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Section 3.6 The Killing Form

There is a natural inner product on any Lie algebra, known as the Killing form.  1  The formal definition of the Killing form \(B\) is

\begin{equation} B(X,Y) = \tr(\ad(X)\circ\ad(Y))\tag{3.6.1} \end{equation}

in terms of the adjoint representation. Let's unpack these symbols.

First of all, the adjoint map \(\ad\) is just the identification of elements \(X\in\gg\) with their matrix representation when acting on \(\gg\) itself via commutation. As discussed in Section 3.5, this map takes basis elements to structure constants, that is

\begin{equation} \ad(e_p) = C_p = (C^n{}_{pm})\tag{3.6.2} \end{equation}

which of course extends linearly to any element \(X\in\gg\text{.}\) Next, The action of \(\ad(X)\) on \(\gg\) is, by definition, commutation, so the composition in the definition of the Killing form refers to a nested commutator, such as

\begin{equation} (\ad(e_p)\circ(\ad(e_q))(e_m) = \bigl[e_p,[e_q,e_m]\bigr] .\tag{3.6.3} \end{equation}

Finally, the trace refers to picking out the coefficient of \(e_m\) in this last expression, then summing over \(m\text{.}\)

Let's work this out. We know that

\begin{equation} \bigl[e_p,[e_q,e_m]\bigr] = [e_p,C^n{}_{qm}\,e_n] = C^a{}_{pn}\,C^n{}_{qm}\,e_a .\tag{3.6.4} \end{equation}

To take the trace, we must first identify the coefficient of \(e_m\) in this expression, then sum over \(m\text{,}\) obtaining finally

\begin{equation} B(e_p,e_q) = C^m{}_{pn}\,C^n{}_{qm}\tag{3.6.5} \end{equation}

where there is of course a double sum over \(m,n\text{.}\) The right-hand side is symmetric in \(p\) and \(q\text{,}\) establishing that \(B\) is a symmetric bilinear form.

We now have two products on elements of a Lie algebra, one of which is symmetric (the Killing form \(B\)), the other of which is antisymmetric (the commutator). What happens if we put them together?

It turns out that the resulting triple product is cyclic, that is,

\begin{equation} B\bigl(X,[Y,Z]\bigr) = B\bigl(Z,[X,Y]\bigr) .\tag{3.6.6} \end{equation}

We can verify this result using components. It is enough to compute

\begin{equation} B\bigl(e_p,[e_m,e_q]\bigr) = C^n{}_{mq}\,B(e_p,e_n) = C^n{}_{mq}\,C^a{}_{pb}\,C^b{}_{na}\tag{3.6.7} \end{equation}

and show that the final expression is cyclic in \(m,p,q\text{.}\) Using the Jacobi identity in the form (3.5.9), we have

\begin{equation} C^n{}_{mq}\,C^b{}_{na} = C^b{}_{mn}\,C^n{}_{qa} - C^b{}_{qn}\,C^n{}_{ma} .\tag{3.6.8} \end{equation}

Inserting (3.6.8) into (3.6.7) now results in

\begin{align} B\bigl(e_p,[e_m,e_q]\bigr) \amp= (C^b{}_{mn}\,C^n{}_{qa} - C^b{}_{qn}\,C^n{}_{ma})\,C^a{}_{pb}\notag\\ \amp= C^b{}_{mn}\,C^n{}_{qa}\,C^a{}_{pb} - C^b{}_{pn}\,C^n{}_{qa}\,C^a{}_{mb}\tag{3.6.9} \end{align}

where we have renamed the dummy indices in the last term. But this expression is clearly antisymmetric in \(m,p\text{,}\) and we have shown that

\begin{equation} B\bigl(e_p,[e_m,e_q]\bigr) = -B\bigl(e_m,[e_p,e_q]\bigr) = +B\bigl(e_m,[e_q,e_p]\bigr)\tag{3.6.10} \end{equation}

as claimed.

Since \(B\) is symmetric, we have therefore also shown that

\begin{equation} B\bigl(e_m,[e_p,e_q]\bigr) + B\bigl(e_p,[e_m,e_q]\bigr) = 0\tag{3.6.11} \end{equation}

which can in turn be rewritten as

\begin{equation} B\bigl([e_q,e_p],e_m\bigr) + B\bigl(e_p,[e_q,e_m]\bigr) = 0 .\tag{3.6.12} \end{equation}

We claim that (3.6.12) establishes that the Killing form is invariant under the action of the Lie group. What does this mean?

For matrix groups, it is straightforward to see how the Lie group \(G\) acts on its Lie algebra \(\gg\text{.}\) If \(M\in G\) and \(X\in\gg\text{,}\) then \(M\) maps \(X\) to \(MXM^{-1}\text{.}\) This operation is called, somewhat confusingly, (matrix) conjugation. So the claim is that

\begin{equation} B(MXM^{-1},MYM^{-1}) = B(X,Y) .\tag{3.6.13} \end{equation}

We ask instead what this invariance means at the Lie algebra level. In effect, we differentiate \(M\text{,}\) replacing it by its infinitesimal version. We must, of course, use the product rule. The result? The claim is now that

\begin{equation} B\bigl([A,X],Y\bigr) + B(\bigl(X,[A,Y]\bigr) = 0 .\tag{3.6.14} \end{equation}

But expanding (3.6.14) in terms of a basis reduces it to expressions of the form (3.6.12), which we have just verified. Thus, the cyclic nature of the triple product is a manifestation of the invariance of the Killing form under the action of the Lie group.

The Killing form is named after mathematician Wilhelm Killing, who independently discovered Lie algebras and worked out many of their properties.