suppose G and H connected Lie groups. Is there $\phi: G \to H$ a morphism of Lie groups such that $d\phi$ is bijective but $\phi$ is not an isomorphism of Lie groups?

I know that $d\phi$ surjective and H connected $\Rightarrow \phi$ surjective. And I know that $d\phi$ injective $\Rightarrow Ker \phi$ is discrete.

So to come up with a potential example, I need to prove that $Ker\phi$ could be non trivial. Is that a possibility?

Many thanks for your hints or help.


Yes. Let $\phi: \mathbb{R} \to \frac{\mathbb{R}}{\mathbb{Z}}$ be the natural map. Then $d\phi: \mathbb{R} \to \mathbb{R}$ is the identity.

  • $\begingroup$ Thank you! Please correct me if I am wrong: $d\phi: Lie(\Bbb R) \to Lie(\Bbb R/\Bbb Z)$. I understand that $Lie(\Bbb R)\cong T\Bbb R = \Bbb R $ but is $Lie(\Bbb R/\Bbb Z) \cong T(\Bbb R/\Bbb Z) = \Bbb R$? $\endgroup$ – PerelMan Apr 9 at 14:20
  • 2
    $\begingroup$ @PerelMan yes. The metaphor is that the tangent space at the identity is located infinitesimally at zero so it can't "see" that you've killed $1$ which is far away, so the tangent spaces are just the same. More rigorously, if you think of a tangent vector as something that eats germs of smooth functions, a smooth function in a neighborhood of $0$ on $\mathbb{R}$ is the same thing as a smooth function in a neighborhood of $0$ on $\mathbb{R}/\mathbb{Z}$ once the neighborhood gets small enough. $\endgroup$ – hunter Apr 9 at 14:26
  • $\begingroup$ Great explanation and metaphor! many thanks @hunter $\endgroup$ – PerelMan Apr 9 at 14:34

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