# What makes differential forms special

There are two natural structures defined on differential manifolds, namely tangent bundle and its dual cotangent bundle. An element of tangent bundle $TM$ at a base point $p\in M$ can be described by a differential operator $X$, also called a tangent vector. If we choose a coordiante for $p=(x_1,\ldots,x_n)$, then we have a natural basis for tangent vector space at $p$ and write $X=X^i\frac{\partial}{\partial x^i}$. Dually, an element of cotangent bundle $T^*M$ at the same base point $p$ can be described by a differential form $\theta$. Take dual basis of tangent vector space and we can write $\theta=\theta_idx^i$.

Everything above concerning tangent bundle and cotangent bundle looks symmetric. However, we can do more on cotangent bundle. We may obtain a second-order differential form $\omega=d\theta=d\theta_i\wedge dx^i$ by exterior differential operating on cotangent bundle. While there's no sort of "exterior tangent" operatation on tangent bundle without imposing extra structure. In addition, there's a distinguished one-form, named "tautology one-form", on cotangent bundle while there's no distinguished tangent vector on tangent bundle. What makes differential form special?

Furthermore, for differential forms, we can define cup product and exterior product on de-Rham cohomology chain. No similar thing can be defined for vectors.

## 1 Answer

One important thing to notice is that the assignment $M \mapsto \Gamma(M, TM)$ does not define a functor on the category of smooth manifolds and smooth maps, because tangent vector fields do not behave well under pushforwards or pullbacks by smooth maps. On the other hand, the assignment $M \mapsto \Gamma(M, \bigwedge^k T^*M)$ defines a contravariant functor on the category of smooth manifolds, since forms can be pulled back by smooth maps.

Also, notice that the functor $C^\infty(-)$ which assigns to each smooth manifold its algebra of smooth functions is the contravariant representable functor $\text{Hom}(-, \mathbb{R})$. So if we want a functorial construction assigning to each smooth manifold some algebra that generalizes the smooth function algebra, it should probably be a contravariant construction.

This fits into the more general duality between space and quantity. The heuristic is that a spacial category can often be thought of as the opposite of an algebraic category, so functors that assign some "algebra of quantities" to a space should generally be contravariant.

• On the one hand, I really like this answer. On the other, it could be read as just emphasizing that the co-case is more well behaved, but not indicating why. I feel the question next question would naturally be "why do vector fields not behave well under pullbacks, what is the obstacle?" – Nikolaj-K May 7 '15 at 9:40
• I added some more explanation to make the idea more clear. – ಠ_ಠ May 7 '15 at 10:00