Given a root system $R\subset {\mathbf{R}}^{n}$
, pick a hyperplane of irrational slope in ${\mathbf{R}}^{n}$
, that is a hyperplane through the origin which doesn't contain any other points of the weight lattice. This divides our roots into two subsets: the positive and negative roots (on either side of the hyperplane). The figure below shows examples of such hyperplanes for the ${A}_{2}$
and ${G}_{2}$
root systems, with positive roots in red and negative roots in blue.

A positive root is called simple if it cannot be written as a sum of two positive roots.

Examples

Example:

The root $\alpha $
in the ${A}_{2}$
root system above can be written as a sum of positive roots: $\alpha =\beta +\varphi $
. Therefore $\alpha $
is not simple. However, $\beta $
and $\varphi $
are simple.

Example:

The roots ${\beta}_{1}$
and ${\alpha}_{6}$
in the ${G}_{2}$
root system are simple and all the other positive roots are not simple. (Exercise)

Properties of simple roots

Simple roots span

Lemma:

Any positive root can be written as a sum of simple roots.

Proof:

If your root $\alpha $
is simple then it's a sum of one simple root and we're done. If not, then it's a sum of two positive roots: $\alpha =\beta +\gamma $
for some positive roots $\beta $
, $\gamma $
. If $\beta $
and $\gamma $
are simple then we're done, if not, we expand one of them as a sum of two positive roots. Continue in this manner. We know that this process will terminate because:

all positive roots have a certain distance from our hyperplane of irrational slope, and they all lie on one side of it;

if we add two positive roots then the result is strictly further away from this hyperplane;

therefore when we write $\alpha =\beta +\gamma $
, the distances of $\beta $
and $\gamma $
from the hyperplane are strictly smaller than that of $\alpha $
.

Since there is a finite number of positive roots, there's a finite number of values this distance can take, so we can't keep splitting up as a sum of positive roots indefinitely or we'd end up with an infinite sequence of positive roots whose distance to the hyperplane was getting strictly smaller and smaller.

Remark:

This tells us that there are at least $n$
simple roots, where $n$
is the dimension of the root system: this is because our positive roots span ${\mathbf{R}}^{n}$
, so by the lemma the simple roots also span ${\mathbf{R}}^{n}$
. In a moment we'll show there are also at most$n$
simple roots (so there are precisely $n$
simple roots).

Weyl chamber

Definition:

The Weyl chamber (with respect to our chosen hyperplane of irrational slope) is the set $$\{v\in {\mathbf{R}}^{n}:\alpha \cdot v\ge 0\text{for all positive roots}\alpha \}.$$

Example:

In the ${A}_{2}$
root system, the red, blue, and orange half-planes are the regions consisting of vectors whose dot product with $\beta $
, $\alpha $
, respectively $\varphi $
is positive. Their intersection is the Weyl chamber shaded on the right.

In this example, you see that we don't need to use $\alpha $
. This is a general fact.

Lemma:

The Weyl chamber is also equal to $$\{v\in {\mathbf{R}}^{n}:\alpha \cdot v\ge 0\text{for all simple roots}\alpha \}.$$

Proof:

By the lemma we've just proved, any positive root is a sum of simple roots (with positive coefficients). Therefore if you have positive dot product with every simple root, you also have positive dot product with every positive root.

When can you add roots to get roots?

Lemma:

If $\alpha $
and $\beta $
are (non-collinear) roots such that $$
then $\alpha +\beta $
is a root.

Example:

In the ${A}_{2}$
example, $$
and $\beta +\varphi =\alpha $
is another root.

Proof:

The proof is an exercise, but here's how the exercise goes. We have two roots $\alpha $
and $\beta $
, so we can focus on the plane ${\mathbf{R}}^{2}$
spanned by $\alpha $
and $\beta $
. The angle between $\alpha $
and $\beta $
is one of $90$
, $60$
, $45$
and $30$
degrees or the complementary angles $120$
, $135$
, $150$
if obtuse (the constraint on angles was about the smallest angle between the corresponding lines). Since the dot product is assumed strictly negative, the only possibilities are $120$
, $135$
or $150$
degrees. This means there are three cases. We'll just see how this works for the case of $120$
degrees.

Rescale everything until $\alpha $
has length 1 (for simplicity) and rotate until $\alpha $
is horizontal. We know that ${P}_{\alpha}(\beta )=\frac{1}{2}{n}_{\beta \alpha}\alpha $
, ${P}_{\beta}(\alpha )=\frac{1}{2}{n}_{\alpha \beta}\beta $
, and $${n}_{\beta \alpha}{n}_{\alpha \beta}=4{\mathrm{cos}}^{2}\varphi =1,$$
because $\varphi =120$
degrees so ${n}_{\alpha \beta}={n}_{\beta \alpha}=\pm 1$
. For positive dot product we get ${n}_{\alpha \beta}={n}_{\beta \alpha}=-1$
, which tells us exactly which vector $\beta $
is: $(-1/2,\sqrt{3}/2)$
.

Now we take the lines ${\pi}_{\alpha}$
orthogonal to $\alpha $
and ${\pi}_{\beta}$
orthogonal to be $\beta $
and reflect e.g. $\beta $
in ${\pi}_{\alpha}$
to construct the root $\alpha +\beta $
(see the figure below).

There are now two more cases to analyse (exercise).

Dot products between simple roots

Lemma:

If $\alpha $
and $\beta $
are simple roots then $\alpha \cdot \beta \le 0$
.

Proof:

Assume $\alpha \cdot \beta >0$
. By the previous result, since $$
, we know that $\alpha -\beta $
is a root.

If $\alpha -\beta $
is a positive root then $$\alpha =(\alpha -\beta )+\beta $$
so $\alpha $
is a sum of two positive roots, which contradicts the assumption that $\alpha $
is simple.

If $\alpha -\beta $
is a negative root then $$\beta =-(\alpha -\beta )+\alpha $$
so $\beta $
is a sum of two positive roots, which contradicts the assumption that $\beta $
is simple.

In either case, we get a contradiction.

Simple roots form a basis

Lemma:

The simple roots are linearly independent.

Remark:

Combined with the earlier fact that they span ${\mathbf{R}}^{n}$
, this means that they form a basis of ${\mathbf{R}}^{n}$
, so there are precisely $n$
of them.

Proof:

The proof uses two facts:

all simple roots lie on one side of our chosen hyperplane of irrational slope;

all simple roots have nonpositive dot products with one another.

Suppose there is a linear dependence between the simple roots: $$\sum {c}_{\alpha}\alpha =0$$
where the sum is over the simple roots with coefficients ${c}_{\alpha}\in \mathbf{R}$
.

If all nonzero coefficients ${c}_{\alpha}$
are positive then we get a contradiction because we're taking a sum (with positive coefficients) of things all lying on one side of a hyperplane, so the result will lie even further from the hyperplane (and hence cannot be zero).

Let's rewrite the linear dependence as $$
that is grouping the terms with positive coefficients on the left-hand side and minus the terms with negative coefficients on the right-hand side. Let's write $v:={\sum}_{{c}_{\alpha}>0}{c}_{\alpha}\alpha $
. By the linear dependence, we have $$
.

The squared length of $v$
is $$
but this is a sum of nonpositive dot products $\alpha \dot{\beta}$
with nonnegative coefficients ${c}_{\alpha}|{c}_{\beta}|$
, so $v\cdot v\le 0$
. Therefore $v=0$
, so $$\sum _{{c}_{\alpha}>0}{c}_{\alpha}\alpha =0$$
and this is a linear dependence with positive coefficients, which we've ruled out already.

Corollary:

There are precisely $n$
simple roots.

Proof:

We've proved that the simple roots form a basis for ${\mathbf{R}}^{n}$
.