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May 29, 2023 at 5:44 comment added Mr. Feynman I'm not quite sure yet. Although my objection sounds reasonable, many authors include $dg$. If we didn't all we'd be left with would be the left translation, which for matrices would mean only $g^{-1}$ instead of $g^{-1}dg$. This PSE question maybe explains it better than I did.
May 29, 2023 at 2:21 comment added Deane Yang @Mr.Feynman, many thanks! I think you're right about this.
May 28, 2023 at 6:34 comment added Mr. Feynman Dear Prof. Yang, I hope you will accept my apologies for commenting a 4-years old post. There is one point in your answer (and elsewhere discussing this) which is not clear to me. Since $dg$ is just the identity map, why do we bother to compose $dL_{g^{-1}}(g)$ with it? I can't see any difference from defining $\omega=dL_{g^{-1}}(g)$ and in fact we have $\omega(v)=dL_{g^{-1}}(g)(v)$ in either case. Could you please clarify that point?
Aug 2, 2019 at 15:31 history answered Deane Yang CC BY-SA 4.0