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Let $\Lambda$ is a constant matrix such that its space derivative is zero $\nabla\Lambda=0$, and $U$ is some element of Lie group such that $UU^{-1}=1$. Will it be possible that $$Tr(\Lambda U\nabla U^{-1})=Tr(\nabla Q)$$

Where $Q=U^{-1}\Lambda U$. The matrix $U$ has space dependence and obviously $\nabla (UU^{-1})=0$.

The reason for my confusion is the following, $$Tr(\nabla Q) = Tr(\nabla[U^{-1}\Lambda U])=Tr[(\nabla U^{-1})(\Lambda U)+(U^{-1}\Lambda)(\nabla U)]$$ Which is not $Tr(\Lambda U\nabla U^{-1})$. The extra term is not zero, or I am missing something?

Mass
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  • Yes, trace(AB)=trace(BA) and you can just rearrange how desired equation. – Nurator Jun 08 '21 at 07:57
  • @Nurator. I got your point, $Tr(\Lambda U\nabla U^{-1})=Tr[(\nabla U^{-1}][\Lambda U])$. But Is not the derivative acts on $\Lambda U$ as well? – Mass Jun 08 '21 at 08:02
  • Sorry yeah you are right. I have no idea how the derivative interacts with the trace. – Nurator Jun 08 '21 at 13:26

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