Killing form for a different representation

Question

In this Math Overflow question, the OP was asking if the Killing form defined for a different representation (than the adjoint) was related to the normal Killing form (defined w.r.t. the adjoint rep.):

$$K_\phi (X,Y):=\text{Tr}\left(\phi(X)\phi(Y)\right)$$

where $\phi$ is a representation of a Lie algebra, and $X,Y\in\mathfrak{g}$ are elements of the Lie algebra. The accepted answer states that if the adjoint representation is irreducible (i.e. the algebra is simple), then indeed the Killing form is proportional to the identity (in any representation).

Here is the accepted answer to the MO question.

They are proportional if $g$ is simple. The form $K_\phi$ defines a homomorphism from the adjoint to the coadjoint representation. If the adjoint representation is irreducible, i.e. $g$ is simple, you know all such homomorphisms are proportional by Schur's lemma.

How do you prove this rigorously? I would equally appreciate a link/reference to somewhere that does show this.

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My thoughts so far.

I can certainly see that the Killing form defines a map from $\mathfrak g$ to $\mathfrak g^*$, where $\mathfrak g^*$ is the dual Lie algebra (i.e. linear functionals over $\mathfrak g$).

$$\begin{align} f: &\mathfrak g \longrightarrow \mathfrak g^*\\ &\\ &X \longrightarrow K_\phi(X,\cdot ) \end{align}$$

Suppose the adjoint representation is irreducible. Then $\text{Ad}$ is an irreducible representation of $\mathfrak g$ over the vector space $\mathfrak g$ (itself). Likewise, the coadjoint representation $\text{Ad}^*$ is an irreducible representation of $\mathfrak g$ over the (dual) vector space $\mathfrak g^*$.

Note: I am actually not so familiar with the coadjoint representation, nor with the precise notion of the dual Lie algebra $\mathfrak g^*$. I am of course familiar with duals of finite-dimensional vector spaces, but, for example, how does the Lie bracket carry over to the dual Lie algebra? Also, is $\mathfrak g \cong \mathfrak g^*$, as it is with finite-dimensional vector spaces?

Now, assuming $\mathfrak g \cong \mathfrak g^*$, we have the following diagram.

If I can simply show that this diagram commutes, i.e. that $\text{Ad}^* \circ f = f \circ \text{Ad}$ as maps, then by Schur's 2nd lemma I can conclude that $f\propto \text{Id}$, where $\text{Id}$ is the identity map between $\mathfrak g$ and itself.

Am I on the right track, or am I going in the wrong direction?

Answer 1

You are on the right track. The fact that $\mathfrak g^*$ is called the "dual of the Lie algebra" does not mean that it is itself a Lie algebra. One just takes the dual of the vector space $\mathfrak g$ and endows it with the representation dual to the adjoint representation. This means that for $\phi\in\mathfrak g^*$ and $X\in\mathfrak g$, you get $(X\cdot\phi)(Y)=-\phi([X,Y])$. If $B$ denotes the Killing form then for $Z\in\mathfrak g$, let us write $B_Z\in\mathfrak g^*$ for the functional $Y\mapsto B(Y,Z)$. Then $(X\cdot B_Z)(Y)=-B([X,Y],Z)$ while $B_{[X,Z]}(Y)=B(Y,[X,Z])$. Thus invariance of the Killing form exactly means that it is a morphism of representations from $\mathfrak g$ to $\mathfrak g^*$.

Answer 2

In a Lie algebra $\mathfrak{g}$ over a field $K$, you have the space of invariant symmetric bilinear forms $B_\mathfrak{g}$, and its subset $B_\mathfrak{g}^{\mathrm{rep}}$ of those that occur as Killing form of some finite-dimensional representation. The latter is an additive submonoid of $B_\mathfrak{g}$.

Suppose now that $K$ has characteristic zero.

If $\mathfrak{g}$ is absolutely simple, $B_\mathfrak{g}$ is 1-dimensional, and its nonzero elements are pairwise proportional (and apart from zero are non-degenerate).

If $\mathfrak{g}$ is semisimple, then

• $B_\mathfrak{g}^{\mathrm{rep}}$ is countable (and actually a finitely generated submonoid)
• the Killing form associated to any finite-dimensional faithful representation is non-degenerate
• $B_\mathfrak{g}$ is 1-dimensional iff $\mathfrak{g}$ is absolutely simple (in particular it's not the case if $\mathfrak{g}$ is simple but not absolutely simple)
• say when $K$ is the real field, and $\mathfrak{g}$ is a complex simple Lie algebra, viewed as a semisimple, simple, not absolutely simple real Lie algebra: then, if its complex Killing form is $b+ib'$ with $b,b'$ real-valued, then $(b,b')$ is a basis of $B_\mathfrak{g}$. In such a case I see no reason why all elements of $B_\mathfrak{g}^{\mathrm{rep}}$ should be proportional, and the argument copied from the MO answer does not prove this, apparently assuming that the Lie algebra is absolutely simple.
• If $\mathfrak{g}/I$ is the largest reductive quotient of $\mathfrak{g}$, then every element of $B_\mathfrak{g}^{\mathrm{rep}}$ contains $I$ in its kernel (i.e., vanishes on $\mathfrak{g}\times I$). For instance, if $\mathfrak{g}$ is solvable, $I=[\mathfrak{g},\mathfrak{g}]$. While, in contrast, $B_\mathfrak{g}$ might be large: there exist finite-dimensional solvable (resp. nilpotent) Lie algebras of arbitrary large solvability (resp. nilpotency) length with non-degenerate symmetric bilinear forms.