Homotopy K-theory

Remarks

The inclusion $R = R[\Delta^0] \to R[\Delta^\bullet]$ gives a natural map $K^B(R) \to KH(R)$. The functor $KH$ is homotopy invariant: $KH_n(R) \cong KH_n(R[t])$ for all $n \in \Z$.

Example: let $R$ be a regular Noetherian ring. Then each $R[\Delta^p]$ is regular, $K_n(R[\Delta^p]) \cong K_n(R)$ by homotopy invariance, and the negative K-groups vanish. So the simplicial spectrum is constant up to equivalence, and $KH_n(R) \cong K_n(R)$ for $n \ge 0$ while $KH_n(R) = 0$ for $n < 0$.

Non-example: ordinary K-theory is not homotopy invariant, so the map $K^B(R) \to KH(R)$ is not always an equivalence. For $R = k[\varepsilon]/(\varepsilon^2)$ over a field $k$, the class of the unit $1 + \varepsilon t$ shows that $K_1(R[t]) \ne K_1(R)$, while $KH_1(R[t]) \cong KH_1(R)$.