An Important Notification

 Due to some technical issues, I decide to move the entire blog to 

kaiwensong.wordpress.com

This means I won't publish further notes on this blog.

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.

Notes on General Topology: Cont. and Homeo.

The idea of topology is extremely useful in capturing properties of a space! According to what Munkres has noted in his book Topology, lots of mathematicians including Hausdorff had searched for an appropriate definition of topology for almost all topological spaces before 1920s. Finally, they generalized the idea as follows:

Def. A topology on a set $X$ is a collection $\tau$ of subsets having

1) $\varnothing,X\in\tau$.

2) $\bigcup_iA_i\in\tau$ where $A_i\in\tau$

3) $\bigcap_j^nB_j\in\tau$ where $n<\infty$ and $B_j\in\tau$.

At first, I was extremely confused about such an abstract idea, because it doesn't seem to be capable to provide any detailed information about an object directly, like the number of handles of your coffee mug or the number of holes in your bread... But you'll see its power in just a second!

In elementary analysis, we've learned the equivalence of "limit" definition of continuity and the $\epsilon-\delta$ condition of continuity. Now, the idea of topology gives us a much general definition called the "open-set" condition.

Def. Let $X$ and $Y$ be two topological spaces connected by a map $f:X\to Y$. $f$ is continuous if $B$ open $\implies f^{-1}(B)$ open. i.e. the preimage of any open set is open. i.e. the preimage of any closed set is closed.

Note that this has to be distinguished from the concept of open map and closed map because the directions of implication are different!!!

It's quite obvious that $\epsilon-\delta$ condition is equivalent to the open set condition. Since the open-set continuity is a generalization of $\epsilon-\delta$ condition. Hence it is also equivalent to the limit definition and the net definition, which I'll be going through.

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There is a special case of continuous map that we always care about ---- Homeomorphism! which requires $f$ to be bijective and continuous at both direction. i.e. Both $f$ and $f^{-1}$ are continuous.

Let's see what this gives us:

 

                           $f(x)=x^3$                                   $g(x)=x$

Well, in the first picture, if we neglect the geometric properties of $f$, then the image is topologically the same as the $x$-axis, and so does $g$. Therefore, Homeomoprhism is used to distinguish the topological property of two spaces. 

For an counterexample of homeomorphisms, see this:

$f(x)=(\sin(2\pi x),\cos(2\pi x)), x\in[0,1)$

The problem with this circle is that we can't find a continuous function between $[0,1)$ and $S^1$ with a continuous inverse function.

And this:

$S^1\vee S^1\not\cong[0,1)$

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However, the reason why mathematicians are sure that "homeomorphisms" preserves topological properties before we actually define (co)homology groups still remains unknown for me. Is it a magic? coincidence? or a consequence of extremely accurate math intuition?

Notes on General Topology (Pre): An Interesting Fact

Today, I'm gonna talk about a common thing in everyone's life --- Counting!

Usually, we set the total sum to be $0$ and add $1$ whenever we meet something new in a set of objects. For example, if we want to count the number of apples in a set, we start by declaring the number to be $0$ and start adding numbers. 

Q: Can we count one object twice?

NO! We can't count one object twice because it's not helpful for us to record the cardinality of this set. Thus, the first restriction of counting is "one-one correspondence", in other word, bijective.

Q: Can we count using fractions?

Usually NO! In my opinion, fractions contain relations which are unnecessary for counting, because we want the answer for "How many", not "How much... in comparison to...". Therefore, the connection between objects and integers is the reason why we need to count.

Def. A set $X$ can be counted by a bijection $f:X\to C\subset\mathbb{Z}_+$.

This definition gives us a magical result if we scrutinize its details, $\mathbb{Q}$ is countably infinite!

From the definition of rational numbers, we know that any rational can be written as the quotient of two integers. Thus, every rational number can be defined by using two integers! Formally, we could consider this process as a function defined on two variables. Let $(a,b)\in\mathbb{Z}^2$ be an ordered pair of integers, then $f:\mathbb{Z}_+^2\to\mathbb{Q}$ is defined by $(a,b)\mapsto\frac{b}{a}$, which can be visualized using the chart below.

Q: How to count them?

Well, obviously you can't finish counting row by row because each row contains infinitely many elements. So, we proceed like the following:

By doing this, we see that $f(n,m)=\frac{m}{n}$ where $\text{gcd}(n,m)=1$ is at the intersection of $n$-th row and $m$-th column. Since $\mathbb{Z}_+^2$ is countable, $\mathbb{Q}$ is also countable and countably infinite.

However, $\mathbb{R}$ is uncountable because $[0,1]\subset\mathbb{R}$ is uncountable, which could be proved using a similar list of real numbers.

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