# E8 math problem solved: please explain in laymen's terms

Don’t try to imagine it, you’ll go blind. Just think of the “248” as the number of parameters you need to specify a point in it. The surface of the earth needs latitude and longitude to specify a point: 2-D. Space needs three numbers: 3-D. To pick an element of E[sub]8[/sub] requires 248 numbers.

Nobody has answered this question of mine yet:

> How does this compare to the classification of all finite simple groups? It sounds
> similar, in that each problem is about characterizing all possible groups of a
> certain sort. In both cases, there were several sporadic or exceptional cases,
> and the complete classification of these things involved identifying all these

The classification of finite simple groups is more analogous to Cartan’s classification of all finite-dimensional simple Lie algebras. Actually, those two are somewhat related. Cartan’s classification is an important tool for the finite simple classification.

What the Atlas project is trying to do is go through and classify all the representations of semisimple Lie groups, which is significantly harder than the Cartan classification, but easier than the finite simple groups. It’s just a monstrously (sorry) huge calculation in places. There’s actually some really clever theoretical mathematics they had to do to figure out how to break up the E[sub]8[/sub] problem into merely colossal chunks so a supercomputer could swallow it.

So, you’re saying this is a sort of mathematical masturbation?

No, that’s memorizing the decimal expansion of pi beyond enough places to calculate the circumference of the universe to within the radius of a proton.

Right. To clear up the confusion, closure is a defining property of subgroups. It got into early definitions of groups because originally all groups were conceived as subgroups of groups of permutations or subgroups of symmetry groups of one sort or another. It has disappeared from modern definitions of groups since the group operation is now conceived as in the quote.

So, how do you know the universe is perfectly round? Otherwise, you wouldn’t need that precision for an approximate circumference, would you? Out of curiosity, how many digits of pi would be needed?

I make it about 42 places. Which means I really need to get memorizing: I only know half that.