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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 $p-q$...

Klaus
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3 Answers3

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I had trouble with exactly the same sentence in Humphrey's book (last sentence before the exercise on page 45 --- there was no "at once" 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.

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 $1 - \langle \beta, \alpha \rangle$.

Paul Orland
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Sid
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Here is a slight refinement of Sid's argument. Using Humphreys' notation we know that $-3 \le \langle \alpha,\beta \rangle \le 3$ for all roots $\alpha,\beta$. Hence, in particular, for the $\alpha$-string through $\beta$ of length $q+r+1$ given by $\beta-r\alpha,\ldots ,\beta+q\alpha$ we find $$ \langle \beta+q\alpha,\alpha\rangle \le 3\mbox{ and} -3\le \langle \beta-r\alpha,\alpha\rangle. $$ But $\langle \alpha,\alpha\rangle=2$ therefore implies $$ 2q\le 3- \langle \beta ,\alpha\rangle \mbox{ and } 2r\le 3+\langle \beta,\alpha\rangle. $$ Thus together we find $q+r+1\le 4$.

MPTuite
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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.

WimC
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  • 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 07 '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 07 '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 07 '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 07 '13 at 19:30
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    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 07 '13 at 19:36