Undergrad How to find the generator of this Lie group?

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The discussion centers on identifying the generator of a Lie group acting on a space with points labeled by continuous parameters. The infinitesimal transformation is expressed in terms of a derivative of the transformed coordinates, leading to the conclusion that the generator is given by a specific formula involving the basis vector. The confusion arises regarding whether the generator should be the function itself or the combination with the partial derivative. It is clarified that the generator is indeed a vector, while the partial derivative serves as the basis vector in differential geometry. Understanding this relationship is crucial for correctly identifying the generator in the context of Lie groups.
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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