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Problem is mostly in the title. I'm a complete amateur when it comes to math. I was doodling on graph paper when I found structure that I'm having a hard time wrapping my head around.

The operation works like this:

  1. $\quad a \ast a = b$

  2. $\quad b \ast b = a$

  3. $\quad a \ast b = a = b \ast a$

The conceptual trouble simply comes from the fact that a and b act a lot like inverses of each other, but there is no identity in the set. If $a^2 = b$ and $a^3 = I,$ it would be much simpler.

The motivation is pretty simple. What a and b originally described were black and white squares in patterns of a set size. By multiplying each square in one pattern with each corresponding square in another, you get a third pattern, e.g. $$■□■ * ■■□ = □■■$$ I was looking for a way to describe these patterns in terms of matrices, but I was unable, or simply not creative enough, to do so with typical matrix multiplication. I then realized that I couldn't even represent one square at a time using matrices (although you can get frustratingly close with k-matrices).

I suppose my question is- are there any notable and/or less abstract representations of this structure? I'm almost certain you cannot define a and b in terms of $\Bbb C$, or $\Bbb C$ modulo n, or complex matrices... I haven't had much luck finding examples of other things that behave like what I'm working with.

Apologies if the question is somewhat vague, but really I'd just like to be pointed in the direction of any work similar to this.

math crab
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  • The things you describe are definitely not inverse semigroups - they're not associative and they're not cancellative. See my answer. – Noah Schweber Oct 29 '16 at 22:09

2 Answers2

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The $*$ works like a NAND for a binary logic, with $a$ the 'True' element, and $b$ the 'False' element.

Bram28
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Your usual sources of examples - $\mathbb{C}$ and its many variations - won't help here: this operation is non-associative. For example, $$a*(a*b)=a*a=b\quad\mbox{but}\quad (a*a)*b=b*b=a.$$ (In particular, I don't understand your claim that you get semigroups out of this.) Indeed it's not even power associative, since we can replace "$b$" with "$(a*a)$" in the above example.

So this isn't even a semigroup. It also isn't a loop, since it doesn't have an identity element. And it's neither left nor right cancellative: $a*b=b*b=b*a$. So it's not a quasigroup either.

Noah Schweber
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