Given a root system R inside R n, pick a hyperplane of irrational slope in 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
A positive root is called
The root alpha in the A_2 root system above can be written as a sum of positive roots: alpha equals beta plus phi. Therefore alpha is not simple. However, beta and phi are simple.
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)
Any positive root can be written as a sum of simple roots.
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 equals beta plus 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 equals beta plus 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.
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 R n, so by the lemma the simple roots also span R n. In a moment we'll show there are also at most n simple roots (so there are precisely n simple roots).
The
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.
The Weyl chamber is also equal to the set of v in R n such that alpha dot v is nonnegative for all simple roots alpha.
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.
If alpha and beta are (non-collinear) roots such that alpha dot beta is strictly negative then alpha plus beta is a root.
In the A_2 example, beta dot phi is negative and beta plus phi equals alpha is another root.
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 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 of beta equals a half n beta alpha times alpha, P beta alpha equals a half n alpha beta times beta, and n beta alpha times n alpha beta equals 4 cos squared phi, which equals 1 because phi is 120 degrees so n alpha beta equals n beta alpha equals minus 1. For positive dot product we get n_{\alpha\beta}=n_{\beta\alpha}=-1, which tells us exactly which vector beta is: minus a half, root 3 over 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 plus beta (see the figure below).
There are now two more cases to analyse (exercise).
If alpha and beta are simple roots then alpha dot beta is nonpositive.
Assume alpha dot beta is strictly positive. By the previous result, since alpha dot minus beta is strictly negative, we know that alpha minus beta is a root.
If alpha minus beta is a positive root then alpha equals (alpha minus beta) plus beta so alpha is a sum of two positive roots, which contradicts the assumption that alpha is simple.
If alpha minus beta is a negative root then beta equals minus (alpha minus beta) plus 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.
The simple roots are linearly independent.
Combined with the earlier fact that they span R n, this means that they form a basis of R n, so there are precisely n of them.
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 of c_alpha times alpha where the sum is over the simple roots with coefficients c_alpha, real numbers.
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 of c_alpha times alpha, summing over alpha such that c_alpha is positive, equals the sum of (absolute value of c_beta) times beta, summing over beta such that c_beta is negative 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 for the sum of c_alpha times alpha, summing over the alpha for which c_alpha is positive. By the linear dependence, we have v is also equals to the sum of (absolute value of c_beta) times beta, summing over beta such that c_beta is negative.
The squared length of v is v dot v equals the sum of c_alpha times alpha over positive c_alphas dotted with the sum of absolute value of c_beta times beta over negative c_betas, which equals the double sum of c_alpha times absolute value of c_beta times alpha dot beta, but this is a sum of nonpositive dot products alpha dot beta with nonnegative coefficients c_alpha times absolute value c_beta, so v dot v less than or equal to zero. Therefore v = 0, so the sum of c_alpha times alpha over positive c_alphas, equals zero and this is a linear dependence with positive coefficients, which we've ruled out already.
There are precisely n simple roots.
We've proved that the simple roots form a basis for R n.