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I am currently reading about differential forms in "Introduction to Smooth Manifolds" by J. M. Lee, and I was wondering exactly how you define the wedge product on the exterior algebra [itex]\Lambda^*(V) = \oplus_{k=0}^n\Lambda^k(V)[/itex] of a vector space [itex]V[/itex]. I understand how the wedge product is defined as a map

[itex]\Lambda^k(V)\times \Lambda^l(V) \rightarrow \Lambda^{k+l}(V)[/itex].

Is the idea simply to extend this definition bilinearly to [itex]\Lambda^*(V)[/itex]? I.e. given elements [itex](T_0, ..., T_n)[/itex] and [itex](S_0, ..., S_n)[/itex] in [itex]\Lambda^*(V)[/itex], do we simply calculate all the combinations [itex]T_j \wedge S_k[/itex], add the obtained forms that have the same degree, and finally let these sums make up the new element of [itex]\Lambda^*(V)[/itex]? This seems to me like an intuitive definition, but Lee wasn't very specific about this, so I thought I should make sure this is correct.

Please let me know if there is something in my post that seems unclear.

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# Definition of the wedge product on the exterior algebra of a vector space

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