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I consider a Lie group $G$, with a group element $g$ parametrised in some manner with parameter $\theta_i$, $i=1,\cdots, \dim G$. Suppose that $K\subset G$. I want to compute the variation of an group element under the left action of $K$ on $G$.

I'm a little confused to how I can do this: Something like $\delta g =g\delta\theta_iT^i$, with $T^i$ the generators of $G$? but restricted to the subgroup $K$?

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    $\begingroup$ As written, this is a pure math question. $\endgroup$ Jun 15, 2015 at 15:33

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Let's say $T^i$ ($i = 1, \dots, \dim G$) are the generators of $G$ chosen in such a way that $T^\mu$ ($\mu = 1, \dots, \dim K$) are the generators of $K \subset G$.

Let $k \in K$ and $g \in G$, then the left action of $k$ on $g$ is just $g' = k g$. Parametrize $k = \exp( \delta \theta_\mu T^\mu )$ then $g' = (1 + \delta \theta_\mu T^\mu) g$, so I would say $\delta g = \delta \theta_\mu T^\mu g$.

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  • $\begingroup$ Thanks, but shouldn't it be $g'=g(1+\delta \theta_\mu T^\mu)$ in first order approximation? $\endgroup$
    – user41746
    Jun 15, 2015 at 14:55
  • $\begingroup$ Sorry, I forgot some brackets. Have edited the post now. Why do you say left action and then always put $k = 1 + \delta \theta_\mu T^\mu$ on the right? $\endgroup$
    – Noiralef
    Jun 15, 2015 at 15:18

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