What is the function space generated by addition and $(a,b)\mapsto (a+b)^{-1}\cdot a\cdot b$ of elements and their inverses?

You are asking for the closure of the vector space of functions spanned by $1, \omega, \omega^{-1}$ closed under addition $h = f + g$ and harmonic mean $\frac{1}{h} = \frac{1}{f} + \frac{1}{g}$.

This will be some subspace of the rational functions $\mathbb{C}(\omega)$ possibly all of them.

• you can get any $a + b\omega + c\omega^{-1}$.

• i'm not even sure you can construct $\omega^2$ this way.

Let me try a little harder. For any function $Z(\omega)$ you can get $$Z + 1 \text{ and } \frac{1}{1 + \frac{1}{Z}} = \frac{Z}{Z+1}$$ These are examples of fractional linear transformations. These correspond to matrices

\[ \left(\begin{array}{cc}1 & 1 \\ 0 & 1 \end{array} \right) \text{ and } \left(\begin{array}{cc}1 & 0 \\ 1 & 1 \end{array} \right)\] These matrices may be easier to fiddle around with than rational functions. I then tried:

\[ \left(\begin{array}{cc}1 & 1 \\ 1 & 0 \end{array} \right) \left(\begin{array}{cc}1 & 0 \\ -1 & 1 \end{array} \right)= \left(\begin{array}{cc}0 & 1 \\ 1 & 0\end{array} \right) \] This would correspond to the function $\frac{1}{Z}$. In terms of rational functions you get the identity

\[ \frac{\frac{z}{-z+1}+1}{\frac{z}{-z}+1} = \frac{z + (-z+1)}{z} = \frac{1}{z} \]

You can imagine the circuit this corresponds to. So your space of rational functions is closed under reciprocals even if you can't multiply.

There are tiny bits of planar algebras and continued fractions in here.

Here's a way to get $\frac{Y^2}{Z}$.

\[ \frac{1}{\frac{1}{y} + \frac{1}{z} } = \frac{zy}{z + y} \]

If we make the replacement $z \mapsto z - y$, the z-transform now reads

\[ \frac{(z-y)y}{z} = \left( 1 - \frac{y}{z} \right) y \text{ and so } -1\left[\left( 1 - \frac{y}{z} \right) y - y\right] = \frac{y^2}{z} \]

You should be able to get $\omega^k$ for any $k \in \mathbb{Z}$ and add to get any Laurent series. Your ring is $\mathbb{C}(\omega)$