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Folding Spaces in Slow Motion
I will (re)start from scratch and this time explain all the intermediary steps toward showing that the imaginary number i occurs naturally in Nature. I have cut the text in different parts. The text is not complicated and it is enough to read it carefully line by line, but be sure that you have correctly understood one part before moving to the next one.

Part 1: The Real Numbers, or a 1-dimensional algebra.

Everyone has got a fairly good intuition about the set of real numbers, at least when we consider its algebraic structure (things are different with its topological structure), so they are a good place to start.

Real numbers are somehow complicated (all those digits after the comma, and we never have enough space to write all of them...), but they are also simple animals: given two of them, that we will call a and b, we can add them and we have learnt to write the resulting element a+b. We can also multiply them. The result of the multiplication of a per b is denoted a*b (or ab).

So rather than focusing on how complicated they can be, let us focus on how easy it is to add them and to multiply them. Adding things and multiplying things are the two core concepts of everything which follows.

Coming back one second to the real numbers, there are two particular real numbers which are called zero and one and which are noted 0 and 1. When we add a number to 0, the result is the same number and when we multiply a number by 1 the result is again the same number.

In the following of this text, I will use the word 'space'. I will say that some structures are spaces. 'Space' is not necessary the word that mathematicians use, but I like it and it will do for the moment. When using this word, what I will actually have in mind is a collection of elements (which can really be anything) and some basic algebraic rules to add them and multiply them.

So the first space than we have talked about in this text is the collection of real numbers together with the two very well known algebraic operations of addition and multiplication.

The title of this part is "The Real Numbers, or a 1-dimensional algebra". What does "1-dimensional" mean ? That's simple. For us today it simply means that it is ok to see the real numbers on a line (with the negatives on the left and the positive on the right in you want), each and everyone of the points of the line represents (or stands for) a real number. Mathematicians would say that the set of real number with the classical operations is a 1-dimensional vectorial space. We will see below some spaces with more than one dimension.

Part 2: Complex numbers.

So in a previous entry I have introduced the space of complex numbers. And another time, by space I mean the collection of them together with the obvious algebraic rules to add them and multiply them.

A complex number is something that we will write a+bi, where a and b are real numbers and i is the imaginary number. We are going to say that the real number a is the real part of the complex number and that b is the imaginary part of the complex number. So basically a complex number is nothing else than a real part plus an imaginary part multiplied by i. For the moment I do not ask you to wonder whether i exists or not, or those kind of things. I simply ask you to accept the fact that I can take a child and teach to him or her how to perform calculus with the complex numbers, and he or she will do that as if this was the most natural thing on Earth.

The algebraic rules to add and multiply complex numbers are simple because they are derived from what we know about the real numbers. For instance, given a complex number a+bi and given another one c+di, I can add them. The result will be the complex number (a+c) + (b+d) i. The result is somehow natural, isn't it ? I have simply added a and c and added b and d, in other words (and this is a good way to remember what is going on), when I add two complex numbers together, the real part of the sum, is the sum of the real parts, and the imaginary part of the sum is the sum of the imaginary parts.

Now we are going to multiply two complex numbers. So let us take a+bi and c+di. I could (if I wanted) define the multiplication of complex numbers is a very simple way, saying that it is enough to multiply the real parts together and multiply the imaginary parts together. I could if I wanted, but I am not going to do this because the resulting structure would not behave the way I want. So I am going to define the product of a+bi and c+di being equal to (ac-bd) + (ad+bc)i. I know, it looks strange but you will get use to it.

We now know how to add complex numbers, and how to multiply them.

Part 3. The truth behind (ac-bd) + (ad+bc) i

In this third part, I will say a bit more about the formula which defines the product of complex numbers. We know that if we have four real numbers x, y, z and u, the following equation is true (x+y) * (z+u) = xz + xu + yz + yu.

let us take x = a, y=bi, z=c and u=di. We then have (x+y) * (z+u) = (a+bi) * (c+di). We also have xz + xu + yz + yu = ac + adi + cbi + bdii. Therefore we have (a+bi) * (c+di) = ac + adi + cbi + bdii. It is now time to remember the fundamental idea behind the creation of i: when multiplied by itself the result is -1. So the term bdii actually equals bd * (-1), or -bd to be short.

In the end we have that (a+bi) * (c+di) = ac + adi + bci - bd. And we can re-write the last term as ac-bd + (ad+bc)i. So (a+bi) * (c+di) = ac-bd + (ad+bc)i.

We then have the same formula that I gave in Part 2, but this time we can see that it was a natural choice, given the fundamental property of i.

Part 4. Homework.

In order for you to be sure that you have understood about complex numbers, you might want to check yourself, by hand, with a paper and a pen, that the following equalities are true

3 * (1+7i) = 3+21i

(1+7i) + (2-6i) = 3+i

(1+7i) * (2-6i) = 44+8i

Part 5. The Space of Polynomials

Here, we are going to play with a quite nice space (which is a rather big one by the way): the space of polynomials. So to be sure that we know what we are talking about, let me remind you that polynomial are folks like 2X^{3}+4X^{2}+X-12.

I decided to use the space of polynomials because most people have already encountered them when they were young (remember when you had to solve equations like 2x+8=0 or 4x^{2}-7x-10=0)

Note how when I write polynomials I use the letter "X "but when I write equations I use the small "x". I actually want to emphasis the difference between writing an equation and considering the polynomial itself, the latter being nothing else than a first class citizen in the space of polynomials.

We know how to multiply a polynomial by a real number to get a new polynomial, for instance 3 * (2X^{3}+4X^{2}+X-12) = 6X^{3}+12X^{2}+3X-36.

We know how to add two polynomials, for instance 2X^{3}+4X^{2}+X-12 + 2X^{2}-X+1 = 2X^{3}+6X^{2}-11.

And we know how to multiply two polynomials together, for instance (2X^{2}+1) * (X-4) = 2X^{3}-8X^{2}+X-4.

Part 6. The beginning.

Note how we now have two spaces: the space of complex numbers and the space of polynomials. In both cases, we can multiply one element of the space by a real number, we can add two elements of the space together to get a new element of the space, and we can multiply two elements of the space together to get a new element of the space.

There is nevertheless a big difference between those two spaces. One of them is of dimension 2 and the other one is of infinite dimension.

Part 7. The dimension of a vectorial space, the case of the complex numbers.

I am not going to define what is a vectorial space because its definition would require me to define what are groups and fields (which you don't want me to do) and write first order logic formulas, but I can tell you that I am going to compare the space of complex numbers and the space of polynomials as vectorial spaces.

Given that we are considering vectorial spaces structures, we can forget in this part that we can always multiply two elements of our spaces. We are only going to use the fact that we can add them and the fact that we can multiply them by a real number.

Let us have a look at the space of complex numbers. So an element of the space is of the form a+bi where a and b are real numbers. I will take two elements of this space and I will call it Unity1 and Unity2.

Unity1 will be the complex number 1 + 0i, that you can write 1 if you want.

Unity2 will be the complex number 0 + 1i, that you can write i if you want.

One important thing about the space of complex numbers is the following: for every complex number c in the space, it is always possible to find two real numbers x and y such that c = x * Unity1 + y * Unity2.

For instance if c = 3+8i, then it is enough to take x=3 and y=8 because 3 * (1 + 0i) + 8 * (0 + 1i) = 3+8i.

I will call the couple (Unity1,Unity2) a base of the vectorial space., and the two real numbers x and y that I mentioned earlier (which I said you can always find), will be called the coordinates of the element c.

Now, there are two things you should agree with. The first one is that given my base, every single complex number has got some coordinates relatively to this base, and the second one is that those coordinates are actually really easy to find because it is enough to see the way the number is written. For instance the coordinates of the complex number -2+9i are -2 and 9.

I could have chosen two other Unities, I could have taken Unity1 = 1+20i and Unity2= -8+2i. In this case the couple would still be a base of the space and it would still be true that every single complex number has got some coordinates for this base. The difference is that you would need a pen and a paper to find the coordinates. How would you find them ? Well, easy, you know that the coordinates are real numbers, right ? Let us call those coordinates x and y and we know that we must have c = x * Unity1 + y * Unity2. This is enough to find x and y ...

If c= -15 + 24i, I let you verify that the coordinates are 1 and 2, for the base Unity1 = 1+20i and Unity2= -8+2i.

One particularity of the space of complex numbers is the fact that all bases of the space always have 2 elements. (If you are a mathematician you will notice that I am simplifying the presentation a bit here, but don't tell anyone...).

From the fact that every base of the space of complex numbers has got 2 elements, we say that the space is of dimension 2. Basically the dimension of a vectorial space is how many elements its bases have got.

Part 8. The dimension of a vectorial space, the case of the polynomials.

Now, let us have a look at the space of polynomials.

Here, the situation is different because it is not possible to find a base made out of a finite number of elements. You can try as hard as you want, but if you come up if a finite collection of elements in the space, (Unity1 to Unity234 for instance), I will always be able to find an element of the space which doesn't have coordinates for your base-wanabe. This fact is not complicated to prove but I will not show it right now (later if someone wants...).

As a result, we say that the space of polynomials is of infinite dimension. The space of polynomials has got a base, but the base has got an infinite number of elements.

Part 9. Light Speed Space Collapse

A bit earlier in the text, we have seen that when we look at their algebraic structure, the space of complex numbers and the space of polynomials are similar, but now we know that their structure of vectorial spaces is different.

What I am going to do now, is that I will take the space of polynomials and fold it until the resulting space is of dimension 2. This is not difficult to do. Remember how I so easily created i yesterday, folding a infinite dimensional space to give it the shape of a piece of paper is no more complicated. I will use my mathematical powers and decide that from now on, the polynomial 2X^{2}+7 and the polynomial 0 are the same element of the space. For this, I simply decide that from now on my space has to verify that the following equation is true 2X^{2}+7=0.

See that was easy, isn't it. And another time, yes, I can. And now you can add on your CV, "I am able to fold an infinite dimensional space into a piece of paper" :-)

Anyway, this poor element 2X^{2}+7 finds itself suddenly equal to 0. Well, this little event will have the effect of collapsing most of the space. For instance the element 4X^{2}+14, which is in fact 2 * (2X^{2}+7) is now 2 * 0, and therefore equal to 0. and this is only the beginning...

The element 2X^{2}+11 which used to be a brave polynomial of degree 2, is in fact equal to 2X^{2}+7 + 4, and therefore equal to 4.

.. and this was only the beginning, the polynomial 2X^{3}+8X+1, which was a brave polynomial of degree 3, is in fact equal to X * ( 2X^{2}+7 ) + X + 1, and therefore equal to X + 1.

What is happening is that the equality 2X^{2}+7=0 acts like a nuclear ignition and from the bottom up contaminates the entire space. Polynomials which used to proudly display their higher degrees, are collapsing towards the floor at the speed of light. After the entire collapse of the space, only polynomials of the form aX+b survive without being put equal to zero.

I will call the collapsed space P' (pronounced "p prime").

Mathematicians call P' a quotient space, and believe me the definition of those doesn't at all involve citizens who die suddenly due to light speed space collapse :-) [Quotient spaces are among my sexiest memories of being an undergraduate because quotient spaces are a particular case of the general idea of equivalence classes which is a very powerful mathematical tool to build things].

So P' has got only members of the form aX+b where a and b are real numbers. The nice thing with P' is that it have not lost its vectorial space structure as you can still multiply members by real numbers and add members together and neither has it lost it algebraic structure as you can still multiply members together, but unlike in the now dead big space of polynomials, this time you have to include the fact that 2X^{2}+7=0. What I mean is the following:

In the big space we had (2X+2)*(X-3) = 2X^{2}-4X-6.

In P' we have (2X+2)*(X-3) = 2X^{2}-4X-6 = 2X^{2}+7-4X-13 = -4X-13.

Part 10.

What have we got in the end ? We have two 2 dimensional vectorial spaces, the space of complex numbers and P', which are also both algebras. Those two spaces tend to behave the same as algebras and also the same as vectorial spaces.

Before carry on let us notice that 2X^{2}+7=0 implies (2/7)X^{2}=-1.

Let us have a look at the polynomial sqrt(2/7)X in P'. What if we multiply it by itsef ? Doing this we have sqrt(2/7)X * sqrt(2/7)X = (2/7)X^{2} = -1.

So in P' we have an element which when you multiply it by itself you get -1. Now some clever people are going to say that -1 in the set of complex numbers of not the same as -1 in P'. This is perfectly true. It is now time to learn something else...

Part 11. Isomorphisms

On one side we have the set of complex numbers and on the other side we have P'. We know that those two vectorial spaces are of dimension 2, right ?. We know that 1+0i and 0+1i are two elements of a base of the first one, so what would a base of P' be ? Well that's easy, a good base of P' is given by the polynomial 1 and the polynomial sqrt(2/7)X.

So now think about it, what if we associated 1 (from the complex numbers) and 1 (from P'), on one hand and on the other hand i (from the complex numbers) and sqrt(2/7)X (from P'). I am going to say it rather than writing more formulas: what is going to happen is that those two spaces, starting from the two bases are going to automatically match more and more elements, until you will find that actually not only every complex number is automatically associated to a polynomial of P', but in fact the two vectorial structures and the two algebraic structures are absolutely identical.

This is not difficult to show, but I don't want right now to define what an isomorphy of vectorial spaces or isomorphy of algebras are.

In fact those two spaces are the same space. When you look at it from one side you see the complex numbers and when you look at it from anther side you see P'.

I have written "from another side" and not "the other side", because this space has got many sides. There are ways to see it under which it looks so strange that you would not recognize it if I don't tell you what you are seeing...

Part 12. Time to eat...

Why have I done all this ? Well to show that if you can get along with P' (which is natural to most people), then there's no mystery about the imaginary number i. That's all :-)