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%opening
\title{Theoretical Minimum\\General Relativity\\Exercises}
\author{Simon Crase (compiler)\\simon@greenweaves.nz}

\begin{document}

\maketitle

\begin{abstract}
	These are some exercises arising from the \emph{General Relativity}\cite{susskind2012general} lectures from Leonard Susskind's \emph{Theoretical Minimum} series\cite{susskind2007theoretical}. Section \ref{sec:schwartzchild:metric} contains a derivation of the Schwarzschild metric. Section \ref{sec:gravitational:waves} examines linearized solutions and gravitational waves.

	Disclaimer: I have created these notes as an aide-m\'emoire for my own use; if you find them useful, you are welcome, but I'd appreciate hearing from you. They are not intended
	as a substitute for listening to the lectures. The intellectual property for all material derived from the lectures belongs, of course, to Professor Susskind; any mistakes, however, are my own.

	The notes were created using TexStudio\cite{TexStudio}, which I recommend for compiling them to a PDF, and the bibliography was created using JabRef\cite{Jabref}.
\end{abstract}

\tableofcontents
\listoffigures
\listoftables
\listoftheorems

\section{Schwarzschild Metric}\label{sec:schwartzchild:metric}

The Schwarzschild Metric is presented, without proof, in \cite[Lecture 6]{susskind2012general}. In this section the coordinates are denoted: $(t,r,\theta,\phi)$. The summation convention is observed for all other indices: e.g. $\Gamma\indices{^\alpha_{\gamma\alpha}}$ denoted summation over the dummy index $\alpha$, but $\Gamma\indices{^\theta_{\gamma\theta}}$ is not summed.

\subsection{Ansatz}
We will seek a solution of the form\cite{Adler1965Introduction}:

\begin{align*}
	dt^2 =& e^{\nu(r)} dt^2 -e^{\lambda(r)} dr^2 -r^2 d \Omega^2 \text{, where} \numberthis \label{eq:schwartzschild:ansatz}\\
	d \Omega^2 =&d\theta^2 + cos^2 \theta d\phi^2
\end{align*}

The  non-zero $g_{\mu\nu}$ are:
\begin{align*}
	g_{tt}=& e^{\nu(r)}\\
	g_{rr}=&-e^{\lambda(r)}\\
	g_{\theta\theta}=&-r^2\\
	g_{\phi\phi}=&-r^2 \cos^2 \theta\\
\end{align*}

The $g^{\mu\nu}$ are the inverse of the $g_{\mu\nu}$, so:
\begin{align*}
	g^{tt}=& e^{-\nu(r)}\\
	g^{rr}=&-e^{-\lambda(r)}\\
	g^{\theta\theta}=&-\frac{1}{r^2}\\
	g^{\phi\phi}=&-\frac{1}{r^2 \cos^2 \theta}
\end{align*}

The only non-zero derivatives of $g^{\mu\nu}$ are:
\begin{align*}
	\partial_r g_{tt}=& \nu^\prime(r) e^{\nu(r)}\\
	\partial_r g_{rr}=& -\lambda^\prime(r) e^{\lambda(r)}\\
	\partial_r g_{\theta\theta}=& -2 r\\
	\partial_r g_{\phi\phi}=& -2 r \cos^2 \theta\\
	\partial_\theta g_{\phi\phi}=& 2 r^2 \cos \theta \sin \theta
\end{align*}

\subsection{Christoffel Symbols}

We start be enquiring which Christoffel are not identically zero. For each $\rho$ we ask whether there are any $(\mu,\nu)$ for which the terms making up $\Gamma\indices{^\rho_{\mu\nu}}$ are non-zero.

\begin{align*}
	\Gamma\indices{^t_{\mu\nu}} =&\frac{1}{2} \big[\underbrace{ \partial_\mu g_{t\nu}}_\text{$\mu=r,\nu=t$} + \underbrace{\partial_\nu g_{\mu t}}_\text{$\mu=t,\nu=r$} - \underbrace{\partial_t g_{\mu\nu}}_\text{$\equiv0$}\big] e^{-\nu(r)}\\
	\Gamma\indices{^t_{rt}}=&\frac{1}{2} \partial_r g_{tt}  e^{-\nu(r)}\\
	=&\frac{1}{2}\nu^\prime(r) \cancel{e^{\nu(r)}}\cancel{e^{-\nu(r)}}\\
	=&\frac{\nu^\prime(r)}{2}\\
	\Gamma\indices{^t_{tr}}=&\frac{\nu^\prime(r)}{2}\\
	\Gamma\indices{^t_{rt}}=&\Gamma\indices{^t_{tr}}=\frac{\nu^\prime(r)}{2} \numberthis \label{eq:Gamma:ttr}
\end{align*}

\begin{align*}
	\Gamma\indices{^r_{\mu\nu}} =& - \frac{1}{2} \big[\underbrace{ \partial_\mu g_{r\nu}}_\text{$\mu=\nu=r$} + \underbrace{\partial_\nu g_{\mu r}}_\text{ $\mu=\nu=r$} - \underbrace{\partial_r g_{\mu\nu}}_\text{$\mu=\nu$}\big] e^{-\lambda(r)}\\
	\Gamma\indices{^r_{tt}} =&-\frac{1}{2} \big[-\partial_r g_{tt}\big]e^{-\lambda(r)}\\
	=&\frac{1}{2} \big[\nu^\prime(r) e^{\nu(r)}\big] e^{-\lambda(r)}\\
	=& \frac{1}{2} \nu^\prime(r) e^{\nu(r)-\lambda(r)} \numberthis \label{eq:Gamma:rtt}\\
	\Gamma\indices{^r_{rr}} =&-\frac{1}{2}\big[\partial_r g_{rr}\big] e^{-\lambda(r)}\\
	=&\frac{1}{2}\big[\lambda^\prime(r) \cancel{e^{\lambda(r)}}\big] \cancel{e^{-\lambda(r)}}\\
	=&\frac{\lambda^\prime(r)}{2} \numberthis \label{eq:Gamma:rrr}\\
	\Gamma\indices{^r_{\theta\theta}} =&-\frac{1}{2}\big[-\partial_r g_{\theta\theta}\big] e^{-\lambda(r)}\\
	=&\frac{1}{\cancel{2}}\big[-\cancel{2} r \big] e^{-\lambda(r)}\\
	=& -r  e^{-\lambda(r)}\numberthis \label{eq:Gamma:rtHtH}\\
	\Gamma\indices{^r_{\phi\phi}} =&-\frac{1}{2} \big[-\partial_r g_{\phi\phi}\big]e^{-\lambda(r)}\\
	=&\frac{1}{\cancel{2}} \big[-\cancel{2} r \cos^2 \theta\big]e^{-\lambda(r)}\\
	=& -r \cos^2 \theta e^{-\lambda(r)} \numberthis \label{eq:Gamma:rpHpH}
\end{align*}

\begin{align*}
	\Gamma\indices{^\theta_{\mu\nu}} =&- \frac{1}{2} \big[\underbrace{ \partial_\mu g_{\theta\nu}}_\text{$\mu=r,\nu=\theta$} + \underbrace{\partial_\nu g_{\mu \theta}}_\text{$\mu=\theta,\nu=r$} - \underbrace{\partial_\theta g_{\mu\nu}}_\text{$\mu=\nu=\phi$}\big] \frac{1}{r^2} \\
	\Gamma\indices{^\theta_{r\theta}} =& - \frac{1}{2} \big[\partial_r g_{\theta\theta} \big] \frac{1}{r^2}\\
	=& - \frac{1}{2} \big[-2r \big] \frac{1}{r^2}\\
	=& \frac{1}{r} \\
	\Gamma\indices{^\theta_{r\theta}} =& \Gamma\indices{^\theta_{\theta r}} =\frac{1}{r} \numberthis \label{eq:Gamma:tHtHr}\\
	\Gamma\indices{^\theta_{\phi\phi}} =&- \frac{1}{2} \big[-\partial_\theta g_{\phi\phi}\big] \frac{1}{r^2}\\
	=& \frac{1}{\cancel{2}} \big[\cancel{2} \bcancel{r^2} \cos \theta \sin \theta\big] \frac{1}{\bcancel{r^2}}\\
	=& \cos \theta \sin \theta \numberthis \label{eq:Gamma:tHpHpH}
\end{align*}

\begin{align*}
	\Gamma\indices{^\phi_{\mu\nu}} =&- \frac{1}{2} \big[\underbrace{ \partial_\mu g_{\phi\nu}}_\text{$\mu\in\{\theta,r\},\nu=\phi$} + \underbrace{\partial_\nu g_{\mu \phi}}_\text{$\mu=\phi,\nu\in\{\theta,r\}$} - \underbrace{\partial_\phi g_{\mu\nu}}_\text{$\equiv0$}\big] \frac{1}{r^2 \cos^2 \theta} \\
	\Gamma\indices{^\phi_{\theta\phi}} =&- \frac{1}{2} \big[ \partial_\theta g_{\phi\phi} \big] \frac{1}{r^2 \cos^2 \theta}\\
	=&- \frac{1}{2} \big[ 2 r^2 \cos \theta \sin \theta \big] \frac{1}{r^2 \cos^2 \theta}\\
	=&-\tan \theta \\
	\Gamma\indices{^\phi_{\theta\phi}} =&\Gamma\indices{^\phi_{\phi\theta}} = -\tan \theta \numberthis \label{eq:Gamma:pHpHtH}\\
	\Gamma\indices{^\phi_{r\phi}} =& - \frac{1}{2} \big[\partial_r g_{\phi\phi}\big] \frac{1}{r^2 \cos^2 \theta}	\\
	=& \bcancel{-} \frac{1}{\cancel{2}} \big[\bcancel{-} \cancel{2} r \xcancel{\cos^2 \theta} \big] \frac{1}{r^2 \xcancel{\cos^2 \theta}}	\\
	=& \frac{1}{r}\\
	\Gamma\indices{^\phi_{r\phi}} =& \Gamma\indices{^\phi_{\phi r}} = \frac{1}{r} \numberthis \label{eq:Gamma:pHrpH}
\end{align*}

\subsection{Riemann \& Ricci Tensors}

From \cite[Lecture II]{akhmedov2016lectures}
\begin{align*}
	R\indices{^\mu_{\nu\alpha\beta}} =& \partial_\alpha \Gamma\indices{^\mu_{\nu\beta}} - \partial_\beta \Gamma\indices{^\mu_{\nu\alpha}} + \Gamma\indices{^\mu_{\gamma\alpha}} \Gamma\indices{^\gamma_{\nu\beta}} - \Gamma\indices{^\mu_{\gamma\beta}} \Gamma\indices{^\gamma_{\nu\alpha}}\\
	R_{\mu\nu} =& R\indices{^\alpha_{\mu\alpha\nu}}\\
	=& \partial_\alpha \Gamma\indices{^\alpha_{\mu\nu}} - \partial_\nu \Gamma\indices{^\alpha_{\mu\alpha}} + \Gamma\indices{^\alpha_{\gamma\alpha}} \Gamma\indices{^\gamma_{\mu\nu}} - \Gamma\indices{^\alpha_{\gamma\nu}} \Gamma\indices{^\gamma_{\mu\alpha}}\\
	R =& R_{\mu\nu} g^{\mu\nu}
\end{align*}

We compute the components of $R_{\mu\nu}$.

\subsubsection{$R_{tt}$ and $R_{rr}$}
Since (\ref{eq:schwartzschild:ansatz}) contains two unknown functions, I expect to be able to determine them from two components of $R_{\mu\nu}$.

\begin{align*}
	R_{tt} =& \partial_\alpha \Gamma\indices{^\alpha_{tt}} - \underbrace{\partial_t \Gamma\indices{^\alpha_{t\alpha}}}_\text{$=0$} + \Gamma\indices{^\alpha_{\gamma\alpha}} \Gamma\indices{^\gamma_{tt}} - \Gamma\indices{^\alpha_{\gamma t}} \Gamma\indices{^\gamma_{t\alpha}} \\
	=& \partial_r \Gamma\indices{^r_{tt}}  + \big[\Gamma\indices{^t_{rt}}+\Gamma\indices{^r_{rr}} + \Gamma\indices{^\theta_{r\theta}}+ \Gamma\indices{^\phi_{r\phi}}\big] \Gamma\indices{^r_{tt}} - 2\Gamma\indices{^t_{r t}} \Gamma\indices{^r_{tt}} \\
	=& \partial_r \Gamma\indices{^r_{tt}}  + \big[-\Gamma\indices{^t_{rt}}+\Gamma\indices{^r_{rr}} + \Gamma\indices{^\theta_{r\theta}}+ \Gamma\indices{^\phi_{r\phi}}\big] \Gamma\indices{^r_{tt}} \\
	=& \partial_r \big[\frac{1}{2} \nu^\prime(r) e^{\nu(r)-\lambda(r)}\big]  + \big[-\frac{\nu^\prime(r)}{2}+\frac{\lambda^\prime(r)}{2} + \frac{1}{r} + \frac{1}{r}\big] \frac{1}{2} \nu^\prime(r) e^{\nu(r)-\lambda(r)}  \\
	=& \frac{1}{2} \big[ \nu^{\prime\prime}(r) + \nu^\prime(r) \big(\nu^\prime(r)-\lambda^\prime(r)\big)\big] e^{\nu(r)-\lambda(r)}+ \big[-\frac{\nu^\prime(r)-\lambda^\prime(r)}{2} + \frac{2}{r}\big] \frac{1}{2} \nu^\prime(r) e^{\nu(r)-\lambda(r)} \\
	=& \frac{1}{2} \bigg[\nu^{\prime\prime}(r) + \frac{1}{2}\nu^\prime(r) \big(\nu^\prime(r)-\lambda^\prime(r)\big)+ \frac{2}{r}\nu^\prime(r)\bigg] e^{\nu(r)-\lambda(r)} \numberthis \label{eq:Rtt}
\end{align*}

\begin{align*}
	R_{rr} =& \partial_\alpha \Gamma\indices{^\alpha_{rr}} - \partial_r \Gamma\indices{^\alpha_{r\alpha}} + \Gamma\indices{^\alpha_{\gamma\alpha}} \Gamma\indices{^\gamma_{rr}} - \Gamma\indices{^\alpha_{\gamma r}} \Gamma\indices{^\gamma_{r\alpha}}\\
	=& \underbrace{\partial_t \Gamma\indices{^t_{rr}}}_\text{$=0$} + \cancel{\partial_r \Gamma\indices{^r_{rr}} } +\underbrace{\partial_\theta \Gamma\indices{^\theta_{rr}}}_\text{$=0$} + \underbrace{\partial_\phi \Gamma\indices{^\phi_{rr}}}_\text{$=0$}\\
	&- \partial_r \Gamma\indices{^t_{rt}} - \cancel{\partial_r \Gamma\indices{^r_{rr}}} - \partial_r \Gamma\indices{^\theta_{r\theta}} - \partial_r \Gamma\indices{^\phi_{r\phi}}\\
	&+ \big[\Gamma\indices{^t_{rt}} + \bcancel{\Gamma\indices{^r_{rr}}} + \Gamma\indices{^\theta_{r\theta}}  + \Gamma\indices{^\phi_{r\phi}}\big] \Gamma\indices{^r_{rr}}\\
	&- \Gamma\indices{^t_{t r}} \Gamma\indices{^t_{rt}} - \bcancel{\Gamma\indices{^r_{r r}} \Gamma\indices{^r_{rr}}} - \Gamma\indices{^\theta_{\theta r}} \Gamma\indices{^\theta_{r\theta}} - \Gamma\indices{^\phi_{\phi r}} \Gamma\indices{^\phi_{r\phi}}\\
	=& - \partial_r \frac{\nu^\prime(r)}{2}  - \partial_r \frac{1}{r} - \partial_r \frac{1}{r} + \big[\frac{\nu^\prime(r)}{2}  + \frac{1}{r}  + \frac{1}{r}\big] \frac{\lambda^\prime(r)}{2}- \big(\frac{\nu^\prime(r)}{2}\big)^2 -  \frac{1}{r^2}  - \frac{1}{r^2}\\
	=& -\frac{\nu^{\prime\prime}(r)}{2} + \cancel{\frac{2}{r^2}} + \bigg(\frac{\nu^\prime(r)}{2}  + \frac{2}{r}  \bigg) \frac{\lambda^\prime(r)}{2}- \big(\frac{\nu^\prime(r)}{2}\big)^2 -  \cancel{\frac{2}{r^2}} \\
	=& -\frac{1}{2}\bigg[\nu^{\prime\prime}(r) - \bigg(\frac{\nu^\prime(r)}{2}  + \frac{2}{r}  \bigg) \lambda^\prime(r) + \frac{\big(\nu^\prime(r)\big)^2}{2}\bigg] \numberthis \label{eq:Rrr}
\end{align*}

\subsubsection{Determination of $\nu$ and $\lambda$}

We now have enough information to determine $\nu$ and $\lambda$. From (\ref{eq:Rtt}) and (\ref{eq:Rrr})

\begin{align*}
	\nu^{\prime\prime}(r) + \frac{1}{2}\nu^\prime(r) \big(\nu^\prime(r)-\lambda^\prime(r)\big)+ \frac{2}{r}\nu^\prime(r)=&0 \numberthis \label{eq:Rtt:a}\\
	\nu^{\prime\prime}(r) - \bigg(\frac{\nu^\prime(r)}{2}  + \frac{2}{r}  \bigg) \lambda^\prime(r) + \frac{\big(\nu^\prime(r)\big)^2}{2}=&0\\
\end{align*}
Subtracting these equations:
\begin{align*}
	\cancel{\nu^{\prime\prime}(r)} + \frac{1}{2}\nu^\prime(r) \big(\nu^\prime(r)-\lambda^\prime(r)\big)+ \frac{2}{r}\nu^\prime(r)-\cancel{\nu^{\prime\prime}(r)} + \bigg(\frac{\nu^\prime(r)}{2}  + \frac{2}{r}  \bigg) \lambda^\prime(r) - \frac{\big(\nu^\prime(r)\big)^2}{2}=&0\\
\end{align*}

\begin{align*}
	 \frac{1}{2}\nu^\prime(r) \big(\cancel{\nu^\prime(r)}-\bcancel{\lambda^\prime(r)}\big)+ \frac{2}{r}\nu^\prime(r) + \bigg(\bcancel{\frac{\nu^\prime(r)}{2}}  + \frac{2}{r}  \bigg) \lambda^\prime(r) - \cancel{\frac{\big(\nu^\prime(r)\big)^2}{2}}=&0\\
	 \frac{2}{r}\big(\nu^\prime(r)+\lambda^\prime(r)\big)=&0\\
	 \nu^\prime(r)+\lambda^\prime(r)=&0 \text{, so for some constant $c$}\\
	 e^\lambda(r) =& \frac{c}{e^\nu(r)}  \numberthis \label{eq:lambda:nu}
\end{align*}

Substituting in (\ref{eq:Rtt:a}):
\begin{align*}
	\nu^{\prime\prime}(r) + \nu^\prime(r)^2 + \frac{2}{r}\nu^\prime(r)=&0 \numberthis \label{eq:nu} \text{. Now}\\
	\big[r e^{\nu(r)}\big]^{\prime\prime}=& \big[e^\nu+r \nu^\prime e^\nu\big]^\prime\\
	=& \big[e^\nu \nu^\prime + e^\nu \nu^\prime + r \nu^{\prime\prime} e^\nu + r (\nu^\prime)^2 e^\nu\big]\\
	=& r e^\nu\big[\frac{2}{r} + \nu^{\prime\prime} + (\nu^\prime)^2\big] \text{, so (\ref{eq:nu}) becomes:}\\
	\big[r e^{\nu(r)}\big]^{\prime\prime}=&0, \text{, or}\\
	r e^{\nu(r)} =&ar + b	\text{, for some constants $a$ and $b$}\\
	e^{\nu(r)} =&a + \frac{b}{r}
\end{align*}

Now we know that the metric (\ref{eq:schwartzschild:ansatz}) should be Minkowskian at infinity, which requires $\nu(r)\rightarrow 1$ as $r \rightarrow \infty$, so from:
\begin{align*}
	dt^2 =& \big(a + \frac{b}{r}\big) dt^2 - \frac{c}{a + \frac{b}{r}} dr^2 -r^2 d  \Omega^2 \text{, for some we see that:}\\
	a =& \frac{c}{a}=1 \text{, using (\ref{eq:lambda:nu}), i.e.} \\
	c=1
\end{align*}

We saw in \cite[Lecture 5]{susskind2012general} that we need $g_{00}\approxeq1-\frac{2MG}{r}$ for small $r$, in order to get agreement with Newtonian gravity, hence $b=-2MG$, and (\ref{eq:schwartzschild:ansatz}) becomes:

\begin{align*}
	dt^2 =& \big(1-\frac{2MG}{R}\big) dt^2 - \frac{dr^2}{1-\frac{2MG}{R}}  -r^2 d  \Omega^2 \numberthis \label{eq:schwartzschild:solved}
\end{align*}

\subsubsection{$R_{\theta\theta}$ and $R_{\phi\phi}$}
We need to establish that the remaining diagonal elements are zero using the metric (\ref{eq:schwartzschild:solved}). We will find the following Lemma useful:
\begin{lemma}$\partial_r \big[r  e^{-\lambda(r)}\big]=1$ \label{lemma:dL}
\end{lemma}

\begin{proof}
	From (\ref{eq:schwartzschild:solved})
	\begin{align*}
	e^{\lambda(r)} =& \frac{1}{1 - \frac{2MG}{r}} \text{, whence}\\
	\partial_r \big[r  e^{-\lambda(r)}\big] =& \partial_r \big[r(1 - \frac{2MG}{r})\big]\\
	=& \partial_r [r-2MG]\\
	=& 1
	\end{align*}
\end{proof}

\begin{align*}
	R_{\theta\theta} =& \partial_\alpha \Gamma\indices{^\alpha_{\theta\theta}} - \partial_\theta \Gamma\indices{^\alpha_{\theta\alpha}} + \Gamma\indices{^\alpha_{\gamma\alpha}} \Gamma\indices{^\gamma_{\theta\theta}} - \Gamma\indices{^\alpha_{\gamma\theta}} \Gamma\indices{^\gamma_{\theta\alpha}}\\
	=& \partial_r \Gamma\indices{^r_{\theta\theta}} - \partial_\theta \big[\Gamma\indices{^\phi_{\theta\phi}}+\Gamma\indices{^r_{\theta r}}\big] + \bigg(\Gamma\indices{^t_{rt}}+\Gamma\indices{^r_{rr}}+\cancel{\Gamma\indices{^\theta_{r\theta}}}+\Gamma\indices{^\phi_{r\phi}}\bigg) \Gamma\indices{^r_{\theta\theta}}\\ &-\cancel{ \Gamma\indices{^r_{\theta\theta}} \Gamma\indices{^\theta_{\theta r}}}  - \Gamma\indices{^\theta_{r\theta}} \Gamma\indices{^r_{\theta\theta}} -\Gamma\indices{^r_{r\theta}} \Gamma\indices{^r_{\theta r}} - \Gamma\indices{^\phi_{\phi\theta}} \Gamma\indices{^\phi_{\phi\theta}}\\
	=& \partial_r \bigg[-r  e^{-\lambda(r)}\bigg] - \partial_\theta \bigg[-\tan \theta +\frac{1}{r}\bigg] \\
	 &+ \bigg(\underbrace{\frac{\nu^\prime(r)}{2}+\frac{\lambda^\prime(r)}{2}}_\text{$=0$ from (\ref{eq:lambda:nu})}+\cancel{\frac{1}{\bcancel{r}}\bigg)\bigg[ -\bcancel{r}  e^{-\lambda(r)}\bigg]}\\
	 & + \cancel{\frac{r e^{-\lambda(r)}}{r}}	 -\frac{1}{r^2} - \tan^2 \theta  \numberthis \label{eq:R:theta:theta}\\
    & - \underbrace{1}_\text{\text{From Lemma \ref{lemma:dL}}} + \underbrace{\sec^2 \theta}_\text{$=1+\tan^2 \theta$} + \cancel{\frac{1}{r^2}} -\cancel{\frac{1}{r^2}} - \tan^2 \theta \\
	=& 0 \text{, as expected.}
\end{align*}

Since the calculation of $R_{\phi\phi}$ has proved complex, I have expanded the terms, and used the cancel symbol to mark vanishing Christoffel symbols.

\begin{align*}
	\partial_\alpha \Gamma\indices{^\alpha_{\phi\phi}} - \partial_\phi \Gamma\indices{^\alpha_{\phi\alpha}} =& \partial_r \big[- \cos^2 \theta r e^{-\lambda(r)}\big]+ \partial_\theta \big[\cos\theta \sin\theta\big]-\underbrace{\partial_\phi \Gamma\indices{^\alpha_{\phi\alpha}}}_\text{= 0}\\
	=& - \cos^2 \theta \cdot \underbrace{\partial_r \big[ r e^{-\lambda(r)}\big]}_\text{$=1$ from Lemma \ref{lemma:dL}}+ \partial_\theta \big[\cos\theta \sin\theta\big]\\
	=& - \bcancel{\cos^2 \theta} -\sin^2\theta +\bcancel{\cos^2 \theta} \numberthis \label{eq:phi:phi:deriv}
\end{align*}

\begin{align*}
	\Gamma\indices{^\alpha_{\gamma\alpha}} \Gamma\indices{^\gamma_{\phi\phi}} =&\Gamma\indices{^\alpha_{t\alpha}} \cancel{\Gamma\indices{^t_{\phi\phi}}}+ \Gamma\indices{^\alpha_{r\alpha}} \Gamma\indices{^r_{\phi\phi}}+\Gamma\indices{^\alpha_{\theta\alpha}} \Gamma\indices{^\theta_{\phi\phi}}+\Gamma\indices{^\alpha_{\phi\alpha}} \cancel{\Gamma\indices{^\phi_{\phi\phi}}}\\
	=&\big[\Gamma\indices{^t_{rt}}+\Gamma\indices{^r_{rr}}+\Gamma\indices{^\theta_{r\theta}}+\Gamma\indices{^\phi_{r\phi}}\big] \Gamma\indices{^r_{\phi\phi}}\\
	+&\big[\cancel{\Gamma\indices{^t_{\theta t}}}+\cancel{\Gamma\indices{^r_{\theta r}}}+\cancel{\Gamma\indices{^\theta_{\theta\theta}}}+\Gamma\indices{^\phi_{\theta\phi}}\big] \Gamma\indices{^\theta_{\phi\phi}}\\
	=& \big[\underbrace{\frac{\nu^\prime}{2} + \frac{\lambda^\prime}{2}}_\text{$=0$}+\frac{1}{r}+\frac{1}{r}\big]\big[-r \cos^2\theta e^{-\lambda}\big] - \tan \theta \cos \theta \sin \theta\\
	=& - 2 \cos^2\theta e^{-\lambda} -\sin^2\theta\numberthis \label{eq:prod:1}
\end{align*}

\begin{align*}
	\Gamma\indices{^\alpha_{\gamma\phi}} \Gamma\indices{^\gamma_{\phi\alpha}}=& \cancel{\Gamma\indices{^t_{t\phi}}} \Gamma\indices{^t_{\phi t}}+\cancel{\Gamma\indices{^t_{r\phi}}} \Gamma\indices{^r_{\phi t}}+\cancel{\Gamma\indices{^t_{\theta\phi}}} \Gamma\indices{^\theta_{\phi t}}+\cancel{\Gamma\indices{^t_{\phi\phi}}} \Gamma\indices{^\phi_{\phi t}}\\
	&+ \cancel{\Gamma\indices{^r_{t\phi}}} \Gamma\indices{^t_{\phi r}} + \cancel{\Gamma\indices{^r_{r\phi}}} \Gamma\indices{^r_{\phi r}}+ \cancel{\Gamma\indices{^r_{\theta\phi}}} \Gamma\indices{^\theta_{\phi r}}+ \Gamma\indices{^r_{\phi\phi}} \Gamma\indices{^\phi_{\phi r}}\\
	&+ \cancel{\Gamma\indices{^\theta_{t\phi}}} \Gamma\indices{^t_{\phi\theta}}+ \cancel{\Gamma\indices{^\theta_{r\phi}}} \Gamma\indices{^r_{\phi\theta}}+ \cancel{\Gamma\indices{^\theta_{\theta\phi}}} \Gamma\indices{^\theta_{\phi\theta}}+ \Gamma\indices{^\theta_{\phi\phi}} \Gamma\indices{^\phi_{\phi\theta}} \\
	&+ \cancel{\Gamma\indices{^\phi_{t\phi}}} \Gamma\indices{^t_{\phi\phi}}+ \Gamma\indices{^\phi_{r\phi}} \Gamma\indices{^r_{\phi\phi}}+ \Gamma\indices{^\phi_{\theta\phi}} \Gamma\indices{^\theta_{\phi\phi}}+ \cancel{\Gamma\indices{^\phi_{\phi\phi}}} \Gamma\indices{^\phi_{\phi\phi}}\\
	=& \Gamma\indices{^r_{\phi\phi}} \Gamma\indices{^\phi_{\phi r}}+  \Gamma\indices{^\theta_{\phi\phi}} \Gamma\indices{^\phi_{\phi\theta}} + \Gamma\indices{^\phi_{r\phi}} \Gamma\indices{^r_{\phi\phi}}+ \Gamma\indices{^\phi_{\theta\phi}} \Gamma\indices{^\theta_{\phi\phi}}\\
	=& 2 \Gamma\indices{^r_{\phi\phi}} \Gamma\indices{^\phi_{\phi r}}+  2 \Gamma\indices{^\theta_{\phi\phi}} \Gamma\indices{^\phi_{\phi\theta}}\\
	=& -2\bcancel{ r} \cos^2 \theta e^{- \lambda(r)} \frac{1}{\bcancel{r}}  - 2 cos\theta \sin\theta \tan\theta\\
	=& -2 \cos^2 \theta e^{- \lambda(r)}   - 2 \bcancel{cos\theta} \sin\theta \frac{\sin\theta}{\bcancel{cos\theta}} \\
	=& -2 \cos^2 \theta e^{- \lambda(r)}   - 2  \sin^2\theta  \numberthis \label{eq:prod:2}
\end{align*}

Combining these three equations:
\begin{align*}
	R_{\phi\phi} =& \partial_\alpha \Gamma\indices{^\alpha_{\phi\phi}} - \partial_\phi \Gamma\indices{^\alpha_{\phi\alpha}} + \Gamma\indices{^\alpha_{\gamma\alpha}} \Gamma\indices{^\gamma_{\phi\phi}} - \Gamma\indices{^\alpha_{\gamma\phi}} \Gamma\indices{^\gamma_{\phi\alpha}}\\
	=&\underbrace{ -\sin^2\theta}_\text{from (\ref{eq:phi:phi:deriv})} - \underbrace{\big[\bcancel{2\cos^2\theta e^{-\lambda}} +\sin^2\theta\big]}_\text{from (\ref{eq:prod:1})} + \underbrace{\bcancel{2 \cos^2 \theta e^{- \lambda(r)}}   + 2  \sin^2\theta}_\text{from (\ref{eq:prod:2})}\\
	=&0
\end{align*}

\subsubsection{Off diagonal elements}
It is easy to show that the off-diagonal terms vanish. With the exception of $R_{r\theta}$ there is no need to expand the $\Gamma$; we merely use our knowledge of which $\Gamma$ and $\partial_\alpha \Gamma$ are non-zero.

\begin{align*}
	R_{tr} =& \partial_\alpha \Gamma\indices{^\alpha_{tr}} - \partial_r \Gamma\indices{^\alpha_{t\alpha}} + \Gamma\indices{^\alpha_{\gamma\alpha}} \Gamma\indices{^\gamma_{tr}} - \Gamma\indices{^\alpha_{\gamma r}} \Gamma\indices{^\gamma_{t\alpha}}\\
	=&\partial_t \Gamma\indices{^t_{tr}} - \underbrace{\partial_r \Gamma\indices{^\alpha_{t\alpha}}}_\text{$=0$} + \underbrace{\Gamma\indices{^\alpha_{t\alpha}}}_\text{$=0$} \Gamma\indices{^t_{tr}} - \underbrace{\Gamma\indices{^t_{r r}}}_\text{$=0$} \Gamma\indices{^r_{tt}} -\underbrace{ \Gamma\indices{^r_{t r}}}_\text{$=0$} \Gamma\indices{^t_{tr}}\\
	=&0
\end{align*}

\begin{align*}
	R_{t\theta} =& \underbrace{\partial_\alpha \Gamma\indices{^\alpha_{t\theta}} - \partial_\theta \Gamma\indices{^\alpha_{t\alpha}} + \Gamma\indices{^\alpha_{\gamma\alpha}} \Gamma\indices{^\gamma_{t\theta}}}_\text{These terms all vanish} - \Gamma\indices{^\alpha_{\gamma\theta}} \Gamma\indices{^\gamma_{t\alpha}}\\
	=&  - \Gamma\indices{^r_{\theta\theta}} \underbrace{\Gamma\indices{^\theta_{tr}}}_\text{$=0$}  - \Gamma\indices{^\theta_{r\theta}} \underbrace{\Gamma\indices{^r_{t\theta}}}_\text{$=0$} \\
	=&0
\end{align*}

\begin{align*}
	R_{t\phi} =& \underbrace{\partial_\alpha \Gamma\indices{^\alpha_{t\phi}} - \partial_\phi \Gamma\indices{^\alpha_{t\alpha}} + \Gamma\indices{^\alpha_{\gamma\alpha}} \Gamma\indices{^\gamma_{t\phi}} }_\text{These terms all vanish}- \Gamma\indices{^\alpha_{\gamma\phi}} \Gamma\indices{^\gamma_{t\alpha}}\\
	=& - \Gamma\indices{^r_{\phi\phi}} \underbrace{\Gamma\indices{^\phi_{tr}}}_\text{$=0$} - \Gamma\indices{^\phi_{\theta\phi}} \underbrace{\Gamma\indices{^\theta_{t\phi}}}_\text{$=0$}- \Gamma\indices{^\phi_{r\phi}} \underbrace{\Gamma\indices{^r_{t\phi}}}_\text{$=0$}\\
	&=0
\end{align*}

 $R_{r\theta}$ does not require a knowledge of $\mu(r)$ or $\nu(r)$.

\begin{align*}
	R_{r\theta} =& \partial_\alpha \Gamma\indices{^\alpha_{r\theta}} - \partial_\theta \Gamma\indices{^\alpha_{r\alpha}} + \Gamma\indices{^\alpha_{\gamma\alpha}} \Gamma\indices{^\gamma_{r\theta}} - \Gamma\indices{^\alpha_{\gamma\theta}} \Gamma\indices{^\gamma_{r\alpha}}\\
	=& \underbrace{\partial_r \Gamma\indices{^r_{r\theta}}}_\text{$=0$} - \underbrace{\partial_\theta \big(\Gamma\indices{^\theta_{r\theta}}+\Gamma\indices{^r_{rr}}\big)}_\text{$=0$} + \Gamma\indices{^\phi_{\theta\phi}} \Gamma\indices{^\theta_{r\theta}} - \Gamma\indices{^r_{\theta\theta}} \underbrace{\Gamma\indices{^\theta_{rr}}}_\text{$=0$} - \Gamma\indices{^\theta_{r\theta}} \underbrace{\Gamma\indices{^r_{r\theta}}}_\text{$=0$}- \Gamma\indices{^\phi_{\phi\theta}} \Gamma\indices{^\phi_{r\phi}}\\
	=&-\frac{1}{r^2} -\cancel{\frac{1}{r}\tan \theta} + \cancel{\frac{1}{r}\tan \theta}\\
	=& 0
\end{align*}

\begin{align*}
	R_{r\phi} =& \partial_\alpha \Gamma\indices{^\alpha_{r\phi}} - \partial_\phi \Gamma\indices{^\alpha_{r\alpha}} + \Gamma\indices{^\alpha_{\gamma\alpha}} \Gamma\indices{^\gamma_{r\phi}} - \Gamma\indices{^\alpha_{\gamma\phi}} \Gamma\indices{^\gamma_{r\alpha}}\\
	=& \underbrace{\partial_\phi \Gamma\indices{^\phi_{r\phi}}}_\text{$=0$} - \underbrace{\partial_\phi \Gamma\indices{^\alpha_{r\alpha}}}_\text{$=0$} + \underbrace{\Gamma\indices{^\alpha_{\phi\alpha}}}_\text{$=0$} \Gamma\indices{^\phi_{r\phi}} - \Gamma\indices{^\phi_{\theta\phi}} \underbrace{\Gamma\indices{^\theta_{r\phi}}}_\text{$=0$} - \Gamma\indices{^\phi_{r\phi}} \underbrace{\Gamma\indices{^r_{r\phi}}}_\text{$=0$}\\
	=&0
\end{align*}

\begin{align*}
	R_{\theta\phi} =& \partial_\alpha \Gamma\indices{^\alpha_{\theta\phi}} - \partial_\phi \Gamma\indices{^\alpha_{\theta\alpha}} + \Gamma\indices{^\alpha_{\gamma\alpha}} \Gamma\indices{^\gamma_{\theta\phi}} - \Gamma\indices{^\alpha_{\gamma\phi}} \Gamma\indices{^\gamma_{\theta\alpha}}\\
	=& \underbrace{\partial_\phi \Gamma\indices{^\phi_{\theta\phi}}}_\text{$=0$} - \underbrace{\partial_\phi \Gamma\indices{^\alpha_{\theta\alpha}}}_\text{$=0$} + \underbrace{\Gamma\indices{^\alpha_{\phi\alpha}}}_\text{$=0$} \Gamma\indices{^\phi_{\theta\phi}} - \underbrace{\Gamma\indices{^\phi_{\phi\phi}}}_\text{$=0$} \Gamma\indices{^\phi_{\theta\phi}}\\
	=&0
\end{align*}

\section{Gravitational waves}\label{sec:gravitational:waves}

Gravitational waves are presented, in \cite[Lecture 10]{susskind2012general}.
In this section the coordinates are denoted: $(t,x,y,x)$. The summation convention is observed for all other indices: e.g. $\Gamma\indices{^\alpha_{\gamma\alpha}}$ denotes summation over the dummy index $\alpha$, but $\Gamma\indices{^x_{\gamma x}}$ is not summed.
\subsection{Linearized Field Equations}

From \cite[Lecture 10]{susskind2012general}:
\begin{align*}
	\Gamma\indices{^{\tau}_{\mu\nu}} =&\frac{1}{2} \big[ \partial_\mu g_{\sigma\nu} + \partial_\nu g_{\mu\sigma} - \partial_\sigma g_{\mu\nu}\big] g^{\sigma\tau}\\
	=& \frac{1}{2} \big[ \partial_\mu h_{\sigma\nu} + \partial_\nu h_{\mu\sigma} - \partial_\sigma h_{\mu\nu}\big] \big[\eta^{\sigma\tau }- O(h)\big]\\
	=& \frac{1}{2} \big[ \partial_\mu h_{\sigma\nu} + \partial_\nu h_{\mu\sigma} - \partial_\sigma h_{\mu\nu}\big] \eta^{\sigma\tau }- O(h^2) \numberthis \label{eq:Gamma:small}
\end{align*}

\subsection{Riemann and Ricci Tensors}
From \cite[Lecture 3]{susskind2012general}:
\begin{align*}
	R\indices{^\alpha_{\beta\gamma\delta}} =& \partial_\gamma\Gamma\indices{^\alpha_{\beta\delta}}-\partial_\delta\Gamma\indices{^\alpha_{\beta\gamma}} + O(\Gamma^2) \text{. Now using $o(\Gamma)=o(h)$ from (\ref{eq:Gamma:small})} \\
	=& \frac{1}{2} \big[ \partial_\gamma\partial_\beta h_{\sigma\delta} + \partial_\gamma\partial_\delta h_{\beta\sigma} - \partial_\gamma\partial_\sigma h_{\beta\delta}\big] \eta^{\sigma\alpha }- \frac{1}{2} \big[ \partial_\delta\partial_\gamma h_{\sigma\beta} + \partial_\delta\partial_\beta h_{\gamma\sigma} - \partial_\delta\partial_\sigma h_{\beta\gamma}\big] \eta^{\sigma\alpha }+ O(h^2)\\
	=& \frac{1}{2} \big[ \partial_\gamma\partial_\beta h_{\sigma\delta} + \partial_\gamma\partial_\delta h_{\beta\sigma} - \partial_\gamma\partial_\sigma h_{\beta\delta} - \partial_\delta\partial_\gamma h_{\sigma\beta} - \partial_\delta\partial_\beta h_{\gamma\sigma} + \partial_\delta\partial_\sigma h_{\beta\gamma}\big] \eta^{\sigma\alpha }+ O(h^2)
\end{align*}
Dropping $O(h^2)$ and contracting
\begin{align*}
	R_{\beta\delta} =& 	R\indices{^\alpha_{\beta\alpha\delta}}\\
	=& \frac{1}{2} \big[ \partial_\alpha\partial_\beta h_{\sigma\delta} + \cancel{\partial_\alpha\partial_\delta h_{\beta\sigma}} - \partial_\alpha\partial_\sigma h_{\beta\delta} - \cancel{\partial_\delta\partial_\alpha h_{\sigma\beta}} - \partial_\delta\partial_\beta h_{\alpha\sigma} + \partial_\delta\partial_\sigma h_{\beta\alpha}\big] \eta^{\sigma\alpha }
	\\
	=& \frac{1}{2} \big[ \partial_\alpha\partial_\beta h_{\sigma\delta} - \partial_\alpha\partial_\sigma h_{\beta\delta} -  \partial_\delta\partial_\beta h_{\alpha\sigma} + \partial_\delta\partial_\sigma h_{\beta\alpha}\big] \eta^{\sigma\alpha } \numberthis \label{eq:linearized}\\
	=& 0 \text{ in empty space}.
\end{align*}
Now $\eta^{00}=-1$, $\eta^{ii}=+1$, and all other values are zero, so Einstein's field equations become:
\begin{align*}
	\sum_i\big[\underbrace{\partial_i\partial_\beta h_{i\delta}}_\text{(a)} - \underbrace{\partial_i\partial_i h_{\beta\delta}}_\text{(b)} -  \underbrace{\partial_\delta\partial_\beta h_{ii}}_\text{(c)} + \underbrace{\partial_\delta\partial_i h_{\beta i}\big]}_\text{(d)}\\
	=\underbrace{\partial_0\partial_\beta h_{0\delta}}_\text{(e)} - \underbrace{\partial_0\partial_0 h_{\beta\delta}}_\text{(f)} -  \underbrace{\partial_\delta\partial_\beta h_{00}}_\text{(g)} + \underbrace{\partial_\delta\partial_0 h_{\beta 0}}_\text{(h)} \numberthis \label{eq:expanded_h}
\end{align*}

\subsection{Ansatz}

We will look for plane wave solutions propagating along the $z$ axis.

\begin{align*}
	h_{\mu\nu}(x,t)=&h^0_{\mu\nu} \sin\big(k(t-z)\big)\\
	\partial_0^2 h_{\mu\nu}=& -k^2 h_{\mu\nu}\\
	\partial_z^2 h_{\mu\nu}=& -k^2 h_{\mu\nu}\\
	\partial_0 \partial_z  h_{\mu\nu}=& k^2 h_{\mu\nu}\\
	\partial_z \partial_0  h_{\mu\nu}=& k^2 h_{\mu\nu}
\end{align*}
All other partial derivatives are zero.

The terms on the left hand side of (\ref{eq:expanded_h}) become:
\begin{align*}
	(a)	\sum_i \partial_i\partial_\beta h_{i\delta} =& \begin{cases}
			\partial_z\partial_\beta h_{z\delta} \text{, $\beta\in \{z,t\}$}\\
			0 \text{, otherwise.}
		\end{cases}\\
	(b)	\sum_i \partial_i\partial_i h_{\beta\delta} =& \partial_z\partial_z h_{\beta\delta}\\
	(c)	\sum_i \partial_\delta \partial_\beta h_{ii} =&\begin{cases}
	\partial_\delta \partial_\beta \sum_i h_{ii} \text{, $\beta\&\delta\in\{z,t\}$}\\
	0 \text{, otherwise}.
	\end{cases}\\
	(d)	\sum_i \partial_\delta\partial_i h_{\beta i} =& \begin{cases}
	\partial_\delta\partial_z h_{\beta z} \text {, $\delta\in\{z,t\}$}\\
	0 \text{, otherwise }
	\end{cases}
\end{align*}

The terms on the right hand side of (\ref{eq:expanded_h}) become:
\begin{align*}
	(e)\;	\partial_0\partial_\beta h_{0\delta}=&0 \text{ unless $\beta\in\{z,t\}$ }\\
	(f)\;	\partial_0\partial_0 h_{\beta\delta}\ne&0\\
	(g)\;	\partial_\delta\partial_\beta h_{00}=&0 \text{ unless $\beta\&\delta\in\{z,t\}$ }\\
	(h)\;	\partial_\delta\partial_0 h_{\beta 0}=&0\text{ unless $\delta\in\{z,t\}$ }&
\end{align*}

\subsection{Detailed Constraints}
Substituting $\beta=0,\delta=0$ in (\ref{eq:expanded_h}):
\begin{align*}
	\sum_i&\big[\underbrace{\partial_i\partial_\beta h_{i\delta}}_\text{(a)} - \underbrace{\partial_i\partial_i h_{\beta\delta}}_\text{(b)} -  \underbrace{\partial_\delta\partial_\beta h_{ii}}_\text{(c)} + \underbrace{\partial_\delta\partial_i h_{\beta i}\big]}_\text{(d)}\\
	&=\underbrace{\partial_0\partial_\beta h_{0\delta}}_\text{(e)}- \underbrace{\partial_0\partial_0 h_{\beta\delta}}_\text{(f)} - \underbrace{\partial_\delta\partial_\beta h_{00}}_\text{(g)} + \underbrace{\partial_\delta\partial_0 h_{\beta 0}}_\text{(h)} \\
	&\underbrace{\partial_z\partial_t h_{zt}}_\text{(a)} - \underbrace{\partial_z\partial_z h_{tt}}_\text{(b)} -  \underbrace{\partial_t\partial_t \sum_i h_{ii}}_\text{(c)} + \underbrace{\partial_t\partial_z h_{tt}}_\text{(d)}\\
	&=\underbrace{\cancel{\partial_t\partial_t h_{tt}}}_\text{(e)}- \underbrace{\cancel{\partial_t\partial_t h_{tt}}}_\text{(f)} - \underbrace{\bcancel{\partial_t\partial_t h_{tt}}}_\text{(g)} + \underbrace{\bcancel{\partial_t\partial_t h_{tt}}}_\text{(h)} \\
	0=&\partial_z\partial_t h_{zt} - \partial_z\partial_z h_{tt} -  \partial_t\partial_t \sum_i h_{ii} + \partial_t\partial_z h_{t z}\\
	=&k^2 \big[h^0_{zt}+h^0_{tt}+\sum_ih^0_{ii}+h^0_{tz}\big] \sin\big(k(t-z)\big)\\
	0=& 2h^0_{zt}+h^0_{tt}+\sum_ih^0_{ii} \numberthis \label{eq:con:00}
	\end{align*}

	Substituting $\beta=0,\delta=1$ in (\ref{eq:expanded_h}):
	\begin{align*}
	\sum_i&\big[\underbrace{\partial_i\partial_\beta h_{i\delta}}_\text{(a)} - \underbrace{\partial_i\partial_i h_{\beta\delta}}_\text{(b)} -  \underbrace{\partial_\delta\partial_\beta h_{ii}}_\text{(c)} + \underbrace{\partial_\delta\partial_i h_{\beta i}}_\text{(d)}\big]\\
	&=\underbrace{\partial_0\partial_\beta h_{0\delta}}_\text{(e)} - \underbrace{\partial_0\partial_0 h_{\beta\delta}}_\text{(f)} -  \underbrace{\partial_\delta\partial_\beta h_{00}}_\text{(g)} + \underbrace{\partial_\delta\partial_0 h_{\beta 0}}_\text{(h)}\\
	&\partial_z\partial_t h_{zx}-\partial_z\partial_z h_{tx}=\cancel{\partial_t\partial_t h_{tx}}-\cancel{\partial_t\partial_t h_{tx}}\\
	0=&k^2 \big[h^0_{zx}+h^0_{tx}\big] \sin\big(k(t-z)\big)\\
	0=&h^0_{zx}+h^0_{tx}  \numberthis \label{eq:con:01}
\end{align*}

Substituting $\beta=0,\delta=2$ in (\ref{eq:expanded_h}):
\begin{align*}
	\sum_i&\big[\underbrace{\partial_i\partial_\beta h_{i\delta}}_\text{(a)} - \underbrace{\partial_i\partial_i h_{\beta\delta}}_\text{(b)} -  \underbrace{\partial_\delta\partial_\beta h_{ii}}_\text{(c)} + \underbrace{\partial_\delta\partial_i h_{\beta i}}_\text{(d)}\big]\\
	&=\underbrace{\partial_0\partial_\beta h_{0\delta}}_\text{(e)} - \underbrace{\partial_0\partial_0 h_{\beta\delta}}_\text{(f)} -  \underbrace{\partial_\delta\partial_\beta h_{00}}_\text{(g)} + \underbrace{\partial_\delta\partial_0 h_{\beta 0}}_\text{(h)}\\
	&\underbrace{\partial_z\partial_t h_{zy}}_\text{(a)} - \underbrace{\partial_z\partial_z h_{ty}}_\text{(b)} =\underbrace{\cancel{\partial_t\partial_t h_{ty}}}_\text{(e)} - \underbrace{\cancel{\partial_t\partial_t h_{ty}}}_\text{(f)}\\
	0=&k^2 \big[h^0_{zy}+h^0_{ty}\big] \sin\big(k(t-z)\big)\\
	0=& h^0_{zy}+h^0_{ty} \numberthis \label{eq:con:02}
\end{align*}

Substituting $\beta=0,\delta=3$ in (\ref{eq:expanded_h}):
\begin{align*}
	\sum_i&\big[\underbrace{\partial_i\partial_\beta h_{i\delta}}_\text{(a)} - \underbrace{\partial_i\partial_i h_{\beta\delta}}_\text{(b)} -  \underbrace{\partial_\delta\partial_\beta h_{ii}}_\text{(c)} + \underbrace{\partial_\delta\partial_i h_{\beta i}}_\text{(d)}\big]\\
	&=\underbrace{\partial_0\partial_\beta h_{0\delta}}_\text{(e)} - \underbrace{\partial_0\partial_0 h_{\beta\delta}}_\text{(f)} -  \underbrace{\partial_\delta\partial_\beta h_{00}}_\text{(g)} + \underbrace{\partial_\delta\partial_0 h_{\beta 0}}_\text{(h)}\\
	&\underbrace{\partial_z\partial_t h_{zz}}_\text{(a)} - \underbrace{\xcancel{\partial_z\partial_z h_{tz}}}_\text{(b)} -  \underbrace{\partial_z\partial_t \sum_i h_{ii}}_\text{(c)} + \underbrace{\xcancel{\partial_z\partial_z h_{tz}}}_\text{(d)}\\
	&=\underbrace{\cancel{\partial_t\partial_t h_{tz}}}_\text{(e)} - \underbrace{\cancel{\partial_t\partial_t h_{tz}}}_\text{(f)} -  \underbrace{\bcancel{\partial_z\partial_t h_{tt}}}_\text{(g)} + \underbrace{\bcancel{\partial_z\partial_t h_{tt}}}_\text{(h)}\\
	0=&k^2 \big[h^0_{xx}+h^0_{yy}\big] \sin\big(k(t-z)\big)\\
	0=&h^0_{xx}+h^0_{yy} \numberthis \label{eq:con:03}
\end{align*}

Substituting $\beta=1,\delta=1$ in (\ref{eq:expanded_h}):
\begin{align*}
	\sum_i&\big[\underbrace{\partial_i\partial_\beta h_{i\delta}}_\text{(a)} - \underbrace{\partial_i\partial_i h_{\beta\delta}}_\text{(b)} -  \underbrace{\partial_\delta\partial_\beta h_{ii}}_\text{(c)} + \underbrace{\partial_\delta\partial_i h_{\beta i}}_\text{(d)}\big]\\
	&=\underbrace{\partial_0\partial_\beta h_{0\delta}}_\text{(e)} - \underbrace{\partial_0\partial_0 h_{\beta\delta}}_\text{(f)} -  \underbrace{\partial_\delta\partial_\beta h_{00}}_\text{(g)} + \underbrace{\partial_\delta\partial_0 h_{\beta 0}}_\text{(h)}\\
	&\underbrace{0}_\text{(a)} - \underbrace{\partial_z\partial_z h_{\beta\delta}}_\text{(b)} -  \underbrace{0}_\text{(c)} + \underbrace{0}_\text{(d)}\\
	&=\underbrace{0}_\text{(e)} - \underbrace{\partial_0\partial_0 h_{\beta\delta}}_\text{(f)} -  \underbrace{0}_\text{(g)} + \underbrace{0}_\text{(h)}\\
	0=&\partial_z\partial_z h_{xx}-\partial_0\partial_0 h_{xx}\\
	0=&k^2 \big[ h^0_{xx} - h^0_{xx}\big] \sin\big(k(t-z)\big) \text{ an empty constraint }
\end{align*}

Substituting $\beta=1,\delta=2$ in (\ref{eq:expanded_h}):
\begin{align*}
	\sum_i&\big[\underbrace{\partial_i\partial_\beta h_{i\delta}}_\text{(a)} - \underbrace{\partial_i\partial_i h_{\beta\delta}}_\text{(b)} -  \underbrace{\partial_\delta\partial_\beta h_{ii}}_\text{(c)} + \underbrace{\partial_\delta\partial_i h_{\beta i}}_\text{(d)}\big]\\
	&=\underbrace{\partial_0\partial_\beta h_{0\delta}}_\text{(e)} - \underbrace{\partial_0\partial_0 h_{\beta\delta}}_\text{(f)} -  \underbrace{\partial_\delta\partial_\beta h_{00}}_\text{(g)} + \underbrace{\partial_\delta\partial_0 h_{\beta 0}}_\text{(h)}\\
	&\underbrace{0}_\text{(a)} - \underbrace{\partial_z\partial_z h_{xy}}_\text{(b)} -  \underbrace{0}_\text{(c)} + \underbrace{0}_\text{(d)}\\
	&=\underbrace{0}_\text{(e)} - \underbrace{\partial_0\partial_0 h_{xy}}_\text{(f)} -  \underbrace{0}_\text{(g)} + \underbrace{0}_\text{(h)}\\
	0=& \partial_z\partial_z h_{\beta\delta} -\partial_0\partial_0 h_{\beta\delta}\\
	0=&k^2 \big[h^0_{xy}-h^0_{xy}\big] \sin\big(k(t-z)\big) \text{ an empty constraint }
\end{align*}

Substituting $\beta=1,\delta=3$ in (\ref{eq:expanded_h}):
\begin{align*}
	\sum_i&\big[\underbrace{\partial_i\partial_\beta h_{i\delta}}_\text{(a)} - \underbrace{\partial_i\partial_i h_{\beta\delta}}_\text{(b)} -  \underbrace{\partial_\delta\partial_\beta h_{ii}}_\text{(c)} + \underbrace{\partial_\delta\partial_i h_{\beta i}}_\text{(d)}\big]\\
	&=\underbrace{\partial_0\partial_\beta h_{0\delta}}_\text{(e)} - \underbrace{\partial_0\partial_0 h_{\beta\delta}}_\text{(f)} -  \underbrace{\partial_\delta\partial_\beta h_{00}}_\text{(g)} + \underbrace{\partial_\delta\partial_0 h_{\beta 0}}_\text{(h)}\\
	&\underbrace{0}_\text{(a)} - \underbrace{\cancel{\partial_z\partial_z h_{xz}}}_\text{(b)} -  \underbrace{0}_\text{(c)} + \underbrace{\cancel{\partial_z\partial_z h_{x z}}}_\text{(d)}\\
	&=\underbrace{0}_\text{(e)} - \underbrace{\partial_t\partial_t h_{xz}}_\text{(f)} -  \underbrace{0}_\text{(g)} + \underbrace{\partial_z\partial_t h_{xt}}_\text{(h)}\\
	0=&\partial_t\partial_t h_{xz}-\partial_z\partial_t h_{xt}\\
	=&k^2 \big[h^0_{xz}+h^0_{xt}\big] \sin\big(k(t-z)\big)\\
	0=&h^0_{xz}+h^0_{xt} \text{, which is the same as (\ref{eq:con:01})}
\end{align*}

Substituting $\beta=2,\delta=2$ in (\ref{eq:expanded_h}):
\begin{align*}
	\sum_i&\big[\underbrace{\partial_i\partial_\beta h_{i\delta}}_\text{(a)} - \underbrace{\partial_i\partial_i h_{\beta\delta}}_\text{(b)} -  \underbrace{\partial_\delta\partial_\beta h_{ii}}_\text{(c)} + \underbrace{\partial_\delta\partial_i h_{\beta i}}_\text{(d)}\big]\\
	&=\underbrace{\partial_0\partial_\beta h_{0\delta}}_\text{(e)} - \underbrace{\partial_0\partial_0 h_{\beta\delta}}_\text{(f)} -  \underbrace{\partial_\delta\partial_\beta h_{00}}_\text{(g)} + \underbrace{\partial_\delta\partial_0 h_{\beta 0}}_\text{(h)}\\
	&\underbrace{0}_\text{(a)} - \underbrace{\partial_z\partial_z h_{yy}}_\text{(b)} -  \underbrace{0}_\text{(c)} + \underbrace{0}_\text{(d)}\\
	&=\underbrace{0}_\text{(e)} - \underbrace{\partial_t\partial_t h_{yy}}_\text{(f)} -  \underbrace{0}_\text{(g)} + \underbrace{0}_\text{(h)}\\
	0=&\partial_z\partial_z h_{yy}-\partial_t\partial_t h_{yy}\\
	=&k^2 \big[h^0_{yy}-h^0_{yy}\big] \sin\big(k(t-z)\big) \\
	0=&h^0_{yy}-h^0_{yy} \text{ an empty constraint}
\end{align*}

Substituting $\beta=2,\delta=3$ in (\ref{eq:expanded_h}):
\begin{align*}
	\sum_i&\big[\underbrace{\partial_i\partial_\beta h_{i\delta}}_\text{(a)} - \underbrace{\partial_i\partial_i h_{\beta\delta}}_\text{(b)} -  \underbrace{\partial_\delta\partial_\beta h_{ii}}_\text{(c)} + \underbrace{\partial_\delta\partial_i h_{\beta i}}_\text{(d)}\big]\\
	&=\underbrace{\partial_0\partial_\beta h_{0\delta}}_\text{(e)} - \underbrace{\partial_0\partial_0 h_{\beta\delta}}_\text{(f)} -  \underbrace{\partial_\delta\partial_\beta h_{00}}_\text{(g)} + \underbrace{\partial_\delta\partial_0 h_{\beta 0}}_\text{(h)}\\
	&\underbrace{0}_\text{(a)} - \underbrace{\cancel{\partial_z\partial_z h_{yz}}}_\text{(b)} -  \underbrace{0}_\text{(c)} + \underbrace{\cancel{\partial_z\partial_z h_{yz}}}_\text{(d)}\\
	&=\underbrace{0}_\text{(e)} - \underbrace{\partial_t\partial_t h_{yz}}_\text{(f)} -  \underbrace{0}_\text{(g)} + \underbrace{\partial_z\partial_t h_{yt}}_\text{(h)}\\
	0=&\partial_t\partial_t h_{yz}-\partial_z\partial_t h_{yt}\\
	=&k^2 \big[ h_{yz}+h_{yt}\big] \sin\big(k(t-z)\big) \\
	0=&  h_{yz}+h_{yt} \text{, which is the same as (\ref{eq:con:02})}
\end{align*}

Substituting $\beta=3,\delta=3$ in (\ref{eq:expanded_h}):
\begin{align*}
	\sum_i&\big[\underbrace{\partial_i\partial_\beta h_{i\delta}}_\text{(a)} - \underbrace{\partial_i\partial_i h_{\beta\delta}}_\text{(b)} -  \underbrace{\partial_\delta\partial_\beta h_{ii}}_\text{(c)} + \underbrace{\partial_\delta\partial_i h_{\beta i}}_\text{(d)}\big]\\
	&=\underbrace{\partial_0\partial_\beta h_{0\delta}}_\text{(e)} - \underbrace{\partial_0\partial_0 h_{\beta\delta}}_\text{(f)} -  \underbrace{\partial_\delta\partial_\beta h_{00}}_\text{(g)} + \underbrace{\partial_\delta\partial_0 h_{\beta 0}}_\text{(h)}\\
	&\underbrace{\cancel{\partial_z\partial_z h_{zz}}}_\text{(a)} - \underbrace{\cancel{\partial_z\partial_z h_{zz}}}_\text{(b)} -  \underbrace{\partial_z\partial_z \sum_i h_{ii}}_\text{(c)} + \underbrace{\partial_z\partial_z h_{zz}}_\text{(d)}\\
	&=\underbrace{\partial_t\partial_z h_{tz}}_\text{(e)} - \underbrace{\partial_t\partial_t h_{zz}}_\text{(f)} -  \underbrace{\partial_z\partial_z h_{tt}}_\text{(g)} + \underbrace{\partial_z\partial_t h_{zt}}_\text{(h)}\\
	0=& \partial_z\partial_z\big(h_{xx}+h_{yy}\big) + \partial_t\partial_z h_{tz} - \partial_t\partial_t h_{zz} - \partial_z\partial_z h_{tt} +\partial_z\partial_t h_{zt}\\
	=&k^2 \big[-h^0_{xx} - h^0_{yy} + h^0_{tz} +  h^0_{zz} + h^0_{tt} + h^0_{zt}\big] \sin\big(k(t-z)\big)l_\delta\partial_\beta h_{00} - \partial_\delta\partial_0 h_{\beta 0} \\
	0=&\underbrace{-h^0_{xx} - h^0_{yy}}_\text{$=0$ from (\ref{eq:con:03})} + h^0_{tz} +  h^0_{zz} + h^0_{tt} + h^0_{zt}\\
	=&  +  h^0_{zz} + h^0_{tt} + 2h^0_{zt} \numberthis \label{eq:con:33}
\end{align*}

\subsection{Summary of Constraints}

Table \ref{table:constraints} shows there are only 5 constraints on the 10 parameters $h^0_{\mu\nu}$, so there are indeed plane wave solutions.

\begin{table}[H]
	\begin{center}
		\caption{Constraints on the parameters $h^0_{\mu\nu}$}\label{table:constraints}
		\begin{tabular}{|c|l|} \hline
			\#&Constraint\\ \hline
			(\ref{eq:con:00})&$2h^0_{zt}+h^0_{tt}+\underbrace{h^0_{xx}+h^0_{yy}}_\text{$=0$ from (\ref{eq:con:03})}+h^0_{zz} =0$\\ \hline
			(\ref{eq:con:01})&$h^0_{zx}+h^0_{tx}=0$\\ \hline
			(\ref{eq:con:02})&$h^0_{zy}+h^0_{ty}=0$\\ \hline
			(\ref{eq:con:03})&$h^0_{xx}+h^0_{yy}=0$\\ \hline
			(\ref{eq:con:33})&$h^0_{zz} + h^0_{tt} + 2h^0_{zt}=0$\\ \hline
		\end{tabular}
	\end{center}
\end{table}

\bibliographystyle{unsrt}
\addcontentsline{toc}{section}{Bibliography}
\raggedright
\bibliography{tm}
\end{document}


\section{Improving Parameter Estimates by Solving Sequences of SIPs}\label{sec:mud-pde-sequence}
When progressing from two dimensions to five in the process of refining our estimate of $g$, we did so using an equal amount of parameter samples despite the volume of the spaces differing.
The measure of $\pspace$ (i.e., $\pmeas(\pspace)$), increased by a factor of $64$ while all else was held constant.
Too many of the functions considered by the initial density are impractical to consider because of their roughness.
In the linear examples of \ref{sec:high-dim-linear-example}, it is shown that initial densities which ascribe higher likelihood to the true parameter lead to MUD estimates that are more accurate.
By making better use of our model-evaluation budget of $1000$ samples for the PDE example, we find that both $\qoi_\text{1D}$ and $\qoi_\text{5D}$ perform significantly better in their ability to resolve $\paramref$.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Motivations for a New Initial Density}
Reducing the volume of the support of the initial density will allow the samples drawn from it to better predict the collected data.
Recall from Example~\ref{subsec:pde-example} that two maps are used to solve the SIP: $\qoi_{1D}$ and $\qoi_{2D}$ and MUD points are shown for representative examples.
We use the solutions from those examples\---namely the ratio which updates the initial density\---to inform the construction of a new initial density in five dimensions.
In Figure~\ref{fig:pde-highd-2d-updated}, we plot the initial densities associated with $\qoi_{1D}$ and $\qoi_{2D}$ and remark that the scalar--valued QoI map identifies a contour which appears to trace a straight line through $\pspace$.
This is helpful for identifying the correlation structure and defining a lower-dimensional subspace to perform rejection sampling. However, the solution that comes from using $\qoi_{2D}$ is better at reducing uncertainty.
Taken together, these observations suggest a number of sampling strategies to generate a new initial density.

\begin{figure}
\centering
  \includegraphics[width=0.45\linewidth]{figures/pde-highd/pde-highd_updated_D2_scalar.png}
  \includegraphics[width=0.45\linewidth]{figures/pde-highd/pde-highd_updated_D2_vector.png}
\caption{
100 measurements are used to solve the SIP for the PDE problem with $\qoi_{1D}$ and $\qoi_{2D}$.
}
\label{fig:pde-highd-2d-updated}
\end{figure}

By considering the relationship between the parameters and the types of functions that are possible given the solution to a 2-D inverse problem, we are able to create a more restricted parameter space in five dimensions.

For a detailed discussion of how a new initial density is constructed for this example, we refer the interested reader to Appendix~\ref{ext:pde-5d-initial}.
We summarize the procedure briefly:
First we generate uniform i.i.d. samples in the three dimensions associated with the new knot points by defining independent bounds for each and taking samples from the cross-product of the directions.\footnote{
The bounds for each are determined by looking at piecewise-linear estimates of $g$ that come from sampling the updated density for the vector--valued solution.
}
We generate $\nsamps=1000$ i.i.d. samples from this 2-D uniform density\footnote{
The (computational) cover is described by a procedure which involves the SVD of samples from the scalar--valued solutions in order to capture the correlation structure.
The use of the $\qoi_\text{1D}$ solution is due to the paucity of samples accepted from the vector--valued solution.
The structure of the latter updated density is more amenable to form a good estimate of this correlation direction in parameter space.
}
, and join them with the three other directions to form the new initial sample set, the functions from which generate the curves shown in Figure~\ref{fig:pde-highd-alt-initial-5d}.

\begin{figure}
\centering
  \includegraphics[width=0.675\linewidth]{figures/pde-highd/pde-highd_init_D5-alt}
\caption{
Initial density constructed for the second attempt at the five--dimensional inverse problem, with the structure of solutions learned from the 2-D example incorporated into the selection of bounds in each direction.
}
\label{fig:pde-highd-alt-initial-5d}
\end{figure}

The new initial curves in \ref{fig:pde-highd-alt-initial-5d}\---especially when contrasted to those in Fig.~\ref{fig:pde-highd-initial-5d}\---represent a far more reasonable set of possibilities.
The slope of the functions considered now all only have a single sign change, a marked improvement over the two or three that many samples from \ref{fig:pde-highd-initial-5d} exhibited.
We note that such considerations of smoothness could be avoided by parameterizing $g$ with a basis of some sort, but that problem is beyond the scope of this work.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{SIP Solutions using New Initial Density}
We now come to the solutions that arise from solving the same five--dimensional inverse problem of interpolating the values of $g$ through equispaced knot points, with both types of maps, in Figure~\ref{fig:pde-highd-5d-alt-mud}.
The difference in comparison to the solutions in Fig.~\ref{fig:pde-highd-5d-mud} is stark: no longer are the estimated functions dramatically under-estimating the local minimum of $g$.
The inadequacy of approximation error is attributable to the choice of knot points imposing a regular structure.
Since $g$'s minimum lies between two knot points, the best approximation of where this minimum is will by definition still be incorrect.

\begin{figure}
\centering
  \includegraphics[width=0.45\linewidth]{figures/pde-highd/pde-highd_pair_D5-alt-5-1_m20.png}
  \includegraphics[width=0.45\linewidth]{figures/pde-highd/pde-highd_pair_D5-alt-5-5_m20.png}
  \includegraphics[width=0.45\linewidth]{figures/pde-highd/pde-highd_pair_D5-alt-5-1_m100.png}
  \includegraphics[width=0.45\linewidth]{figures/pde-highd/pde-highd_pair_D5-alt-5-5_m100.png}
\caption{Solutions to the SIP using one hundred measurements for $\ndata = 20$ (top) and $100$ (bottom).
(Left): Scalar-valued solutions for alternative approach to the five-dimensional problem.
(Right): Vector-valued solutions.
}
\label{fig:pde-highd-5d-alt-mud}
\end{figure}

Even when only $20$ measurements are incorporated into constructing the QoI maps, there is a considerable improvement in the predicted boundary conditions when using a better initial density, as seen by comparing the solutions in Fig.~\ref{fig:pde-highd-5d-alt-mud} to Fig.~\ref{fig:pde-highd-5d-mud}.
Owing to the reduced volume of support for the initial density, both QoI maps resolve the residuals similarly, especially as more data are incorporated (shown in the bottom of \ref{fig:pde-highd-5d-alt-mud}).
Since ``unreasonable'' functions are no longer being considered, both maps produce qualitatively similar estimates.

\subsection{Demonstration of Reduction in Uncertainty}
As a final note on this experiment, we contrast the resulting $L^2$-errors to $g$\footnote{derived from computational approximation with the trapezoidal rule} of these MUD solutions, to the previous two examples in Figure~\ref{fig:pde-highd-5d-hist}.
With each successive problem, our uncertainty is reduced and the MUD solutions have lower variance and improved accuracy.
Note that they appear to be moving towards a value away from zero, which represents a fixed bias (five equispaced knots can only approximate this particular $g$ so well).

\begin{figure}
\centering
  \includegraphics[width=0.675\linewidth]{figures/pde-highd/pde-highd_hist_D5_t5-0E-01}
\caption{
Comparison of the 2D initial errors to the 5D ones, as well as the reduction of uncertainty that solving a SIP problem for each provides.
}
\label{fig:pde-highd-5d-hist}
\end{figure}


%auto-ignore
\providecommand{\MainFolder}{..}
\documentclass[\MainFolder/Text.tex]{subfiles}

\begin{document}
\section{Poincar\'e DGA's and Poincar\'e duality models}\label{SubSec:PoincModel}
\allowdisplaybreaks
\Correct[noline,caption={DONE Hom to weak Hom}]{Homotopy to weak homotopy}
\Correct[noline,caption={DONE Add orientation to the data of PDGA}]{Add orientation to the data of PDGA}
\Correct[noline,caption={DONE Change integral to or}]{Change integral to or}

In this section, we restrict to non-negatively graded unital commutative $\DGA$'s, which we denote by $\nnuCDGA$. In this case, the notions of orientation and of cyclic structure agree by Proposition~\ref{Prop:OrAndCyc}.

We modify and combine definitions from~\cite{Van2019} and~\cite{Lambrechts2007} as follows.

\begin{Definition}[Dif.~Poincar\'e duality algebra, $\PDGA$ and formality]\label{Def:PDGA}
A \emph{differential Poincar\'e duality algebra of degree $n$} is a $\nnuCDGA$ $(V,\Dd,\wedge)$ of finite type with orientation~$\Or$ in degree~$n$ such that the induced pairing on $V$ satisfies Poincar\'e duality. If $\Dd = 0$, we call it just \emph{Poincar\'e duality algebra.}

A \emph{Poincar\'e~$\DGA$ (shortly $\PDGA$) of degree $n$} is a $\nnuCDGA$ $(V,\Dd,\wedge)$ together with an orientation $\Or^\H: \H(V) \rightarrow \R$  of degree $n$ which makes $\H(V)$ into a Poincar\'e duality algebra.

A \emph{morphism of $\PDGA$'s} $(V_1,\Dd_1,\wedge_1,\Or^\H_1)$ and $(V_2,\Dd_2,\wedge_2,\Or^\H_2)$ is a $\DGA$-morphism $f: V_1 \rightarrow V_2$ such that the induced map $f_*: \H(V_1) \rightarrow \H(V_2)$ preserves orientation, i.e., it holds $\Or_2^\H \circ f_* = \Or_1^\H$.

A \emph{quasi-isomorphism (or weak equivalence)} of $\PDGA$'s is a morphism of $\PDGA$'s $f: V_1 \rightarrow V_2$ such that $f_*: \H(V_1) \rightarrow \H(V_2)$ is an isomorphism of oriented $\DGA$'s.

Two $\PDGA$'s are \emph{weakly homotopy equivalent (or isomorphic in the homotopy category)} if they are connected by a zig-zag of $\PDGA$-quasi-isomorphisms.

A $\PDGA$ is \emph{formal} if it is weakly homotopy equivalent (as a $\PDGA$) to its homology.
\end{Definition}

It follows from Proposition~\ref{Prop:OrAndCyc} that a differential Poincar\'e duality algebra according to Definition~\ref{Def:PDGA} is precisely a cyclic $\DGA$ from Part~I. In particular, it is finite dimensional. However, if we relax finite type, unitality or commutativity, we obtain a different notion.

\begin{Remark}[Frobenius algebra]
A differential Poincar\'e duality algebra, resp.~a cyclic $\DGA$, is precisely a finite-dimensional symmetric dg-Frobenius algebra from \cite[p.~13]{Vallette2012} or \cite[Theorem~1.1]{Cohen2006}.
\end{Remark}
%Notice that $\DGA$-morphisms of differential Poincar\'e duality algebras preserve $\Or^V$ if and only if they preserve $\Or^\H$, and in this case, they are injective by \eqref{Lem:AutomaticInjectivity}.

\begin{Definition}[Poincar\'e duality model]\label{Def:PDModel}
A \emph{Poincar\'e duality model} of a $\PDGA$ $(V,\Dd,\wedge,\Or^\H)$ is a differential Poincar\'e duality algebra $(\Model,\Dd^\Model,\wedge^\Model,\Or^\Model)$ which is weakly homotopy equivalent to $V$ as a $\PDGA$.

We call a Poincar\'e duality model \emph{small} if $\VansQuotient(\VansSmall(\Model)) \simeq \Model$ for every Hodge decomposition.
\end{Definition}

\begin{Remark}[On Poincar\'e duality models]
\begin{RemarkList}
\item The definition of a Poincar\'e duality model in~\cite{Lambrechts2007} requires only weak homotopy equivalence of $\DGA$'s, i.e., it does not require quasi-isomorphisms to preserve orientation on homology.

\item In general, a Sullivan minimal model $\Lambda U$ fails easily to be a Poincar\'e duality model because it often has non-zero elements in degree $>n$, e.g., powers of even generators; see Example~\ref{Ex:SphereModel} for $\Sph{2}$.

For a compact connected Lie group $G$, the subalgebra of harmonic forms $\Harm$ for any biinvariant Riemannian metric is isomorphic to a free algebra on odd generators, see \cite[Chapter~1]{Felix2008}. Therefore, $\Harm$ with zero differential and the induced cyclic structure is at the same time the Sullivan minimal model and a Poincar\'e duality model for $\DR(G)$.

\item Poincar\'e duality models are not ``strongly unique'' in the sense that two Poincar\'e duality models of the same algebra must not be isomorphic; see Example~\ref{Ex:SUsix} for $\mathrm{SU}(6)$. However, there is a ``weak uniqueness'' statement in Proposition~\ref{Prop:LambrechtUnique} below. The situation is similar to the situation with Sullivan and minimal Sullivan models; see \cite{Felix2008}. We introduced ``smallness'' in Definition~\ref{Def:PDModel} as a candidate for a minimality condition on a Poincar\'e duality model which might imply its ``strong uniqueness''; see Question~\ref{Q:QuestionsPonc}.
\qedhere
\end{RemarkList}
\end{Remark}

Consider the functor from $\PDGA$'s to $\DGA$'s which forgets the orientation on homology. We have the following trivial yet somewhat surprising observation.

\begin{Proposition}[$\PDGA$-formality is the same as $\DGA$-formality]\label{Prop:PoincModelOfFormal}
A Poincar\'e $\DGA$ $(V,\Dd,\wedge,\Or^\H)$ is formal (as a $\PDGA$) if and only if it is formal as a $\DGA$.
%In this case, $(\H(V),\wedge,\int)$ is its minimal Poincar\'e duality model.
\end{Proposition}
\begin{proof}
The ``only if'' part is clear.

As for the ``if'' part, let
\begin{equation}\label{Eq:ZZ}
V \longleftarrow \bullet \dotsb \bullet \longrightarrow \H(V)
\end{equation}
be a weak homotopy equivalence of $\DGA$'s. Denote by $f: \H(V) \rightarrow \H(V)$ the isomorphism on homology induced by \eqref{Eq:ZZ} from the left to the right. We adjoin $\H(V)$ to the right of~\eqref{Eq:ZZ} to obtain the homotopy
\begin{equation}\label{Eq:ZZII}
V \longleftarrow \bullet \dotsb \bullet \longrightarrow \H(V) \xrightarrow{f^{-1}} \H(V)
\end{equation}
whose induced map on homology from the left to the right is the identity. Therefore, we can orient homologies of the inner nodes of~\eqref{Eq:ZZII} so that all maps preserve orientation on homology.
\end{proof}

The next proposition will be used to show the existence of Poincar\'e duality models.

\begin{Proposition}[Extension of Hodge type]\label{Prop:ExtensionOfHodgeType}
Let $V$ be a $\PDGA$ of degree $n\ge 5$ which is of finite type and satisfies $V^0=\Span\{1\}$ and $V^1 = 0$. Then it is a retract of an oriented $\DGA$ $\LambrechtsExtension(V)$ of Hodge type in the category of $\PDGA$'s.
\end{Proposition}

\begin{proof}
Pick an arbitrary harmonic subspace $\Harm$ and an arbitrary complement $C$ of $\ker \Dd$ in $V$.
If $C$ is not perpendicular to $\Harm$, replace it with $\{c - \pi(c)\mid c\in C\}$, where $\pi: V \rightarrow \Harm$ is the orthogonal projection.
We start with $l=\lceil \frac{n}{2} \rceil$ and apply Lemma~\ref{Lemma:Exte} inductively to get an extension $\hat{V} = V \otimes \Lambda$ which admits a decomposition
\begin{equation}\label{Eq:HatDecomp}
\hat{V} = \hat{\Harm}\oplus\Dd\hat{V}\oplus\hat{C}
\end{equation}
of type \eqref{Eq:DecompOfV} such that there is a complement $\hat{E}$ of $\hat{C}^\perp$ in $\hat{C}$ and a linear map $\hat{\rho}: \hat{E}^{\lceil n/2\rceil}\oplus\dotsb\oplus\hat{E}^n\rightarrow\Dd\hat{V}$ such that \eqref{Eq:ConditionTemp} and \eqref{Eq:ConditionTempII} hold.
We consider a linear map
\[
\kappa: \hat{C} \longrightarrow \Dd \hat{V}
\]
such that
\[
\kappa(e)=
\begin{cases}
	0 & \text{for }e\in \hat{E}^i\text{ with }i<\lceil\frac{n}{2}\rceil,\\
	\hat{\rho}(e) & \text{for }e\in \hat{E}^i\text{ with } i>\lceil\frac{n}{2}\rceil, \\
\end{cases}\quad\text{and}\quad \kappa(c^\perp) = 0\quad\text{for }c^\perp\in\hat{C}^\perp.
\]
The case $n = 2k$ and $e\in \hat{E}^k$ is specified as follows.
If $k$ is even, then $\langle \cdot,\cdot \rangle: \hat{E}^{k}\otimes \hat{E}^{k} \rightarrow \R$ is an inner product, and there is an orthonormal basis $\eta_1$, $\dotsc$, $\eta_m$ for some $m\in\N$.
We require
\begin{equation}\label{Eq:InnerProdCase}
 \kappa(\eta_i) = \frac{1}{2}\hat{\rho}(\eta_i)\quad\text{for all }i=1, \dotsc, m.
\end{equation}
If $k$ is odd, then $\langle \cdot,\cdot \rangle: \hat{E}^{k}\otimes \hat{E}^{k} \rightarrow \R$ is a symplectic form, and there is a symplectic basis $\eta_1$, $\theta_1$, $\dotsc$, $\eta_m$, $\theta_m$ for some $m\in\N$.
We use the convention $\langle \theta_i,\eta_j\rangle = \delta_{ij}$ for $i$, $j=1$,~$\dotsc$, $m$.
We require
\begin{equation}\label{Eq:SymplCase}
 \kappa(\eta_i) = \hat{\rho}(\eta_i)\quad\text{and}\quad\kappa(\theta_i)= 0\quad\text{for }i=1,\dotsc,m.
\end{equation}
Let
\[
\hat{C}' \coloneqq \{c - \kappa(c) \mid c\in \hat{C}\}.
\]
This is a complement of $\ker \Dd$ in $\hat{V}$ perpendicular to $\hat{\Harm}$ because $\hat{C}$ is and $\im \kappa \subset\Dd\hat{V}$.
Given homogenous $c_1$, $c_2\in \hat{C}$ with $\deg c_1 + \deg c_2 = n$ and $\deg c_1 \le \deg c_2$, write $c_1 = c^\perp_1 + e_1$ and $c_2 = c^\perp_2 + e_2$ for $c_1^\perp$, $c_2^\perp \in \hat{C}^\perp$ and $e_1$, $e_2\in\hat{E}$, and compute
\begin{align*}
\langle c_1 - \kappa(c_1), c_2 - \kappa(c_2) \rangle &= \langle c_1, c_2 \rangle - \langle \kappa(c_1), c_2 \rangle - \langle c_1, \kappa(c_2) \rangle\\
&=\begin{aligned}[t]
&\underbrace{\langle e_1, e_2 \rangle - \langle \kappa(e_1), e_2 \rangle - \langle e_1, \kappa(e_2) \rangle}_{\eqqcolon(*)} \\ &{}-\underbrace{\langle\kappa(e_1),c_2^\perp\rangle - \langle c_1^\perp, \kappa(e_2)\rangle}_{\eqqcolon(**)}.
\end{aligned}
\end{align*}
Now, $(**)=0$ because of \eqref{Eq:ConditionTempII}.
As for $(*)$, if $\deg c_1 < \deg c_2$, then
\begin{align*}
(**) &= \langle e_1, e_2 \rangle - \langle e_1,\kappa(e_2)\rangle \\
     &= \langle e_1, e_2 \rangle - \langle e_1,\hat{\rho}(e_2)\rangle \\
     &= 0
\end{align*}
because of \eqref{Eq:ConditionTemp}.
If $\deg c_1 = \deg c_2 = k$ and $k$ is even, we plug in the orthonormal basis and get using \eqref{Eq:InnerProdCase} that
\begin{align*}
e_1 = \eta_i,\ e_2 = \eta_j:  && (**) &= \langle \eta_i, \eta_j \rangle - \langle \kappa(\eta_i),\eta_j\rangle - \langle \eta_i,\kappa(\eta_j)\rangle \\
&& & = \langle \eta_i, \eta_j\rangle - \langle \eta_j, \kappa(\eta_i)\rangle - \langle \eta_i, \kappa(\eta_j)\rangle \\
&& & = \langle \eta_i, \eta_j \rangle - \frac{1}{2}\langle \eta_j, \hat{\rho}(\eta_i)\rangle - \frac{1}{2}\langle\eta_i,\hat{\rho}(\eta_j)\rangle\\
&& & = \langle \eta_i, \eta_j \rangle - \frac{1}{2}\langle \eta_j, \eta_i \rangle - \frac{1}{2}\langle \eta_i, \eta_j \rangle \\
&& & = 0.
\end{align*}
If $k$ is odd, we plug in the symplectic basis and get using \eqref{Eq:SymplCase} that
\begin{align*}
e_1 = \eta_i,\ e_2 = \eta_j: && (**) &= \langle \eta_j, \kappa(\eta_i) \rangle - \langle \eta_i, \kappa(\eta_j) \rangle \\
&& &= \langle \eta_j, \hat{\rho}(\eta_j) \rangle - \langle \eta_i, \hat{\rho}(\eta_j) \rangle \\
&& &= 0, \\
e_1 = \theta_i,\ e_2 = \eta_j: && (**) &= \langle \theta_i, \eta_j\rangle - \langle \theta_i, \kappa(\eta_j) \rangle \\
&& &= \langle \theta_i, \eta_j\rangle - \langle \theta_i, \hat{\rho}(\eta_j) \rangle \\
&& &= 0, \\
e_1 = \theta_i,\ e_2 = \theta_j: && (**) &= 0.
\end{align*}
This shows that $\hat{C}\perp\hat{C}$, and hence \eqref{Eq:HatDecomp} is a Hodge decomposition.

Finally, because $\hat{V} = V \otimes \Lambda$ as a $\DGA$, both the inclusion $\iota: V \rightarrow \hat{V}$ of $V$ into $\hat{V}_0$ and the projection $\pi: \hat{V} \rightarrow V$ from $\hat{V}_0$ onto $V$ are $\DGA$ morphisms.
Because $\pi \circ \iota = \Id$ and because $\iota_*$ is an orientation preserving isomorphism, $\pi_*$ is an orientation preserving isomorphism as well.
Therefore, $V$ is a retract of $\LambrechtsExtension(V)\coloneqq\hat{V}$ in the category of $\PDGA$'s.
\end{proof}

The next example shows that the Sullivan minimal model is sometimes of Hodge type.

\begin{Example}[Sullivan minimal model of $\Sph{2}$ is of Hodge type]\label{Ex:SphereModel}
The Sullivan minimal model of $\Sph{2}$ is the free graded commutative algebra $\Model\coloneqq\Lambda(\eta_2,\eta_3)$ with $\Abs{\eta_2}=2$, $\Abs{\eta_3}=3$, $\Dd \eta_2 = 0$ and $\Dd\eta_3=\eta_2\wedge\eta_2$.
From degree reasons, it holds $\Model = \Span\{\eta_2^k \eta_3^l \mid k\ge 0, l \in \{0,1\}\}$ as a graded vector space.
We have a canonical decomposition $\Model = \Harm \oplus \Dd \Model \oplus C$, where $\Harm = \Span\{\eta_2\}$, $\Dd \Model = \Span\{\eta_2^k \mid k \ge 2 \}$ and $C = \Span\{\eta_2^{k}\eta_3 \mid k \ge 0\}$.
We define an orientation $\Or : \Model \rightarrow \R$ in degree $2$ by $\Or(\eta_2) \coloneqq 1$ on $\Harm$ and by $0$ on $\Dd \Model$ and $C$.
It is easy to see that $C\perp \Harm$ and $C\perp C$ with respect to the induced cyclic structure $\langle \cdot,\cdot\rangle$.
Consider the $\DGA$-quasi-isomorphism $f: \Model \rightarrow \DR(\Sph{2})$ defined by $f(\eta_2)\coloneqq \Vol$ and $f(\eta_3)\coloneqq 0$.
Clearly, it is orientation preserving.
We have $\Model/\Model^\perp\simeq \Lambda(\eta_2)$
\end{Example}

The following proposition about the existence of a Poincar\'e dualiy model of a Poincar\'e $\DGA$ $V$ with $\H^1(V)=0$ was originally proven in \cite[Theorem~1.1]{Lambrechts2007}.
It was formulated in the category of $\DGA$'s, i.e., not checking whether the arrows are orientation preserving on homology.
The idea was to construct an extension $\LambrechtsExtension(\Lambda U )$ of the Sullivan minimal model $\Lambda U$ of~$V$ and an orientation on it such that the degenerate subspace is acyclic; the Poincar\'e duality model is then obtained by taking the quotient.
The extension is constructed by adding elements which kill the so called orphans.

By Proposition~\ref{Prop:HodgeAcyc}, we know that $V^\perp$ is acyclic if and only if $V$ is of Hodge type ($V$ needs to be of finite type for the direct implication).
Based on this, we give a new construction of $\LambrechtsExtension(\Lambda U)$ using Lemma~\ref{Lemma:Exte}, i.e., by adding exact partners to non-degenerates.
Our construction works for $n\ge 5$, whereas the assumption of \cite{Lambrechts2007} is $n\ge 7$.
We also do not need $\Dd (\Lambda U)^2 = 0$, although it follows from $\H^1(V) = 0$.
It is also clear from our construction that the arrows preserve orientation on homology.
However, this can be checked for the construction of \cite{Lambrechts2007} as well.

\begin{Proposition}[Existence of Poincar\'e duality model for $\H^1 = 0$]\label{Prop:ExOfLambrStan}
A~Poincar\'e $\DGA$ $V$ with $\H^0(V) = \Span\{1\}$ and $\H^1(V)=0$ admits a Poincar\'e duality model $\Model$.
%If moreover $\H^2 = \H^3 = 0$, then for any two finite dimensional Poincar\'e duality models $M_1$ and $M_2$ there is another finite dimensional Poincar\'e duality model $M_3$ and the zig-zag of orientation preserving quasi-isomorphisms
%$$\begin{tikzcd}
%& M_3 & \\
%M_1\arrow{ur} & & M_2.\arrow{ul}
%\end{tikzcd}$$
\end{Proposition}
\begin{proof}
If $\Or^\H$ comes from a pairing on $V$ which is of Hodge type, then we can take $\Model=\VansQuotient(\VansSmall(V))$. The weak homotopy equivalence of $\PDGA$'s looks like
\begin{equation}\label{Eq:ModOne}
\begin{tikzcd}
 &  \VansSmall(V) \arrow[two heads]{dl}\arrow[hook]{dr} & \\
 \VansQuotient(\VansSmall(V)) & & V.
\end{tikzcd}
\end{equation}
If $V$ is not of Hodge type, we proceed as follows. Let $n$ be the degree of the orientation on $\H(V)$.
If $n\le 6$, then $V$ is formal as a $\DGA$ by~\cite{Miller1979}, and we can take $\Model=\H(V)$ by Proposition~\ref{Prop:PoincModelOfFormal}.
The weak homotopy equivalence of $\PDGA$'s looks like
\begin{equation}\label{Eq:ModTwo}
\begin{tikzcd}
 & \Lambda U \arrow{dl}\arrow{dr} & \\
 \H(V) & & V,
\end{tikzcd}
\end{equation}
where $\Lambda U$ is the Sullivan minimal model of $V$.
The Sullivan minimal model $W\coloneqq \Lambda U$ is of finite type and satisfies $W^0 = \R$, $W^1 = 0$ and $\Dd W^2 = 0$.
Suppose that $n\ge 7$.
Let $\LambrechtsExtension(W)$ be the extension of $W$ of Hodge type either from Proposition~\ref{Prop:ExtensionOfHodgeType} or from \cite[Section~4]{Lambrechts2007}.
This extension is of finite type, the inclusion $W\hookrightarrow\LambrechtsExtension(W)$ is a quasi-isomorphism of $\DGA$'s and $\LambrechtsExtension(W)^\perp$ is acyclic.
Moreover, $W\hookrightarrow\LambrechtsExtension(W)$ preserves orientation on homology.
We take $\Model = \VansQuotient(\LambrechtsExtension(\Lambda U))$ and obtain the following weak homotopy equivalence of $\PDGA$'s:
\begin{equation}\label{Eq:ModThree}\begin{tikzcd}
 & \Lambda U \arrow{dl}\arrow{dr} & \\
\VansQuotient(\LambrechtsExtension(\Lambda V))& & V.
\end{tikzcd}\end{equation}
This proves the proposition.
\end{proof}

The following is mostly \cite[Theorem~7.1]{Lambrechts2007}.
In addition, we check that the orientation on homology is preserved.
Also, by using our extension of Hodge type, we can improve from $n\ge 7$ to $n\ge 5$.

\begin{Proposition}[``Weak uniqueness'' of Poincar\'e duality model]\label{Prop:LambrechtUnique}
Let $V_1$ and $V_2$ be differential Poincar\'e duality algebras of degree $n$ which are weakly homotopy equivalent as $\PDGA$'s.
Suppose that $\H^0(V_1)=\H^0(V_2)=\Span\{1\}$ and $\H^1(V_1) = \H^1(V_2) = 0$.
In addition, suppose that $\H^2(V_1) = \H^2(V_2) = 0$, $V_1^1 = V_2^1 = 0$ and $n\ge 5$.
Then there is a differential Poincar\'e duality algebra $V_3$ and $\PDGA$-quasi-isomorphisms\footnote{These are automatically injective and orientation preserving.}
\begin{equation}\label{Eq:LambrechtsZigZag}
\begin{tikzcd}
& V_3 & \\
V_1\arrow{ur}& & \arrow{ul}V_2.
\end{tikzcd}
\end{equation}
%Moreover, $V_3$ is of finite type, $V_3^0 = \R$, $V_3^1 = 0$ and $\Dd V^3 = 0$.
\end{Proposition}
\begin{proof}
By the assumption, there is $k\ge 1$ and a zig-zag of $\PDGA$-quasi-isomorphisms
\begin{equation}\label{Eq:ZigZag}
V_1 \longleftarrow Z_1 \longrightarrow Z_2 \longleftarrow Z_3 \longrightarrow Z_4 \longleftarrow \dotsb \longleftarrow Z_k \longrightarrow V_2.
\end{equation}
Consider the Sullivan minimal model $\Lambda U \rightarrow Z_2$ and use the Lifting Lemma \cite[Lemma~2.15]{Felix2008} to construct $\DGA$-quasi-isomorphisms $\Lambda U \rightarrow Z_1$ and $\Lambda U \rightarrow Z_3$ such that the diagram
$$\begin{tikzcd}
& \Lambda U \arrow{d} \arrow{ld} \arrow{rd} & \\
Z_1 \arrow{r} & Z_2 & \arrow{l} Z_3
\end{tikzcd}$$
commutes up to homotopy of $\DGA$'s. It is easy to see that there is an orientation on $\H(\Lambda U)$ such that all morphisms preserve orientation on homology. Therefore, we can replace the segment $V_1 \longleftarrow Z_1 \longrightarrow Z_2 \longleftarrow Z_3 \longrightarrow Z_4$ in \eqref{Eq:ZigZag} by $V_1 \longleftarrow \Lambda U \longrightarrow Z_4$. Repeating this process, we can shorten \eqref{Eq:ZigZag} to
\begin{equation}\label{Eq:DiagDiag}
\begin{tikzcd}
& \Lambda U \arrow{dr}{f_2} \arrow[swap]{dl}{f_1} & \\
V_1 & & V_2,
\end{tikzcd}
\end{equation}
where $f_1$ and $f_2$ are $\PDGA$-quasi-isomorphisms. In order to take $\VansQuotient(\LambrechtsExtension(\Lambda U))$ and obtain a zig-zag with three terms, we have to revert the arrows in \eqref{Eq:DiagDiag}. The trick from~\cite{Lambrechts2007} is the following:

Consider the relative minimal model of the multiplication $\mu: \Lambda U \otimes \Lambda U \rightarrow \Lambda U$; from \cite[Example~2.48]{Felix2008}, it is given by
\begin{equation}\label{Eq:RelMinMod}
\begin{tikzcd}
\Lambda U \otimes \Lambda U\arrow{r}{\mu} \arrow{rd}{i} & \Lambda U \\
& M(\mu)\coloneqq \Lambda U \otimes \Lambda U \otimes \Lambda(U[1]), \arrow{u}{p}
\end{tikzcd}
\end{equation}
where $i$ is the inclusion into the first two factors, which is a cofibration, and $p$ is a surjective quasi-isomorphism. Let $\iota_i : \Lambda U \rightarrow \Lambda U \otimes \Lambda U$ for $i=1$, $2$ be the inclusions to the first and the second factor, respectively. Because $\mu \circ \iota_i = \Id$ and $p$ is a quasi-isomorphism, the maps $i \circ \iota_i : \Lambda U \rightarrow M(\mu)$ for $i=1$, $2$ are quasi-isomorphisms. Moreover, it is easy to see that $\H(M(\mu))$ inherits an orientation such that $p_*$ and $(i\circ \iota_i)_*$ are orientation preserving. To transfer this situation to $V_i$, we use the diagram
\begin{equation}\label{Eq:Pushout}
\begin{tikzcd}
\Lambda U \arrow{r}{f_i}\arrow{d}{\iota_i}& V_1 \arrow{d}{\iota_i^V} \\
\Lambda U \otimes \Lambda U \arrow{r}{f_1\otimes f_2} \arrow{d}{i} & V_1 \otimes V_2 \arrow{d}{g_2} \\
M(\mu) \arrow{r}{g_1} & \tilde{V}_3 \coloneqq  M(\mu) \otimes_{\Lambda U \otimes \Lambda U} (V_1\otimes V_2),
\end{tikzcd}
\end{equation}
where $\iota_i^V : V_i \rightarrow V_1 \otimes V_2$ for $i=1$, $2$ are inclusions. The lower square, i.e., the maps $g_1$, $g_2$ and the $\DGA$ $\tilde{V}_3$, is a pushout diagram (see \cite[Example~1.4]{LoopSpaces}). According to~\cite{MO204414}, the model category of $\nnuCDGA$ is proper, and hence pushouts along cofibrations preserve quasi-isomorphisms. Therefore, $f_1\otimes f_2$ being a quasi-isomorphism implies that $g_1$ is a quasi-isomorphism. We push the orientation to~$\H(\tilde{V}_3)$ via $g_{1*}$. Since $i\circ \iota_i$, $g_1$ and $f_i$ are quasi-isomorphisms preserving orientation on homology, it follows that $h_i\coloneqq g_2\circ \iota_i^V:  V_i \rightarrow V_3$ are quasi-isomorphisms preserving orientation of homology as well. It holds
\[
\tilde{V}_3 \simeq \Lambda(U[1]) \otimes V_1 \otimes V_2.
\]
Clearly, $\tilde{V}_3$ is of finite type.
Using $\H^1(V_i) = \H^2(V_i) = 0$, we have $U^1 = U^2 = 0$, and hence $(\Lambda(U[1]))^1 = 0$.
This together with $V_i^1 = 0$ implies that $\tilde{V}_3^1 = 0$.
%$\tilde{V}_3^1 = 0$ and $\tilde{V}_3^2 = \tilde{V}_1^0 \otimes \tilde{V}_2^0 \otimes U^{3} = 0$ hold due to the additional assumptions.
Therefore, all conditions for an application of the Hodge extension $\LambrechtsExtension$ from Proposition~\ref{Prop:ExtensionOfHodgeType} are satisfied, and we can set
$$ V_3\coloneqq \VansQuotient(\LambrechtsExtension(\tilde{V}_3)). $$
This finishes the proof.
%
%I could have started with Poincar\'e $\DGA$'s (i.e., Poincar\'e duality algebra on homology) and obtain a Poincar\'e DGA $\tilde{V}_3$ and subsequently a Poincar\'e model. Hence quasi-isomorphic $\PDGA$'s, then there exists a Poincar\'e model $V_3$ and the arrows to them.
%
%Suppose we have started with differential Poincar\'e duality algebras, then we obtain isomorphisms of small algebras.
\end{proof}

\begin{Conjecture}\label{Conj:PDGALST}
The additional assumptions of Proposition~\ref{Prop:LambrechtUnique} can be dropped.
\end{Conjecture}

\begin{Remark}[Weak uniqueness in the case of $\H^1(V)=0$ and $n\le 3$]
 For $n=1$, there is no differential Poincar\'e duality algebra $V$ with $\H^1(V) = 0$.

 For $n=2$, a general differential Poincar\'e duality algebra can be written in terms of its Hodge decomposition as
 \begin{align*}
 	V^2 & = \Span\{\Vol\}\oplus \Dd C^1\\
	V^1 & = \Dd C^0 \oplus C^1\\
	V^0 & = \Span\{1\} \oplus C^0.
 \end{align*}
 Now, $\Harm = \Span\{1\}\oplus \Span\{\Vol\}$ is a dg-subalgebra which is itself a Poincar\'e duality algebra. Therefore, two differential Poincar\'e duality algebras with $\H(V_1)\simeq \H(V_2)$ and $\H^1(V_i) = 0$ are connected via the zig-zag
\[
\begin{tikzcd}
& \Harm \arrow{dr}{} \arrow[swap]{dl}{} & \\
V_1 & & V_2.
\end{tikzcd}
\]
For $n=3$, we have
\begin{align*}
	V^3 & = \Span\{\Vol\} \oplus \Dd C^2\\
 	V^2 & =\Dd C^1 \oplus C^2 \\
	V^1 & = \Dd C^0 \oplus C^1\\
	V^0 & = \Span\{1\} \oplus C^0,
\end{align*}
and the same situation as for $n=2$ occurs.

For $n=4$ and $V^1 = 0$, we have $V \simeq \Harm$.
\end{Remark}

We would like to define a ``minimal Poincar\'e duality model''. We motivate this notion in the following remark.

\begin{Remark}[Model and minimal model]\label{Rem:Models}
We shall understand models and minimal models in terms of model categories and their homotopy categories.

Let us illustrate this on Sullivan models. A Sullivan $\DGA$ is a free graded commutative algebra $\Lambda U$ over a positively graded vector space $U$ which admits a well-ordered homogenous basis $(v_\alpha)$ such that $\Dd v_\alpha \in \Lambda(v_\beta \mid \beta < \alpha)$ ($\coloneqq$\,the subalgebra of $\Lambda U$ generated by the $v_\beta$'s) for all $\alpha$. A Sullivan $\DGA$ is called minimal if $\im \Dd \subset \Lambda_{\ge 2} U$ ($\coloneqq$\,the set of decomposable elements).

According to \cite[Theorem~4.3]{Bousfield1976}, the category $\nnuCDGA$ is a model category with weak equivalences being $\DGA$-quasi-isomorphisms, fibrations being degree-wise surjective $\DGA$-morphisms and cofibrations being retracts of relative Sullivan algebras (see \cite[Proposition~2.22 and Proposition~2.28]{Felix2008}). Cofibrant objects are then precisely Sullivan algebras.

So, the homotopy extension property holds already for Sullivan $\DGA$'s. To see the role of minimality, we shall descent to the homotopy category. The homotopy category is constructed from a model category by localizing morphisms at weak equivalences. An isomorphism in the homotopy category, called weak homotopy equivalence, corresponds to a zig-zag of weak equivalences. If $V_1$ is weakly homotopy equivalent to $V_2$, we say that $V_2$ is a model of $V_1$. If there is a weak equivalence $V_2 \rightarrow V_1$, we say that $V_2$ is a resolution of~$V_1$. We understand minimality as a condition which is in each weak homotopy equivalence class satisfied by at most one object up to isomorphism in the model category. If minimal models exist, they form a skeleton of the homotopy category. This is precisely the case of $\nnuCDGA$ and minimal Sullivan algebras. Indeed, by \cite[Theorem~2.24]{Felix2008}, every connected $\nnuCDGA$ is resolved by a minimal Sullivan algebra. Next, by \cite[Proposition~2.26]{Felix2008}, a $\DGA$-morphism lifts to resolutions by Sullivan algebras, and by \cite[Corollary~2.13]{Felix2008}, quasi-isomorphic minimal Sullivan $\DGA$'s are isomorphic. Finally, this implies that weakly homotopy equivalent minimal Sullivan algebras are isomorphic, and hence are minimal in the sense above.

As another example, for an operad (or properad) $\Operad$, one wants to construct a dg-operad~$\Operad_\infty$ which is a quasi-free resolution of $\Operad$ (see \cite{Vallette2012}). Quasi-free means that after forgetting the differential, the operad $\Operad_\infty$ is free over $\Perm$-bimodules. This is similar to Sullivan models which are free over vector spaces. For quadratic operads, $\Operad_\infty$ is often constructed as the cobar construction $\Omega$ of the Koszul dual cooperad $\Operad^{\mbox{!`}}$. This is the case of $\AInfty$, $\LInfty$ or of the properad $\IBLInfty$. The differential on $\Omega \Operad^{\mbox{!`}}$ is the extension of the decomposition on~$\Operad^{\mbox{!`}}$ to a derivative, and hence it has decomposable image (c.f., the explicit formula~\cite[Formula~(2)]{Peksova2018}). This is similar to minimal Sullivan models. It would be interesting to know whether $\Operad_\infty$ can be constructed using the same inductive method of ``killing'' and ``adding'' generators of homology as Sullivan minimal models.

Finally, let us note that in \cite{Cirici2019}, they use the inductive method to construct (minimal) models of $\Operad$-algebras for a wide class of operads $\Operad$. Note that cyclic $\AInfty$-algebras can be formulated in the language of cyclic operads. However, in the case of $\PDGA$'s, we have the non-degenerate pairing on homology and an operadic description is not clear.
\end{Remark}
%\begin{Lemma}
%Let $f: V_1 \rightarrow V_2$ be an orientation preserving quasi-isomorphism of differential Poincar\'e duality algebras $(V_1,\Dd_1,\wedge_1,\Or_1)$ and $(V_2,\Dd_2,\wedge_2,\Or_2)$. Then $f$ is injective and... \todo{What more properties? Dies there exist a dg-ideal $Z$ s.t. $V_2 = Z \oplus f(V_1)$?}
%\end{Lemma}
%\begin{proof}
%It follows easily that $f$ is injective. Now we have $V = W \oplus W_\perp$. From the cyclicity of $\Dd$ we get $\Dd W_\perp \subset V$. We would like to construct a complementary differential graded ideal. We still haven't used the fact that $W$ is a subalgebra and the cyclicity.
%\end{proof}
%
%\begin{Lemma}
%A cyclic cochain complex which is of finite-type (degreewise finite dimensional) and satisfies Poincar\'e duality is of Hodge type.
%\end{Lemma}
%\begin{proof}
%
%\end{proof}
%
%\begin{Lemma}
%Let $V$ be of finite type. Then Poincar\'e duality is equivalent to non-degeneracy.
%\end{Lemma}
%
%Therefore, Poincar\'e duality algebras are contained in degrees $0$ $\dots$ $n$.
%
%
%
%\begin{Lemma}
%Any two small algebras are isomorphic.
%\end{Lemma}
%\begin{proof}
%\end{proof}
%
%\begin{Lemma}
%If $V$ is differential Poincar\'e duality algebra of finite type, then $Q(V_{\text{small}})$ is a weakly equivalent differential Poincar\'e duality algebra of finite type.
%\end{Lemma}
%
%\begin{Lemma}
%Let $V$ be a differential Poincar\'e duality algebra of finite type. Then the sequence of taking $Q(\bullet_{smallal}$ stabilizes. I.e. there are going to be isomorphic.
%\end{Lemma}
%\begin{proof}
%Clear for finite dimensional.
%\end{proof}
%
%\begin{Lemma}
%Any two minimal Poincar\'e duality models are isomorphic.
%\end{Lemma}

%Consider small Poincar\'e duality models. Let $(V,\Dd,\wedge,\Or^\H)$ be a $\PDGA$ with $\H^0(V) = \R$ and $\H^1(V)=0$. Proposition~\ref{Prop:ExOfLambrStan} gives a Poincar\'e duality model $\Model$. Proposition~\ref{Prop:PropPropertiessd} asserts that $\VansQuotient(\VansSmall(\Model))$ is small and Proposition~\ref{Prop:HodgeAcyc} asserts that it is weakly homotopy equivalent to $\Model$ and hence to $V$ as a $\PDGA$. Therefore, small Poincar\'e duality models exist. Let $\Model_1$ and $\Model_2$ be two small Poincar\'e duality models of $V$. Pick any of their Hodge decompositions. Proposition~\ref{Prop:LambrechtUnique}, more generally Conjecture~\ref{Conj:PDGALST}, gives the diagram \eqref{Eq:LambrechtsZigZag}, where $V_1 = \Model_1$, $V_2 = \Model_2$ and $V_3$ are differential Poincar\'e duality algebras and the maps, which we denote by $f_1: V_1 \rightarrow V_3$ and $f_2: V_2\rightarrow V_3$, are $\PDGA$-quasi-isomorphisms. Lemma~\ref{Eq:LemSmallSub} asserts that there are Hodge decompositions of $V_3$ with small subalgebras $\VansSmall_1(V_3)$ and $\VansSmall_2(V_3)$ such that $f_1$ and $f_2$ induce pairing preserving isomorphisms $\VansSmall(V_1) \simeq \VansSmall_1(V_3)$ and $\VansSmall(V_2) \simeq \VansSmall_2(V_3)$, respectively. Conjecture~\ref{Conj:UnieqSmal} asserts that $\VansQuotient(\VansSmall_1(V_3)) \simeq \VansQuotient(\VansSmall_2(V_3))$. In total, we have
%$$ \Model_1 \simeq \VansQuotient(\VansSmall(\Model_1)) \simeq \VansQuotient(\VansSmall_1(V_3))\simeq \VansQuotient(\VansSmall_2(V_3)) \simeq \VansQuotient(\VansSmall(\Model_2)) \simeq \Model_2 $$
%as differential Poincar\'e duality algebras.
%\begin{Corollary}[Small Poincar\'e duality models are minimal]\label{Cor:SmalPoinc}
%Small Poincar\'e duality models exist when $\H^0 = \R$ and $\H^1 = 0$ and are unique up to an isomorphism provided that the conjectures from Sections~\ref{SubSec:CycStr} and~\ref{SubSec:PoincModel} are true.
%\end{Corollary}

The following example shows that the zig-zag of a Poincar\'e duality model can not always be shortened to one arrow.

\begin{Example}[No Poincar\'e dualiy model with one arrow]\phantomsection\label{Ex:NoOneArrow}
\begin{ExampleList}
\item The closed genus $2$ surface $\Sigma_2$ does not admit a Poincar\'e duality model $A$ with just one arrow $A \rightarrow \DR(\Sigma_2)$. Suppose the contrary. We consider the quasi-isomorphism $f:A\rightarrow \DR(\Sigma_2)$ and compute for homogenous $v_1$, $v_2\in A$ the following:
\begin{align*}
\langle f(v_1),f(v_2)\rangle &= \pm \langle f(v_1)\wedge f(v_2),1 \rangle\\
&= \pm \langle f(v_1\wedge v_2),1 \rangle\\
&= \pm \langle [f(v_1\wedge v_2)],[1]\rangle\qquad\text{(*)}\\
&= \pm \langle f_*[v_1\wedge v_2], f^*[1] \rangle\\
&= \pm \langle [v_1\wedge v_2],[1]\rangle\\
&= \pm \langle v_1\wedge v_2, 1 \rangle\qquad(*) \\
&= \langle v_1, v_2 \rangle.
\end{align*}
Stars hold because the only non-zero case is when $\deg(v_1) + \deg(v_2) = \deg(\langle\cdot,\cdot\rangle)$, and hence $\Dd (v_1\wedge v_2) = 0$ because non-degeneracy implies vanishing of higher degrees. We see that $f$ preserves the pairing on the chain-level and is injective by non-degeneracy. We can thus assume that $A\subset\DR(\Sigma_2)$ is a dg-subalgebra equipped with the restriction of the intersection pairing. Since $\dim(A)<\infty$, there is a Hodge decomposition
\begin{align*}
	A^2 &= \Harm^2 \oplus \Dd C^1\\
	A^1 &= \Harm^1 \oplus \Dd C^0 \oplus C^1\\
	A^0 &= \Harm^0 \oplus C^0.
\end{align*}
If $1\neq f\in C^{\infty}(\Sigma_2)$, then $f^k$, $k\in \N_0$ are linearly independent over $\R$. Therefore, it must hold $\Harm^0 = \Span\{1\}$ and $C^0 = 0$. Duality implies $\Dd C^1 = 0$, and hence $\Harm^2$ is spanned by $\omega \in \DR^2(\Sigma_2)$ with $\int_{\Sigma_2}\omega = 1$. It holds even $C^1=0$ as $\ker \Dd \cap C = 0$ in a Hodge decomposition. Since $A\simeq \H(\Sigma_2)$ as a $\DGA$, there are closed $\alpha_1$, $\beta_1$, $\alpha_2$, $\beta_2\in\DR^1(\Sigma_2)$ such that $\Harm^1 = \Span\{\alpha_1, \beta_1, \alpha_2, \beta_2\}$ and such that for all $x\in\Sigma_2$ the following holds:
\begin{align*}
	\alpha_1(x) \wedge \alpha_2(x) &=  \alpha_1(x) \wedge \beta_2(x) = 0\\
	\quad\alpha_1(x)\wedge\beta_1(x) &= \alpha_2(x)\wedge\beta_2(x) = \omega(x).
\end{align*}
Taking an $x\in \Sigma_2$ with $\omega(x) \neq 0$ gives a contradiction.

\item  The simply-connected $4$-manifold $\CP^{2\# 7}$, where $\#$ denotes the connected sum, does not admit a Poincar\'e duality model $A$ with just one arrow $A \rightarrow \DR(M)$. Similarly as in the proof for $\Sigma_2$, we can restrict to the case $A\subset\DR(\CP^{2 \# 7})$. We obtain $A^4 = \Harm^4 = \langle \omega \rangle$ for a somewhere non-vanishing $4$-form $\omega$ and $\Harm^2 = \langle K_0, \dotsc, K_6 \rangle$ such that for all $x\in \CP^{2\#7}$ the following holds:
\begin{align*}
	K_i(x)\wedge K_i(x) &= \pm \omega(x) \\
	K_i(x) \wedge K_j(x) &= 0.
\end{align*}
We now view $K_i(x)$ as vectors in $\R^6$ so that $K_0(x)\wedge K_j(x) = 0$ corresponds to taking the scalar product. If $\sum_{i=0}^6 \lambda_i K_i(x) = 0$ for some $\lambda_i\in \R$ with $\lambda_{i_0} \neq 0$, then
\[
0 = K_{i_0}(x) \wedge \Bigl(\sum_{i=0}^6 \lambda_i K_i(x)\Bigr) = \pm \lambda_{i_0} \omega(x).
\]
Therefore, $\omega(x) \neq 0$ implies that $K_i(x)$ are linearly independent. Hence, in this case, $K_0(x) \wedge K_i(x) = 0$ for all $i=1$, $\dotsc$, $6$ implies $K_0(x) = 0$, which is a contradiction with $K_0(x) \wedge K_0(x) = \omega(x)$.\qedhere
\end{ExampleList}
\end{Example}

\begin{Questions}\phantomsection\label{Q:QuestionsPonc}
\begin{RemarkList}
%\item Is it possible to prove Proposition~\ref{Prop:LambrechtUnique} for $n\le 6$ when formality holds in an easier way?
\item Give an example of a Sullivan minimal model $f: \Lambda U \rightarrow V$ of a $\PDGA$ $V$ which is not of Hodge type for any orientation such that $f_*: \H(\Lambda U) \rightarrow \H(V)$ is orientation preserving.
\item Can the additional assumptions in Lemma~\ref{Lemma:Exte} and Proposition~\ref{Prop:ExtensionOfHodgeType} be dropped?
%\item Is t a model category structure on Poincar\'e $\DGA$'s ?
%\item Sullivan's inductive construction gives a minimal model which resolves $V$. The construction of Proposition~\ref{Prop:ExOfLambrStan} does not resolve $V$ in any case \eqref{Eq:ModOne}, \eqref{Eq:ModTwo} and \eqref{Eq:ModThree}. Is it possible to resolve a $\PDGA$ by a Poincar\'e duality model? It is clearly possible in the geometrically formal case, i.e., when there is a $\DGA$-quasi-isomorphism $\H(V)\rightarrow V$.
%For the sake of comparison with Sullivan's inductive construction, it might be interesting to know whether $\VansQuotient(\VansSmall(\cdot))$ is free over $\DGA$'s of Hodge type and whether its differential has decomposable image.
\item Is smallness or minimal dimension the correct notion of ``minimality'' of a Poincar\'e duality model?  \qedhere
\end{RemarkList}
\end{Questions}

\end{document}


% !TeX root = ../main.tex
\documentclass[../main.tex]{subfiles}
\begin{document}
\section{The Model}
\stepcounter{Counter}
\begin{ModelGroupElement}
For any parameter \(\kappa, n\),
\begin{align*}
\elements&\coloneqq M\\
\groupoperation{\elements}&\coloneqq\cdot\\
\charts{\elements}&\coloneqq\tensor{X}{}\mapsto\tensor{\theta}{}\\
\innerprod{\elements}&\coloneqq g\\
\points{\elements}&\equiv R\\
\transformations{\elements}&\equiv M\\
\end{align*}
for injective smooth function \(K\colon\kappa\mapsto\lambda=\sign{\kappa}\sqrt{\abs{\kappa}}\in\R\to\R\).
\end{ModelGroupElement}
\begin{ModelGroupAssertion}
\end{ModelGroupAssertion}
\begin{ModelCurvatureAssertion}
\end{ModelCurvatureAssertion}
\end{document}


\subsection{Homomorphism as a tensor product}