General Discussion on De Rham Cohomology Groups (base on my experience in 2020 Yau International Mathcamp)

 In my opinion, de Rham cohomology group can be thought of as a start point of learning these kind of algebraic object after studying some basics about fundamental groups (to build up a sense in algebraic topology).

From a geometric point of view, I think de Rham cohomology groups is defined for differentiable manifold (a Hausdorff, locally Euclidean space), which is easy to visualize in low dimensions as long as one has a solid background of (multivariable) calculus. From an algebraic perspective, (co)homology groups is defined for a "complex", i.e., a sequence of Abelian groups connected by homomorphisms with special properties. The whole complex can be denoted by $\{\mathcal{A},\partial\}$, which is the same as

$$\ldots\to A_n\overset{\partial_n}{\to}A_{n-1}\overset{n-1}{\to}\ldots$$

Dualizing each Abelian group by applying the "Hom" functor, we get a dual complex, i.e.,

$$\ldots\leftarrow \text{Hom}(A_n,G)\overset{d_{n-1}}{\leftarrow}\text{Hom}(A_{n-1},G)\overset{d_{n-2}}{\leftarrow}\ldots$$

where $G$ is called the coefficient group. Use my knowledge of some homological algebra, the existence of non-trivial (co)homology group is the lack of exactness.

Therefore, you can see that the concept of (co)homology group is not only geometric but also algebraic to some extent.

In de Rham cohomology group, we often set the coefficient group to be $\mathbb{R}$ or $\mathbb{C}$, and we form a complex by connecting the set of differential forms using the differential operator $d:\Omega^k(M^n)\to\Omega^{k+1}(M^n)$. Recall that a differential $k$-form is an expression of the form

$$\omega^k= \sum_{i_1,...,i_k} P_{i_1i_2...i_k}(\vec{x}) dx_{i_1} \wedge dx_{i_2} \wedge \cdot \cdot \cdot \wedge dx_{i_k}$$

and 

$$d\omega^k=\underbrace{\sum_{i_1,...,i_k} dP_{i_1i_2...i_k}(\vec{x})dx_{i_1} \wedge dx_{i_2} \wedge \cdot \cdot \cdot \wedge dx_{i_k}}_{\text{(k+1)-form}}$$

(I was really confused about the wedge product when the professor introduced this definition the first time, but after drawing and researching, I found that the this wedge can still be regarded as wedge product of vectors, which can be directly observed by the following graph

The wedge product $\vec{u}\wedge\vec{v}$ is just the region enclosed by that trapezium, which can be considered as a two dimensional vector. Now, it becomes geometrically clear that $d$ maps a $k$-form to $(k+1)$ form, which raises dimension by $1$.)

However, this is not enough for us to assert that sets of differential forms build a complex because we have to prove that $d^2=0$. (i.e. if we apply the differential operator twice on a differential forms, it get mapped to $0$) This is our homework for week 1 in Analysis & Topology course.

Proof of $d^2=0$:

In order to provide a brief argument, I think it suffices to consider a simple case, where $\omega$ only has the following form.

Suppose $\omega^k=fdx^*$ is a differentiable $k$-form. Then, apply the differential operator twice, we get

$$d\omega^k=\sum_{j=1}^n(\sum_{i=1}^n\dfrac{\partial^2 f}{\partial x^j\partial x^i}dx^j\wedge dx^i\wedge dx^*)$$

It's obvious that this sum is "symmetric" because partial derivatives commutes and because of the properties of wedge product. Therefore, the term

$$\dfrac{\partial^2f}{\partial x^j\partial x^i}dx^j\wedge dx^j\wedge dx^i\wedge dx^*$$

cancels with

$$\dfrac{\partial^2f}{\partial x^i\partial x^j}dx^i\wedge dx^j\wedge dx^*$$

So, we conclude that $d^2=0$.

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Now, we can make use of the standard construction of (co)homology groups to generate de Rham cohomology. For a smooth manifold $M$, its $k$-th de Rham cohomology group is defined as

$$H^k(M)=\text{ker}(\Omega^k(M)\to\Omega^{k+1}(M))/{\text{im}(\Omega^{k-1}(M)\to\Omega^{k}(M))}$$

To describe $H^k(M)$ in a clearer, shorter way, we need two more definitions:

Definition (Closed forms and Exact forms)

A differential $k$-form $\omega^k$ is said to be closed if $d\omega^k=0$, and exact if $\omega^k=d(\eta^{k-1})$, where $\eta^{k-1}$ is a differential $(k-1)$-form.

With new terms, we can rewrite the previous expression.

$$H^k(M)=Z^k(M)/B^k(M)$$

where $Z^k(M)$ is the set of all closed $k$-forms and $B^k(M)$ is the set of all exact $k$-forms. Note that this expression makes sense because $d^2=0$, which makes $B^k(M)\subset Z^k(M)$.

I'll present further thought in my next post regarding this topic.