Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\Phi$ be a root system. In his Introduction to Lie algebras and Representation Theory, J. Humphreys proves that if $$\beta-p\alpha,\dots,\beta,\dots,\beta+q\alpha$$ is the $\alpha$-root string through $\beta$, then $$\frac{2(\beta,\alpha)}{(\alpha,\alpha)}=p-q.$$ Then, he states that it follows "at once" that root strings are of length at most four.

I dont get why. Of course, we know that $-4\le 2(\beta,\alpha)/(\alpha,\alpha)\le 4$, but then length of the root string is $p+q+1$, not $p-q$...

share|cite|improve this question
up vote 2 down vote accepted

By positive definiteness $0 \leq (\alpha, \beta)^2 < (\alpha, \alpha)( \beta, \beta )$ or, equivalently, $0 \leq (\alpha, \beta^\vee) (\beta, \alpha^\vee ) < 4$. Note the strict inequality.

Edit: I replaced $\langle \, , \, \rangle$ by $(\, , \,)$ to denote the inner product because of their conflicting interpretations.

share|cite|improve this answer
What does $\beta^\vee$ mean ? Moreover, I don't see how/why this helps ? And by $\langle\alpha,\beta\rangle$, you mean $2(\alpha,\beta)/(\alpha,\alpha)$ ? – Klaus Jan 7 '13 at 18:38
Thank you for the precision. However, I still don't understand why this gives $p+q+1\le 4$... Could you be more explicit ? – Klaus Jan 7 '13 at 19:04
What $\alpha$-root string do you get for $\beta + q \alpha$ instead? What inequality can you derive in that case? – WimC Jan 7 '13 at 19:15
Sorry, I must be very stupid today. I had this idea before, but the $\alpha$-root string for $\beta+q\alpha$ is $$\beta+q\alpha-(q+p)\alpha,\dots,\beta+q\alpha,\dots,\beta+q\alpha+0\alpha,$$ so we find $q+p=2(\beta,\alpha)/(\alpha,\alpha)+2q$, which we already knew... – Klaus Jan 7 '13 at 19:30
Or in other words $q+p = (\beta + q\alpha, \alpha^\vee) \in \{0, 1, 2, 3 \}$. (Note that $\beta + q\alpha$ is itself a root.) – WimC Jan 7 '13 at 19:36

\noindent I had trouble with exactly the same sentence in Humphrey's book (last sentence before the exercise on page 45 --- there was no \lq at once\rq\ for me). After googling, I ended up on this webpage, but I found the exchange of remarks tricky to follow. For the record, an easy way to prove that a root string has length at most 4 is to note that the sequence $$ \langle \beta + i \alpha, \alpha \rangle, \enspace \hbox {where $i$ is an integer}, $$ forms an AP with difference 2. If $\phi$ and~$\psi$ are any two linearly independent roots, then $$ \langle \psi, \phi \rangle =-3, -2, -1, 0, 1, 2 \hbox { or } 3, $$ from which it follows that the root string has length at most 4. \medskip

\noindent The formula $r - q = \langle \beta, \alpha \rangle$ becomes really useful slightly later when constructing the Hasse diagram of the positive roots from a base of simple roots using root strings: $$ \hbox {string length} = r + q + 1 = 2r + 1 - \langle \beta, \alpha \rangle. $$ In particular, when $r=0$ (that is, when $\beta - \alpha$ is not a root, as happens when $\alpha$ and $\beta$ are simple roots), then the string length is $1 - \langle \beta, \alpha \rangle$.

share|cite|improve this answer
The site is not entirely TeX... You don't need the \noindent's – Emily Aug 2 '13 at 3:16

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.