\subsection{Computation}
In this section we prove Theorem \ref{thm:compRFH}. In other words, we will calculate the full Rabinowitz Floer homology of $H$ using the isomorphism of the positive Rabinowitz Floer homologies proven in Theorem \ref{thm:iso} and the following long exact sequences (see \eqref{ActionLongExSeq}):
\begin{gather}
\dots \to H_{*+n-1}(\Sigma) \to RFH_{*}^{\geq 0}(H) \to RFH^+_*(H)\to \dots \label{exact1}\\
\dots \to RFH^-_{*}(H) \to RFH_{*}(H) \to RFH^{\geq 0}_*(H)\to \dots \label{exact2}
\end{gather}

We begin by collecting the properties of the Rabinowitz Floer homology for the compact hypersurface $ \Sigma_0$ that we need\footnote{ We remind the reader (Remark \ref{rem:onlyonsigma}) that in the compact case the Rabinowitz Floer homology only depends on $H_0$ through its zero level set $ \Sigma_0$. Therefore instead of $RFH_*(H_0)$ we could write $RFH_*(\Sigma_0)$, and similarly for the other variants. For consistency with the non-compact case, however, we will not do this.}. 
\begin{prop}
\label{compact}
For the compact hypersurface $ \Sigma_0 \subseteq \mathbb{R}^{2k}$ we have
\begin{equation}\label{plusIminus1}
RFH^+_*(H_0) =  \left\lbrace
\begin{array}{c c c}
\Z_2 &  & *= k+1,\\
0 & & *\neq  k+1.
\end{array}
\right.
\qquad
RFH^-_*(H_0)= \left\lbrace
\begin{array}{c c c}
\Z_2 & & *=  -k,\\
0 & & *\neq  -k.
\end{array}
\right.
\end{equation}
\end{prop}
\begin{proof}
Since $\Sigma_0$ is compact, it is displaceable in $\mathbb{R}^{2k}$. Thus by \cite[Thm.\ 1.2]{CieliebakFrauenfelder2009} we have $RFH(H_0)=0$. Thus also the symplectic homology $SH_*(\Sigma_0)$ and cohomology $SH^*(\Sigma_0)$ vanish by \cite[Thm.\ 13.3]{ritter_2013}. Therefore the positive symplectic homology $SH^+_*(\Sigma_0)$ agrees\footnote{This can also easily be proved directly.} with $H^{k-1-*}( \mathbb{R}^{2k})$ by \cite[Lem.\ 2.1]{CieliebakFrauenfelderOancea2010}. Then by \cite[Thm.\ 1.4]{CieliebakFrauenfelderOancea2010} we have $RFH^+_*(H_0) \cong SH^+_*( \Sigma_0)$. Now \eqref{plusIminus1} follows from \eqref{exact1} and \eqref{exact2}.
\end{proof}


That, together with the isomorphism from Theorem \ref{thm:iso} will allow us to calculate the full Rabinowitz Floer homology of $\Sigma$:

\vspace*{0.5cm}
\paragraph{\textit{Proof of Theorem \ref{thm:compRFH}:}}
Recall from Remark \ref{H2} that $\Sigma \simeq S^{n+k-1}\times\R^{n-k}$, hence
\begin{equation}\label{homSigma}
H_*(\Sigma) =  \begin{cases} \Z_2 & *=0, n+k-1,\\ 0 &\textrm{otherwise.}\end{cases}\quad\Rightarrow\quad H_{*+n-1}(\Sigma)=\begin{cases} \Z_2 & *=1-n, k,\\ 0 &\textrm{otherwise.}\end{cases}
\end{equation}
We now use the natural map (all squares commute!) between the long exact sequences proven in Theorem \ref{thm:iso}:
$$
\xymatrix{
 \dots \ar[r] &H_{*+n-1}(\Sigma) \ar[r] \ar[d]_{\Psi^0}
                 & RFH_{*}^{\geq 0}(H) \ar[r] \ar[d] &    RFH^+_*(H) \ar[r] \ar[d]_{\Psi^+} & \dots \\
\dots \ar[r]  & H_{*+k-1}(\Sigma_0)  \ar[r]   & RFH_{*}^{\geq 0}(H_0) \ar[r] & RFH^+_*(H_0) \ar[r]  & \dots}
$$
By Theorem \ref{thm:iso} the map $\Psi^+ \colon  RFH^+_*(H) \to  RFH^+_*(H_0)$ is an isomorphism. Hence, we obtain from \eqref{plusIminus1} and \eqref{homSigma} that $RFH_{*}^{\geq 0}(H)= 0$ for all $* \notin \{1-n, k+1, k\}$. In case $*= 1-n$ we have
\[
\equalto{RFH^+_{ 2-n}(H)}{0}\to \equalto{H_{0}(\Sigma)}{\Z_2} \to RFH_{ 1-n}^{\geq 0}(H) \to \equalto{RFH^+_{ 1-n}(H)}{0},
\]
which implies $RFH_{ 1-n}^{\geq 0}(H)=\Z_2$. In case $* \in \{ k+1, k\}$ we have 
$$
\xymatrix{
 0 \rightarrow RFH_{k +1}^{\geq 0}(H) \ar[r] \ar[d] &  RFH^+_{k  +1}(H) \ar[r] \ar[d]_{\Psi^+} &H_{n+k-1}(\Sigma) \ar[r] \ar[d]_{\Psi^0}
                 & RFH_{k}^{\geq 0}(H) \ar[d] \rightarrow 0\\
0 \rightarrow \equalto{RFH_{k +1}^{\geq 0}(H_0)}{0} \ar[r] &\equalto{RFH^+_{k  +1}(H_0)}{\Z_2} \ar[r] & \equalto{H_{2k-1}(\Sigma_0) }{\Z_2} \ar[r]   & \equalto{RFH_k^{\geq 0}(H_0)}{0} \rightarrow 0}
$$
Observe that the maps $RFH^+_{k  +1}(H_0)\rightarrow H_{2k-1}(\Sigma_0), \Psi^+$ and $\Psi^0$ are isomorphisms (cf.\ Theorem \ref{thm:iso}), implying that the map $RFH^+_{k  +1}(H)\rightarrow H_{n+k-1}(\Sigma)$ is also an isomorphism. Consequently, $RFH_*^{\geq 0}(H)=0$ for $* \in \{ k+1, k\}$. We conclude that
\[
RFH_*^{\geq 0}(H) \coloneqq  \left\lbrace
\begin{array}{c c l }
\Z_2 & & *= 1-n,\\
0 & & \textrm{otherwise}.
\end{array}
\right.
\]
Having determined $RFH^{\geq 0}_*(H)$ we plug it into \eqref{exact2} to determine $RFH(H)$. Combining it with \eqref{plusIminus1} and  Remark \ref{rem:Sigma-} we obtain
$RFH_\ast(H)=0$ for $*\notin \{1{-}n, -k\}$. In the other two cases we have the following exact sequences, which depend on the relation between $n$ and $k$. In case $n\neq k{+}1$ we have
\[
\equalto{RFH^-_{ 1-n}(H)}{0} \to RFH_{ 1-n}(H) \to \equalto{RFH^{\geq 0}_{ 1-n}(H)}{\Z_2}\to 0,
\]
and
\[
0 \to \equalto{RFH^-_{-k}(H)}{\Z_2} \to RFH_{-k}(H) \to \equalto{RFH^{\geq 0}_{-k}(H)}{0},
\]
which gives $RFH_{-k}(H)=RFH_{1-n}(H)=\Z_2$. However, if $n=k{+}1$ we have
\[
0 \to \equalto{RFH^-_{ 1-n}(H)}{\Z_2} \to RFH_{1-n}(H) \to \equalto{RFH^{\geq 0}_{ 1-n}(H)}{\Z_2}\to 0,
\]
which gives $RFH_{ 1-n}(H)=\Z_2\oplus \Z_2$. \hfill $\square$
{"config": "arxiv", "file": "2006.16839/comp.tex"}