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Folding Spaces
In my previous entry I introduced the complex imaginary number i, out of a magician's hat. To me that's fine, but people with no mathematical education may still wonder about the validity of all this. So I will show here, that complex numbers are not unfamiliar and that they were already there (we simply could not see them). To this, I will take a familiar space, the space of polynomials. It is an classic regular algebra over the field of real numbers and behaves pacefully. It is of infinite dimention (as a vectorial space).

An element of this space is, for instance, 3X^{4}+23X^{2}-4X+12, another one is X^{10}+23. I can multiply a member by a real number and I can add members together (vectorial space structure at play), I can also multiply two members (algebraic structure at play); and the expression 3X^{4}+23X^{2}-4X+12 + 3 * ( X^{10}+23 ) is actually equal to 3X^{10} + 3X^{4}+23X^{2}-4X+81.

For instance the expression (X^2+9) * (X^2+X+1) equals to X^{4} + X^{3} + 10X^{2} + 9X + 1

Note that I have not properly defined what is an algebra, a vectorial space, a field and the dimention of a vectorial space. I would if I was giving a lecture, but for now we can avoid it. I chose the algebra of polynomial which is familiar enough.

The title is "Folding Spaces", what am I actually having in mind ? Well, our current algrebra is of infinite dimention over the field of real numbers, right ? I am going to write down an equality and our algebra is going to fold into itself and will become an algebra of dimention 2.

The equation is 2X^{2}+7=0.

No, I am not asking you to resolve this, I have simply written a decision about this space. I have written that the element 2X^{2}+7 is now equal to the neutral element of the space. So you may not have seen this happening (it happened very fast), but in fact our space has instantly stopped being a space of infinite dimention to become a space of dimention 2.

To see what I mean, we take two members of the space, 2X^{2}+1 and X^{3}-1, and we are going to multiply them.

So have ( 2X^{2}+1 ) * ( X^{3}-1 ) = 2X^{5} + X^{3} - 2X^{2} -1 = 2X^{5} +7X^{3} -6X^{3} - 2X^{2} -1 = X^{3} ( 2X^{2} + 7) - 6X^{3} - 2X^{2} -1 = 6X^{3} +21X - 2X^{2} -21X -1 = 3X * (2X^{2} +7) - 2X^{2} -21X -1 = -1 * (2X^{2} + 7) -21X -1 + 7 = -21X +6

Note how I have used our fundamental equation to reduce the degree of the product to 1, while it should have been 5 in the unfolded space. So in our folded space the product of 2X^{2}+1 and X^{3}-1 is -21X +6.

Where am I going with this? Well, you may not have noticed, but our folded space is like the space of complex numbers, a brave algebra which is also a vectorial space of dimention 2 over the set of real numbers.

No, no, I am not joking. In the previous entry we were dealing with numbers, and now we are dealing with polynomials, but the two spaces are absolutely isomorphic (meaning that mathematically speaking there's no way you could distinguish them).

So a good question now would be "Ok, you claim that your folded space is like the complex numbers, so where is i ?"

That a good question and we are going to answer to this using the fundamental equality that i should respect,i.e. i^{2}=-1.

So let aX+b be a member of our space and it has to be such that (aX+b)^{2} = -1, in other words 7 * (aX+b)^{2} = -7 =2X^{2} (from 2X^{2}+7=0)

The equation 7 * (aX+b)^{2} =2X^{2} has the solution a = sqrt(2/7) and b=0. So now you have it, the polynomial sqrt(2/7)X is our imaginary number. Is checks ( sqrt(2/7) X )^{2} = (2/7)X^{2} = -1, in the folded space.

Now, I think I am going to stop, because for sure I have lost Aubrey, but remember that when others would see only boring calculus on polynomials, I see spaces with infinite dimentions which suddenly fold themselves and restructure in a way which make them indistinguisable to our algebra of complex numbers, and we even have found what I introduced in the previous entry as an alien number coming from another world, it was just there, waiting for us, and you would not believe how many times we have looked at it in the eyes without recognizing it:-)

The end.
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