Let T be a torus (namely, topology homeomorphic to quotient topology $\mathbb{R}/\mathbb{Z}$). Intuitively it makes sense to say that tangent space of torus at identity (i.e. lie algebra of 1D torus) is $\mathbb{R}$ since we can geometrically think of tangent space as straight line passing through the torus at identity. However, how could we formally prove this fact? It seems like if I formally want to prove it, I need to consider all derivation maps (or curve passing through origin) then somehow need to show that the set of derivations correspond to the reals, but I am not very sure how the argument works here. Also, how is tangent space of k-dimensional torus like? Could anyone help me to figure out this? Thank you a lot for all of your support in advance!
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2This seems a little confused. Firstly the 1-dimensional torus is just a circle so we don't really need to mention tori. Then the tangent spaces to any 1-dimensional real manifold are automatically isomorphic to $\mathbb{R}$ so there isn't really anything to prove. Similarly the tangent spaces to a k-dimensional manifold are isomorphic to $\mathbb{R}^k$ – Callum Feb 21 '24 at 07:56
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1Think of partial derivatives as telling us directions along which we probe changes of functions on the manifold. Since partial derivatives can be linearly combined they form a vector space. Functions on the one-dimensional torus eat only one variable and there is only one relevant derivative to take. How many dimensions has then the space of partial derivatives? That space we call tangent space. Why all this and not simpler? My favourite answer: it leads to the correct coordinate transformation formulas simply by applying – Kurt G. Feb 21 '24 at 08:10
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the chain rule. – Kurt G. Feb 21 '24 at 08:10