Let $G$ be a Lie group and define the conjugation map to be $c_g(x) = gxg^{-1}$. I have often seen the claim that the differential $(c_g)_*$ is smooth (for example, in the proof of the adjoint representation being smooth), but this is not obvious to me. I know that $c_g$ itself is a smooth map.

How would one prove this? More generally, is the differential of a smooth map always smooth?

  • $\begingroup$ You need to make your question precise. $(c_g)_*$ is a linear map of vector spaces, so obviously smooth. Are you asking if it varies smoothly with $g$? To make sense of this you have to talk about tangent bundles (which are trivial in the case of a Lie group). $\endgroup$ Jun 3 at 23:09

1 Answer 1


This is basically by definition of smoothness: smooth means infinitely differentiable, so the derivative is still smooth. More precisely, suppose $M$ and $N$ are smooth manifolds and $f:M\to N$ is smooth and consider the differential $df:TM\to TN$. Picking local coordinates on $M$ and $N$, $f$ locally looks like a map $f:\mathbb{R}^m\to\mathbb{R}^n$ and then $df$ locally is the map $\mathbb{R}^m\times\mathbb{R}^m\to\mathbb{R}^n\times\mathbb{R}^n$ which sends $(x,v)$ to $(f(x),D_xv)$ where $D_x$ is the $n\times m$ matrix whose entries are the partial derivatives of the components of $f$ at $x$. Since $f$ is infinitely differentiable, so are all these partial derivatives, so $(x,v)\mapsto (f(x),D_xv)$ is infinitely differentiable.

  • $\begingroup$ Thank you very much! So this would imply the differential of a smooth map is always smooth? $\endgroup$
    – CBBAM
    Jun 3 at 18:05
  • 2
    $\begingroup$ That's correct. $\endgroup$ Jun 3 at 18:12

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