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Exterior differentiation

(This section is intended to introduce the background. Feel free to skip if you already know exterior differentiation.)

There are several useful tools for vector calculus on $\mathbb{R}^3,$ namely gradient, curl, and divergence. It is possible to treat the gradient of a differentiable function $f$ on $\mathbb{R}^3$ at a point $x_0$ as the Fréchet derivative at $x_0$. But it does not work for curl and divergence at all. Fortunately there is another abstraction that works for all of them. It comes from differential forms.

Let $x_1,\cdots,x_n$ be the linear coordinates on $\mathbb{R}^n$ as usual. We define an algebra $\Omega^{\ast}$ over $\mathbb{R}$ generated by $dx_1,\cdots,dx_n$ with the following relations:

$$\begin{cases} dx_idx_i=0 \\ dx_idx_j = -dx_jdx_i \quad i \neq j \end{cases}$$

This is a vector space as well, and it’s easy to derive that it has a basis by

$$1,dx_i,dx_idx_j,dx_idx_jdx_k,\cdots,dx_1\dots dx_n$$

where $i<j<k$. The $C^{\infty}$ differential forms on $\mathbb{R}^n$ are defined to be the tensor product

$$\Omega^*(\mathbb{R}^n)=\{C^{\infty}\text{ functions on }\mathbb{R}^n\} \otimes_\mathbb{R}\Omega^*.$$

As is can be shown, for $\omega \in \Omega^{\ast}(\mathbb{R}^n)$, we have a unique representation by

$$\omega=\sum f_{i_1\cdots i_k}dx_{i_1}\dots dx_{i_k},$$

and in this case we also say $\omega$ is a $C^{\infty}$ $k$-form on $\mathbb{R}^n$ (for simplicity we also write $\omega=\sum f_Idx_I$). The algebra of all $k$-forms will be denoted by $\Omega^k(\mathbb{R}^n)$. And naturally we have $\Omega^{\ast}(\mathbb{R}^n)$ to be graded since

$$\Omega^{*}(\mathbb{R}^n)=\bigoplus_{k=0}^{n}\Omega^k(\mathbb{R}^n).$$

The operator $d$

But if we have $\omega \in \Omega^0(\mathbb{R}^n)$, we see $\omega$ is merely a $C^{\infty}$ function. As taught in multivariable calculus course, for the differential of $\omega$ we have

$$d\omega=\sum_{i}\partial\omega/\partial x_idx_i$$

and it turns out that $d\omega\in\Omega^{1}(\mathbb{R}^n)$. This inspires us to obtain a generalization onto the differential operator $d$:

$$\begin{aligned} d:\Omega^{k}(\mathbb{R}^n) &\to \Omega^{k+1}(\mathbb{R}^n) \\ \omega &\mapsto d\omega \end{aligned}$$

and $d\omega$ is defined as follows. The case when $k=0$ is defined as usual (just the one above). For $k>0$ and $\omega=\sum f_I dx_I,$ $d\omega$ is defined ‘inductively’ by

$$d\omega=\sum df_I dx_I.$$

This $d$ is the so-called exterior differentiation, which serves as the ultimate abstract extension of gradient, curl, divergence, etc. If we restrict ourself to $\mathbb{R}^3$, we see these vector calculus tools comes up in the nature of things.

Functions

$$df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial z}dz.$$

$1$-forms

$$d(f_1dx+f_2dy+f_3dz)=\left(\frac{\partial f_3}{\partial y}-\frac{\partial f_2}{\partial z}\right)dydz-\left(\frac{\partial f_1}{\partial z}-\frac{\partial f_3}{\partial x}\right)dxdz+\left(\frac{\partial f_2}{\partial x}-\frac{\partial f_1}{\partial y}\right)dxdy.$$

$2$-forms

$$d(f_1dydz-f_2dxdz+f_3dxdy)=\left(\frac{\partial f_1}{\partial x}+\frac{\partial f_2}{\partial y}+ \frac{\partial f_3}{\partial z}\right)dxdydz.$$

The calculation is tedious but a nice exercise to understand the definition of $d$ and $\Omega^{\ast}$.

Conservative field - on the kernel and image of $d$

By elementary computation we are also able to show that $d^2\omega=0$ for all $\omega \in \Omega^{\ast}(\mathbb{R}^n)$ (Hint: $\frac{\partial^2 f}{\partial x_i \partial x_j}=\frac{\partial^2 f}{\partial x_j \partial x_i}$ but $dx_idx_j=-dx_idx_j$). Now we consider a vector field $\overrightarrow{v}=(v_1,v_2)$ of dimension $2$. If $C$ is an arbitrary simply closed smooth curve in $\mathbb{R}^2$, then we expect

$$\oint_C\overrightarrow{v}d\overrightarrow{r}=\oint_C v_1dx+v_2dy$$

to be $0$. If this happens (note the arbitrary of $C$), we say $\overrightarrow{v}$ to be a conservative field (path independent).

So when conservative? It happens when there is a function $f$ such that

$$\nabla f=\overrightarrow{v}=(v_1,v_2)=(\partial{f}/\partial{x},\partial{f}/\partial{y}).$$

This is equivalent to say that

$$df=v_1dx+v_2dy.$$

If we use $C^{\ast}$ to denote the area enclosed by $C$, by Green’s theorem, we have

$$\begin{aligned} \oint_C v_1dx+v_2dy&=\iint_{C^*}\left(\frac{\partial{v_2}}{\partial{x}}-\frac{\partial{v_1}}{\partial{y}}\right)dxdy \\ &=\iint_{C^*}d(v_1dx+v_2dy) \\ &=\iint_{C^*}d^2f \\ &=0 \end{aligned}$$

If you translate what you’ve learned in multivariable calculus course (path independence) into the language of differential form, you will see that the set of all conservative fields is precisely the image of $d_0:\Omega^0(\mathbb{R}^2) \to \Omega^1(\mathbb{R}^2)$. Also, they are in the kernel of the next $d_1:\Omega^1(\mathbb{R}^2) \to \Omega^2(\mathbb{R}^2)$. These $d$’s are naturally homomorphism, so it’s natural to discuss the factor group. But before that, we need some terminologies.

de Rham complex and de Rham cohomology group

The complex $\Omega^{\ast}(\mathbb{R}^n)$ together with $d$ is called the de Rham complex on $\mathbb{R}^n$. Now consider the sequence

$$\Omega^0(\mathbb{R}^n)\xrightarrow{d_0}\Omega^1(\mathbb{R}^n)\xrightarrow{d_1}\cdots\xrightarrow{d_{n-2}}\Omega^{n-1}(\mathbb{R}^n)\xrightarrow{d_{n-1}}\Omega^{n}(\mathbb{R^n}).$$

We say $\omega \in \Omega^k(\mathbb{R}^n)$ is closed if $d_k\omega=0$, or equivalently, $\omega \in \ker d_k$. Dually, we say $\omega$ is exact if there exists some $\mu \in \Omega^{k-1}(\mathbb{R}^n)$ such that $d\mu=\omega$, that is, $\omega \in \operatorname{im}d_{k-1}$. Of course all $d_k$’s can be written as $d$ but the index makes it easier to understand. Instead of doing integration or differentiation, which is ‘uninteresting’, we are going to discuss the abstract structure of it.

The $k$-th de Rham cohomology in $\mathbb{R}^n$ is defined to be the factor space

$$H_{DR}^{k}(\mathbb{R}^n)=\frac{\ker d_k}{\operatorname{im} d_{k-1}}.$$

As an example, note that by the fundamental theorem of calculus, every $1$-form is exact, therefore $H_{DR}^1(\mathbb{R})=0$.

Since de Rham complex is a special case of differential complex, and other restrictions of de Rham complex plays no critical role thereafter, we are going discuss the algebraic structure of differential complex directly.

The long exact sequence of cohomology groups

We are going to show that, there exists a long exact sequence of cohomology groups after a short exact sequence is defined. For the convenience let’s recall here some basic definitions

Exact sequence

A sequence of vector spaces (or groups)

$$\cdots \rightarrow G_{k-1} \xrightarrow{f_{k-1}} G_k \xrightarrow{f_k} G_{k+1} \xrightarrow{f_{k+1}}\cdots$$

is said to be exact if the image of $f_{k-1}$ is the kernel of $f_k$ for all $k$. Sometimes we need to discuss a extremely short one by

$$0 \rightarrow A \xrightarrow{f} B \xrightarrow{g} C \rightarrow 0.$$

As one can see, $f$ is injective and $g$ is surjective.

Differential complex

A direct sum of vector spaces $C=\oplus_{k \in \mathbb{Z}}C^k$ is called a differential complex if there are homomorphisms by

$$\cdots \rightarrow C^{k-1} \xrightarrow{d_{k-1}} C^k \xrightarrow{d_k} C^{k+1} \xrightarrow{d_{k+1}}\cdots$$

such that $d_{k-1}d_k=0$. Sometimes we write $d$ instead of $d_{k}$ since this differential operator of $C$ is universal. Therefore we may also say that $d^2=0$. The cohomology of $C$ is the direct sum of vector spaces $H(C)=\oplus_{k \in \mathbb{Z}}H^k(C) $ where

$$H^k(C)=\frac{\ker d_{k}}{\operatorname{im}d_{k-1}}.$$

A map $f: A \to B$ where $A$ and $B$ are differential complexes, is called a chain map if we have $fd_A=d_Bf$.

The sequence

Now consider a short exact sequence of differential complexes

$$0 \rightarrow A \xrightarrow{f} B \xrightarrow{g} C \rightarrow 0$$

where both $f$ and $g$ are chain maps (this is important). Then there exists a long exact sequence by

$$\cdots\rightarrow H^q(A) \xrightarrow{f^*} H^{q}(B) \xrightarrow{g^*} H^q(C)\xrightarrow{d^{*}}H^{q+1}(A) \xrightarrow{f^*}\cdots.$$

Here, $f^{\ast}$ and $g^{\ast}$ are the naturally induced maps. For $c \in C^q$, $d^{\ast}[c]$ is defined to be the cohomology class $[a]$ where $a \in A^{q+1}$, and that $f(a)=db$, and that $g(b)=c$. The sequence can be described using the two-layer commutative diagram below.

The long exact sequence is actually the purple one (you see why people may call this zig-zag lemma). This sequence is ‘based on’ the blue diagram, which can be considered naturally as an expansion of the short exact sequence. The method that will be used in the following proof is called diagram-chasing, whose importance has already been described by Professor James Munkres: master this. We will be abusing the properties of almost every homomorphism and group appeared in this commutative diagram to trace the elements.

Proof

First, we give a precise definition of $d^{\ast}$. For a closed $c \in C^q$, by the surjectivity of $g$ (note this sequence is exact), there exists some $b \in B^q$ such that $g(b)=c$. But $g(db)=d(g(b))=dc=0$, we see for $db \in B^{q+1}$ we have $db \in \ker g$. By the exactness of the sequence, we see $db \in \operatorname{im}{f}$, that is, there exists some $a \in A^{q+1}$ such that $f(a)=db$. Further, $a$ is closed since

$$f(da)=d(f(a))=d^2b=0$$

and we already know that $f$ has trivial kernel (which contains $da$).

$d^{\ast}$ is therefore defined by

$$d^*[c]=[a],$$

where $[\cdot]$ means “the homology class of”.

But it is expected that $d^{\ast}$ is a well-defined homomorphism. Let $c_q$ and $c_q’$ be two closed forms in $C^q$. To show $d^{\ast}$ is well-defined, we suppose $[c_q]=[c_q’]$ (i.e. they are homologous). Choose $b_q$ and $b_q’$ so that $g(b_q)=c_q$ and $g(b_q’)=c_q’$. Accordingly, we also pick $a_{q+1}$ and $a_{q+1}’$ such that $f(a_{q+1})=db_q$ and $f(a_{q+1}’)=db_q’$. By definition of $d^{\ast}$, we need to show that $[a_{q+1}]=[a_{q+1}’]$.

Recall the properties of factor group. $[c_q]=[c_q’]$ if and only if $c_q-c_q’ \in \operatorname{im}d$. Therefore we can pick some $c_{q-1} \in C^{q-1}$ such that $c_q-c_q’=dc_{q-1}$. Again, by the surjectivity of $g$, there is some $b_{q-1}$ such that $g(b_{q-1})=c_{q-1}$.

Note that

$$\begin{aligned} g(b_q-b_q'-db_{q-1})&=c_q-c_{q}'-g(db_{q-1}) \\ &=dc_{q-1}-d(g(b_{q-1})) \\ &=dc_{q-1}-dc_{q-1} \\ &= 0. \end{aligned}$$

Therefore $b_q-b_q’-db_{q-1} \in \operatorname{im} f$. We are able to pick some $a_q \in A^{q}$ such that $f(a_q)=b_q-b_q’-db_{q-1}$. But now we have

$$\begin{aligned} f(da_q)=df(a_q)&=d(b_q-b_q'-db_{q-1}) \\ &=db_q-db_q'-d^2b_{q-1} \\ &=db_q-db_q' \\ &=f(a_{q+1}-a_{q+1}'). \end{aligned}$$

Since $f$ is injective, we have $da_q=a_{q+1}-a_{q+1}’$, which implies that $a_{q+1}-a_{q+1}’ \in \operatorname{im}d$. Hence $[a_{q+1}]=[a_{q+1}’]$.

To show that $d^{\ast}$ is a homomorphism, note that $g(b_q+b_q’)=c_q+c_q’$ and $f(a_{q+1}+a_{q+1}’)=d(b_q+b_q’)$. Thus we have

$$d^*[c_q+c_q']=[a_{q+1}+a_{q+1}'].$$

The latter equals $[a_{q+1}]+[a_{q+1}’]$ since the canonical map is a homomorphism. Therefore we have

$$d^*[c_q+c_q']=d^*[c_q]+d^*[c_q'].$$

Therefore the long sequence exists. It remains to prove exactness. Firstly we need to prove exactness at $H^q(B)$. Pick $[b] \in H^q(B)$. If there is some $a \in A^q$ such that $f(a)=b$, then $g(f(a))=0$. Therefore $g^{\ast}[b]=g^{\ast}[f(a)]=[g(f(a))]=[0]$; hence $\operatorname{im}f \subset \ker g$.

Conversely, suppose now $g^{\ast}[b]=[0]$, we shall show that there exists some $[a] \in H^q(A)$ such that $f^{\ast}[a]=[b]$. Note $g^{\ast}[b]=\operatorname{im}d$ where $d$ is the differential operator of $C$ (why?). Therefore there exists some $c_{q-1} \in C^{q-1}$ such that $g(b)=dc_{q-1}$. Pick some $b_{q-1}$ such that $g(b_{q-1})=c_{q-1}$. Then we have

$$g(b-db_{q-1})=g(b)-d(g(b_{q-1}))=g(b)-dc_{q-1}=0.$$

Therefore $f(a)=b-db_{q-1}$ for some $a \in A^q$. Note $a$ is closed since

$$f(da)=df(a)=d(b-db_{q-1})=db-d^2b_{q-1}=db=0$$

and $f$ is injective. $db=0$ since we have

$$g(db)=d(g(b))=d(dc_{q-1})=0.$$

Furthermore,

$$f^*[a]=[f(a)]=[b-dc_{q-1}]=[b]-[0]=[b].$$

Therefore $\ker g^{\ast} \subset \operatorname{im} f$ as desired.

Now we prove exactness at $H^q(C)$. (Notation:) pick $[c_q] \in H^q(C)$, there exists some $b_q$ such that $g(b_q)=c_q$; choose $a_{q+1}$ such that $f(a_{q+1})=db_q$. Then $d^{\ast}[c_q]=[a_{q+1}]$ by definition.

If $[c_q] \in \operatorname{im}g^{\ast}$, we see $[c_q]=[g(b_q)]=g^{\ast}[b_q]$. But $b_q$ is closed since $[b_q] \in H^q(B)$, we see $f(a_{q+1})=db_q=0$, therefore $d^{\ast}[c_q]=[a_{q+1}]=[0]$ since $f$ is injective. Therefore $\operatorname{im}g^{\ast} \subset \ker d^{\ast}$.

Conversely, suppose $d^{\ast}[c^q]=[0]$. By definition of $H^{q+1}(A)$, there is some $a_q \in A$ such that $da_q = a_{q+1}$ (can you see why?). We claim that $b_q-f(a_q)$ is closed and we have $[c_q]=g^{\ast}[b_q-f(a_q)]$.

By direct computation,

$$d(b_q-f(a_q))=db_q-d(f(a_q))=db_q-f(d(a_q))=db_q-f(a_{q+1})=0.$$

Meanwhile

$$g^*[b_q-f(a_q)]=[g(b_q)]-[g(f(a_q))]=[c_q].$$

Therefore $\ker d^{\ast} \subset \operatorname{im}g^{\ast}$. Note that $g(f(a_q))=0$ by exactness.

Finally, we prove exactness at $H^{q+1}(A)$. Pick $\alpha \in H^{q+1}(A)$. If $\alpha \in \operatorname{im}d^{\ast}$, then $\alpha=[a_{q+1}]$ where $f(a_{q+1})=db_q$ by definition. Then

$$f^*(\alpha)=[f(a_{q+1})]=[db_q]=[0].$$

Therefore $\alpha \in \ker f^{\ast}$. Conversely, if we have $f^{\ast}(\alpha)=[0]$, pick the representative element of $\alpha$, namely we write $\alpha=[a]$; then $[f(a)]=[0]$. But this implies that $f(a) \in \operatorname{im}d$ where $d$ denotes the differential operator of $B$. There exists some $b_{q+1} \in B^{q+1}$ and $b_q \in B^q$ such that $db_{q}=b_{q+1}$. Suppose now $c_q=g(b_q)$. $c_q$ is closed since $dc_q=g(db_q)=g(b_{q+1})=g(f(a))=0$. By definition, $\alpha=d^{\ast}[c_q]$. Therefore $\ker f^{\ast} \subset \operatorname{im}d^{\ast}$.

Remarks

As you may see, almost every property of the diagram has been used. The exactness at $B^q$ ensures that $g(f(a))=0$. The definition of $H^q(A)$ ensures that we can simplify the meaning of $[0]$. We even use the injectivity of $f$ and the surjectivity of $g$.

This proof is also a demonstration of diagram-chasing technique. As you have seen, we keep running through the diagram to ensure that there is “someone waiting” at the destination.

This long exact group is useful. Here is an example.

Application: Mayer-Vietoris Sequence

By differential forms on a open set $U \subset \mathbb{R}^n$, we mean

$$\Omega^*(U)=\{C^{\infty}\text{ functions on }U\}\otimes_\mathbb{R}\Omega^*.$$

And the de Rham cohomology of $U$ comes up in the nature of things.

We are able to compute the cohomology of the union of two open sets. Suppose $M=U \cup V$ is a manifold with $U$ and $V$ open, and $U \amalg V$ is the disjoint union of $U$ and $V$ (the coproduct in the category of sets). $\partial_0$ and $\partial_1$ are inclusions of $U \cap V$ in $U$ and $V$ respectively. We have a natural sequence of inclusions

$$M \leftarrow U\amalg V \leftleftarrows^{\partial_0}_{\partial_1}\leftleftarrows U \cap V.$$

Since $\Omega^{*}$ can also be treated as a contravariant functor from the category of Euclidean spaces with smooth maps to the category of commutative differential graded algebras and their homomorphisms, we have

$$\Omega^*(M) \rightarrow \Omega^*(U) \oplus \Omega^*(V) \rightrightarrows^{\partial^*_0}_{\partial^*_1}\rightrightarrows\Omega^*({U \cap V}).$$

By taking the difference of the last two maps, we have

$$\begin{aligned} 0 \rightarrow \Omega^*(M) \rightarrow \Omega^*(U) \oplus \Omega^*(V) &\rightarrow \Omega^*(U \cap V) \rightarrow 0 \\ (\omega,\tau) &\mapsto \tau-\omega \end{aligned}$$

The sequence above is a short exact sequence. Therefore we may use the zig-zag lemma to find a long exact sequence (which is also called the Mayer-Vietoris sequence) by

$$\cdots\to H^q(M) \to H^q(U) \oplus H^q(V) \to H^q(U \cap V) \xrightarrow{d^*} H^{q+1}(M) \to \cdots$$

An example

This sequence allows one to compute the cohomology of two union of two open sets. For example, for $H^{*}_{DR}(\mathbb{R}^2-P-Q)$, where $P(x_p,y_p)$ and $Q(x_q,y_q)$ are two distinct points in $\mathbb{R}^2$, we may write

$$(\mathbb{R}^2-P)\cap(\mathbb{R}^2-Q)=\mathbb{R}^2-P-Q$$

and

$$(\mathbb{R}^2-P)\cup(\mathbb{R}^2-Q)=\mathbb{R}^2.$$

Therefore we may write $M=\mathbb{R}^2$, $U=\mathbb{R}^2-P$ and $V=\mathbb{R}^2-Q$. For $U$ and $V$, we have another decomposition by

$$\mathbb{R}^2-P=(\mathbb{R}^2-P_x)\cup(\mathbb{R}^2-P_y)$$

where

$$P_x=\{(x,y_p):x \in \mathbb{R}\}.$$

But

$$(\mathbb{R}^2-P_x)\cap(\mathbb{R}^2-P_y)$$

is a four-time (homeomorphic) copy of $\mathbb{R}^2$. So things become clear after we compute $H^{\ast}_{DR}(\mathbb{R}^2)$.

References / Further reading

• Raoul Bott, Loring W. Tu, Differential Forms in Algebraic Topology
• Munkres J. R., Elements of Algebraic Topology
• Micheal Spivak, Calculus on Manifolds
• Serge Lang, Algebra