nLab biaction




Monoid theory



A ternary function which simultaneously exhibits an action on a set from both the left and the right side.

Sets with biactions are the bimodule objects internal to Set.


Given a set SS and monoids (M,e M,μ M)(M, e_M, \mu_M) and (N,e N,μ N)(N, e_N, \mu_N), a MM-NN-biaction or two-sided action is a ternary function α:M×S×NS\alpha:M \times S \times N \to S such that

  • for all sSs \in S, α(e M,s,e N)=s\alpha(e_M, s, e_N) = s

  • for all sSs \in S, aMa \in M, bMb \in M, cNc \in N, and dNd \in N, α(a,α(b,s,c),d)=α(μ M(a,b),s,μ N(c,d))\alpha\big(a, \alpha(b, s, c), d\big) = \alpha\big(\mu_M(a, b), s, \mu_N(c, d)\big)

Left and right actions

The left M M -action is defined as

α M(a,s)α(a,s,e N)\alpha_M(a, s) \coloneqq \alpha(a, s, e_N)

for all aMa \in M and sSs \in S. It is a left action because

α M(e M,s)=α(e M,s,e N)=s\alpha_M(e_M, s) = \alpha(e_M, s, e_N) = s
α M(a,α L(b,s))=α(a,α(b,s,e N),e N)=α(μ M(a,b),s,μ N(e N,e N))=α(μ M(a,b),s,e N)=α M(μ M(a,b),s)\alpha_M\big(a, \alpha_L(b, s)\big) = \alpha\big(a, \alpha(b, s, e_N), e_N\big) = \alpha\big(\mu_M(a, b), s, \mu_N(e_N, e_N)\big) = \alpha\big(\mu_M(a, b), s, e_N\big) = \alpha_M\big(\mu_M(a, b), s\big)

The right N N -action is defined as

α N(s,c)α(e M,s,c)\alpha_N(s, c) \coloneqq \alpha(e_M, s, c)

for all cNc \in N and sSs \in S. It is a right action because

α N(s,e N)=α(e M,s,e N)=s\alpha_N(s, e_N) = \alpha(e_M, s, e_N) = s
α N(α N(s,c),d)=α(e M,α(e M,s,c),d)=α(μ M(e M,e M),s,μ N(c,d))=α(e M,s,μ N(c,d))=α N(s,μ N(c,d))\alpha_N\big(\alpha_N(s, c), d\big) = \alpha\big(e_M, \alpha(e_M, s, c), d\big) = \alpha\big(\mu_M(e_M, e_M), s, \mu_N(c, d)\big) = \alpha\big(e_M, s, \mu_N(c, d)\big) = \alpha_N\big(s, \mu_N(c, d)\big)

The left MM-action and right NN-action satisfy the following identity:

  • for all sSs \in S, aMa \in M and cNc \in N, α M(a,α N(s,c))=α N(α M(a,s),c)\alpha_M\big(a, \alpha_N(s, c)\big) = \alpha_N\big(\alpha_M(a, s), c\big).

This is because when expanded out, the identity becomes:

α(a,α(e M,s,c),e N)=α(e M,α(a,s,e N),c)\alpha\big(a, \alpha(e_M, s, c), e_N\big) = \alpha\big(e_M, \alpha(a, s, e_N), c\big)
α(μ M(a,e M),s,μ N(c,e N))=α(μ M(e M,a),s,μ N(e N,c))\alpha\big(\mu_M(a, e_M), s, \mu_N(c, e_N)\big) = \alpha\big(\mu_M(e_M, a), s, \mu_N(e_N, c)\big)
α(a,s,c)=α(a,s,c)\alpha(a, s, c) = \alpha(a, s, c)

See also

Last revised on May 25, 2022 at 02:17:35. See the history of this page for a list of all contributions to it.