How to find the generator of this Lie group?

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SUMMARY

The discussion focuses on identifying the generator of a Lie group operating in a space defined by points ##X^\iota##. The infinitesimal transformation is expressed as $$\gamma[\delta N^i]X^\iota=X^\iota+\xi^\iota_i(X^\kappa) \delta N^i$$, leading to the conclusion that the generator is given by the formula ##X_i=\xi^\iota_i (X^\kappa) \frac \partial {\partial X^\iota}##. This indicates that the generator is a vector field, with ##\frac \partial {\partial X^\iota}## serving as the basis vector in differential geometry. The discussion clarifies the relationship between the generator and the basis vector, emphasizing the need for ##n## coordinates over all ##\iota##.

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Haorong Wu
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How to find the generator of this Lie group?
Hello, there. Consider a Lie group operating in a space with points ##X^\iota##. Its elements ##\gamma [ N^i]## are labeled by continuous parameters ##N^i##. Let the action of the group on the space be ##\gamma [N^i] X^\iota=\bar X^\iota (X^\kappa, N^i)##. Then the infinitesimal transformation is given by $$\gamma[\delta N^i]X^\iota=X^\iota+\left . \frac {\partial \bar X^\iota (X^\kappa, N^i)}{\partial N^i} \right |_{N^i=0}\delta N^i=X^\iota+\xi^\iota_i(X^\kappa) \delta N^i$$ in the neighborhood of the identity ##N^i=0##. According to the paper, the generators of the Lie group is ##X_i=\xi^\iota_i (X^\kappa) \frac \partial {\partial X^\iota}## I do not see how to get this conclusion. I thought the genrator should be ##\xi^\iota_i(X^\kappa) ## itself. Is it related to the fact that the genrator is a vector, and ##\frac \partial {\partial X^\iota}## is merely the basis vector as in the differential geometry?

Thanks!
 
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Haorong Wu said:
Summary:: How to find the generator of this Lie group?

Is it related to the fact that the generator is a vector, and ##\frac \partial {\partial X^\iota}## is merely the basis vector as in the differential geometry?
Yes. We have to have ##n## coordinates over all ##\iota##.
 
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