Nilpotent ideal is in nullspace for Killing form

Statement

Suppose $F$ is a field, $L$ is a Lie algebra over $F$, $\kappa$ is the Killing form (?) on $L$, and $A$ is a Nilpotent ideal (?) of $L$. Then, for $x\in A,y\in L$, $\!\kappa (x,y)=0$.

Related facts

Weaker facts

• Abelian ideal is in nullspace for Killing form

Other related facts

• Killing form on ideal equals restriction of Killing form
• Cartan's first criterion
• Cartan's second criterion

Facts used

1. Lower central series members are derivation-invariant
2. Derivation-invariant subring of ideal implies ideal

Proof

Given: A field $F$, a Lie algebra $L$ over $F$, a class $c$-nilpotent ideal $A$ of $L$. $\kappa$ is the Killing form on $L$.

To prove: For $x\in A,y\in L$, $\kappa (x,y)=0$.

Proof: Consider:

$\operatorname {ad} (x)\circ (\operatorname {ad} (y)\circ \operatorname {ad} (x))^{c}$.

The right-most $\operatorname {ad} (x)$ sends any element of $L$ inside $A$, since $A$ is an ideal. $\operatorname {ad} (y)$ again sends this inside $A$, since $A$ is an ideal.

The next $\operatorname {ad} (x)$ now sends the element inside $[A,A]$. Since $[A,A]$ is a derivation-invariant subring of $A$ (fact (1)) which is an ideal in $L$, $[A,A]$ is an ideal in $L$ (fact (2)). So $\operatorname {ad} (y)$ sends the element within $[A,A]$.

Inductively, after $d$ steps, the element is in $[\dots [A,A],A],\dots ,A]$ with $d$ $A$s. Applying $\operatorname {ad} (x)$ sends it to $[\dots [A,A],\dots A]$ with $d+1$ $A$s. This is a derivation-invariant subring of $A$ which is an ideal of $L$, so it is an ideal of $L$. So $\operatorname {ad} (y)$ preserves it.

Thus, $(\operatorname {ad} (y)\circ \operatorname {ad} (x))$ is in $A_{c}=[\dots [A,A],\dots ,A]$ where $A$ is repeated $c$ times. Applying $\operatorname {ad} (x)$ to this sends it to $A_{c+1}$, which is zero since $A$ has class $c$. Thus:

$\operatorname {ad} (x)\circ (\operatorname {ad} (y)\circ \operatorname {ad} (x))^{c}=0$

From this, we get that $(\operatorname {ad} (x)\circ \operatorname {ad} (y))^{c+1}=0$. Thus, $\operatorname {ad} (x)\circ \operatorname {ad} (y)$ is nilpotent, so it has trace zero, so $\kappa (x,y)=0$.