Killing form

Killing form

The goal of this video is to explain why the weight lattice of S ⁢ U ⁢ ( 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 × 𝔤 → 𝐂 defined by K ⁢ ( X , Y ) = Tr ⁢ ( ad X ⁢ ad Y ) .

Remember that ad X : 𝔤 → 𝔤 is the map ad X ⁢ ( Z ) = [ X , Z ] . Therefore ad X ⁢ ad Y : 𝔤 → 𝔤 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 ⁢ ( a 1 ⁢ X 1 + a 2 ⁢ X 2 , b 1 ⁢ Y 1 + b 2 ⁢ Y 2 ) = ∑ i = 1 2 ∑ j = 1 2 a i ⁢ b j ⁢ K ⁢ ( X i , Y j ) .

Proof:

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

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

Example

Example:

Let 𝔤 = 𝔰 ⁢ 𝔩 ⁢ ( 3 , 𝐂 ) and let H 13 = ( 1 0 0 0 0 0 0 0 - 1 ) and H 23 = ( 0 0 0 0 1 0 0 0 - 1 ) . Together, these span the subspace 𝔥 𝐑 = { H θ := θ 1 ⁢ H 13 + θ 2 ⁢ H 23 = ( θ 1 0 0 0 θ 2 0 0 0 - θ 1 - θ 2 ) : θ 1 , θ 2 ∈ 𝐑 } 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 ) .