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I'm watching a lecture series on differential geometry and a Lie-algebra valued one-form $A$ on a principle $G-$bundle $(P, \pi, M)$ is written to belong to the space $$ A \in \Omega^1(M) \otimes T_e G $$ Where $\Omega^1(M)$ is the space of one-forms over $M$ and $T_e G$ is the lie algebra of the lie group $G$.

Can someone explain why the tensor product shows up here?

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Remember from linear algebra that when working with finite-dimensional vector spaces (bundles), the space (bundle) of linear maps $L(V, W)$ is isomorphic to the tensor product $V^* \otimes W.$ Since the usual ($\mathbb R$-valued) differential one-forms are sections of the bundle $$TM^* = L(TM, \mathbb R),$$ we thus see that if $E$ is some other vector space or bundle, the $E$-valued one-forms on $M$ should be sections of $$L(TM,E) = TM^* \otimes E.$$

In this particular case, $E$ is the vector space $T_e G$ (the Lie algebra of $G$), so the $E$-valued one-forms are sections of $TM^*\otimes T_e G.$ (If you want to get pedantic about this being a tensor product of vector bundles over $M$, $T_e G$ here is really shorthand for the trivial bundle $M \otimes T_e G \to M.$)

The space of sections of this bundle is $$\Gamma(TM^* \otimes T_e G) = \Gamma(TM^*) \otimes T_e G = \Omega^1(M) \otimes T_eG.$$

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A one-form eats tangent vectors and returns scalars, so a $G$-valued one-form ought to eat tangent vectors and return elements of $G$.

An element of $\Omega \otimes G$ does exactly this, viewing $\omega \otimes g$ as the thing that eats $v$ and returns $\omega(v) \cdot g$.

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