## Idea ## A ternary function which behaves as simultaneous actions on both the left and the right side: ## Definition ## ### A single action monoid ### Given a set $S$ and a monoid $(M, e, \mu)$, a **$M$-biaction** is a ternary function $\alpha:M \times S \times M \to S$ such that * for all $s:S$, $\alpha(e, s, e) = s$ * for all $s:S$, $a:M$, $b:M$, $c:M$, and $d:M$, $\alpha(a, \alpha(b, s, c), d) = \alpha(\mu(a, b), s, \mu(c, d))$ ### Two action monoids ### Given a set $S$ and monoids $(M, e_M, \mu_M)$ and $(N, e_N, \mu_N)$, a **$M$-$N$-biaction** is a ternary function $\alpha:M \times S \times N \to S$ such that * for all $s:S$, $\alpha(e_M, s, e_N) = s$ * for all $s:S$, $a:M$, $b:M$, $c:N$, and $d:N$, $\alpha(a, \alpha(b, s, c), d) = \alpha(\mu_M(a, b), s, \mu_N(c, d))$ ## Left and right actions The [[left action|left $M$-action]] is defined as $$\alpha_M(a, s) \coloneqq \alpha(a, s, e_N)$$ for all $a:M$ and $s:S$. It is a left action because $$\alpha_M(e_M, s) = \alpha(e_M, s, e_N) = s$$ $$\alpha_M(a, \alpha_L(b, s)) = \alpha(a, \alpha(b, s, e_N), e_N) = \alpha(\mu_M(a, b), s, \mu_N(e_N, e_N)) = \alpha(\mu_M(a, b), s, e_N) = \alpha_M(\mu_M(a, b), s)$$ The [[right action|right $N$-action]] is defined as $$\alpha_N(s, c) \coloneqq \alpha(e_M, s, c)$$ for all $c:N$ and $s:S$. It is a right action because $$\alpha_N(s, e_N) = \alpha(e_M, s, e_N) = s$$ $$\alpha_N(\alpha_N(s, c), d) = \alpha(e_M, \alpha(e_M, s, c), d) = \alpha(\mu_M(e_M, e_M), s, \mu_N(c, d)) = \alpha(e_M, s, \mu_N(c, d)) = \alpha_N(s, \mu_N(c, d))$$ The left $M$-action and right $N$-action satisfy the following identity: * for all $s:S$, $a:M$ and $c:N$, $\alpha_M(a, \alpha_N(s, c)) = \alpha_N(\alpha_M(a, s), c)$. This is because when expanded out, the identity becomes: $$\alpha(a, \alpha(e_M, s, c), e_N) = \alpha(e_M, \alpha(a, s, e_N), c)$$ $$\alpha(\mu_M(a, e_M), s, \mu_N(c, e_N)) = \alpha(\mu_M(e_M, a), s, \mu_N(e_N, c))$$ $$\alpha(a, s, c) = \alpha(a, s, c)$$ ## See also ## * [[action]] * [[bimodule]] * [[biaction homomorphism]]