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Let $G$ be a linear algebraic group over a field $k$. I think that the tangent bundle should be the sheaf $Der(\mathcal{O}_G,\mathcal{O}_G)$, which is isomorphic to the dual of the differentials. However, there is also the set $\displaystyle G\left(\frac{k[\epsilon]}{(\epsilon^2)}\right)$, which seems to deserve the name "tangent bundle" in some way, even though it is not even a sheaf. Can anyone clarify the relationship between these two objects?

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up vote 2 down vote accepted

The canonical map $k[\epsilon]\to k$ induces a map $ G(k[\epsilon])\to G(k)$, and we have an exact sequence $$ 0\to \mathrm{Lie}(G) \to G(k[\epsilon])\to G(k)\to 1.$$ This can be taken as a definition of $\mathrm{Lie}(G)$. It is known that $\mathrm{Lie}(G) $ has canonically a structure of $k$-vector space and is isomorphic to the tangent space $T_{G, e}$ of $G$ at the unit $e\in G$.

On the other hands, the tangent bundle $T_{G/k}$ (the dual of the differentials) is free (as $O_G$-module) and satisfies $$T_{G/k}\simeq \mathrm{Lie}(G)\otimes_k O_G \simeq T_{G,e}\otimes_k O_G.$$

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