The Unapologetic Mathematician

Mathematics for the interested outsider

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.

February 4, 2010 Posted by | Geometry, Root Systems | 8 Comments

   

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