Associated to any vector space $V$ is its exterior algebra $\Lambda(V)$ which has the direct sum decomposition $\Lambda(V) = \bigoplus_{i=0}^n\Lambda^i(V)$ where $n = \dim V$.
My first interaction with the concept of an exterior algebra was in differential geometry where one defines a $k$-form on a smooth manifold $M$ to be an element of $\Gamma(M, \Lambda^k(T^*M))$. Here $\Lambda^k(T^*M)$ is a vector bundle; in particular, $\Lambda^k(T^*M) = \bigsqcup_{m\in M}\Lambda^k(T_m^*M)$. A little bit further into my differential geometry studies, I encountered the concept of an $E$-valued $k$-form, where $E$ is a vector bundle on $M$, which is defined to be an element of $\Gamma(M, \Lambda^k(M)\otimes E)$.
Before seeing the definition of an $E$-valued form (or truly understanding the concept), I was under the impression that the exterior algebra of $E$ would appear in the definition. I now know why it doesn't, but I have come to realise that I have never seen the exterior algebra associated to a vector bundle other than the cotangent bundle. Therefore, I ask the following question:
Are there any situations in which one wants/needs to consider the exterior algebra of a vector bundle other than the cotangent bundle?
Best Answer
If $E$ is a complex vector bundle of rank $r$, its first Chern class is equal to the first chern class of its top exterior product: $$c_1(E)=c_1(\wedge ^r E)$$
This is extremely useful since the first (and only!) chern class of a line bundle is generally easy to compute.
For example on a compact Riemann surface or on a smooth projective curve, the line bundle $L=\mathcal O(D)$ associated to a divisor $D$ has its first chern class equal to the degree of the divisor: $$c_1(L)=\text {deg} D $$