## Root systems

In order to study Lie algebras, one needs a reasonable command of root systems. Root systems are closely associated with groups generated by reflections in $\mathbb{R}^n.$ We equip $\mathbb{R}^n$ with the Euclidean scalar product $(u,v)$ (w.l.o.g. $(u,v)=\sum_{i=1}^n u_i v_i$). Then for

a non-zero $a\in \mathbb{R}^n$ we have the linear transformation, called reflection $\sigma_a$ of $\mathbb{R}^n$ defined by $\sigma_a(v)=v-2\frac{(v,a)}{(a,a)}a.$ Note $\sigma_a(a)=-a$ and $\sigma_a(u)=u$ for any $u$ such that $(u,a)=0.$

In the basis consisting of $n-1$ linearly independent vectors $u_i$ satisfying $(u_i,a)=0$ and the vector $a$ one has $\sigma_a$ as the diagonal matrix $\mathrm{diag}(1,\dots,1,-1).$ It shows that in particular $\sigma_a^2=1.$

A subset $\Phi$ of $\mathbb{R}^n$ equipped with the Euclidean scalar product $(u,v)$ is called
a root system if

The elements of $\Phi$ are called roots, and $n$ is called the rank.

An orthogonal transformation of $\mathbb{R}^n,$ i.e. a linear transformation $g$ satisfying $gg^T=1,$ and thus $(gu,gv)=(u,v)$ for all $u,v\in \mathbb{R}^n,$ maps $\Phi$ to an isomorphic root system. However, there is even coarser equivalence relation: we say that $\Phi$ is isomorphic to $\Phi'$ is there is a linear transformation $g$ such that $g(\Phi)=\Phi'$ and $(u,v)/(v,v)=(\Phi(u),\Phi(v))/(\Phi(u),\Phi(u))$ for all $u,v\in\Phi.$ In particular, multiplying every $u\in\Phi$ by a scalar does not change the isomorphism class of $\Phi.$

Let us consider few examples first.

• $n=1.$ Then there is just one example, denoted $A_1,$ so that $\Phi=\{a,-a\},$ with, say, $(a,a)=2.$
• $n=2,$ $\Phi=\{a,-a,b,-b\}$ with $(a,b)=0.$ This is denoted $A_1\times A_1,$ and we can choose $(a,a)=(b,b)=2.$
• $n=2,$ $\Phi=\{a,-a,b,-b,a+b,-a-b\},$ where $a=(\sqrt{2},0), b=(\frac{-\sqrt{2}}{2},\frac{\sqrt{6}}{2}),$ so $(a,b)=-1.$ This is denoted by $A_2.$

A system like $A_1\times A_1$ is called reducible, as it consists of two completely independent root systems living in orthogonal to each other subspaces. It is obvious that in this way it is always possible to construct the root system $\Phi\times\Psi\subset\mathbb{R}^{n+m},$ given a root system $\Phi\in \mathbb{R}^n$ and a root system $\Psi\in \mathbb{R}^m.$ On the other hand, $A_2$ is an example of irreducible, i.e. not reducible, root system.
It is more convenient to represent $A_2$ in the plane $x_1+x_2+x_3=0$ in $\mathbb{R}^3.$ We take $a=(1,-1,0)$ and $b=(0,1,-1).$ A generalisation of this construction is easy, and gives one of the examples of an infinite series $A_n$ of irreducible root systems.

the root system $A_n\subset \mathbb{R}^{n+1}$, with the roots $\pm (1,-1,0,\dots,0),$ $\pm (0,1,-1,0,\dots,0),$$\pm (0,\dots,0,1,-1).$ It is not so hard to see that $|\Phi|=n(n+1).$

It turns out that a complete classification of root systems is possible. Certainly, it suffices to classify irreducible ones. The idea is that one settles the case of rank $n=2$ and then uses it to rule out almost all possibilities for pairs of roots to appear in a root system of a bigger rank. This results in 3 infinite families, $A_n,$ $B_n,$ and $D_n,$ and few exceptions.

We present here a straightforward, albeit somewhat messy, treatment for root systems of rank at most 3.

Rank 2 case
We can assume that in a rank 2 system $\Phi$ there are roots $a=(\sqrt{2},0)$ and $b=(b_1,b_2)\in\Phi-\{a,-a\}$ such that $(b,a)\neq 0$ (otherwise we have a reducible system, $A_1\times A_1$). As $\sigma_a(b)=(-b_1,b_2)\in\Phi$ by 3), w.l.o.g. $b_1>0.$ Using 4) twice, we have $b_1\sqrt{2}, \frac{2b_1\sqrt{2}}{b_1^2+b_2^2}\in \mathbb{Z}_{+}.$ Thus $b_1=\sqrt{2}k,$ with $2k\in \mathbb{Z},$ and $4k\geq 2k^2 +b_2^2.$ Thus $b_2^2\leq 2k(2-k),$ and so $k\in\{1/2,1\}.$

1. $k=1.$ Then $\frac{4}{2+b_2^2}\in \mathbb{Z}_{+}.$ As $0 < \frac{b_2^2}{2}\leq 1,$ we derive that $b_2=\pm\sqrt{2}.$ Then $\sigma_b(a)=(\pm\sqrt{2},0)=\pm\gamma,$ and so we obtain (applying the reflection $\sigma_\gamma$) the root system $B_2=\{\pm a,\pm\gamma, \pm(\alpha+\gamma),\pm(\alpha-\gamma)\}.$ Note that there are 2 copies of $A_1\times A_1$ sitting inside.
2. $k=1/2.$ Then $\frac{4}{1+2b_2^2}\in \mathbb{Z}_{+}.$ We derive that $b_2^2\in\{1/2,3/2\}.$
1. if $b_2^2=3/2$ then one obtains (a system isomorphic to) $A_2.$
2. if $b_2^2=1/2$ then one obtains $G_2,$ certain root system with 12 roots of two different squared norms (we chose them to be 1 and 2). Note that one can find two copies of $A_2$ inside, one for each norm.

We are not quite done yet; it remains to see whether we can add more roots to each of the irreducible systems we constructed. One can certainly add more roots to $A_2$ and obtain $G_2,$ but is there any other missed option? (The answer is no.)

Rank 3 case

If for an irreducible root system $\Phi\supseteq G_2$ then $G_2=\Phi.$

In other words, $G_2\cong\Delta=\{\pm a,\pm b, \pm (a+b), \pm (2a+b), \pm(a+2b),\pm(b-2a)\},$ where $a$ is as above and $b=\frac{1}{2}(-\sqrt{2},\sqrt{6}),$ cannot get extended. Observe that the squared lengths of the roots in $G_2$ are $2=(a,a)$ and $6=(2a+b,2a+b)=4(a,a)+4(a,b)+(b,b)=8-4+2.$
Let $r\in\Phi-\Delta$ be such that $(r,r)=2$ and $(a,r)\neq 0.$ Replacing, if necessary, $r$ by $-r$, and considering the possibilities for $a, r$ to generate a root system, we can assume $(a,r)=1.$
Similarly, we conclude that $\gamma=(b,r)\in \{0,\pm 1\}.$ We compute $(r,2a+b)=2+\gamma\in\{1,2,3\}.$ On the other hand $r,2a+b$ generate a $G_2,$ and so $(r,2a+b)=3,$ implying $\gamma=1.$ As well, $r,a+2b$ generate a $G_2,$ but $(r,b-2a)=1-2=-1,$ a contradiction.
Thus $(a,r)=(b,r)=0.$ By irreducibility, either the claim holds, or
there exists $r\in\Phi-\Delta$ such that $(a,r)\neq 0.$ (Again, we can assume $(a,r)>0.$) So there are 2 possibilities: either $a,r$ generate a $G_2,$ or a $B_2.$ In the 1st case $(r,r)=6,$ $(a,r)=3,$ $\gamma=(b,r)\in\{0,\pm 3\}.$ As $(r,a+b)\leq 3,$ one has $\gamma\in\{-3,0\}.$ As $r\neq 2a+b,$ one has $(r,2a+b)=6+\gamma=3,$ implying $\gamma=-3.$ But then $(r,2a-b)=9,$ a nonsense.
In the 2nd case $(r,r)=4,$ $(a,r)=2,$ $(b,r)\in\{0,\pm 2\}.$ Thus $(r,2a+b)> 0.$ But the squared length ratio of two roots in an irreducible root system in $\mathbb{R}^2$ is either 1, or 2, or 3, but for $2a+b$ and $r$ it is 3/2, the final contradiction.

Next, we shall prove the following.

Let $a,b,c\in\Phi,$ a root system, be linearly independent, and $(a,b)\neq 0.$ Then there is a root $r$ in the rank 2 system generated by $a,b$ such that $(r,c)=0.$ Moreover, the systems generated by $c$ and roots that are not proportional to $r$ cannot both be equal to $B_2.$

Consider first the case when $a,b$ generate a subsystem $B_2\cong\Delta\subset\Phi.$ Say, $2(a,a)=(b,b)=4.$ Then, without loss in generality, we may assume that $(a,b)=2.$ Applying an orthogonal transformation to the space, we may assume that
$a=(110\dots 0),$ and $b=(020\dots 0).$

Suppose first $(c,c)=4.$ W.l.o.g. $(a,c)\neq 0$ (for otherwise, if $(a,c)=(b-a,c)=0$ then we have a reducible system). Then $a,c$ generate $B_2$, so, changing, if necessary, the sign of $c$, we have $(a,c)=2.$
Thus $c=(x,2-x,z_1,\dots)$ and so $2>x>0,$ as follows from $x^2+(2-x)^2\leq 4=(c,c).$ If $(b,c)\neq 0$ then $(b,c)=4-2x=\pm 2,$ and the inequality above implies $x=1,$ i.e. $c=(1,1,z_1,\dots).$ Now we can take $r=b-a.$
(We constructed the system known as $C_3.$ It has 12 long roots and 6 short ones.)

Now, suppose $(c,c)=2.$ Then $a,c$ generate, w.l.o.g., $A_2$
(for if neither $a,c$ nor $b-a,c$ generate $A_2$ then we have a reducible system again). Thus w.l.o.g. $(a,c)=1$ and so $c=(x,1-x,z_1,\dots),$ and $x^2+(1-x)^2\leq 2=(c,c).$ So $2x^2-2x\leq 1.$
Assume that $b, c$ generate $B_2.$ Then $\pm 2=(b,c)=2(1-x),$ i.e. either $x=0,$ or $x=-2.$ The latter case is ruled out by the inequality above. So $c=(0,1,z_1,\dots).$ So we can take $r=2a-b=(20\dots 0).$ Otherwise, $(b,c)=0,$ and we can take $r=b.$
(In both cases, we constructed the system known as $B_3.$ It has 6 long roots and 12 short ones.)

Now, by changing the roles and signs of $a,b,c$ if necessary, we may assume that $a,b$ and $a,c$ generate $A_2$‘s and that $a=(1,-1,0,\dots 0),$ $b=(0,1,-1,0\dots 0),$ $c=(x,x+1,z,\dots).$ If $(b,c)\neq 0$ then $(b,c)=x+1-z=\pm 1.$ In the first case $x=z,$ and so $r=a+b,$ and we obtain the root system $A_3.$ In the second case $z=x+2,$ and $(a+b,c)=-2,$ a contradiction.

It is not so hard to see that we actually showed that

An irreducible root system $\Phi,$ of rank 3 is isomorphic to one of the following systems: $A_3,$ $B_3,$ $C_3.$