So one-parameter group $G$ is defined with a continuous group homomorphism $\phi: \mathbb{R} \rightarrow G$.

But according to the texts I read, they say that $G$ must be distinguished from groups as $G$ is not a group. So my wonder is, but there is group homomorphism there... So what is going on?

  • $\begingroup$ @anon I think you might as well have posted that as an answer. $\endgroup$ – Hagen von Eitzen Apr 7 '13 at 11:26

What is going on is that we are referring to the continuous group homomorphism itself as a "one-parameter (sub)group." Much like how a curve, in differential geometry (or topology more generally), may be described as a continuous map from an interval into a space. Different functions can have the same image, and a single group may be parametrized in different ways even though the image the two parametrizations inhabit inside the codomain are identical. One may think of the parametrization as extra data attached to a bona fide (sub)group.

  • 1
    $\begingroup$ For example, $\phi\colon\mathbb R\to \mathbb C^\times$, $t\mapsto e^{it}$ describes (has as image) the subgroup $S^1$, but so does $t\mapsto e^{i\omega t}$ for any nonzero real $\omega$. For the image $S^1$, $\omega$ does not play a role, but we do have a different one-parameter subgorup for each possible coice of $\omega$. $\endgroup$ – Hagen von Eitzen Apr 7 '13 at 12:03

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.