What are all the continuous and associative binary maps $M: \mathbb{R}^2 \times \mathbb{R}^2 \rightarrow \mathbb{R}^2$ that satisfy $M(x,y+z) = M(x,y) + M(x,z)$?

Besides the 0 constant function, and the identity multiplicative map $M((x_0,x_1),(y_0,y_1)) = (x_0y_0, x_1y_1)$ The theory of the complex numbers gives our first non trivial specimen: $M((x_0,x_1),(y_0, y_1)) = (x_0y_0-x_1y_1, x_0y_1+x_1y_0)$

How to find the rest? If there are any? Posing this as a system of two functional equations (associativity, distributivity over addition) seems like a natural direction but i lack machinery to turn that into a general solution.

Along the functional equation approach: we are looking for solutions to the following system of equations

$$M(x,M(y,z)) = M(M(x,y),z)$$ $$M(x,y+z) = M(x,y)+M(x,z)$$

  • 2
    $\begingroup$ Do you mean $M(x,y+z)=M(x,y)+M(x,z)$? Do you want the operation to be commutative as well, or are you happy with noncommutative multiplications as well (since you do not mention additivity in the first argument)? $\endgroup$ – Jeppe Stig Nielsen Sep 16 '18 at 14:52
  • $\begingroup$ Yep, I corrected it, and yep, dont care for commutativity $\endgroup$ – frogeyedpeas Sep 16 '18 at 15:24
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    $\begingroup$ Why not include $M(x+y,z)=M(x,z)+M(y,z)$ as well, then? By the way, constants do not seem to obey this, unless that constant satisfies $c=c+c$ which happens only for $c=(0,0)$. $\endgroup$ – Jeppe Stig Nielsen Sep 16 '18 at 15:49
  • $\begingroup$ I just want to view the M’s as honomorphisms for the additive structure on R^2, if the space is very rich and complex then I might add your condition as well, and if it still has a lot going on then perhaps commutativity too $\endgroup$ – frogeyedpeas Sep 16 '18 at 15:51

Let $z=a+bi$ with $b\ne0$. Then $\mathbb C=\mathbb R[z]$. In particular, $1,z$ is a basis for $\mathbb C$ over $\mathbb R$. In this basis, multiplication is given by $$ \begin{align} (x_1,y_1)(x_2,y_2) &=(x_1+y_1 z)(x_2+y_2 z) \\&=x_1 x_2 + y_1 y_2 z^2 +(x_1 y_2 +x_2 y_1)z \\&=(x_1 x_2 - y_1 y_2(a^2+b^2))+(x_1 y_2 +x_2 y_1+2a)z \\&=(x_1 x_2 - y_1 y_2(a^2+b^2),x_1 y_2 +x_2 y_1+2a) \end{align} $$


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