I'm acquainting myself with the world of Lie groups, and have seen that since there is a bijection between left invariant vector fields and a tangent vector at the identity. And thus one defines the Lie algebra of a Lie group as $\mathfrak{g}=T_{1_G}(G)$. However in the book I am reading they discuss the exponential map and say that it is continuous, but with no mention of the topology on $\mathfrak{g}$.

My question is given such a Lie algebra $\mathfrak{g}$ is the topology induced from the tangent bundle? Is it simply the topology on $\mathbb{R}^d$ ($d$ being the dimension of the Lie group)? Or is it the topology such that Lie algebra is a topological vector space and the Lie bracket is smooth?

  • $\begingroup$ Over non-discrete locally compact fields $K$ (e.g. $K= \Bbb R$ or $K = \Bbb C$), finite-dim. vector spaces have kind of a natural topology, namely, the product topology when viewed as $K^n$ after choice of some basis. That is the unique topology which makes them locally compact Hausdorff top. vect. spaces. See e.g. terrytao.wordpress.com/2011/05/24 and mathoverflow.net/questions/264812/…). So I would think this is the one to go with, and I assume (but have not checked) that it is the same as all the options you mention. $\endgroup$ – Torsten Schoeneberg Dec 19 '18 at 20:18

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