Let $G$ be a Lie group endowed with a bi-invariant metric $Q$. Then $$\mathrm{Ric}(X,Y)=\mathrm{tr\ }\frac{1}{4}[[X, ·],Y]=-\frac{1}{4}B(X,Y),\qquad B(X,Y):=\mathrm{tr\ }(\mathrm{ad}(X)\circ \mathrm{ad}(Y)).$$ Where $B$ is a symmetric bilinear form called Killing form of $G$.
Claim: Ricci tensor $\mathrm{Ric}$ of Lie group $G$ is independent of the bi-invariant metric $Q$.
Is it really this clime true? Because $B$ is defined using trace operator and trace isn't independent of metric. Isn't?