Let's break the proof into two steps as requested: --- ### **Step 1: Recall Definitions and Use Engel's Theorem** **Recall:** - A Lie algebra \( L \) is **nilpotent** if the lower central series terminates in \( \{0\} \) after finitely many steps. - **Engel's theorem** states: If all elements of \( L \) act nilpotently via the adjoint representation (\(\operatorname{ad}_x\)), then \(L\) is nilpotent. **Observation:** - If \(L\) is nilpotent, then for every \(x \in L\), the linear map \(\operatorname{ad}_x: L \to L\) defined by \(\operatorname{ad}_x(y) = [x, y]\) is a nilpotent linear transformation. **Explanation:** This means for each \( x \in L \), there exists some power \( n \) such that \( (\operatorname{ad}_x)^n = 0 \). --- ### **Step 2: Killing Form and Trace of Nilpotent Operators** The **Killing form** \( \kappa \) on \(L\) is defined as: \[ \kappa(x, y) = \operatorname{Tr}\left(\operatorname{ad}_x \circ \operatorname{ad}_y\right) \] for all \(x, y \in L\), where \( \operatorname{Tr} \) denotes the trace of a linear operator. **Since \( \operatorname{ad}_x \) is nilpotent for all \( x \in L \), the composition \( \operatorname{ad}_x \circ \operatorname{ad}_y \) is also nilpotent.** **The trace of any nilpotent linear transformation is zero.** **Explanation:** For any nilpotent endomorphism \( T \), all its eigenvalues are zero, so \( \operatorname{Tr}(T) = 0 \). Thus, \( \kappa(x, y) = 0 \) for all \( x, y \in L \). --- ### **Final Answer:** **If \( L \) is nilpotent, then the Killing form of \( L \) is identically zero.**