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Suppose that $G$ is a connected Lie group of unitary matrices and $U(t)\in G$ depends continuously differentiable on a real parameter $t$ and has no real eigenvalue -1. Then the principal value of the logarithm $W(t):=\log U(t)$ is also continuously differentiable in $t$.

If $G$ is abelian, we may conclude (with dot denoting differentiation by $t$) the formula $$\dot W(t)=U(t)^{-1}\dot U(t).$$ I am looking for an explicit formula (with proof) generalizing this that expresses $\dot W(t)$ in terms of $U(t)$ and $\dot U(t)$ when $G$ is nonabelian.

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  • $\begingroup$ If $U(t)$ commutes with $\dot{U}(t)$ (at any one value of time $t$) then you get the same as above, by differentiating the Taylor series for the exponential. But if they don't commute, then I think you don't have an elementary expression for the result. $\endgroup$
    – Ben McKay
    Jul 29, 2019 at 13:07
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    $\begingroup$ There is a concise formula involving integral on stackexchange: math.stackexchange.com/a/2085547/27752 $\endgroup$ Jul 29, 2019 at 13:53
  • $\begingroup$ Can't you diagonalise your one-parameter subgroup and get the same result? $\endgroup$
    – LSpice
    Jul 29, 2019 at 14:08
  • $\begingroup$ @LSpiceL No, since the transform depends on $t$. $\endgroup$ Jul 29, 2019 at 14:13

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The comment by Vít Tuček leads to the formula $$\frac{d}{dt}\log(1+A(t))=\int_0^1 (1+sA(t))^{-1}\dot A(t)(1+sA(t))^{-1} ds.$$ This can be verified by Taylor expanding both sides if the norm of $A(t)$ is less than 1, and follows in general by analytic continuation when no real eigenvalue of $A(t)$ is $\le -1$. (The proof in https://math.stackexchange.com/a/2085547/27752 only covers the first situation.)

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