Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
Are you distinguishing considering the entire exterior algebra from considering individual exterior powers? –  Qiaochu Yuan Jul 15 '13 at 20:21
add comment

3 Answers

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: $$c_1(L)=\text {deg} D $$

share|improve this answer
add comment

Tractor bundles are certain vector bundles that are more suitable than tensor bundles for the construction of invariant differential operators on some (so called parabolic) geometries.

For instance, a conformal structure $c = [g]$ on a smooth manifold $M$ defines a parabolic geometry in this sense (conformal geometry), and there exist so called (standard conformal) tractor bundle which in any choice of a metric $g \in c$ from the conformal class is just the direct sum $$ \Bbb T = \Omega^0 \oplus \Omega^1 \oplus \Omega^0 $$ but when the other metric $\hat{g} \in c$ is chosen this direct sum decomposition transforms nicely so that the (standard conformal) tractor metric and the (standard conformal) tractor connection are well-defined (invariant) on the tractor bundle.

In "Conformally invariant operators, differential forms, cohomology and a generalisation of Q-curvature" of T.Branson and A.R. Gover, see e.g. here, exterior powers $$\Bbb T^k = \underbrace{\Bbb T \wedge \dots \wedge \Bbb T}_{\text{k times}}$$ of the tractor bundles were introduced (under the name of $k$-form tractors, see p. 24 there). They have a number of applications in the theory of conformally invariant differential operators.

share|improve this answer
add comment

In Poisson geometry and related fields (e.g., deformation quantisation à la Kontsevich), one does actually consider the bundle $\wedge TM$ of multivector fields together with a generalisation of the Lie bracket on $\Gamma(TM)$ to $\Gamma(\wedge TM)$ called the Schouten--Nijenhuis bracket. In particular, specifying a Poisson bracket $\{\cdot,\cdot\}_M$ on a manifold $M$ is equivalent to specifying a Poisson bivector, i.e., a section $\eta \in \Gamma(\wedge^2 TM)$ such that $[\eta,\eta]=0$, via $\{f,g\}_M = (df \otimes dg)(\eta)$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.