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

  1. |\Phi|<\infty, 0\not\in\Phi, and \Phi spans \mathbb{R}^n
  2. if v\in\Phi then \{\alpha v\mid \alpha\in \mathbb{R}\}=\{v,-v\}.
  3. for any a,b\in\Phi one has \sigma_a(b)\in\Phi, i.e. \sigma_a leaves \Phi invariant.
  4. for any a,b\in\Phi one has 2\frac{(b,a)}{(a,a)}\in \mathbb{Z}.

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.


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