0
$\begingroup$

I am wondering how to show that the Lie functor taking Lie groups to Lie algebras is faithful? In particular, I am looking for a constructive proof, since I am working in the context of synthetic differential geometry (and thus restricted to intuitionistic logic).

Sorry if this is an obvious question, but my algebra background isn't that strong, and I mostly taught myself what little Lie theory I know.

$\endgroup$
1
  • $\begingroup$ This is false unless you restrict to connected Lie groups. I am not sure what's constructive, but what proofs have you seen? There's two that I know: one that first passes through formal group laws (in Serre's book on Lie groups) and one that passes through foliations (you can find this in Lee's book). $\endgroup$ – user98602 Jan 4 '16 at 14:34
1
$\begingroup$

There is a reference with a proof given here. It says "the second Lie theorem establishes that this functor is fully faithful". See also here.

For non-connected Lie groups the Lie functor is not faithful, see page $35$ here for "explicit counterexamples".

$\endgroup$
5
  • 1
    $\begingroup$ "Faithful" is not synonymous with "injective on objects"... In fact these two properties are quite unrelated. Did I misunderstand your answer? $\endgroup$ – Najib Idrissi Jan 4 '16 at 14:38
  • 1
    $\begingroup$ Sorry, this was not correct. I included the nLab reference now. $\endgroup$ – Dietrich Burde Jan 4 '16 at 14:48
  • $\begingroup$ Thanks! For some reason I had thought that the Lie functor was faithful for the entire category of Lie groups. Do you happen to know of an explicit counterexample by the way (i.e. a pair of distinct Lie group homomorphisms which map to the same Lie algebra homomorphism under the Lie functor?) $\endgroup$ – ಠ_ಠ Jan 5 '16 at 0:16
  • 1
    $\begingroup$ Yes, there are many explicit counterexamples. Just consider any two non-connected Lie groups $G$ and $H$ such that their respective identity connected components $G_I$ and $H_I$ are Lie groups themselves - see my edit. $\endgroup$ – Dietrich Burde Jan 5 '16 at 11:10
  • 1
    $\begingroup$ To be rather silly: $G = \Bbb Z/2$, $H = *$. $\endgroup$ – user98602 Jan 6 '16 at 18:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.