Killing form

Killing form

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 𝔤 be a Lie algebra. The Killing form of 𝔤 is the map K:𝔤×𝔤𝐂 defined by K(X,Y)=Tr(adXadY).

Remember that adX:𝔤𝔤 is the map adX(Z)=[X,Z]. Therefore adXadY:𝔤𝔤 is the map Z[X,[Y,Z]] . This is a linear map, so if we pick a basis of 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(a1X1+a2X2,b1Y1+b2Y2)=2i=12j=1aibjK(Xi,Yj) .

Proof:

It's symmetric because Tr(MN)=Tr(NM) for any two matrices M and N .

It's bilinear because (a) adX is linear in X , (b) adY 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 𝔤 . We will discuss this later. For now, let's work out an example.

Example

Example:

Let 𝔤=𝔰𝔩(3,𝐂) and let H13=(10000000-1) and H23=(00001000-1) . Together, these span the subspace 𝔥𝐑={Hθ:= of real diagonal matrices in 𝔰 𝔩 ( 3 , 𝐂 ) . I want to compute all possible Killing "dot products" between these two. We will see that K ( H 13 , H 13 ) = K ( H 23 , H 23 ) = 12 and K ( H 13 , H 23 ) = 6 .

If we think of this as a dot product, then this would be telling us | H 13 | = | H 23 | = 12 and the cosine of the angle between them would be 60 degrees.

Let's proceed with the calculation. What is ad H 13 : 𝔰 𝔩 ( 3 , 𝐂 ) 𝔰 𝔩 ( 3 , 𝐂 ) ? Let's calculate the matrix of this linear map with respect to the basis: H 13 , H 23 , E 12 , E 21 , E 13 , E 31 , E 23 , E 32 . We have ad H 13 ( H 13 ) = [ H 13 , H 13 ] = 0 ad H 13 ( H 23 ) = [ H 13 , H 23 ] = 0 because the matrices H i j are diagonal and commute with one another. We also have ad H θ ( E i j ) = ( θ i - θ j ) E i j and θ = ( 1 , 0 , - 1 ) for H 13 , so ad H 13 ( E 12 ) = E 12 , ad H 13 ( E 21 ) = - E 21 ad H 13 ( E 13 ) = 2 E 13 , ad H 13 ( E 31 ) = - 2 E 31 ad H 13 ( E 23 ) = E 23 , ad H 13 ( E 32 ) = - E 32 .

As a matrix, therefore, ad H 13 is the diagonal matrix: ad H 13 = ( 0 0 1 - 1 2 - 2 1 - 1 ) .

Similarly, we find that ad H 23 = ( 0 0 - 1 1 1 - 1 2 - 2 ) .

Therefore K ( H 13 , H 13 ) is the trace of ( 0 0 1 - 1 2 - 2 1 - 1 ) 2 = ( 0 0 1 1 4 4 1 1 ) which is 12 . Similarly for K ( H 23 , H 23 ) .

Finally, we have K ( H 13 , H 23 ) = Tr ( ( 0 0 1 - 1 2 - 2 1 - 1 ) ( 0 0 - 1 1 1 - 1 2 - 2 ) ) = - 1 - 1 + 2 + 2 + 2 + 2 = 6 .

Pre-class exercise

Exercise:

Consider the basis of 𝔰 𝔲 ( 2 ) given by the Pauli matrices σ 1 = ( i 0 0 - i ) , σ 2 = ( 0 1 - 1 0 ) , σ 3 = ( 0 i i 0 ) Find all possible Killing dot products K ( σ i , σ j ) .