The goal of this video is to explain why the weight lattice of SU(3) is drawn as the triangular lattice. The reason is that the weight lattice has an intrinsic notion of geometry (dot product, lengths, angles) associated to it, coming from a bilinear form called the Killing form

Definition:

Let little g be a Lie algebra. The Killing form of little g is the map K from little g times little g to C defined by K of X, Y equals the trace of little ad_X times little ad_Y.

Remember that little ad_X from little g to little g is the map little ad_X of Z equals X bracket Z. Therefore little ad_X times little ad_Y from little g to little g is the map Z maps to X bracket (Y bracket Z). This is a linear map, so if we pick a basis of little g we can think of it as a matrix. The Killing form of X with Y is the trace of this matrix.

Remark:

The choice of basis doesn't affect the answer: if we change basis, this conjugates the matrix of the linear map by a change of basis matrix, and trace is invariant under conjugation.

Lemma:

The Killing form is a symmetric bilinear form. This means that K(X, Y) = K(Y, X) and that K of a_1 X_1 + a_2 X_2, b_1 Y_1 + b_2 Y_2 equals the sum over i and j from 1 to 2 of a_i b_j K(X_i, Y_j).

Proof:

It's symmetric because trace of M N equals trace of N M for any two matrices M and N.

It's bilinear because (a) little ad_X is linear in X, (b) little ad_Y is linear in Y, and (c) the trace is linear on matrices.

Remark:

A more famous symmetric bilinear form is the dot product on Euclidean space. So I want to think of K as like a dot product, and use it to define lengths of and angles between vectors in the same way (using the same formulas) that we do for dot product. That's what I mean when I say that certain roots make a 120 degree angle with one another, for example. For this to be true, we will need the Killing form to have certain positivity properties when restricted to subspaces of little g. We will discuss this later. For now, let's work out an example.

Example

Example:

Let little g be little s l 3 C and let H_{1 3} be the 3-by-3 diagonal matrix with diagonal entries 1, 0, minus 1 and H_{2 3} be the 3-by-3 diagonal matrix with diagonal entries 0, 1, minus 1. Together, these span the subspace little h R, that is the set of matrices H_(theta) equals theta_1 H_{1 3} plus theta_2 H_{2 3}, or equivalently diagonal theta_1, theta_2, minus theta_1 minus theta_2, as theta_1 and theta_2 vary in R of real diagonal matrices in little s l 3 C. I want to compute all possible Killing "dot products" between these two. We will see that K(H_{1 3}, H_{1 3}) = K(H_{2 3}, H_{2 3}) = 12 and K(H_{1 3}, H_{2 3}) = 6.

If we think of this as a dot product, then this would be telling us that the length of H_{1 3} and H_{2 3} is square root of 12 and the cosine of the angle between them would be 60 degrees.

Let's proceed with the calculation. What is little ad of H_{1 3}, as a linear map from little s l 3 C to itself? Let's calculate the matrix of this linear map with respect to the basis: H_{1 3}, H_{2 3}, E_{1 2}, E_{2 1}, E_{1 3}, E_{3 1}, E_{2 3}, E_{3 2}. We have little ad of H_{1 3} applied to H_{1 3} equals H_{1 3} bracket with itself, which is zero, and little ad of H_{1 3} applied to H_{2 3} equals H_{1 3} bracket with H_{2 3}, which is zero because the matrices H_{ij} are diagonal and commute with one another. We also have little ad of H_{theta} applied to E_{i j} equals (theta_i minus theta_j) times E_{i j} and theta equals 1, 0, minus 1 for H_{1 3}, so little ad of H_{1 3} applied to E_{1 2} equals E_{1 2}, applied to E_{2 1} equals minus E_{2 1} applied to E_{1 3} equals 2 E_{1 3}, applied to E_{3 1} equals minus 2 E_{3 1} applied to E_{2 3} equals E_{2 3} and applied to E_{3 2} equals minus E_{3 2}.

As a matrix, therefore, little ad of H_{1 3} is the diagonal matrix: with diagonal entries 0, 0, 1, minus 1, 2, minus 2, 1, minus 1

Similarly, we find that little ad of H_{2 3} is the diagonal matrix with diagonal entries 0, 0, minus 1, 1, 1, minus 1, 2, minus 2.

Therefore K(H_{1 3}, H_{1 3}) is the trace of little ad of H_{1 3} squared equals the diagonal matrix with diagonal entries 0, 0, 1, 1, 4, 4, 1, 1 which is 12. Similarly for K(H_{2 3}, H_{2 3}).

Finally, we have K(H_{1 3}, H_{2 3}) equals the trace of little ad of H_{1 3} times little ad of H_{2 3}, which is the diagonal matrix with diagonal entries 0, 0, minus 1, minus 1, 2, 2, 2, 2, which has trace 6.

Pre-class exercise

Exercise:

Consider the basis of little s u 2 given by the Pauli matrices sigma_1 equals the matrix i, 0, 0, minus i; sigma_2 equals the matrix 0, 1, minus 1, 0; and sigma_3 equals the matrix 0, i, i, 0 Find all possible Killing dot products K(\sigma_i, \sigma_j).