any larger values of l will put smaller circles further in than any line can touch. Thus, all possible values of \beta which satisfy both conditions are the twenty highlighted points in this image, six examples of which are indicated in the diagram.

The three examples towards the top of the diagram correspond to the second, fourth, and sixth lines in the above table. The three examples toward the bottom are similar, but have \lVert\beta\rVert\leq\lVert\alpha\rVert. We can see that if we project any of the examples onto \alpha, the projection is a half-integral multiple of \alpha. Conversely, projecting \alpha onto any of the examples gives a half-integral multiple of that vector.

So we can say a lot about how any two vectors in \Phi are related if they aren’t parallel or orthogonal. The question now is how we can pick a whole collection of vectors so that each pair relates to each other in one of these very particular ways.

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January 28, 2010 - Posted by | Geometry, Root Systems

5 Comments »

  1. Clear, fun, and good graphics to help.

    Comment by Jonathan Vos Post | January 28, 2010 | Reply

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