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 We equip with the Euclidean scalar product (w.l.o.g. ). Then for

a non-zero we have the linear transformation, called

reflectionof defined by Note and for any such that

In the basis consisting of linearly independent vectors satisfying and the vector one has as the diagonal matrix It shows that in particular

A subset of equipped with the Euclidean scalar product is called

a root system ifThe elements of are called

roots, and is called therank.

An orthogonal transformation of i.e. a linear transformation satisfying and thus for all maps to an isomorphic root system. However, there is even coarser equivalence relation: we say that is isomorphic to is there is a linear transformation such that and for all In particular, multiplying every by a scalar does not change the isomorphism class of

Let us consider few examples first.

- Then there is just one example, denoted so that with, say,
- with This is denoted and we can choose
- where so This is denoted by

A system like 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 given a root system and a root system On the other hand, is an example of *irreducible*, i.e. not reducible, root system.

It is more convenient to represent in the plane in We take and A generalisation of this construction is easy, and gives one of the examples of an infinite series of irreducible root systems.

the root system , with the roots … It is not so hard to see that

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 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, and 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 there are roots and such that (otherwise we have a reducible system, ). As by 3), w.l.o.g. Using 4) twice, we have Thus with and Thus and so

- Then As we derive that Then and so we obtain (applying the reflection ) the root system Note that there are 2 copies of sitting inside.
- Then We derive that

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 and obtain but is there any other missed option? (The answer is *no*.)

If for an irreducible root system then

In other words, where is as above and cannot get extended. Observe that the squared lengths of the roots in are and

Let be such that and Replacing, if necessary, by , and considering the possibilities for to generate a root system, we can assume

Similarly, we conclude that We compute On the other hand generate a and so implying As well, generate a but a contradiction.

Thus By irreducibility, either the claim holds, or

there exists such that (Again, we can assume ) So there are 2 possibilities: either generate a or a In the 1st case As one has As one has implying But then a nonsense.

In the 2nd case Thus But the squared length ratio of two roots in an irreducible root system in is either 1, or 2, or 3, but for and it is 3/2, the final contradiction.

Next, we shall prove the following.

Let a root system, be linearly independent, and Then there is a root in the rank 2 system generated by such that Moreover, the systems generated by and roots that are not proportional to cannot both be equal to

Consider first the case when generate a subsystem Say, Then, without loss in generality, we may assume that Applying an orthogonal transformation to the space, we may assume that

and

Suppose first W.l.o.g. (for otherwise, if then we have a reducible system). Then generate , so, changing, if necessary, the sign of , we have

Thus and so as follows from If then and the inequality above implies i.e. Now we can take

(We constructed the system known as It has 12 long roots and 6 short ones.)

Now, suppose Then generate, w.l.o.g.,

(for if neither nor generate then we have a reducible system again). Thus w.l.o.g. and so and So

Assume that generate Then i.e. either or The latter case is ruled out by the inequality above. So So we can take Otherwise, and we can take

(In both cases, we constructed the system known as It has 6 long roots and 12 short ones.)

Now, by changing the roles and signs of if necessary, we may assume that and generate ‘s and that If then In the first case and so and we obtain the root system In the second case and a contradiction.

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

An irreducible root system of rank 3 is isomorphic to one of the following systems:

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