I'll describe to you in the following a procedure to construct all the roots of a simple Lie algebra. This procedure is based on Slansky's review article. It is different from the method described in the Wikipedia page.
I think that it needs as little as possible input data.
Actually, the only inputs you need is the Dynkin diagram (Given in table 5 of Slansky), the highest weights of the adjoint representation (Given in table 8) and a few rules described in the sequel to make the construction.
- Construction of the Cartan matrix from the Dynkin diagram.The elements of the cartan matrix are the scalar product of the roots divided by the second root squared length.
1.1. Unconnected roots are orthogonal, the angle betwen two roots connected by one line is 120 degrees, 2 lines 135 degrees and three lines 150 degrees.
1.2. Only the relative lengths matter, it is conventional to take the squared length of the short roots (empty dots) as 1 and of the long roots (full dots) as 2 except for $G_2$ which is taken as 3.
1.3 Now that you have constructed the Cartan matrix, you can read the primitive roots which are given by the rows of the cartan matrix.
Remarks: These are the primitive root components in the weight basis in which the primitive weights have only one nonvanishing unit component).One must remenber that the weight space is not Eucledian. It has a metric that must be used in scalar products. The metric tensor can be constructed from the inverse of the Cartan matrix according to equation 4.11 in Slansky.
- Now that you have the primitive roots, one method to construct the whole root system is to find the weights of the adjoint representation. For this purpose, one can construct the weight diagram starting from the highest weight given in table 8 of Slansky using the method described on page 31, 32, the positive roots are just the positive weights of the adjoint representation.
2.1. The highest weight is picked from table 8.
2.2. From the highest weight and from any intermediate weight the n-th primitive root is subtracted a number of times equal to the n-th component of the weight if it is positive and not subtracted it it is zero or negative.
For example, in $A_3$ the roots $\alpha_1 = (2, -1, 0)$ and $\alpha_3 = (0, -1, 2)$ are subtracted each once from the highest weight $(1, 0, 1)$ in order to obtain the second level weights: $(-1 , 1, 1)$ and $(1, 1, -1)$.
Remark: This method allows you to construct the weight diagram but not to know the weight multiplicity, but in our case of the adjoint representation, we know that the multiplicity of all the weights is one because they are roots except for the zero weight whose multiplicity is equal to the rank of the Lie algebra.
This post imported from StackExchange Mathematics at 2015-06-18 19:07 (UTC), posted by SE-user David Bar Moshe