Last time I had asked for intuitions about associativity, @JasonHise64 suggested that associativity is commutativity squared. In retrospect, his suggestion seems much clearer than I had initially been able to see. But I think I got it now. Let's take a look together... (thread)

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(ab)c=a(bc)

That's how we ususally write associativity. How can we rewrite this, such that it looks like commutativity? I couldn't get it to work with a single operation L_x(y), but when I add another notation for the same operation with arguments flipped R_y(x) it worked!

With these, we can write

R_c(L_a(b)) = L_a(R_c(b))

or even only talk about L and R like so:

R(L(b)) = L(R(b))

This is in loose analogy to adjunctions, often seen in the context of Lie algebras. There, you'd write Ad_a(b) to obtain an operator Ad_a one can apply to elements b.

So what did we just do? Instead of our multiplication (·) we used two isomorphic notations L_a(b) = R_b(a) and then added a commutation law like so

L·R = R·L

If we had required these operatiors to be identical instead L = R we would have obtained commutativity instead!

I wonder if there's a further level three commutativity, and what it would look like. I suppose there might be an even nicer way to write down these laws, maybe using commutative diagrams from category theory, which would make these generalizations more obvious.

Arguably, this might settle my question from last time, how to recognize associativity looking at multiplication tables: we don't, we look at higher multiplication tables instead. Or not, as it's cheating, and I should spend some time playing with multiplication tables instead.

But I have more questions! What's the associative analog for the geometrical intuition for a commutator? Remember, it measures the failure of commutativity [A,B] = AB-BA You can look at it as the difference of one way of summing arrows versus the other way.

There's also an associator [A,B,C] = (AB)C - A(BC) about which I know next to nothing. But I know a little more about commutators. For these the Jacobi identity holds [A,[B,C]] + [B,[C,A]] + [C,[A,B]] = 0 which is kind of very similar to associativity when written like this:

[[A,B],C] - [A,[B,C]] = [[A,C],B]

Oh no, now it starts all over again! What's the symmetry group of that, and how can we read it off a multiplication table ...

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