# Simple roots

## Simple roots

### Definition

Definition:

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: $\alpha=\beta+\phi$ . Therefore $\alpha$ is not simple. However, $\beta$ and $\phi$ 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 $\{v\in\mathbf{R}^{n}\ :\ \alpha\cdot v\geq 0\mbox{ for all positive roots }% \alpha\}.$

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 $\phi$ 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 $\{v\in\mathbf{R}^{n}\ :\ \alpha\cdot v\geq 0\mbox{ for all simple roots }% \alpha\}.$

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 $\alpha\cdot\beta<0$ then $\alpha+\beta$ is a root.

Example:

In the $A_{2}$ example, $\beta\cdot\phi<0$ and $\beta+\phi=\alpha$ 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 $n_{\beta\alpha}n_{\alpha\beta}=4\cos^{2}\phi=1,$ because $\phi=120$ 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 $\alpha\cdot\beta\leq 0$ .

Proof:

Assume $\alpha\cdot\beta>0$ . By the previous result, since $\alpha\cdot(-\beta)<0$ , 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 $\sum_{c_{\alpha}>0}c_{\alpha}\alpha=\sum_{c_{\beta}<0}|c_{\beta}|\beta,$ 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 $v=\sum_{c_{\beta}<0}|c_{\beta}|\beta$ .

The squared length of $v$ is $v\cdot v=\left(\sum_{c_{\alpha}>0}c_{\alpha}\alpha\right)\cdot\left(\sum_{c_{% \beta}<0}|c_{\beta}|\beta\right)=\sum\sum c_{\alpha}|c_{\beta}|\alpha\cdot\beta,$ but this is a sum of nonpositive dot products $\alpha\dot{\beta}$ with nonnegative coefficients $c_{\alpha}|c_{\beta}|$ , so $v\cdot v\leq 0$ . 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 $\mathbf{R}^{n}$ .