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Let $G$ be a Lie Group end let us consider the Cartesian product of $n$ copies of $G$ ($N\geq 2$): $G\times G\times \ldots \times G$. What is the Lie Algebra of this group? Is the Lie Algebra the direct sum g+g...+g (where g is the Lie algebra of $G$) ?

Ivo Terek
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mary satollo
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1 Answers1

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Of course it is.

Generally, when $M_1,\ldots,M_n$ are manifolds and for every $i$, $p_i\in M_i$ we have the identity $$T_{(p_1,\ldots,p_n)}M_1\times\ldots\times M_n=\bigoplus_iT_{p_i}M_i.$$Now given a Lie group $G$, its Lie algebra can be identified with $T_eG$, and we're done.

Amitai Yuval
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  • You need to know that this identification is natural in order to know that it respects the Lie bracket; a priori this argument only tells you that you get an identification as vector spaces. – Qiaochu Yuan Sep 02 '14 at 04:19