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2 $begingroup$ Let $U:text{Hom}(mathbb{R}^d,mathbb{R}^d) to mathbb{R}$ be a smooth function . If $U$ is orthogonally-invariant, i.e. $U(QA)=U(A)$ for every $Q in text{SO}(n),A in text{Hom}(mathbb{R}^d,mathbb{R}^d)$ , then $$ dU_{QA}(QB)=dU_{A}(B) tag{1}$$ Does the converse hold? That is, suppose $U$ is a $C^{infty}$ function whose differential satisfies equation $(1) $ for every $A,Bin text{Hom}(mathbb{R}^d,mathbb{R}^d),Qin text{SO}(n)$ . Is it true that $U(QA)=U(A)$ for every $Q in text{SO}(n)$ ? lie-groups symmetry orthogonal-matrices symmetric-functions share | cite | improve this question edited Dec 10 '18 at 9:27 ...