# Differential of conjugation map is smooth

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?

• 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). Jun 3 at 23:09

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.