## M Theory Lesson 28

I distinctly remember learning about the complex number \$i\$ when I was about 13. The teacher alluded briefly to some new class of number for solving equations, which she then promptly told us we shouldn’t worry our little heads over. Of course I approached her after class, having put away the book I was reading during her lesson, and I begged her to tell me about these new numbers. She told me that it really wasn’t a good idea, even though I was very smart, to jump too far ahead of what I was supposed to know. But I whined and whined until she relented and told me about the square root of -1.

At first I was a little disappointed. Only when I found out about Euler’s relation did I decide that it was pretty cool. But even now, the geometry enters as a tool, rather than as a means of defining the number itself. Numbers should come from Euler characteristics of categories. But as Khovanov homology tells us, even Euler characteristics are derived from higher dimensional (categorified) invariants.

The Jones polynomial is an Euler characteristic for Khovanov homology. The parameter \$q\$ of the Jones polynomial is usually taken to be complex valued in quantum group machinations. We saw that the factor \$q + q^{-1}\$ arose naturally as a label for a single loop. One way to view the number \$i\$ is as a solution to the equation \$i + i^{-1} = 0\$. What if we needed a zero to arise in this manner? First we would need a concept of imaginary number.

## 6 Responses so far »

1. 1

### Mahndisa S. Rigmaiden said,

03 17 07

Hello Kea:
Nice post. Interesting statement below:

“One way to view the number i is as a solution to the equation i+(i^-1)=0. What if we needed a zero to arise in this manner? First we would need a concept of imaginary number.”

Well I think a broad way to look at the equation for ‘i’ is that ‘i’ is a transformation of idempotent quantity:
i + (i^-1)= 0.
=> i= -(i^-1)

Recall that i^2=-1 and insert:
=> i= (i^2) (i^-1)

Recall that Idempotents have the property that P=P^2 and exploit:

=> i= (i^2) (1/i)

Now we have a geometric interpretation of i:
1. It is defined in terms of itself=> sense of trancendental
2.It is defined as an idempotent combined with an inversion on itself!

I hope this makes sense, aside from the obvious there may be richness in a diagrammatical representation of i…

2. 2

### Mahndisa S. Rigmaiden said,

03 17 0

Hello Kea:
Thanks for stopping by. Now the thoughts are making more sense:

1. i is a number formed by taking the composition of two transformations: doubling and inversin.

2. My diagram tried to illustrate the symmetric aspects of i by having the two triangles, but the inversion point at the bubble was what I should have also explained. The bubble around the i means that one can rotate i about that axis without changing it, structure preserving morphisms are about… Mere thoughts…thanks for your input and pls continue to correct any problems conceptually I may have, cuz that is how we learn:)

Now how might you draw the picture?

3. 3

### Kea said,

If we remember that for the number 1 there are many representations: a one element set (0-category), a simple 2 arrow 1-category, etc. then we should expect many possibilities for i, but there will be a minimal one, like the 1 object picture for 1. I can’t help thinking the square picture for -1 (with 5 arrows) is the right starting point, but it isn’t actually a category until we add the extra arrow to make a tetrahedron (2-sphere).

4. 4

### kneemo said,

I’ve been looking at this in terms of morphisms. In this sense, -i^-1=i is a special case of morphisms that satisfy -f^-1 o (id) = f o (id). For brevity let’s call these Heron morphisms, after Heron of Alexandria. Given any two distinct Heron morphisms, say g and h, and assuming g^-1 o h^-1 = (h o g)^-1, one can show (g o h)o(h o g)=id. Thus, (h o g)=(g o h)^-1, quickly leading us to (g o h)=-(g o h)^-1, making the composition (g o h) a Heron morphism as well.

If you build a composition table using id, g, h, and (g o h), you’ll see its just a fancy way of writing down the quaternion unit multiplication table for 1, i, j and k.

Constructing the octonions isn’t much more difficult. We merely start with three distinct Heron morphisms g, h, and f and write out the corresponding composition table.

5. 5

### Kea said,

kneemo, cool. Yes, this simple defining relation seems to be right. Heron is a good choice of name. Given a Heronian (rational) triangle, one can split it into two right-angled triangles, whose sidelengths form Pythagorean triples with rational entries.

6. 6

### L. Riofrio said,

Great that you talk about i. the concept of “imaginary time” was used by Einstein but forgotten in today’s “relativity” textbooks.