Some Lemmas on Simple Roots

If $\Delta$ is some fixed base of a root system $\Phi$ , we call the roots $\alpha\in\Delta$ “simple”. Simple roots have a number of nice properties, some of which we’ll run through now.

First off, if $\alpha\in\Phi^+$ is positive but not simple, then $\alpha-\beta$ is a (positive) root for some simple $\beta\in\Delta$ . If $\langle\alpha,\beta\rangle\leq0$ for all $\beta\in\Delta$ , then the same argument we used when we showed $\Delta(\gamma)$ is linearly independent would show that $\Delta\cup\{\alpha\}$ is linearly independent. But this is impossible because $\Delta$ is already a basis.

So $\langle\alpha,\beta\rangle>0$ for some $\beta\in\Delta$ , and thus $\alpha-\beta\in\Phi$ . It must be positive, since the height of $\alpha$ must be at least $2$ . That is, at least one coefficient of $\alpha-\beta$ with respect to $\Delta$ must be positive, and so they all are.

In fact, every $\alpha\in\Phi^+$ can be written (not uniquely) as the sum $\beta_1+\dots+\beta_{\mathrm{ht}(\alpha)}$ for a bunch of $\beta_i\in\Delta$ , and in such a way that each partial sum $\beta_1+\dots+\beta_k$ is itself a positive root. This is a great proof by induction on $\mathrm{ht}(\alpha)$ , for if $\mathrm{ht}(\alpha)=1$ then $\alpha$ is in fact simple itself. If $\alpha$ is not simple, then our argument above gives a $\beta_{\mathrm{ht}(\alpha)}$ so that $\alpha=\alpha'+\beta_{\mathrm{ht}(\alpha)}$ for some $\alpha'\in\Phi^+$ with $\mathrm{ht}(\alpha')=\mathrm{ht}(\alpha)-1$ . And so on, by induction.

If $\alpha$ is simple, then the reflection $\sigma_\alpha$ permutes the positive roots other than $\alpha$ . That is, if $\alpha\neq\beta\in\Phi^+$ , then $\sigma_\alpha(\beta)\in\Phi^+$ as well. Indeed, we write

$\displaystyle\beta=\sum\limits_{\gamma\in\Delta}k_\gamma\gamma$

with all $k_\gamma$ nonnegative. Clearly $k_\gamma\neq0$ for some $\gamma\neq\alpha$ (otherwise $\beta=\alpha$ ). But the coefficient of $\gamma$ in $\sigma_\alpha(\beta)=\beta-(\beta\rtimes\alpha)\alpha$ must still be $k_\gamma$ . Since this is positive, all the coefficients in the decomposition of $\sigma_\alpha(\beta)$ are positive, and so $\sigma_\alpha(\beta)\in\Phi^+$ . Further, it can’t be $\alpha$ itself, because $\alpha$ is the image $\sigma_\alpha(-\alpha)$ .

In fact, this leads to a particularly useful little trick. Let $\delta$ be the half-sum of all the positive roots. That is,

$\displaystyle\delta=\frac{1}{2}\sum\limits_{\beta\in\Phi^+}\beta$

then $\sigma_\alpha(\delta)=\delta-\alpha$ for all simple roots $\alpha$ . The reflection shuffles around all the positive roots other than $\alpha$ itself, which it sends to $-\alpha$ . This is a difference in the sum of $-2\alpha$ , which the $\frac{1}{2}$ turns into $-\alpha$ .

Now take a bunch of $\alpha_1,\dots,\alpha_t\in\Delta$ (not necessarily distinct) and write $\sigma_i=\sigma_{\alpha_i}$ . If $\sigma_1\dots\sigma_{t-1}(\alpha_t)\prec0$ , then there is some index $1\leq s<t$ that we can skip. That is,

$\displaystyle\sigma_1\dots\sigma_t=\sigma_1\dots\sigma_{s-1}\sigma_{s+1}\dots\sigma_{t-1}$

Write $\beta_i=\sigma_{i+1}\dots\sigma_{t-1}(\alpha_t)$ for every $i$ from ${0}$ to $t-2$ , and $\beta_{t-1}=\alpha_t$ . By our assumption, $\beta_0\prec0$ and $\beta_{t-1}\succ0$ . Thus there is some smallest index $s$ so that $\beta_s\succ0$ . Then $\sigma_s(\beta_s)\prec0$ , and we must have $\beta_s=\alpha_s$ . But we know that $\sigma_{\tau(\alpha)}=\tau\sigma_\alpha\tau^{-1}$ . In particular,

$\displaystyle\sigma_s=\left(\sigma_{s+1}\dots\sigma_{t-1}\right)\sigma_t\left(\sigma_{s+1}\dots\sigma_{t-1}\right)^{-1}$

And then we can write

$\displaystyle\begin{aligned}\sigma_1\dots\sigma_{s-1}(\sigma_s\sigma_{s+1}\dots\sigma_{t-1})\sigma_t&=\sigma_1\dots\sigma_{s-1}(\sigma_{s+1}\dots\sigma_{t-1}\sigma_t)\sigma_t\\&=\sigma_1\dots\sigma_{s-1}\sigma_{s+1}\dots\sigma_{t-1}\end{aligned}$

From this we can conclude that if $\sigma=\sigma_1\dots\sigma_t$ is an expression in terms of the basic reflections with $t$ as small as possible, then $\sigma(\alpha_t)\prec0$ . Indeed, if $\sigma(\alpha_t)\succ0$ , then

$\displaystyle\sigma_1\dots\sigma_{t-1}(\alpha_t)=\sigma_1\dots\sigma_{t-1}(\sigma_t(-\alpha_t))=-\sigma(\alpha_t)\prec0$

and we’ve just seen that in this case we can leave off $\sigma_t$ as well as some $\sigma_s$ in the expression for $\sigma$ .

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February 4, 2010 - Posted by John Armstrong | Geometry, Root Systems

8 Comments »

The following doesn’t seem to parse for me: if $\sigma_1\hdots\sigma_{t-1}(\alpha_t)$ ,

is this vector supposed to equal something?

Comment by Gilbert Bernstein | February 5, 2010 | Reply
Sorry, yes, I must have accidentally cut something..

Comment by John Armstrong | February 5, 2010 | Reply
[...] Action of the Weyl Group on Weyl Chambers With our latest lemmas in hand, we’re ready to describe the action of the Weyl group of a root system on the set [...]

Pingback by The Action of the Weyl Group on Weyl Chambers « The Unapologetic Mathematician | February 5, 2010 | Reply
Just noticed another typo:

“then $\sigma_\alpha(\delta)=\delta - \alpha$ for all simple roots $\delta$.”

The final $\delta$ should be $\alpha$ ?

Comment by Gilbert Bernstein | February 7, 2010 | Reply
Yes, thanks.

Comment by John Armstrong | February 7, 2010 | Reply
[...] of Weyl Group Elements With our theorem from last time about the Weyl group action, and the lemmas from earlier about simple roots and reflections, we can define a few notions that make discussing Weyl groups [...]

Pingback by Lengths of Weyl Group Elements « The Unapologetic Mathematician | February 8, 2010 | Reply
[...] many roots, there can be only finitely many heights, and so there is some largest height. And we know that we can get to any positive root of any height by adding more and more simple roots. So we will [...]

Pingback by From Cartan Matrix to Root System « The Unapologetic Mathematician | February 17, 2010 | Reply
[...] with a positive sign as a positive sum of the two vectors in . For example, in accordance with an earlier lemma, we can [...]

Pingback by Construction of the G2 Root System « The Unapologetic Mathematician | March 8, 2010 | Reply

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