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.

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

8 Comments »

  1. 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

  2. Sorry, yes, I must have accidentally cut something..

    Comment by John Armstrong | February 5, 2010 | Reply

  3. [...] 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

  4. 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

  5. Yes, thanks.

    Comment by John Armstrong | February 7, 2010 | Reply

  6. [...] 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

  7. [...] 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

  8. [...] with a positive sign as a positive sum of the two vectors in . For example, in accordance with an earlier lemma, we can [...]

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