# The Unapologetic Mathematician

## 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 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 $\sigma$.