## Root Strings

First, a lemma: given a root system , let and be nonproportional roots. That is, . Then if — if the angle between the vectors is strictly acute — then is also a root. On the other hand, if then is also a root. We know that is a root. If we replace with we can see that the second of these assertions follows from the first, so we just need to prove that one.

We know that is positive if and only if is. So let’s look at the table we worked up last time. We see that when is positive, then either that or equals . If it’s , then

is a root. On the other hand, if then is a root.

So given two nonproportional and nonorthogonal roots and we’re guaranteed to have more than one vector of the form for some integer in the root system . We call the collection of all such vectors in the -string through .

Let be the largest integer so that , and let be the largest integer so that . I say that the root string is unbroken. That is, for all .

Indeed, if there’s some integer so that , then we can find and with so that

But then the above lemma tells us that , while . Subtracting, we find

The two inequalities tell us that this difference should be positive, but is positive and is negative. Thus we have a contradiction, and the root string must be unbroken from to .

We can also tell that just adds a positive or negative multiple of to any root. Then it’s clear from the geometry that just reverses a root string end to end. That is, . But we can also calculate

Thus . And so the length of the string through can be no more than .

Cool! Some of this was new to me. Thank you!

Comment by Jonathan Vos Post | January 30, 2010 |

[…] are two vectors in a base , then we know that and . Indeed, our lemma tells us that if then would be in . But this is impossible, because every vector in can only be […]

Pingback by Bases for Root Systems « The Unapologetic Mathematician | February 1, 2010 |

[…] if and are distinct vectors in then . Indeed, by our lemma if then . And so either or lies in . In the first case, we can write , so is decomposable. In […]

Pingback by The Existence of Bases for Root Systems « The Unapologetic Mathematician | February 2, 2010 |

[…] is irreducible, there must be at least one and so that , and so we must have for this . This proves that must also be a root, which contradicts the maximality of . And so we conclude that is empty, […]

Pingback by Properties of Irreducible Root Systems I « The Unapologetic Mathematician | February 10, 2010 |

[…] But beyond that, we don’t know which Cartan matrices actually arise, and that’s the whole point of our project. For now, though, we will assume that our matrix does in fact arise from a real root system, and see how to use it to construct a root system whose Cartan matrix is the given one. And our method will hinge on considering root strings. […]

Pingback by From Cartan Matrix to Root System « The Unapologetic Mathematician | February 17, 2010 |

this is great thanks – helping me finally understand some of this root system stuff, just in time for my exams!

Comment by twentyfifteen | May 21, 2012 |