# Definition of Exterior Algebra of Differential Forms

I've read that we form the exterior algebra $$\Omega^*(M)$$ of differential forms on a smooth $$n$$-dimensional manifold $$M$$ by $$\Omega^*(M) = \bigoplus_{k=0}^n \Omega^k(M),$$ so it is a direct sum of the spaces of differential $$k$$-forms for all $$k$$. What if we approached this from a slightly different angle, and created a vector bundle over $$M$$ where the fiber over $$p \in M$$ is $$\Lambda(T^*_pM)$$ (so we create the graded algebras over each point first and then sew them together, instead of vice versa.) Will this create the same object?

Your proposed construction is consistent to taking the direct sum of all the differential $$k$$-forms. Note to construct $$\Lambda(T_p^*M)$$ we need to construct each fiber as a Whitney sum anyway:
$$\Lambda(T_p^*M) \;\; =\;\; \bigoplus_{k=0}^n \Lambda^k(T_p^*M).$$
I suppose the one key difference is to just remember the difference between the graded alternating tensor bundle $$\Lambda(T^*M)$$ and differential forms $$\Omega^*(M)$$. Elements $$\sigma \in \Omega^*(M)$$ are smooth sections $$\sigma: M \to \Lambda(T^*M)$$.