If $G$ acts linearly and isometrically on $V$ then the unit disk $D(V)$ and its boundary $S(V)$ are an appealing kind of equivariant disk and sphere. You can try taking the pairs $(D(V),S(V))$ as the building blocks for equivariant complexes, but you won't get all the equivariant homotopy types that you want, not even the manifolds with smooth $G$-action.

You will get them all if you consider also representations of subgroups. If $H$ acts on $V$ then look at $(G\times_HD(V),G\times_HS(V))$, where $G\times_HX$ means the quotient of $G\times X$ with $(g,hx)$ identified with $(gh,x)$. (This converts an $H$-action into a $G$-action.)

On the other hand, although you also get everything you want if you only use trivial representations of arbitrary subgroups rather than arbitrary representations of arbitrary subgroups.

If $G$ acts linearly and isometrically on $V$ then the unit disk $D(V)$ and its boundary $S(V)$ are an appealing kind of equivariant disk and sphere. You can try taking the pairs $(D(V),S(V))$ as the building blocks for equivariant complexes, but you won't get all the equivariant homotopy types that you want, not even the manifolds with smooth $G$-action.

You will get them all if you consider also representations of subgroups. If $H$ acts on $V$ then look at $(G\times_HD(V),G\times_HS(V))$, where $G\times_HX$ means the quotient of $G\times X$ with $(g,hx)$ identified with $(gh,x)$. (This converts an $H$-action into a $G$-action.)

On the other hand, you also get everything you want if you only use trivial representations of arbitrary subgroups rather than arbitrary representations of arbitrary subgroups.