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I have a question regarding a remark in the book "Reflection Groups And Coxeter Groups" by James E. Humphreys (unfortunately the book is not to be found as a whole on google books or such).

In chapter 2 we encounter the definition of a crystallographic root system ($\varPhi$ crystallographic iff $\frac{2(a,b)}{(b,b)} \in \mathbb{Z}\forall a,b \in \varPhi$) and its associated Weyl group $W$ (the group generated by all reflections belonging to elements of $\varPhi$).

After the definition Humphreys follows up with some facts about crystallographic root systems. It is quite easy to see that in the case where $\varPhi$ is irreducible (i.e. the Coxeter graph is connected) there can occur at most two squared length of roots (the roots are then called long and short roots respectively) and $W$ acts transitively on the long and the short roots. After fixing a simple system $\Delta \subset \varPhi$ a partial ordering is introduced on $\varPhi$ via $a \leq b$ iff $b-a$ is a non-negative linear combination of $\Delta$ and the author writes that there is a unique greatest element of $\varPhi$ with respect to this ordering which is a long root.

I can show (using transitivity and a fundamental domain given earlier in the book) that there can be at most two greatest elements one being a long root and one a short root. However I am unable to see why the long root should be greater than the short root. Is there a proof for this fact (preferably one that does not use Lie theory) that does not use the classification and construction of all possible root systems (this is at the moment the only way I could find)?

Any help would be appreciated. Thanks in advance.

Edit: As requested the definition of a simple system: A subset $\Delta \subset \varPhi$ of a rootsystem $\varPhi$ is called simple if $\Delta$ is a basis for the span of $\varPhi$ which satisfies the additional condition that any element of $\varPhi$ is either a non-negative or a non-positive linear combination of $\Delta$.

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To be able to show this with as little extra as possible, could you please give the exact definition of simple system that the author uses? – Hagen von Eitzen Nov 17 '12 at 10:17
I added the definition as by your request. – Sebastian Schoennenbeck Nov 17 '12 at 11:52

2 Answers 2

up vote 3 down vote accepted

Assume there is a maximal short root $a$ and there exist long roots. By switching to a different simple system, we may assume (using transitivity) that $a$ corresponds to a vertex of the Dynkin graph and is adjacent in it to a long root $b$. We need not walk through the complete classification, but can quite easily show that $a$ and $b$ span a root system of type $B_2$ or $G_2$, where the claim can be verified by inspection.

For example, in $G_2$ there are six short and six long roots. If $a$ is our maximal short root and $b$ is another positive root, then $a+b$ or $a+2b$ is a long root (since $b\ne -a$) and is $>a$.

And in $B_2$ we have our maximal short root $a$ and an orthogonal positive short root $b$. Then $a+b$ is a long root $>a$.

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My first intuition was to do something like that but I don't think it solves the problem. Switching to a different system means we are also changing the partial ordering we are looking at. Your proof shows that a short root belonging to a simple system and adjacent in its coxeter graph to a long root can never be the maximal element with regard to the partial ordering derived form that system. But this is more or less trivial since $(a,b) \leq 0 \forall a,b \in \Delta$ so the maximal element will never be contained in $\Delta$. – Sebastian Schoennenbeck Nov 17 '12 at 14:39
Furthermore the ordering is not respected by elements of $W$ and this is basically what we would need to get this proof to work in my opinion. – Sebastian Schoennenbeck Nov 17 '12 at 14:40
The scalar product is left invariant by $W$ and I hoped that would be enough – Hagen von Eitzen Nov 17 '12 at 15:28
Using your answer (basically filling in some calculations and extending some of the shorter arguments) I was able to completely prove the result. I will add a separate answer later so that the complete proof is given in this thread. Thank you very much for your help. – Sebastian Schoennenbeck Nov 17 '12 at 17:12

As promised here the complete proof: There are two cases we need two consider:

Case A: One of the labels of the edges of the coxeter graph is "4". Then the ratio of squared lengths of long roots and short roots is $2$. Wlog we will assume that $(a,a)=2$ for long roots and $(b,b)=1$ for short roots. Let $b$ be a positive short root. By transitivity there exists a long root $a$ such that $(a,b)=-1$ and we have $a+b=s_a(b)$ is a short root orthogonal to $b$. Hence we have two subcases.

subcase A.1: $a+b$ is negative. Then $-a-b$ is positive and we can conclude that $-a=-a-b+b$ is a root (since $a$ was a root) which is greater in our partial ordering than $b$.

subcase A.2: $a+b$ is positive. Then $a+b+b=a+2b=s_b(a)$ is in fact a root and in fact again greater than $b$.

Case B: One of the labels of the edges of the coxeter graph is "6". Then the ratio of squared lengths of long roots and short roots is $3$. Wlog we will assume that $(a,a)=3$ for long roots and $(b,b)=1$ for short roots. Let $b$ be a positive short root. By transitivity there exists a long root $a$ such that $(a,b)=-\frac 3 2$. We have $a+b=s_a(b)$ is a short root and $(a+b,b)=-\frac 1 2$. The same subcases arise:

subcase B.1: b+a is negative. Then $-b-a$ is positive and $-a=-b-a+b$ is a root greater than $b$.

subcase B.2: b+a is positive. Then $2b+a=b+a+b=s_{a+b}(b)$ is a root which is greater than $b$.

Therefore we find an even greater root for every positive short root $b$ in our irreducible system $\varPhi$ and a greatest element with respect to our partial ordering has to be a long root. (This long root $a$ is uniquely determined since $W$ acts transitively on the long roots, $(a,c) \geq 0 \forall c \in \Delta$ and the fact that the subset of our vector space defined by this condition forms a fundamental domain for the action of $W$ on the vector space.)

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How did you get the statement "By transitivity there exists a long root a such that (a,b)=-1"? – user Jun 26 at 1:10
We assume that $\Phi$ is irreducible i.e. the Coxeter-graph is connected. Hence there are a long root $a'$ and a short root $b'$ in $\Delta$ which are connected in the Coxeter-graph (which implies $(a',b')=-1$). The Weyl group acts transitively on the short roots so there is a $w \in W$ with $wb'=b$. Furthermore $W$ acts orthogonally and hence $wa'=:a$ is a long root with $(a,b)=(wa',wb')=(a',b')=-1$. – Sebastian Schoennenbeck Jun 26 at 6:50
See my original post. All of these facts can be found or are easy corollaries to facts in Humphreys book. – Sebastian Schoennenbeck Jun 27 at 11:26
The usual, $W$ respects the inner product. – Sebastian Schoennenbeck Jun 27 at 16:17
Yes, otherwise they wouldn't even appear in the Coxeter-graph. – Sebastian Schoennenbeck Jun 27 at 16:36

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