Let $V$ be a chain complex, which is either $Z$ or $Z/2$ graded. A circle action on $V$ is by definition an action of the dga $H_\ast(S^1)$. This consists of a map $D : V → V$ , which is of square zero, commutes with $d$, and in the $Z$ graded case is of degree 1. the homotopy invariant is $V^{hS^1}= V [[t]]$ with differential $d+tD$. That is if $vf(t)$ is an element $V [[t]]$, then $d(vf(t)) = (dv)f(t) + (Dv)tf(t)$

Could anyone explain to me how this works? why homotopy invariant is that? I am not quite familiar with homotopy invariant.(but I know what it is)

Let $V$ be a chain complex, which is either $Z$ or $Z/2$ graded. A circle action on $V$ is by definition an action of the dga $H_\ast(S^1)$. This consists of a map $D : V → V$ , which is of square zero, commutes with $d$, and in the $Z$ graded case is of degree 1. the homotopy invariant is $V^{hS^1}= V [[t]]$ with differential $d+tD$. That is if $vf(t)$ is an element $V [[t]]$, then $d(vf(t)) = (dv)f(t) + (Dv)tf(t)$

The Tate complex is $V_{Tate} = V ((t))$ again with differential $d+tD.$

The homotopy coinvariants is the space $V_{hS^1} = t^−1V [t^−1] = V_{Tate}/V^{hS^1}$ with differential induced from that on $V_{Tate}$

Could anyone explain to me how this works? why homotopy (co)invariant is that? I am not quite familiar with homotopy (co)invariant.(but I know what it is)