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 .