Dual Killing form

Killing form

0.00 Recall that any Lie algebra admits a symmetric bilinear form called the Killing form $K(X,Y)=\mathrm{Tr}(\mathrm{ad}_{X}\mathrm{ad}_{Y}).$

Example:

0.25 Suppose $\mathfrak{g}$ is abelian, so the bracket is zero. Then $\mathrm{ad}_{X}=0$ for all $X$ , so $K(X,Y)=0$ for all $X$ and $Y$ . The Killing form was introduced to equip $\mathfrak{g}$ with some interesting geometry, and defining the dot product of two vectors to be identically zero isn't an interesting geometry, so we want to focus on examples where this doesn't happen.

Definition:

1.05 A symmetric bilinear form $K$ is called nondegenerate if for all $X\neq 0$ there exists $Y$ such that $K(X,Y)\neq 0$ .

1.45 We say that $\mathfrak{g}$ is semisimple if its Killing form is nondegenerate.

Remark:

I'm being immoral here. The correct definition of semisimplicity is actually quite different from this, involving diagonalisability of certain matrices. It's then a theorem ("Cartan's criterion") that semisimplicity is equivalent to nondegeneracy of the Killing form. However, the way I've set things up, we don't need to know the usual definition of semisimplicity, but we do need to know about nondegeneracy of the Killing form.

Compact groups

Here is a theorem (which we won't prove; it will be an exercise) which gives a large class of examples of semisimple Lie algebras.

Theorem:

2.50 If $G$ is a compact group then the Killing form $K\colon\mathfrak{g}\times\mathfrak{g}\to\mathbf{R}$ is (real-valued and) negative semidefinite, i.e. $K(X,X)\leq 0$ for all $X\in\mathfrak{g}$ . Moreover $K(X,X)=0$ if and only if $\mathrm{ad}_{X}=0$ , i.e. if $X$ commutes with everything in the Lie algebra.

Definition:

3.55 The centre of a Lie algebra $\mathfrak{g}$ is defined to be the set $Z(\mathfrak{g})$ of $X\in\mathfrak{g}$ such that $[X,Y]=0$ for all $Y\in\mathfrak{g}$ .

Remark:

Recall that the centre $Z(G)$ of a group $G$ is the set of all elements $g$ which commute with every other element of the group. It's possible to have Lie groups with nontrivial centre but where the Lie algebra has trivial centre, for example $SU(2)$ has $Z(SU(2))=\{\pm I\}$ while $Z(\mathfrak{su}(2))=\{0\}$ .

Corollary:

5.10 If $G$ is compact and $Z(\mathfrak{g})=0$ then $\mathfrak{g}$ is semisimple.

Proof:

Since $Z(\mathfrak{g})=0$ , $K(X,X)<0$ for all $X\in\mathfrak{g}\setminus\{0\}$ , so $\mathfrak{g}$ is semisimple.

Corollary:

6.00 Assume $Z(\mathfrak{g})=0$ . Let $T\subset G$ be a maximal torus. Let $\mathfrak{t}$ be the Lie algebra of $T$ . Let $\mathfrak{h}=\mathfrak{t}\otimes\mathbf{C}$ and $\mathfrak{h}_{\mathbf{R}}=i\mathfrak{t}$ . Then $K|_{\mathfrak{h}_{\mathbf{R}}}$ is positive definite.

Proof:

$K(X,X)<0$ , so $K(iX,iX)=-K(X,X)>0$ (unless $X=0$ ).

7.20 Another name for a positive-definite symmetric bilinear form $K\colon V\times V\to\mathbf{R}$ is a Euclidean inner product. If we pick an orthonormal basis with respect to $K$ then the change of basis matrix gives an isomorphism from $K$ to the usual Euclidean dot product $\cdot\colon V\times V\to\mathbf{R}$ . In other words, we can really think of $\mathfrak{h}_{\mathbf{R}}$ equipped with the Killing form as ordinary Euclidean space, provided our Lie algebra is the Lie algebra of a compact semisimple group.

The dual

Definition

9.25 Remember that our weight diagram lives in $\mathfrak{h}_{\mathbf{R}}^{*}$ , not in $\mathfrak{h}_{\mathbf{R}}$ .

Lemma:

If $K\colon V\times V\to\mathbf{C}$ is a nondegenerate symmetric bilinear form then $V^{*}$ inherits a symmetric bilinear form $K^{*}\colon V^{*}\times V^{*}\to\mathbf{C}$ .

Proof:

There is an isomorphism $\sharp\colon V^{*}\to V$ (its inverse is called $\flat\colon V\to V^{*}$ ) defined in the following way: $K(\alpha^{\sharp},v)=\alpha(v)$ for all $\alpha\in V^{*}$ and $v\in V$ .

11.30 This uniquely determines the map $\sharp$ because $K$ is nondegenerate (it will be an exercise to see this).
Define $K^{*}(\alpha,\beta)=K(\alpha^{\sharp},\beta^{\sharp})$ .

More concretely...

12.45 To make this a little less abstract, let's write out what this means in terms of matrices. Pick a basis $e_{1},\ldots,e_{n}$ of $V$ and define a matrix $K$ whose $i, j$ entry is $K_{ij}=K(e_{i},e_{j})$ . Then $K(v,w)=K(\sum_{i}v_{i}e_{i},\sum_{j}w_{j}e_{j})=\sum_{i,j}v_{i}w_{j}K_{ij}.$

14.05 Given $\alpha\in V^{*}$ , we can think of $\alpha$ as a row vector $\begin{pmatrix}\alpha_{1}&\cdots&\alpha_{n}\end{pmatrix}$ and we want to produce a column vector $\alpha^{\sharp}=\begin{pmatrix}\alpha_{1}^{\sharp}\\ \vdots\\ \alpha_{n}^{\sharp}\end{pmatrix}$ such that $K(\alpha^{\sharp},v)=\alpha(v)$ for all $v\in V$ .

14.50 We have $\alpha(v)=\sum\alpha_{j}v_{j}\quad K(\alpha^{\sharp},v)=\sum K_{ij}\alpha^{% \sharp}_{i}v_{j}.$ Now taking coefficients of $v_{j}$ on both sides we get $\alpha_{j}=\sum_{i}K_{ij}\alpha^{\sharp}_{i}.$

15.40 Since $K_{ij}$ is symmetric, this implies $\alpha_{j}=\sum_{i}K_{ji}\alpha^{\sharp}_{i}$ , which we can write as $\alpha^{T}=K\alpha^{\sharp}.$ Hence $\alpha^{\sharp}=K^{-1}\alpha^{T}.$

Remark:

16.30 This only makes sense if $K_{ij}$ is invertible, and this turns out to be equivalent to nondegeneracy. The flat map is then given by $v^{\flat}=(Kv)^{T}$ , which makes sense even if $K$ is degenerate.

16.50 Finally, let's define $K^{*}_{ij}$ by $K^{*}(\alpha,\beta)=\sum K^{*}_{ij}\alpha_{i}\beta_{j}$ . Since $K^{*}(\alpha,\beta):=K(\alpha^{\sharp},\beta^{\sharp})$ , we have $\sum K^{*}_{ij}\alpha_{i}\beta_{j}=\sum K_{ij}K^{-1}_{ip}\alpha_{p}K^{-1}_{jq}% \beta_{q}=\sum K^{-1}_{pq}\alpha_{p}\beta_{q}.$

18.30 Therefore $K^{*}_{ij}=K^{-1}_{ij}$ . So if we want to compute the matrix of the dual symmetric bilinear form $K^{*}$ (with respect to the dual basis), we first compute the matrix of $K$ , then we invert it.

Example

Example:

19.05 For $SU(3)$ , we calculated $K(H_{13},H_{13})=K(H_{23},H_{23})=12$ and $K(H_{13},H_{23})=6$ for the Killing form on $\mathfrak{h}_{\mathbf{R}}$ (spanned by $e_{1}=H_{13}$ and $e_{2}=H_{23}$ ). So, as a matrix $K_{ij}=K(e_{i},e_{j})$ , we get $K=\begin{pmatrix}12&6\\ 6&12\end{pmatrix}.$

20.00 On the dual space $\mathfrak{h}_{\mathbf{R}}^{*}$ , we therefore get $K^{*}=\frac{1}{108}\begin{pmatrix}12&-6\\ -6&12\end{pmatrix}.$ Let $e_{1}^{\flat}$ and $e_{2}^{\flat}$ be the dual basis of $\mathfrak{h}_{\mathbf{R}}^{*}$ . Then $|e_{i}^{\flat}|=1/3,\quad e_{1}^{\flat}\cdot e_{2}^{\flat}=-6/108,$ so if $\phi$ is the angle between these two vectors then $\cos\phi=-6\times 9/108=-1/2$ , and $\phi$ is $120$ degrees.

Remark:

21.10 This is why I was drawing my $k$ and $\ell$ axes at 120 degrees to one another in $SU(3)$ weight diagrams.