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Denis Nardin
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First of all, note that as proven at the beginning of page 38 of the paper:right before example 3.9 they prove that $$H^G_*(S^V;\underline{\mathbb{Z}})=H_*(C^{cell}_*(S^V)^G)\,,$$ where $C^{cell}_*(S^V)$ is the cellular complex for some $G$-CW-structure on $S^V$ (and so levelwise is just a sum of permutation modules). In particular, since $S^V$ is $n$-dimensional $$H^G_n(S^V;\underline{\mathbb{Z}})=\ker\left(C^{cell}_n(S^V)^G\to C^{cell}_{n-1}(S^V)^G\right)\,.$$ Hence $$H^G_n(S^V;\underline{\mathbb{Z}})=\ker\left(C^{cell}_n(S^V)\to C^{cell}_{n-1}(S^V)\right)^G=H^u_n(S^V;\mathbb{Z})^G\,,$$ since taking fixed points commute with taking kernels.

Now, $V$ is orientable iff the action of $G$ preserves the orientation iff $G$ acts trivially on $H^u_n(S^V;\mathbb{Z})$ (recall that an orientation of $V$ is the same thing as a generator of $H^u_n(S^V;\mathbb{Z})$). So the restriction map $H^G_n(S^V;\underline{\mathbb{Z}})\to H^u_n(S^V;\mathbb{Z})$ is an isomorphism iff $V$ is orientable.

WARNING: In general it is not possible to find a $G$-CW structure such that $G$ acts trivially on $C^{cell}_n(S^V)$, even if $V$ is orientable.

First of all, note that as proven at the beginning of page 38 of the paper: $$H^G_*(S^V;\underline{\mathbb{Z}})=H_*(C^{cell}_*(S^V)^G)\,,$$ where $C^{cell}_*(S^V)$ is the cellular complex for some $G$-CW-structure on $S^V$ (and so levelwise is just a sum of permutation modules). In particular, since $S^V$ is $n$-dimensional $$H^G_n(S^V;\underline{\mathbb{Z}})=\ker\left(C^{cell}_n(S^V)^G\to C^{cell}_{n-1}(S^V)^G\right)\,.$$ Hence $$H^G_n(S^V;\underline{\mathbb{Z}})=\ker\left(C^{cell}_n(S^V)\to C^{cell}_{n-1}(S^V)\right)^G=H^u_n(S^V;\mathbb{Z})^G\,,$$ since taking fixed points commute with taking kernels.

Now, $V$ is orientable iff the action of $G$ preserves the orientation iff $G$ acts trivially on $H^u_n(S^V;\mathbb{Z})$ (recall that an orientation of $V$ is the same thing as a generator of $H^u_n(S^V;\mathbb{Z})$). So the restriction map $H^G_n(S^V;\underline{\mathbb{Z}})\to H^u_n(S^V;\mathbb{Z})$ is an isomorphism iff $V$ is orientable.

WARNING: In general it is not possible to find a $G$-CW structure such that $G$ acts trivially on $C^{cell}_n(S^V)$, even if $V$ is orientable.

First of all, note that right before example 3.9 they prove that $$H^G_*(S^V;\underline{\mathbb{Z}})=H_*(C^{cell}_*(S^V)^G)\,,$$ where $C^{cell}_*(S^V)$ is the cellular complex for some $G$-CW-structure on $S^V$ (and so levelwise is just a sum of permutation modules). In particular, since $S^V$ is $n$-dimensional $$H^G_n(S^V;\underline{\mathbb{Z}})=\ker\left(C^{cell}_n(S^V)^G\to C^{cell}_{n-1}(S^V)^G\right)\,.$$ Hence $$H^G_n(S^V;\underline{\mathbb{Z}})=\ker\left(C^{cell}_n(S^V)\to C^{cell}_{n-1}(S^V)\right)^G=H^u_n(S^V;\mathbb{Z})^G\,,$$ since taking fixed points commute with taking kernels.

Now, $V$ is orientable iff the action of $G$ preserves the orientation iff $G$ acts trivially on $H^u_n(S^V;\mathbb{Z})$ (recall that an orientation of $V$ is the same thing as a generator of $H^u_n(S^V;\mathbb{Z})$). So the restriction map $H^G_n(S^V;\underline{\mathbb{Z}})\to H^u_n(S^V;\mathbb{Z})$ is an isomorphism iff $V$ is orientable.

WARNING: In general it is not possible to find a $G$-CW structure such that $G$ acts trivially on $C^{cell}_n(S^V)$, even if $V$ is orientable.

Source Link
Denis Nardin
  • 16.5k
  • 2
  • 69
  • 103

First of all, note that as proven at the beginning of page 38 of the paper: $$H^G_*(S^V;\underline{\mathbb{Z}})=H_*(C^{cell}_*(S^V)^G)\,,$$ where $C^{cell}_*(S^V)$ is the cellular complex for some $G$-CW-structure on $S^V$ (and so levelwise is just a sum of permutation modules). In particular, since $S^V$ is $n$-dimensional $$H^G_n(S^V;\underline{\mathbb{Z}})=\ker\left(C^{cell}_n(S^V)^G\to C^{cell}_{n-1}(S^V)^G\right)\,.$$ Hence $$H^G_n(S^V;\underline{\mathbb{Z}})=\ker\left(C^{cell}_n(S^V)\to C^{cell}_{n-1}(S^V)\right)^G=H^u_n(S^V;\mathbb{Z})^G\,,$$ since taking fixed points commute with taking kernels.

Now, $V$ is orientable iff the action of $G$ preserves the orientation iff $G$ acts trivially on $H^u_n(S^V;\mathbb{Z})$ (recall that an orientation of $V$ is the same thing as a generator of $H^u_n(S^V;\mathbb{Z})$). So the restriction map $H^G_n(S^V;\underline{\mathbb{Z}})\to H^u_n(S^V;\mathbb{Z})$ is an isomorphism iff $V$ is orientable.

WARNING: In general it is not possible to find a $G$-CW structure such that $G$ acts trivially on $C^{cell}_n(S^V)$, even if $V$ is orientable.