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 
.
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January 29, 2010 -
Posted by John Armstrong |
Geometry, Root Systems
Cool! Some of this was new to me. Thank you!
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this is great thanks – helping me finally understand some of this root system stuff, just in time for my exams!