For a finite-dimensional <Lie algebra representation> $\rho$ on $V$, the <Trace form of a Lie algebra representation> is
$$
(x,y)_V=\operatorname{tr}\bigl(\rho(x)\rho(y)\bigr).
$$
Write again $A=\rho(a)$, $B=\rho(b)$, and $C=\rho(c)=[A,B]$. The operator $C$ commutes with both $A$ and $B$. Direct use of the <cyclic property of the trace> gives
$$
\begin{aligned}
\operatorname{tr}(CA)
&=\operatorname{tr}(ABA-BA^2)=0,\\
\operatorname{tr}(CB)
&=\operatorname{tr}(AB^2-BAB)=0,\\
\operatorname{tr}(C^2)
&=\operatorname{tr}(ABC-BAC)=0.
\end{aligned}
$$
In the last line, cyclicity and $AC=CA$ turn $\operatorname{tr}(ABC)$ into $\operatorname{tr}(BAC)$. Thus the nonzero vector $c\in\mathfrak g$ is orthogonal to the basis $a,b,c$, and hence to all of $\mathfrak g$. The <bilinear form> is therefore <degenerate bilinear form>[degenerate].

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 102 / 1 / iii / Solution

{wiki=Bilinear_form#Definitions}

A bilinear form is degenerate when its <radical of a bilinear form>[radical] contains a nonzero vector.

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