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\section{Orientation, Poincar\'e duality, Hodge decomposition}
\label{SubSec:CycStr}

In this section, a $\DGA$ $V$ is a triple $(V,\Dd,\wedge)$, where
\begin{itemize}
\item $V = \bigoplus_{n\in \Z} V^n$ is a $\Z$-graded vector space,
\item $\Dd: V\rightarrow V$ a differential of degree $1$, and
\item $\wedge: V\otimes V \rightarrow V$ an associative product of degree $0$ such that $\Dd$ and $\wedge$ satisfy the Leibnitz identity $\Dd(v_1 \wedge v_2) = \Dd v_1 \wedge v_2 + (-1)^{v_1} v_1 \wedge\Dd v_2$ for all homogenous $v_1$, $v_2\in V$.
%\[ \Dd(v_1 \wedge v_2) = \Dd v_1 \wedge v_2 + (-1)^{v_1} v_1 \wedge \Dd v_2. \]
\end{itemize}
In other words, we consider general, possibly non-unital and non-commutative, $\Z$-graded $\DGA$'s. We denote by $\deg(v)$ the degree of a homogenous element $v\in V$ and write $(-1)^v$ in the exponent.
%We will consider two closely related notions on a $\DGA$ --- an orientation and a cyclic structure.

%We will not need degree shifts, and hence we will use $\Abs{v}$ and $\Deg(v)$ (and~$v$ in the exponent) to denote the degree of $v$ interchangeably.

\begin{Definition}[Orientations and cyclic structures]\label{Def:CycStr}
Let $(V,\Dd)$ be a $\Z$-graded cochain complex. We define the following:
\begin{itemize}
\item An \emph{orientation in degree $n$} is a linear function $\Or : V^n \rightarrow \R$ such that
\[\Or \neq 0\quad\text{and}\quad\Or \circ \Dd = 0. \]
In other words, it is a surjective chain map $\Or : (V,\Dd) \rightarrow (\R[-n],0)$.
\item A~\emph{cyclic structure of degree $n$} is a homogenous bilinear form $\langle \cdot, \cdot \rangle : V \otimes V \rightarrow \R$ of degree $-n$ as a map such that for all homogenous $v_1$, $v_2\in V$, we have
\begin{equation}\label{Eq:CycStr}
\langle \Dd v_1, v_2\rangle = (-1)^{1+v_1 v_2} \langle \Dd v_2, v_1 \rangle.
\end{equation}
\end{itemize}
A cyclic structure on a $\DGA$ $(V,\Dd,\wedge)$ is additionally required to satisfy
\begin{equation}\label{Eq:CycStrII}
\langle v_1 \wedge v_2, v_3 \rangle = (-1)^{v_3(v_1 + v_2)}\langle v_3\wedge v_1, v_2 \rangle
\end{equation}
for all homogenous $v_1$, $v_2$, $v_3\in V$.
%Morphisms of such objects are morphisms of $\DGA$'s preserving the cyclic structure, resp.~the orientation.
\end{Definition}

Analogously, one can define a cyclic structure on an $\AInfty$-algebra.

\begin{Remark}[Cyclic $\DGA$]\label{Rem:Eq}
A cyclic $\DGA$ $(V,\Pair,\mu_1,\mu_2)$ of degree $n$ from Definition~\ref{Def:CyclicAinfty} is the same as a $\DGA$ equipped with a cyclic structure~$\langle\cdot,\cdot\rangle$ of degree $n$ which is symmetric, i.e.,
\[ \langle v_1, v_2 \rangle = (-1)^{v_1 v_2} \langle v_2, v_1\rangle \]
for all homogenous $v_1$, $v_2\in V$, and non-degenerate (see Definition~\ref{Def:PoincDual}). The correspondence is via the degree shift
\begin{align*}
\mu_1(\SuspU v) &= \SuspU \Dd(v),\\
\mu_2(\SuspU v_1, \SuspU v_2) &= (-1)^{v_1}\SuspU(v_1\wedge v_2),\\
\Pair(\SuspU v_1, \SuspU v_2) & = (-1)^{v_1} \langle v_1,v_2\rangle,
\end{align*}
where $\SuspU$ is a formal symbol of degree $-1$. From this reason, we sometimes call a $\DGA$ equipped with a non-degenerate symmetric cyclic structure a cyclic $\DGA$, although there are degree shifts involved.
\end{Remark}

The homology $\H(V)\coloneqq \H(V,\Dd)$ of a $\DGA$ $(V,\Dd,\wedge)$ is also a $\DGA$ with the induced product $\wedge$ and with zero differential. Given an orientation $\Or$ or a cyclic structure $\langle\cdot,\cdot\rangle$ on~$V$, we define the maps $\Or^\H: \H(V) \rightarrow \R$ and $\langle\cdot,\cdot\rangle^\H: \H(V)\otimes\H(V)\rightarrow\R$ for all closed $h_1$, $h_2\in V$ by
\begin{equation}\label{Eq:IndOnHom}
\begin{aligned}
\Or^\H([h_1]) &\coloneqq \Or(h_1)\text{ and}\\
\langle[h_1],[h_2]\rangle^\H &\coloneqq \langle h_1, h_2\rangle,
\end{aligned}
\end{equation}
respectively, where $[\cdot]$ denotes the cohomology class. It is easy to see that $\langle\cdot,\cdot\rangle^\H$ is a cyclic structure on $\H(V)$ and that $\Or^\H$ is an orientation on $\H(V)$ provided that $\Restr{\Or}{\Ker \Dd} \neq 0$.

\begin{Proposition}[Orientation on homology]\label{Prop:OrOnHomG}
We have the following:
\begin{ClaimList}
\item Let $(V,\Dd)$ be a $\Z$-graded cochain complex, and let $n\in \Z$. Given an orientation $\tilde{\Or}: \H^n(V) \rightarrow \R$, there is an orientation $\Or^V: V^n \rightarrow \R$ such that~$\tilde{\Or} = \Or^\H$. If $\Dd V^n = 0$, then we have the correspondence
\[\text{Orientations on }V\text{ in degree }n\ \overset{1:1}{\simeq}\ \text{Orientations on }\H(V)\text{ in degree }n. \]
\item If $(V_1,\Dd_1,\Or_1)$ and $(V_2,\Dd_2,\Or_2)$ are $\Z$-graded cochain complexes oriented in degree $n$ with $\Dd V_1^n = 0 = \Dd V_2^n$, then a chain map $f: V_1 \rightarrow V_2$ preserves the orientation if and only if the induced map $f_*: \H(V_1) \rightarrow \H(V_2)$ preserves the induced orientation.
\end{ClaimList}
\end{Proposition}
\begin{proof}
\begin{ProofList}
\item We define $\Or^V(v) \coloneqq \tilde{\Or}([v])$ for all closed $v\in V$ and extend it by $0$ to a complement of $\Ker \Dd$ in~$V$. It is obvious that $\Or^\H = \tilde{\Or}$. If $\Dd V^n = 0$, any complement is trivial and $\Or^V$ is uniquely determined by $\Or^\H$.

\item Let $v\in V_1^n$. Because $\Dd v = 0$, we have
\[ \Or_2(f(v)) = \Or^{\H}_2([f(v)]) = \Or^{\H}_2(f_*[v]) = \Or^{\H}_1([v]) = \Or_1(v). \]
This finishes the proof.\qedhere
\end{ProofList}
\end{proof}

Suppose that $1$ is a unit for $(V,\Dd,\wedge)$, i.e., $1\in V^0$, $\Dd 1 = 0$ and $1\wedge v = v \wedge 1 = v$ for all $v\in V$, and let $\langle \cdot,\cdot\rangle$ by a cyclic structure on $V$. For all homogenous $v_1$, $v_2\in V$, we have
\begin{align*}
\langle v_1, v_2 \rangle &= \langle v_1\wedge 1,v_2\rangle \\
& = (-1)^{v_1 v_2}\langle 1\wedge v_2, v_1 \rangle \\
& = (-1)^{v_1 v_2}\langle v_2, v_1 \rangle,
\end{align*}
and hence $\langle \cdot,\cdot\rangle$ is automatically symmetric.

Recall that a $\DGA$ is called commutative if $v_1 \wedge v_2 = (-1)^{v_1 v_2} v_2\wedge v_1$ for all homogenous $v_1$, $v_2\in V$. Commutativity of a $\DGA$ and symmetry of a general cyclic structure on it seem to be unrelated.

%The induced cyclic structure on the $\DGA$ $(\H(V), 0, \wedge)$ is denoted by the same symbol and defined by
%\begin{equation}\label{Eq:IndHom}
%\langle [v_1], [v_2] \rangle = \langle v_1, v_2 \rangle\quad\text{for all closed }v_1, v_2\in V.
%\end{equation}
%
%The (induced) orientation in degree $n$ on $\H(V)$ is denoted by $\int: \H^n(V) \rightarrow \R$; by definition, we have
%\begin{equation}\label{Eq:IndOr}
%\int [v] = \Or(v)\quad\text{for all closed }in V.
%\end{equation}

%
%It turns out that, e.g., for unital $\DGA$'s with the degree bounded from above by~$n$, cyclic structures $\langle\cdot,\cdot\rangle$ of degree $n$ on $V$ are in one-to-one correspondence with orientations $\int: \H^n(V) \rightarrow \R$ on the homology. The precise relation is stated in the following lemma.
\begin{Proposition}[Correspondence of orientations and cyclic structures on $\DGA$'s]\label{Prop:OrAndCyc}
Let $(V,\Dd,\wedge)$ be a $\DGA$. Then the following holds:
\begin{ClaimList}
\item If $\wedge$ is commutative, then any orientation $\Or$ in degree $n$ induces a cyclic structure~$\langle\cdot,\cdot\rangle$ of degree $n$ which is given for all homogenous $v_1$, $v_2\in V$ by
\begin{equation}\label{Eq:PairForm}
\langle v_1, v_2 \rangle \coloneqq \begin{cases}
    \Or(v_1 \wedge v_2) & \text{if }\deg(v_1) + \deg(v_2) = n, \\
    0 & \text{otherwise}. \end{cases}
\end{equation}
\item If $1$ is a unit, then any non-zero cyclic structure $\langle\cdot,\cdot\rangle$ of degree $n$ induces an orientation~$\Or$ in degree $n$ by defining
\begin{equation}\label{Eq:OrForm}
\Or(v) \coloneqq \begin{cases}
 \langle v, 1 \rangle & \text{for }v\in V^n,\\
 0 & \text{otherwise.}
 \end{cases}
\end{equation}
\item For a unital commutative $\DGA$ $(V,\Dd,\wedge,1)$, formulas \eqref{Eq:PairForm} and \eqref{Eq:OrForm} define the correspondence
\[ \text{Orientations in degree }n\ \overset{1:1}{\simeq}\ \text{Non-zero cyclic structures of degree }n.\]
%If moreover $(V,\Dd,\wedge,1)$ is unital, then there is a canonical correspondence
%\[ \text{cyc.~str.~on }V\text{ of degree }n\ \overset{1:1}{\simeq}\  \text{ cyc.~str.~on }\H(V)\text{ of degree }n. \]
%%In particular, a DGA-morphism $f: V \rightarrow V'$ preserves the corresponding structure on the chain level if and only if $f_*: \H(V) \rightarrow \H(V')$ preserves it on homology.
\end{ClaimList}
\end{Proposition}
\begin{proof}
\begin{ProofList}
\item  Using the Leibnitz identity, properties of an orientation and commutativity, we check that
\begin{align*}
\langle \Dd v_1, v_2 \rangle &= \Or(\Dd v_1 \wedge v_2) \\
&= \Or\bigl(\Dd(v_1\wedge v_2) - (-1)^{v_1} v_1 \wedge \Dd v_2\bigr) \\
&= (-1)^{1+v_1}\Or(v_1\wedge\Dd v_2)\\
&= (-1)^{1+v_1 v_2} \Or(\Dd v_2 \wedge v_1) \\
&= (-1)^{1+v_1 v_2} \langle \Dd v_2, v_1\rangle
\end{align*}
and
\begin{align*}
\langle v_1\wedge v_2, v_3\rangle &= \Or(v_1\wedge v_2\wedge v_3)\\
&= (-1)^{v_3(v_1+v_2)} \Or(v_3 \wedge v_1 \wedge v_2)\\
&= (-1)^{v_3(v_1+v_2)} \langle v_3\wedge v_1, v_2\rangle
\end{align*}
for all homogenous $v_1$, $v_2$, $v_3\in V$.
\item For all $v\in V$, we have
\begin{align*}
\Or(\Dd v) = \langle \Dd v, 1 \rangle = - \langle \Dd 1, v \rangle = 0.
\end{align*}
From $\langle\cdot,\cdot\rangle \neq 0$ it follows that there are homogenous $v_1$, $v_2\in V$ with $\deg(v_1) + \deg(v_2) = n$ such that $\langle v_1, v_2 \rangle \neq 0$. Then $v_1\wedge v_2\in V^n$ and
\begin{align*}
\Or(v_1 \wedge v_2) &= \langle v_1\wedge v_2,1\rangle\\
&= \langle 1 \wedge v_1, v_2\rangle\\
&=\langle v_1,v_2\rangle \neq 0.
\end{align*}
Therefore, $\Or$ is an orientation on $V$.
\item This is a combination of (a) and (c) plus the uniqueness, which is easy to check.\qedhere
\end{ProofList}
\end{proof}

\begin{Remark}[Volume forms]\label{Rem:VolForms}
If $\H^n(V) \simeq \R$, then orientations $\Or: \H^n(V)\rightarrow\R$  and elements $0\neq [\Vol] \in \H^n(V)$ called \emph{volume forms} are in one-to-one correspondence via
\[ \Or([\Vol]) = 1. \]
A consequence is the following:
Suppose that $(V_1,\Dd_1,\Or_1)$ and $(V_2,\Dd_2,\Or_2)$ are cochain complexes oriented in degree $n$ which satisfy
\begin{equation}\label{Eq:SDDFG}
\H^n(V_1)\simeq\H^n(V_2)\simeq\R\quad\text{and}\quad\Dd_1 V_1^n \simeq \Dd_2 V_2^n = 0,
\end{equation}
so that Proposition~\ref{Prop:OrOnHomG} applies. Then a chain map $f: V_1 \rightarrow V_2$ preserves orientation if and only if the induced map $f_*: \H^n(V_1) \rightarrow \H^n(V_2)$ maps $[\Vol_1]$ to $[\Vol_2]$. In the category of unital commutative $\DGA$'s satisfying \eqref{Eq:SDDFG}, so that also Proposition~\ref{Prop:OrAndCyc} holds, if the orientations come from cyclic structures, then $f$ preserves cyclic structure if and only if~$f_*$ preserves volume form.
%If $V$ is an oriented augmented unital $\DGA$ if the pairing is non-deg, one defines $\Vol$ uniquely by requiring that $\Vol \perp \bar{V}$ and $\Or(\Vol) = 1$.
\todo[noline,caption={DONE Augmented case}]{In the augmented unital case, if the pairing is non-deg, one defines $\Vol$ uniquely by requiring that $\Vol \perp \bar{V}$ and $\Or(\Vol) = 1$. Let us not write it.}
\end{Remark}

%Observe that if the cyclic structure $\langle\cdot,\cdot\rangle$ comes from an orientation, which is always the case when the $\DGA$ is unital by Proposition~\ref{Prop:OrAndCyc}, then it is \emph{graded symmetric} (shortly symmetric), i.e., it holds
%\[ \langle v_1,v_2\rangle = (-1)^{v_1 v_2}\langle v_2,v_1\rangle\quad\text{for all homogenous }v_1, v_2\in V. \]

\begin{Definition}[Non-degeneracy and Poincar\'e duality]\label{Def:PoincDual}
%An orientation $\Or: V \rightarrow \R$ on a graded vector space~$V$ is called \emph{non-degenerate} if for every $v\in V$, the following implication holds:
%\[ \Or(v\wedge w)= 0\quad\text{for all }w\in V\quad\Implies\quad v=0. \]
Given a graded vector space~$V$, a homogenous bilinear form ($\eqqcolon$\,pairing) $\langle\cdot,\cdot\rangle: V\otimes V \rightarrow \R$ of degree $-n$ as a map which is graded symmetric is called \emph{non-degenerate} if for every $v\in V$, the following implication holds:
\[ \langle v,w \rangle= 0\quad\text{for all }w\in V\quad\Implies\quad v=0. \]
We say that $\langle\cdot,\cdot\rangle$ satisfies \emph{Poincar\'e duality} if the map $\flat: V \rightarrow V^{\GD}$ (graded dual) defined by
\[\flat(v)(w) \coloneqq \langle v,w\rangle\quad \text{for all }v,w\in V\]
is a graded isomorphism (it has degree $-n$ as a map) of graded vector spaces.
\end{Definition}

\begin{Remark}[On non-degeneracy and Poincar\'e duality]\phantomsection\label{Rem:NonDegPD}
\begin{RemarkList}
\item Clearly, Poincar\'e duality implies non-degeneracy. If the degree $k$ component $V^k$ of $V$ is finite-dimensional for every~$k\in \Z$ --- we say that $V$ is of \emph{finite type} --- then the opposite is true as well. If $n=0$, then Poincar\'e duality implies that $V$ is of finite type.
%In this case, we obtain the well-known Poincar\'e duality $V^k \simeq V^{n-k}$.
\item If $V$ is non-negatively graded, then non-degeneracy of $\langle\cdot,\cdot\rangle: V\otimes V \rightarrow \R$ implies $V = V^0 \oplus \dotsb \oplus V^n$. Therefore, non-negatively graded vector spaces of finite type which admit a non-degenerate homogenous bilinear form are finite-dimensional.
\item The de Rham complex $(\DR(M),\Dd,\wedge)$ of an oriented closed $n$-manifold~$M$ with the orientation $\int: \DR^n(M) \rightarrow \R$ is an oriented $\DGA$ whose cyclic structure is non-degenerate but does not satisfy Poincar\'e duality. On the other hand, the induced structure on homology $\H(\DR(M))$ satisfies Poincar\'e duality.
\qedhere
\end{RemarkList}
\end{Remark}
%
%We now relate similar definitions from \cite{Cieliebak2015}, \cite{VanLe2019} and \cite{Lambrechts2007} to our definitions:
%\begin{itemize}
%\item A ``cyclic $\DGA$'' as defined in \cite{Cieliebak2015} is a $\DGA$ $(V,\Dd,\wedge)$ with a symmetric non-degenerate cyclic structure $\langle \cdot,\cdot \rangle : V \otimes V \rightarrow \R$. Different signs in their definition occur because they consider the operations on the degree shift $V[1]$ (c.f., Definition~\ref{Def:})
%\item A ``Poincar\'e-DGCA of degree $n$'' as defined in \cite{VanLe2019} is a unital commutative $\DGA$ (shortly $\uCDGA$) $(V,\Dd,\wedge,1)$ with $V=V^0\oplus \dotsb\oplus V^n$ and with an orientation $\int: \H^n(V) \rightarrow \R$ such that $(\H(V),\wedge,\int)$ is a finite-dimensional Poincar\'e algebra.
%\item An ``oriented Poincar\'e-CDGA'' as defined in \cite{Lambrechts2007} is a $\uCDGA$ with non-negatively graded $V$ of finite type such that $\H(V)$ is a Poincar\'e algebra.
%\end{itemize}
%For unital cyclic $\DGA$'s, cyclicity implies \emph{graded symmetry} of $\langle\cdot,\cdot\rangle$. Indeed, for all homogenous $v_1$, $v_2\in V$, we have
%\[ \langle v_1, v_2 \rangle = \langle 1 \wedge v_1, v_2 \rangle = (-1)^{v_1 v_2} \langle v_2 \wedge 1, v_1 \rangle = (-1)^{v_1 v_2} \langle v_2, v_1\rangle. \]
%Graded symmetry is also automatic for $\langle\cdot,\cdot\rangle$ coming from an orientation.
%%For simplicity, we will assume from now on that $\langle\cdot,\cdot\rangle$ is symmetric.

An analog of the following definition is used in \cite{Van2019} and also in \cite{Lambrechts2007} (under the name ``set of orphans'').

\begin{Definition}[Degenerate subspace and non-degenerate quotient]\label{Def:NonDegQ}
Given a symmetric pairing $\langle\cdot,\cdot\rangle: V\otimes V\rightarrow\R$ on a graded vector space~$V$, we define the \emph{degenerate subspace}~$V^\perp\subset V$ by
\begin{align*}
V^\perp\coloneqq \{ v\in V \mid \langle w,v \rangle = 0\text{ for all }w\in V\}.
\end{align*}
If $(V,\Dd,\wedge)$ is a $\DGA$ and $\langle\cdot,\cdot\rangle$ a cyclic structure, then \eqref{Eq:CycStr} implies that $V^\perp$ is a differential graded ideal in $V$, and thus we obtain the short exact sequence of $\DGA$'s
\begin{equation}\label{Eq:ImportantSES}
\begin{tikzcd}
0 \arrow{r} & V^\perp \arrow[hook]{r}{\iota} & V \arrow[two heads]{r}{\pi^\VansQuotient} & \VansQuotient(V) \coloneqq V / V^\perp \arrow{r} & 0,
\end{tikzcd}
\end{equation}
where $\iota$ is the inclusion and $\pi^\VansQuotient$ the canonical projection. We call the $\DGA$ $\VansQuotient(V)$ together with the induced non-degenerate cyclic structure $\langle\cdot,\cdot\rangle^\VansQuotient$ such that $\langle\pi^\VansQuotient(\cdot),\pi^\VansQuotient(\cdot)\rangle^\VansQuotient = \langle\cdot,\cdot\rangle$ the \emph{non-degenerate quotient.}
\end{Definition}

It was observed in~\cite{Van2019} that the question whether $(V^\perp,\Dd)$ is acyclic, and hence~$\pi^\VansQuotient$ is a quasi-isomorphism, turns out to be related to the existence of Hodge decomposition.

\begin{Definition}[Hodge decomposition]\label{Def:HodgeDecomp}
A cochain complex $(V,\Dd)$ with a symmetric cyclic structure $\langle\cdot,\cdot\rangle: V\otimes V \rightarrow \R$ is \emph{of Hodge type} if there exist subspaces $\Harm \subset \Ker \Dd$ and $C\subset V$ such that
\begin{equation}\label{Eq:HodgeDecomp}
V = \Ker \Dd \oplus C, \quad \Ker \Dd = \Im \Dd \oplus \Harm\quad\text{and}\quad C \perp \Harm \oplus C,
\end{equation}
where $\perp$ denotes the relation of being perpendicular with respect to $\langle\cdot,\cdot\rangle$. Such decomposition is called a \emph{Hodge decomposition.} We call $\Harm$ the \emph{harmonic subspace} and $C$ the \emph{coexact part}.

Given a Hodge decomposition, we define the \emph{standard Hodge homotopy} $\HtpStd: V \rightarrow V$~by
\[\HtpStd \coloneqq \begin{cases}
    -(\Restr{\Dd}{C})^{-1} & \text{on }\Im \Dd, \\
    0 & \text{on } \Harm \oplus C.
   \end{cases}\]
Then we have $\Dd \HtpStd = - \pi_{\Im \Dd}$, $\HtpStd \Dd = -\pi_{C}$, and hence
\[ [\Dd,\HtpStd] = \Dd \HtpStd + \HtpStd \Dd = \pi_\Harm - \Id.\]
 We call $(\Harm,\HtpStd)$ the \emph{Hodge pair} associated to the Hodge decomposition \eqref{Eq:HodgeDecomp}.
\end{Definition}

\begin{Proposition}[Non-deg., fin.~type implies Hodge type]\label{Prop:NDegFin}
Any cochain complex of finite type with a non-degenerate symmetric cyclic structure is of Hodge type.
\end{Proposition}
\begin{proof}
This is \cite[Lemma~11.1]{Cieliebak2015}, and the proof uses formal Hodge theory.
\end{proof}

\begin{Remark}[Harmonic subspaces]\label{Rem:RemarkHarm}
In the situation of Proposition~\ref{Prop:NDegFin}, it was shown in \cite[Remark~2.6]{Van2019} that for any complement $\Harm$ of $\im\Dd$ in $\ker\Dd$ (in other words, the image of a section $\H(V)\rightarrow \Ker\Dd$) there is a coexact part $C$ such that $V=\Harm \oplus \Im \Dd \oplus C$ is a Hodge decomposition. From this reason, we call any complement of $\im \Dd$ in $\ker \Dd$ a \emph{harmonic subspace.}\footnote{Given a Hodge decomposition $V=\Harm\oplus\im\Dd\oplus C$ and a harmonic subspace $\Harm'$, then it holds $\Harm' = \Graph(\alpha: \Harm \rightarrow \Dd V)$ because $\Harm \oplus \Dd V = \Harm' \oplus \Dd V$, and one can take $C'=\Graph(- \alpha^\dagger - \frac{1}{2}\alpha\alpha^\dagger: C \rightarrow \Harm\oplus\Dd V)$.}
\end{Remark}

The following lemma will be used in the proof of Proposition~\ref{Prop:HodgeAcyc}.

\begin{Lemma}[Complement of acyclic subcomplex over $\R$]\label{Lem:Pom}
Let $f: V_1 \rightarrow V_2$ be an injective chain map of cochain complexes $(V_1,\Dd_1)$ and $(V_2,\Dd_2)$ over $\R$ such that $(V_1,\Dd_1)$ is acyclic. Then there is a chain map $g: V_2 \rightarrow V_1$ such that $g\circ f=\Id$.\footnote{This lemma can be used to prove that over $\R$, every surjective quasi-isomorphism is a deformation retraction and every injective quasi-isomorphism is a section of a deformation retraction.}
\end{Lemma}

\begin{proof}
For every $i\in \Z$, consider the diagram
\[\begin{tikzcd}
\Ker \Dd_1^i \oplus C^i_1 \arrow{r}{f^i} \arrow{d}{\Dd_1^i} & \Ker \Dd_2^i \oplus C^i_2 \arrow{d}{\Dd_2^i} \\
\Ker \Dd^{i+1}_1 \oplus C^{i+1}_1 \arrow{r}{f^{i+1}} & \Ker \Dd_2^{i+1} \oplus C^{i+1}_2,
\end{tikzcd}\]
where $C^i_1$ is a complement of $\Ker \Dd_1^i$ in $V_1^i$ and $C^i_2$ is a complement of $\Ker \Dd_2^i$ in~$V_2^i$. With respect to this decomposition, we write
\begin{align*}
f^i & = \begin{pmatrix}
f^{i}_{11} & f^i_{12} \\
f^{i}_{21} & f^i_{22}
\end{pmatrix},  & g^i &= \begin{pmatrix}
g^{i}_{11} & g^i_{12} \\
g^{i}_{21} & g^i_{22}
\end{pmatrix}, \\
\Dd^i_1 &= \begin{pmatrix}
0 & d^i_1 \\
0 & 0
\end{pmatrix},& \Dd^i_2 &= \begin{pmatrix}
0 & d^i_2 \\
0 & 0
\end{pmatrix}.
\end{align*}

The assumption $\H(V_1)=0$ implies that $d^i_1$ is an isomorphism. The fact that~$f$ is a chain map translates to
\begin{equation}\label{Eq:EqEqEq}
d_2^i f^i_{21} = 0,\quad f^{i+1}_{21} d_1^i = 0,\quad f^{i+1}_{11} d^i_1 = d^i_2 f^i_{22}.
\end{equation}
From the second relation and surjectivity of $d_1^i$, we get that $f_{21}^{i+1} = 0$. Now,~$f^{i+1}_{11}$ has to be injective because it is the only possibly non-zero part of $f$ on $\Ker d_1^{i+1}$. From the third relation of \eqref{Eq:EqEqEq} and the fact that $d_1^i$ is injective, we get that $f_{22}^i$ is injective as well.
Relations \eqref{Eq:EqEqEq} hold also for $g$ with $d_1$ and $d_2$ switched. In particular, we have $g_{21}^{i}=0$. The relation $g \circ f = \Id$ translates using $f^i_{21} = g^i_{21} = 0$ to
\begin{equation}\label{Eq:EqEq}
g^i_{11} f^i_{11} = \Id, \quad g^i_{11} f^i_{12} + g^i_{12}f^i_{22}=0, \quad g^i_{22} f^i_{22} = \Id.
\end{equation}
Because $d_1^i$ is an isomorphism, the last equation is equivalent to
\[ \Id = d_1^i g^i_{22}f^i_{22} (d_1^i)^{-1} = g^{i+1}_{11} d_2^i f^i_{22} (d_1^i)^{-1} = g^{i+1}_{11} f^{i+1}_{11} d_1^i (d_1^i)^{-1} = g^{i+1}_{11} f^{i+1}_{11}. \]

We see that $g$ can be constructed as follows. For all $i\in \Z$, let $g_i^{11}$ be an arbitrary left inverse of $f_i^{11}$. Set $g_{22}^i \coloneqq (d_1^i)^{-1}g_{11}^{i+1} d_2^i$ and $g_i^{21}\coloneqq0$. Finally, $g_{12}^i$ has to be chosen such that the second equation of \eqref{Eq:EqEq} is satisfied. This is possible since we can first define~$g_{12}^i$ on $\Im f^i_{22}$ because~$f^i_{22}$ injective and then extend it by $0$ to a complement.
\end{proof}

Claim (b) of the following proposition corresponds to \cite[Lemma~2.8]{Van2019}. Claim (c) was suggested by Prof.~Hông Vân Lê via e-mail correspondence.

\begin{Proposition}[Hodge decomposition and acyclicity of $V^\perp$]\label{Prop:HodgeAcyc}
Let $(V,\Dd)$ be a cochain complex with a symmetric cyclic structure $\langle\cdot,\cdot\rangle: V\otimes V\rightarrow\R$. If $V$ is of Hodge type, then the following implications hold:
\begin{ClaimList}
\item If $\langle \cdot,\cdot \rangle$ is non-degenerate, then $\langle \cdot,\cdot \rangle^\H$ is non-degenerate.
\item If $\langle \cdot,\cdot\rangle^\H$ is non-degenerate, then $(V^\perp,\Dd)$ is acyclic.
\end{ClaimList}
Moreover, the following reverse implication holds:
\begin{ClaimList}[resume]
\item If $V$ is of finite type and $(V^\perp,\Dd)$ is acyclic, then $V$ is of Hodge type.
\end{ClaimList}
\end{Proposition}
\begin{proof}
\begin{ProofList}
\item Let $V = \Im \Dd \oplus \Harm \oplus C$ be a Hodge decomposition. Then $\Im \Dd \oplus C \subset \Harm^\perp$, and hence
\[ \langle \Dd \eta + b + c, b' \rangle = \langle b,b'\rangle\quad\text{for all }\eta\in V, c\in C\text{ and }b, b'\in \Harm. \]
The claim follows easily. Notice that having a Hodge decomposition, it holds
\[ \Restr{\langle \cdot,\cdot\rangle}{\Harm\otimes \Harm}\text{ non-degenerate}\quad\Equiv\quad \Im\Dd \oplus C=\Harm^\perp. \]
\item Consider a Hodge decomposition as above, and let $v\in V^\perp \cap \Ker \Dd$ be a non-zero vector. Suppose that $v\not\in \Im\Dd$. Then $[v] \neq 0$ in $\H(V)$, and hence there is a $b\in \Harm$ such that $\langle v,b \rangle \neq 0$ by non-degeneracy of $\langle\cdot,\cdot\rangle^\H$. This is a contradiction with $v\in V^\perp$. Therefore, it holds $V^\perp \cap \Ker \Dd = V^\perp \cap \Im \Dd$. In particular, there is an $\eta \in C$ such that $v = \Dd \eta$. Now, for any $\eta'\in C$, $b\in \Harm$ and $c\in C$, we have using $C\perp \Harm\oplus C$ and $v\in V^\perp$ the following:
\begin{align*}
\langle \Dd \eta' + b + c, \eta \rangle &= \langle \Dd \eta',\eta \rangle \\
&=(-1)^{1+\eta\eta'}\langle\Dd\eta,\eta'\rangle\\
&= (-1)^{1+\eta\eta'}\langle v,\eta'\rangle\\
&= 0.
\end{align*}
Therefore, it holds $\eta\in V^\perp$, and we have shown that $V^\perp \cap \Im\Dd = \Dd V^\perp$. The claim follows.
\item Because $V^\perp \subset V$ is an acyclic subcomplex and we work over $\R$, there is a complementary subcomplex $Z\subset V$; i.e., $\Dd Z\subset Z$ and $V = V^\perp\oplus Z$. This follows from Lemma~\ref{Lem:Pom} by setting $V^\perp = \Im f$ and $Z=\Ker g$. \Add[caption={Deformation retract over $\R$ add},noline]{Add here deformation retract somewhere.}Now, the restriction of $\langle\cdot,\cdot\rangle$ to $Z$ is non-degenerate, and Proposition~\ref{Prop:NDegFin} provides its Hodge decomposition $Z=\Dd Z \oplus \Harm \oplus D$. Let $E\subset V^\perp$ be a graded vector space which is complementary to $\Dd V^\perp$ in $V^\perp$; i.e., $V^\perp = \Dd V^\perp \oplus E$. It is easy to check that $V=\Im\Dd\oplus\Harm\oplus C$ with $C\coloneqq D \oplus E$ is a Hodge decomposition.
\qedhere
\end{ProofList}
\end{proof}
\todo[noline,caption={PD implies Hodge without fin type}]{Does the following hold without assuming finite type? I.e. does Poincar\'e duality imply Hodge type?}
The following notions were taken from \cite{Van2019}.

\begin{Definition}[Hodge subalgebra and small subalgebra]\label{Def:SmallSubalg}
Consider a $\DGA$ $(V,\Dd,\wedge)$ with a symmetric cyclic structure $\langle\cdot,\cdot\rangle$. Suppose that it admits a Hodge decomposition with the Hodge pair $(\Harm,\HtpStd)$. A \emph{Hodge subalgebra} is a differential graded subalgebra $W\subset V$ which satisfies
\[ \Harm\subset W \quad\text{and}\quad \HtpStd W \subset W. \]
We denote the smallest Hodge subalgebra of $V$ by $\VansSmall(V)$ and call it the \emph{small subalgebra}. We stress that the definition of $\VansSmall(V)$ depends on $(\Harm,\HtpStd)$!
\end{Definition}
%In the situation of Definition~\ref{Def:SmallSubalg}, it holds
%\[ \VansSmall(V) = \bigcap_{\substack{W\subset V\\\text{Hodge subalgebra}}} W. \]

The following is a version of \cite[Proposition~3.3]{Van2019} which generalizes to the non-simply-connected case (see (iii) of Remark~\ref{Rem:OnHodgeSubalg} below for the comparison).

\begin{Proposition}[Description of small subalgebra]\label{Prop:SmallDescription}
Consider the situation of Definition~\ref{Def:SmallSubalg}. The small subalgebra $\VansSmall(V)$ is generated as a graded vector space by Kontsevich-Soibelman--like evaluations of rooted binary trees with $k\ge 1$ leaves labeled with homogenous elements of~$\Harm$, interior vertices labeled with $\wedge$ and interior edges labeled either with~$\StdHtp$ or with~$\Id$.
\end{Proposition}
\begin{proof}
We abbreviate $\VansSmall \coloneqq \VansSmall(V)$, denote the set of labeled trees by $\Trees$ and denote the vector space generated by evaluations of elements of $\Trees$ by $\langle \Trees\rangle$.

Clearly, we have $\langle \Trees\rangle \subset \VansSmall$. For ``$=$'', it suffices to check that for any $T$, $T_1$, $T_2\in\Trees$, it holds $\Dd T$, $\HtpStd T$, $T_1 \wedge T_2 \in \langle\Trees\rangle$, i.e., that $\langle \Trees\rangle$ is a Hodge subalgebra.

As for $\Dd T$, we imagine $\Dd$ propagating from the root to the leaves. When it encounters $L\wedge R$, where $L$ stands for the left and $R$ for the right sub-branch, it duplicates the tree and continues propagating in $L$ and $R$, respectively (we take the sum of the two copies in the end). This is justified by the Leibnitz identity $\Dd(L\wedge R) = \Dd L\wedge R + (-1)^L L \wedge \Dd R$. When it encouners $\HtpStd$, it triples the tree and either goes past $\HtpStd$ and continues propagating, or exchanges $\HtpStd$ for $\Id$ and stops, or exchanges $\HtpStd$ for $\pi_\Harm$ and stops. This is justified by $\Dd \HtpStd = -\HtpStd \Dd - \Id + \pi_\Harm$. If it encounters $\Id$, nothing happens and it keeps propagating. If it reaches a leaf with $h\in \Harm$, then the corresponding tree evaluates to $0$. We see that we alway obtain an element of $\langle\Trees\rangle$.

As for $\HtpStd T$, we have either $\HtpStd T = 0$ if the interior edge adjacent to the root ($\eqqcolon$ the root edge) is labeled with $\HtpStd$, or $\HtpStd T$ is a new tree in $\Trees$ which arises from $T$ by replacing $\Id$ by $\HtpStd$ on the root edge.

As for $T_1 \wedge T_2$, using $\Id = \pi_\Harm - \Dd \HtpStd - \HtpStd \Dd$ and the Leibnitz identity, we get
\begin{align*}
T_1 \wedge T_2 &= \pi_\Harm(T_1 \wedge T_2) - (\Dd \HtpStd)(T_1 \wedge T_2) - (\HtpStd \Dd)(T_1 \wedge T_2) \\
&= \pi_\Harm(T_1 \wedge T_2) - \Dd\bigl(\HtpStd(T_1 \wedge T_2)\bigr) - \HtpStd(\Dd T_1 \wedge T_2)
- (-1)^{T_1} \HtpStd(T_1 \wedge \Dd T_2).
\end{align*}
We see that $T_1\wedge T_2 \in \langle \Trees\rangle$.
\end{proof}

\begin{Remark}[On Hodge subalgebra and small subalgebra]\phantomsection\label{Rem:OnHodgeSubalg}
\begin{RemarkList}
\item Any Hodge subalgebra~$W$ inherits the Hodge decomposition
\begin{equation}\label{Eq:InducedHodgeDecomp}
W = \Harm \oplus \Dd W \oplus \HtpStd W
\end{equation}
with the Hodge pair $(\Harm,\Restr{\HtpStd}{W})$.
\item We will be in the situation of Definition~\ref{Def:SmallSubalg} and use the notation of the proof of Proposition~\ref{Prop:SmallDescription}. In addition, we suppose that $V$ is non-negatively graded. Having established $\VansSmall = \langle \Trees \rangle$, consider the Hodge decomposition \eqref{Eq:InducedHodgeDecomp} for $W = \VansSmall$. Let $\widebar{\VansSmall} = \bigoplus_{k\ge 1} \VansSmall^k$ denote the reduced part of $\VansSmall$, and let $\langle\widebar{\VansSmall} \wedge \widebar{\VansSmall}\rangle$ denote the graded vector space generated by products $v_1 \wedge v_2$ for $v_1$, $v_2\in \widebar{\VansSmall}$. We have
\begin{align*}
\HtpStd\langle\Trees\rangle &= \bigl\langle\{ T\in \Trees \text{ with }\HtpStd\text{ on the root edge}\}\bigr\rangle \\
&= \HtpStd\langle\widebar{\VansSmall} \wedge \widebar{\VansSmall}\rangle
\end{align*}
and
\begin{align*}
 \Dd\langle\Trees\rangle &= \Dd (\pi_\Harm - \Dd \HtpStd - \HtpStd\Dd)\langle\Trees\rangle \\
 &= \Dd \HtpStd \Dd \langle\Trees\rangle \\
 &\subset \Dd \HtpStd \langle\Trees\rangle \\
 &\subset \Dd \HtpStd\langle\widebar{\VansSmall}\wedge\widebar{\VansSmall}\rangle.
\end{align*}
It holds even $\Dd\langle\Trees\rangle = \Dd\HtpStd\langle\widebar{\VansSmall}\wedge\widebar{\VansSmall}\rangle$ due to \eqref{Eq:InducedHodgeDecomp}. We can now write \eqref{Eq:InducedHodgeDecomp} as
\begin{equation}\label{Eq:VansFormula}
\VansSmall^k = \Harm^k \oplus \Dd \HtpStd \langle\widebar{\VansSmall}\wedge\widebar{\VansSmall}\rangle^k \oplus \HtpStd \langle\widebar{\VansSmall}\wedge\widebar{\VansSmall}\rangle^{k+1}.
\end{equation}
If $\H^1 = 0$, this agrees with the formula from \cite[Proposition~3.3]{Van2019}. In this case, it holds $\VansSmall^1 = 0$, and hence $\langle\widebar{\VansSmall}\wedge\widebar{\VansSmall}\rangle^{k+1}$ depends only on $\VansSmall^i$ for $i<k$. Therefore, we can compute~$\VansSmall^k$ from \eqref{Eq:VansFormula} inductively starting with $\VansSmall^0 = \langle 1\rangle$.
\item The previous remark implies the following: Let $(V,\Dd,\wedge)$ be a non-negatively graded $\DGA$ with cyclic structure $\langle\cdot,\cdot\rangle$ of Hodge type. Suppose that $\H^1(V)=0$. If $\H(V)$ is of finite type, then so is $\VansSmall(V)$.
%\item The subalgebra $\VansSmall(V)$ ``resolves'' $V$; i.e., there is a $\DGA$-quasi-isomorphism $\VansSmall(V) \xhookrightarrow{} V$ (the inclusion). Consider the smallest subalgebra $\langle\Harm\rangle^\wedge$ containing~$\Harm$; similarly as $\VansSmall(V)$, it is generated as graded vector space by evaluations of rooted binary trees with $\Id$ on interior edges. The $\DGA$ $\langle\Harm\rangle^\wedge$ also ``resolves'' $V$.
\item The advantage of $\VansSmall(V)$ is that we have the diagram of pairing preserving $\DGA$-quasi-isomorphisms
\[\begin{tikzcd}
 V & \VansSmall(V)\arrow[hook']{l} \arrow[two heads]{r} & \VansQuotient(\VansSmall(V)).
\end{tikzcd}\]
Moreover, in the case of (iii), the non-degenerate quotient is finite-dimensional, and hence a Poincar\'e duality model of $V$ (see the next section).
\item Notice that if there is a quasi-isomorphism $f:\H(V)\rightarrow V$, then $\VansSmall(V) = V$ for any Hodge decomposition with harmonic subspace $\im f$.  \qedhere
\end{RemarkList}
\end{Remark}

\begin{Proposition}[Properties of $\VansSmall$ and $\VansQuotient$]\label{Prop:PropPropertiessd}
Let $(V,\Dd,\wedge)$ be a $\DGA$ with a symmetric cyclic structure $\langle\cdot,\cdot\rangle$. Suppose that it admits a Hodge decomposition with Hodge pair $(\Harm,\HtpStd)$. Then $\VansSmall(V)$ and $\VansQuotient(V)$ admit Hodge decompositions with the Hodge pairs $(\Harm,\HtpStd^\VansSmall\coloneqq\Restr{\HtpStd}{\VansSmall(V)})$ and $(\pi^\VansQuotient(\Harm),\HtpStd^\VansQuotient)$, respectively, where $\pi^\VansQuotient: V \rightarrow \VansQuotient(V)$ is the canonical projection and $\HtpStd^\VansQuotient$ the unique map on $\VansQuotient(V)$ satisfying $\HtpStd^\VansQuotient \circ \pi^\VansQuotient=\pi^\VansQuotient\circ \HtpStd$. Furthermore, with respect to the induced Hodge decompositions, we have
\begin{equation}\label{Eq:QSRelations}
\VansSmall(\VansSmall(V)) = \VansSmall(V), \quad \VansQuotient(\VansQuotient(V)) = \VansQuotient(V)\quad\text{and}\quad \VansSmall(\VansQuotient(\VansSmall(V))) = \VansQuotient(\VansSmall(V)).\end{equation}
\end{Proposition}
\begin{proof}
The fact that $\VansSmall\coloneqq\VansSmall(V)$ admits a Hodge decomposition with Hodge pair $(\Harm,\HtpStd^\VansSmall)$ was stated in (i) of Remark~\ref{Rem:OnHodgeSubalg} and is easy to check.

We prove that $\VansQuotient\coloneqq \VansQuotient(V)$ has a Hodge decomposition with Hodge pair $(\pi^\VansQuotient(\Harm),\HtpStd^\VansQuotient)$. Because $\pi^\VansQuotient$ is a quasi-isomorphism, $\pi^\VansQuotient(\Harm)$ is a harmonic subspace of $\VansQuotient$, and we have $\Ker \Dd^\VansQuotient = \pi^\VansQuotient(\Harm)\oplus \Im \Dd^\VansQuotient$. Let $c\in C$ with $\Dd c \in V^\perp$ such that $\pi^\VansQuotient(c)\in\Ker\Dd^\VansQuotient \cap \pi^\VansQuotient(C)$. From the cyclicity of $\langle\cdot,\cdot\rangle$ with respect to $\Dd$ and from $C\perp\Harm\oplus C$, it follows that $c\in V^\perp$, and thus $\pi^\VansQuotient(c)=0$. Together with surjectivity of $\pi^\VansQuotient$ this implies that $\VansQuotient = \ker \Dd^\VansQuotient \oplus \pi^\VansQuotient(C)$. From $\langle \pi^\VansQuotient(\cdot),\pi^\VansQuotient(\cdot)\rangle^\VansQuotient$, we see that $\pi^\VansQuotient(C) \perp \pi^\VansQuotient(\Harm) \oplus \pi^\VansQuotient(C)$. Therefore, $\VansQuotient = \pi^\VansQuotient(\Harm)\oplus\Im\Dd^\VansQuotient\oplus\pi^\VansQuotient(C)$ is a Hodge decomposition, and it is easy to see that its standard Hodge homotopy $\HtpStd^\VansQuotient$ satisfies $\HtpStd^\VansQuotient \circ \pi^\VansQuotient=\pi^\VansQuotient\circ \HtpStd$. This defines $\HtpStd^\VansQuotient$ uniquely because $\pi^\VansQuotient$ is surjective.

As for the relations \eqref{Eq:QSRelations}, the first two are clear. The third can be seen as follows. Since $\pi^\VansQuotient: V \rightarrow \VansQuotient$ is a $\DGA$-morphism mapping the harmonic subspaces isomorphically onto each other and commuting with $\Dd$ and $\HtpStd$, the assignments $Y\subset V \mapsto \pi^\VansQuotient(Y)\subset \VansQuotient$ and $Z\subset \VansQuotient \mapsto (\pi^{\VansQuotient})^{-1}(Z)\subset V$ preserve Hodge subalgebras. Therefore, if $Z\subset \VansQuotient(\VansSmall)$ is a Hodge subalgebra, then $(\pi^\VansQuotient)^{-1}(Z)\subset\VansSmall$ is a Hodge subalgebra. It holds even $(\pi^\VansQuotient)^{-1}(Z)=\VansSmall$ by minimality of $\VansSmall$, and hence $Z=\VansQuotient(\VansSmall)$ by surjectivity of~$\pi^\VansQuotient$.
\end{proof}

%In the ideal world, the isomorphism class of $\VansQuotient(\VansSmall(V))$ would not depend on the chosen Hodge decomposition of $V$. We collect some lemmas which shall point us to a uniqueness statement for $\VansQuotient(\VansSmall(V))$ below.

%Notice that solely the fact that $\langle\cdot,\cdot\rangle$ is non-degenerate on $V$ does not imply that $\VansSmall(V) \simeq V$. A counterexample is $V = \DR(M)$, where $M$ is a compact oriented manifold with $\HDR^1(M) = 0$; indeed, by (iii) of Remark~\ref{Rem:OnHodgeSubalg}, $\VansSmall(\DR(M))$ is of finite type because $\HDR(M)$ is, but $\DR(M)$ is not of finite type.

A natural question is, how does $\VansQuotient(\VansSmall(V))$ depend on the chosen Hodge pair and how does it behave under quasi-isomorphisms? The following lemmas might be useful.

\begin{Lemma}[Kernel of pairing-preserving morphism]\label{Lem:PomLemma}
Let $V_1$ and $V_2$ be vector spaces with symmetric bilinear forms $\langle \cdot,\cdot \rangle_1: V_1 \otimes V_1 \rightarrow \R$ and $\langle \cdot,\cdot\rangle_2: V_2\otimes V_2 \rightarrow \R$, respectively. Let $f: V_1 \rightarrow V_2$ be a linear map such that
\begin{equation}\label{Eq:Isometryyy}
\langle v_1, v_2 \rangle_1 = \langle f(v_1), f(v_2) \rangle_2 \quad \text{for all }v_1, v_2 \in V_1.
\end{equation}
Then it holds
\[ \Ker f \subset V_1^\perp\quad\text{and}\quad f(V_1^\perp)\subset f(V_1)^\perp. \]
In particular, the following statements are true:
\begin{ClaimList}
\item If $\langle\cdot,\cdot\rangle_1$ is non-degenerate, then $f$ is injective.
\item If $\langle\cdot,\cdot\rangle_2$ is non-degenerate and $f$ is surjective, then $\Ker f = V_1^\perp$.
\end{ClaimList}
\end{Lemma}
\begin{proof}
Clear.
\end{proof}

\begin{Lemma}[Injectivity on domain with non-degenerate orientation]\label{Lem:AutomaticInjectivity}
Let $(V_1,\Dd_1,\wedge_1)$ be a non-negatively graded commutative $\DGA$ with an orientation $\Or_1$ in degree $n$ such that the induced cyclic structure is non-degenerate. Let $(V_2,\Dd_2,\wedge_2)$ be any $\DGA$, and let $f: V_1 \rightarrow V_2$ be a morphism of $\DGA$'s. Then injectivity of $f_*: \H^n(V_1) \rightarrow \H^n(V_2)$ implies injectivity of $f: V_1 \rightarrow V_2$.
\end{Lemma}
\begin{proof}
Firstly, injectivity of a homogenous map is equivalent to degree-wise injectivity. Secondly, because $V_1$ is non-negatively graded and the cyclic structure of degree $n$ is non-degenerate, we have $V_1 = V_1^0\oplus\dotsb\oplus V_1^n$. Now, suppose that $v\in V_1^k$ for some $k=0$,~$\dotsc$, $n$ satisfies $f(v) = 0$. For any $w\in V^{n-k}$, the product $v\wedge w$ lies in $V^n$, and thus $\Dd(v\wedge w)=0$. We compute
\begin{align*}
f_* [v\wedge w] = [f(v\wedge w)] = [f(v)\wedge f(w)] = 0,
\end{align*}
and hence $v\wedge w = \Dd \eta$ for some $\eta \in V_1^{n-1}$ by injectivity of $f_*$. Consequently, we have
\begin{align*}
\Or_1(v\wedge w) &= \Or_1(\Dd \eta) =0,
\end{align*}
and hence $v = 0$ by non-degeneracy of $\Or_1$.
\end{proof}

\begin{Lemma}[Small subalgebra of cyclic $\DGA$ and quasi-iso.]\label{Eq:LemSmallSub}
Let $(V_1,\Dd_1,\wedge_1,\langle\cdot,\cdot\rangle_1)$ and~$(V_2,\Dd_2,\wedge_2,\langle\cdot,\cdot\rangle_2)$ be non-negatively graded unital commutative $\DGA$'s of finite type with non-degenerate cyclic structures of degree $n$ (hence finite-dimensional). Let $f: V_1 \rightarrow V_2$ be a $\DGA$-morphism such that $f_*: (\H(V_1),\Or_1^\H)\rightarrow (\H(V_2),\Or_2^\H)$ is an isomorphism. Then a Hodge decomposition of $V_1$ with Hodge pair $(\Harm_1,\HtpStd^1)$ induces a Hodge decomposition of~$V_2$ with Hodge pair $(f(\Harm_1),\HtpStd^2)$, where~$\HtpStd^2$ satisfies $\HtpStd^2 \circ f = f\circ \HtpStd^1$. Consequently,~$f$ induces an isomorphism $\VansSmall(V_1) \simeq \VansSmall(V_2)$.
\end{Lemma}
\begin{proof}
Because $f_*$ preserves orientation and it holds $V_1^{n+1} = 0 = V_2^{n+1}$, Proposition~\ref{Prop:OrOnHomG} implies that $f$ preserves cyclic structure and hence is injective by Lemma~\ref{Lem:PomLemma}. Let $V_1 = \Harm_1 \oplus \Im \Dd_1 \oplus C_1$ be a Hodge decomposition. From the injectivity of $f$, it follows that
\[ f(V_1) = f(\Harm_1) \oplus \Dd f(V_1) \oplus f(C_1) \]
and that the restriction of $\langle\cdot,\cdot\rangle_2$ to $f(V_1)\otimes f(V_1)$ is non-degenerate. Because $V_1$ is of finite type, $f(V_1)$ is of finite type, and so non-degeneracy implies Poincar\'e duality $f(V_1)^{\GD}\simeq f(V_1)$. It follows that
\[ V_2 = f(V_1) \oplus f(V_1)^\perp.\]
Cyclicity of $\Dd_2$ with respect to $\langle\cdot,\cdot\rangle_2$ implies that $f(V_1)^\perp\subset V_2$ is a subcomplex. Because~$f$ is a quasi-isomorphism, we have $\H(V_1)\simeq \H(f(V_1))\simeq \H(V_2)$; because the homology is additive, we have $\H(V_2)\simeq\H(f(V_1))\oplus\H(f(V_1)^\perp)\simeq \H(V_2)\oplus\H(f(V_1)^\perp)$; finally, because $\H(V_2)$ is of finite type, we have $\H(f(V_1)^\perp) = 0$. Because $V_2 = f(V_1) \oplus f(V_1)^\perp$ and $\langle\cdot,\cdot\rangle_2$ is non-degenerate, its restriction to $f(V_1)^\perp$ is non-degenerate too. As $V_2$ and hence $f(V_1)^\perp$ is of finite type, Proposition~\ref{Prop:NDegFin} gives a Hodge decomposition $f(V_1)^\perp = \Dd f(V_1)^\perp \oplus C_2'$. It is easy to check that
\[ V_2 = f(\Harm_1)\oplus\underbrace{\bigl(\Dd f(V_1) \oplus \Dd f(V_1)^\perp)}_{\displaystyle=\Im\Dd_2}\oplus(\underbrace{f(C_1)\oplus C_2'}_{\displaystyle\eqqcolon C_2})\]
is a Hodge decomposition of $V_2$. The corresponding standard Hodge homotopy clearly satisfies $\HtpStd^2 \circ f = f\circ \HtpStd^1$, and it holds $\HtpStd^2(f(V_1)) \subset f(V_1)$, so that $f(V_1)$ is a Hodge subalgebra. By injectivity, it follows that $\VansSmall(V_1) \simeq \VansSmall(V_2)$.
\end{proof}
%
%Application of this theory is possible since .
%where the other arrows are the canonical inclusions and projections. How to get it?
%
%\end{proof}
%
%\begin{Lemma}
%A small algebra is an image of a Sullivan's model. I.e., whenever $A$ is small and we have a quasi-isomorphism from the minimal model into $A$, then it has to be surjective --- this would be cool.
%
%Every orientation preserving quasi-isomorphism inside a small algebra has to be surjective. Then I have it!
%\end{Lemma}
%
%
%\begin{Example}
%\begin{ExampleList}
%\item The de Rham complex $\DR(M)$ of a closed oriented manifold~$M$ is non-degenerate, of Hodge type and does not satisfy Poincar\'e duality. Proposition~\ref{Prop:HodgePModel}
%applies and gives a weakly equivalent Poincar\'e model.
%\item Example of finite-dimensional and not of Hodge type (Hence necessary degenerate).
%\item Example of infinite-dimensional non-degenerate and not of Hodge type. Maybe even Poincar\'e duality?
%\item An example where two small subalgebras are not isomorphic. They do not even have to homotopy equivalent because image of homotopy equivalent morphisms do not have to.
%\end{ExampleList}
%\end{Example}
%
%\begin{Proposition}[Small subalgebra in geometrically formal case]\label{Prop:SmalGeomForm}
%Let $(V,\Dd,\wedge)$ be a non-negatively graded unital commutative $\DGA$ with a cyclic structure $\langle\cdot,\cdot\rangle$ of Hodge type such that the induced cyclic structure on $\H(V)$ is non-degenerate. Suppose that there is $\DGA$-quasi-isomorphism $f: \H(V) \rightarrow V$ preserving orientation on homology. Then there is a Hodge decomposition of $V$ with $\Harm = \im f$ and $f$ induces an isomorphism $\H(V) \simeq \VansQuotient(\VansSmall(V))$.
%\end{Proposition}
%
%\begin{proof}
%By Remark~\ref{Rem:RemarkHarm}, there is a Hodge decomposition with $\Harm = \Im f$. Therefore, we can restrict the codomain of $f$ and obtain the $\DGA$-quasi-isomorphism $\tilde{f}: \H(V) \rightarrow \VansSmall(V) \rightarrow \VansQuotient(\VansSmall(V))$ which preserves orientation on homology. Lemma~\ref{Eq:LemSmallSub} implies that there is a Hodge decomposition of $V$ such that $\tilde{f}$ induces an isomorphism $\VansSmall(\H(V))\simeq \VansSmall(\VansQuotient(\VansSmall(V)))$. However, $\VansSmall(\H(V))=\H(V)$ and $\VansSmall(\VansQuotient(\VansSmall(V)))=\VansQuotient(\VansSmall(V))$ by Proposition~\ref{Prop:PropPropertiessd}.
%\end{proof}

Since $\VansSmall=\VansSmall(V)$ is ``small'', a question whether it can be fit inside the image of a Sullivan's minimal model arose. This would imply, under some additional assumptions, that for any two Hodge decompositions of $V$, the non-degenerate quotients $\VansQuotient_1$, resp.~$\VansQuotient_2$ of the corresponding small subalgebras $\VansSmall_1$, resp.~$\VansSmall_2$ would be isomorphic as Poincar\'e duality algebras. Let us sketch the idea of this construction assuming that $f_1: \Lambda U_1 \twoheadrightarrow \VansSmall_1$ and $f_2: \Lambda U_2 \twoheadrightarrow \VansSmall_2$ are surjective Sullivan's minimal models. Uniqueness from \cite[Theorem~2.24]{Felix2008} gives an isomorphism $\Lambda U \coloneqq \Lambda U_1 \simeq \Lambda U_2$ such that the following diagram commutes up to homotopy of $\DGA$'s:
\begin{equation}\label{Eq:HpyCommutDiag}
\begin{tikzcd}
& & \VansSmall_1 \arrow[hook]{ld}\arrow{r} & \VansQuotient_1 \\
\Lambda U \arrow[bend left,two heads]{rru}{f_1}\arrow[bend right,two heads]{rrd}[below]{f_2}  &  V & &  \\
& & \VansSmall_2 \arrow[hook]{lu}\arrow{r}& \VansQuotient_2.
\end{tikzcd}
\end{equation}
Now, $\Lambda U \xrightarrow{f_1} \VansSmall_1 \xhookrightarrow{} V$ and $\Lambda U \xrightarrow{f_2} \VansSmall_2 \xhookrightarrow{} V$ induce the same isomorphism on homology; this can be used to pullback the orientation~$\Or^\H$ on $\H(V)$ to an orientation $\tilde{\Or}$ on $\H(\Lambda U)$, so that all maps in \eqref{Eq:HpyCommutDiag} will preserve the orientation on homology. Under the assumptions $\Dd V^n = 0$ and $\Dd (\Lambda U)^n = 0$, Proposition~\ref{Prop:OrOnHomG} applies, and we obtain a cyclic structure $\langle\cdot,\cdot\rangle^{\Lambda U}$ on $\Lambda U$ which is preserved by both $f_1$ and~$f_2$ on the chain level. We denote $f_i^\VansQuotient\coloneqq \pi^{\VansQuotient}_i \circ f_i$ and write down the following diagram with pairing-preserving $\DGA$-quasi-isomorphisms:
\begin{equation}\label{Eq:Diagram}
\begin{tikzcd}
& \arrow[two heads,swap]{ld}{f_1^\VansQuotient} \bigl(\Lambda U,\langle\cdot,\cdot\rangle^{\Lambda U}\bigr)\arrow[two heads]{rd}{f_2^\VansQuotient}& \\
\bigl(\VansQuotient_1,\langle\cdot,\cdot\rangle^{\VansQuotient}_1\bigr) & & \bigl(\VansQuotient_2,\langle\cdot,\cdot\rangle^{\VansQuotient}_2 \bigr).
\end{tikzcd}
\end{equation}
Claim (b) of Lemma~\ref{Lem:PomLemma} implies that $\ker f_1^\VansQuotient = \ker f_2^\VansQuotient = (\Lambda U)^\perp$, and hence
\[ \VansQuotient_1 \simeq  \Lambda U / (\Lambda U)^\perp\simeq \VansQuotient_2. \]
Unfortunately, the next two examples show that one can not, in general, expect~$\VansSmall_1$ and~$\VansSmall_2$, or even $\VansQuotient_1$ and $\VansQuotient_2$ to be isomorphic and $\VansSmall$ to fit inside the image of the minimal model.\footnote{Note that it is always possible to construct a surjective (non-minimal) Sullivan model $f: \Lambda U \twoheadrightarrow V$ by taking the minimal Sullivan model and inductively adding generators $\xi_i$ and $\mu_i$ with $\Dd \mu_i = 0$ and $\Dd \xi_i = \mu_i$ (the minimality condition on a Sullivan's algebra would require $\Dd \xi_i$ to be decomposable), and mapping them to $v\in V$ and $\Dd v$, respectively. Nevertheless, the uniqueness property of non-minimal Sullivan models is much weaker, see \cite[Lemma~2.20]{Felix2008}.}

\begin{Example}[Small algebras are, in general, not unique and not contained in images of Sullivan minimal models]\label{Ex:NonUniqueSmall}
Consider $M=\CP^2$, and let $\Kaehler$ be the Fubini--Study form on~$M$.
For $\alpha\in\DR^1(M)$, set $\Kaehler_\alpha \coloneqq \Kaehler + \Dd\alpha$.
Then $K_\alpha \wedge K_\alpha = K \wedge K + \Dd(2\alpha\wedge K + \alpha \wedge \Dd\alpha)$.
We can choose $\alpha$ such that $\Dd(2\alpha\wedge K + \alpha\wedge\Dd\alpha)\neq 0$.
Consider the Riemannian Hodge decomposition of $\DR(M)$ with $\Harm = \langle 1, \Kaehler, \Kaehler\wedge \Kaehler\rangle$.
According to Remark~\ref{Rem:RemarkHarm}, there is also a ``twisted'' Hodge decomposition with the harmonic subspace $\Harm_\alpha \coloneqq \langle 1, \Kaehler_\alpha, \Kaehler \wedge \Kaehler\rangle$.

The small subalgebra $\VansSmall$ for the Riemann Hodge decomposition is $\VansSmall=\langle 1, \Kaehler, \Kaehler\wedge\Kaehler\rangle$.
The small subalgebra $\VansSmall_\alpha$ for the twisted Hodge decomposition must contain $1$, $\Kaehler_\alpha$ and both $\Kaehler_\alpha\wedge\Kaehler_\alpha$ and $\Kaehler\wedge\Kaehler$ (we require $\Harm_\alpha\subset\VansSmall_\alpha$ by definition).
Therefore, it contains
\[
\Kaehler_\alpha \wedge \Kaehler_\alpha - \Kaehler\wedge\Kaehler = \Dd(2\alpha\wedge K + \alpha\wedge\Dd\alpha)
\]
and also
\[
\StdPrpg^\alpha\Dd(2\alpha\wedge K + \alpha\wedge\Dd\alpha)=\pi_{C_\alpha}(2\alpha\wedge K + \alpha\wedge\Dd\alpha).
\]
Proposition~\ref{Prop:SmallDescription} asserts that these vectors, together with $1$ and $\Kaehler_\alpha$, generate $\VansSmall_\alpha$ as a graded vector space.
Clearly, $\VansSmall_\alpha$ is not isomorphic to $\VansSmall$ for generic $\alpha$, but
\[
\VansQuotient(\VansSmall_\alpha)=\langle1,\Kaehler_\alpha,\Kaehler_\alpha\wedge\Kaehler_\alpha\rangle\simeq \langle 1, \Kaehler, \Kaehler\wedge\Kaehler\rangle= \VansQuotient(\VansSmall).
\]
In fact, the previous argument works in general and implies that for $\CP^2$, the non-degenerate quotients of two small subalgebras are isomorphic as Poincar\'e duality algebras (this does not hold for any $M$, see Example~\ref{Ex:SUsix}).

The Sullivan minimal model of $M$ is the free $\DGA$ $\Model \coloneqq \Lambda(\eta,\mu)$ with $\Abs{\eta}=2$, $\Abs{\mu}=5$, $\Dd\eta = 0$ and $\Dd\mu=\eta\wedge\eta$.
A $\DGA$-quasi-isomorphism $f: \Model\rightarrow\DR(M)$ is specified by its values on~$\eta$ and~$\mu$; for example, $f(\eta)\coloneqq\Kaehler$ and $f(\mu)\coloneqq 0$.
We see that neither $\VansSmall_\alpha$ nor $\Harm_\alpha$ can lie in $\im f$ because $\dim (\im f)^{4}=1$ for any $f$.
\end{Example}

\begin{Example}[Non-degenerate quotients of small subalgebras for different Hodge decompositions are, in general, not isomorphic]\label{Ex:SUsix}
Consider $M=\mathrm{SU}(6)$.
It is a compact simply-connected Lie group of dimension~$35$ ($=n^2 - 1$ for $n=6$) whose cohomology ring is freely generated by single elements in degrees  $3$, $5$,~$\dotsc$,~$11$; see \cite[Corollary~3.11]{Mimura1991}.
There is a biinvariant Riemannian metric and there are biinvariant differential forms $x_3$, $x_5$,~$\dotsc$, $x_{11}$ in the corresponding degrees such that
\[
\Harm \coloneqq \Lambda(x_3, \dotsc, x_{11})\subset \DR(M)
\]
is the algebra of harmonic forms, see \cite[Chapter~1]{Felix2008}.

For $\eta_6\in \DR^6(M)$ and $\eta_8\in\DR^8(M)$, which are going to be specified later, set
\[
\tilde{x}_7 \coloneqq x_7 + \Dd \eta_6\quad\text{and}\quad\tilde{x}_9 \coloneqq x_9 + \Dd \eta_8,
\]
and let $\Harm'$ denote the graded vector space obtained from $\Harm$ by replacing the vectors $x_{7}$ and $x_{9}$ with $\tilde{x}_7$ and $\tilde{x}_9$, respectively.
We emphasize that we are replacing just the vectors, not the products; e.g., $x_7\wedge x_9$ is an element of $\Harm'$ but $\tilde{x}_7\wedge \tilde{x}_9$ might not be.
Let $\VansSmall = \Harm$ be the small subalgebra corresponding to the Riemannian Hodge decomposition, and let~$\VansSmall'$ be the small subalgebra corresponding to a Hodge decomposition based on~$\Harm'$ (such always exists by Remark~\ref{Rem:RemarkHarm}).
The following elements in degrees $15$, resp.~$20$ must be contained in $\VansSmall'$:
\begin{align*}
y&\coloneqq\StdPrpg'(\tilde{x}_7 \wedge x_9 -x_7 \wedge x_9)\\
 &=\StdPrpg'\Dd(\eta_6 \wedge x_9),\\
z&\coloneqq\tilde{x}_9 \wedge x_{11} - x_9 \wedge x_{11}\\
 &=\Dd(\eta_8 \wedge x_{11}).
\end{align*}
Using Stokes theorem and $\Dd\circ\StdPrpg' = \pi_{\Im \Dd}$, we get
\begin{align*}
\langle y,z\rangle & = \pm \int_{M} \StdPrpg'\Dd(\eta_6 \wedge x_9) \wedge\Dd(\eta_8 \wedge x_{11})\\
&= \pm \int_{M} \Dd\eta_6\wedge \eta_8 \wedge x_9 \wedge x_{11}.
\end{align*}
We claim that the integral can be made non-zero by a choice of $\eta_6$ and $\eta_8$.
Indeed, because $x_9\wedge x_{11}$ generates non-zero homology, there is an $m\in M$ such that $x_9(m)\wedge x_{11}(m)\neq 0$.
Pick local coordinates $(x^i)$ centered at $m$, and let $\alpha^I\Diff{x}^I$ be a non-zero coefficient in $x_9(m)\wedge x_{11}(m)$. Consider the complement $J = I^C$ and decompose $J= J_1 \cup J_2$ into two parts with $\Abs{J}_1 = 7$ and $\Abs{J}_2 = 8$.
For some $0\neq c\in\R$, set locally
\[
\eta_6 \coloneqq ((x^{J_{11}}+c)\Diff{x}^{J_1\backslash\{J_{11}\}})\quad\text{and}\quad\eta_8=\Diff{x}^{J_2},
\]
and extend them to the whole of $M$ by multiplying with a bump function which is constant non-zero in a neighborhood of $m$.
We have achieved that the integrand $\omega \coloneqq \Dd\eta_6\wedge \eta_8 \wedge x_9 \wedge x_{11}$ is non-zero around $m$.
Because the intersection pairing is non-degenerate and because $\omega \neq 0$, there is a function $f\in C^\infty(M)$ such that $\int_M f\omega \neq 0$.
We can now just rescale $\eta_8$ by $f$.

We have shown that $z$ induces a non-zero element $\pi_{\VansQuotient}'(z)\in \VansQuotient(\VansSmall')^{20}$, where $\pi_{\VansQuotient}': \VansSmall' \rightarrow \VansSmall'/{\VansSmall'}^\perp$ is the canonical projection.
Now, $\pi_{\VansQuotient}'$ is a $\DGA$-morphism, and hence $\pi_{\VansQuotient}'(z)$ is exact.
Because $\VansSmall'$ is of Hodge type, $\pi_{\VansQuotient}'$ is also a quasi-isomorphism, and hence $\pi_{\VansQuotient}'(x_{9}\wedge x_{11})$ generates non-trivial homology.
It follows that $\pi_{\VansQuotient}'(z)$ is not a multiple of $\pi_{\VansQuotient}'(x_9\wedge x_{11})$, and thus $\dim \VansQuotient(\VansSmall')^{20} \ge 2$.
However, we have $\VansQuotient(\VansSmall)^{20} = \Harm^{20} = \langle x_9\wedge x_{11} \rangle$.
This shows that $\VansQuotient(\VansSmall)$ and $\VansQuotient(\VansSmall')$ can not be isomorphic as vector spaces.
\end{Example}

%The following conjecture is based on a remark of Prof.~Hông Vân Lê, communicated in an e-mail correspondence, that ``the small subalgebra is the image of a minimal model''.

%\begin{Conjecture}[Hodge-Sullivan minimal model]\label{Conj:HodgeSullMin}
%Let $(V,\Dd,\wedge)$ be a non-negatively graded unital commutative $\DGA$ with a cyclic structure $\langle\cdot,\cdot\rangle$ of Hodge type. Suppose that~$V$ is of finite type, $\H^0(V) = \R$ and $\H^1(V) = 0$. Then for any Hodge decomposition of~$V$, there is a Sullivan minimal model $\Lambda U \rightarrow V$ (see \cite{Felix2008} or Remark~\ref{Rem:Models} later) whose image contains a Hodge subalgebra.
%\end{Conjecture}
%\begin{proof}[Idea of proof] The author tried to modify the well-known inductive construction of $\Lambda U \rightarrow V$ to achieve that its image is $\HtpStd$-invariant; he observed that vanishing of higher homology groups might be required. In any case, it sufficient (and necessary) to prove that the image contains at least the trees from Proposition~\ref{Prop:SmallDescription}.
%\end{proof}
%If Conjecture~\ref{Conj:HodgeSullMin} is true, then we have the following:
%\begin{Conjecture}[Uniqueness of non-degenerate small subalgebra]\label{Conj:UnieqSmal}
%Let $(V,\Dd,\wedge)$ be a non-negatively graded unital commutative $\DGA$ with a cyclic structure $\langle\cdot,\cdot\rangle$ of degree $n$ of Hodge type. Suppose that~$\H(V)$ is of finite type, $\H^0(V) = \R$ and $\H^1(V) = 0$. Suppose, in addition, that $\Dd V^n = 0$ and $\Dd (\Lambda U)^n = 0$ for the Sullivan minimal model $\Lambda U$ of $V$. Then for any two Hodge decompositions with small subalgebras $\VansSmall_1(V)$ and $\VansSmall_2(V)$, there is a pairing-preserving isomorphism $\VansQuotient(\VansSmall_1(V))\simeq\VansQuotient(\VansSmall_2(V))$.
%\end{Conjecture}
%\begin{proof}
%\end{proof}

It is not hard to come up with artificial examples of oriented $\DGA$'s which are not of Hodge type.\Add[noline,caption={Not of Hodge type}]{Example of an oriented dga without a Hodge decomposition.}

The following lemma and proposition show that in some cases it is possible to extend a $\DGA$ to a $\DGA$ of Hodge type.

\begin{Lemma}[Giving partners to non-degenerates]\label{Lemma:Exte}
 Let $(V,\Dd,\wedge,\Or)$ be a unital commutative $\DGA$ which is non-negatively graded and oriented in degree $n$.
 For $k=\lceil\frac{n}{2}\rceil$, $\dotsc$, $n$, consider the following property $(P_k)$ of a direct sum decomposition%(``almost Hodge decomposition'')
\begin{equation}\label{Eq:DecompOfV}
V=\Harm\oplus \Dd V \oplus C,
\end{equation}
where $\Harm$ is a harmonic subspace and $C$ a complement of $\ker \Dd$ in $V$ perpendicular to~$\Harm$ with respect to the induced cyclic structure $\langle\cdot,\cdot\rangle$:
\begin{description}
\item[$(P_{k})$] There is a complement $E$ of
\[
C^\perp\coloneqq \{ c^\perp \in C \mid \langle c^\perp,c\rangle=0\text{ for all }c\in C\}
\]
in $C$ and a homogenous linear map
\[
\rho:  E^{\lceil n/2\rceil}\oplus \dotsb \oplus E^{k} \longrightarrow \Dd V
\]
such that for all $e'\in E$ and $c^\perp\in C^\perp$, the following holds:
\begin{align}
\langle e', \rho(e) \rangle &=\langle e', e \rangle, \label{Eq:ConditionTemp} \\
\langle c^\perp, \rho(e) \rangle &=  0.\label{Eq:ConditionTempII}
\end{align}
\end{description}
Suppose that $V$ is non-negatively graded, of finite type and satisfies $V^0=\Span\{1\}$ and $V^1 = 0$.
Suppose that $n \ge 5$ and that $(\H(V), \wedge, \Or^\H)$ is a Poincar\'e duality algebra.
Given $\lceil\frac{n}{2}\rceil\le l\le n$, suppose that $V$ admits a decomposition of type \eqref{Eq:DecompOfV} such that either $l=\lceil\frac{n}{2}\rceil$ or $l>\lceil\frac{n}{2}\rceil$ and $(P_{l-1})$ holds.
Then there is an $m\in \N_0$ and a Sullivan $\DGA$
 \begin{equation}\label{Eq:SullAlg}
 \Lambda \coloneqq \Lambda(w_1,\dotsc,w_m,z_1,\dotsc,z_m)
 \end{equation}
 specified by $\deg w_i = l-1$, $\deg z_i = l$, $\Dd z_i = 0$ and $\Dd w_i = z_i$ for all $i=1$, $\dotsc$, $m$ such that the tensor product $\DGA$
\[
 \hat{V}\coloneqq V\otimes \Lambda
 \]
 admits an orientation $\hat{\Or}: \hat{V}\to\R$ which extends $\Or: V \rightarrow \R$ on the canonical inclusion $V \hookrightarrow\hat{V}$, and there is a decomposition
\begin{equation}\label{Eq:DevompOfVHat}
\hat{V} = \hat{\Harm}\oplus \Dd \hat{V}\oplus \hat{C}
\end{equation}
of type \eqref{Eq:DecompOfV} for which $(P_{l})$ holds.
\end{Lemma}

\begin{proof}
The proof consists of a construction of $\hat{\Or}: \hat{V}\rightarrow\R$, a construction of a harmonic subspace $\hat{\Harm}$ in $\hat{V}$, a construction of a complement $\hat{C}$ of $\ker \Dd$ in $\hat{V}$, a degreewise description of $\hat{C}$ and $\hat{C}^\perp$, a degreewise construction of a complement $\hat{E}$ of $\hat{C}^\perp$ in $\hat{C}$ and a proof of the property $(P_l)$ for the constructed decomposition.
\begin{description}[leftmargin=0pt,font=\normalfont\itshape]
\item[Construction of $\hat{\Or}$:]
Consider the decomposition \eqref{Eq:DecompOfV}.
Because $\Dd : C \rightarrow \Dd V$ is an isomorphism, we can write
\begin{equation}\label{Eq:VDecomp}
V = \Harm \oplus \underbrace{\Dd E \oplus \Dd C^\perp}_{\displaystyle \Dd V}\oplus \underbrace{E\oplus C^\perp}_{\displaystyle C}.
\end{equation}
The restriction of $\langle\cdot,\cdot\rangle$ to $E$ is non-degenerate, and $V$ is of finite type by assumption; hence, $E$ is finite-dimensional.
Set
\[
m \coloneqq \dim E^l.
\]
Let $\xi_1$, $\dotsc$, $\xi_m$ be a basis of $E^{l}$ and $\xi^1$, $\dotsc$, $\xi^m$ its dual basis in $E^{n-l}$.
The Sullivan algebra~$\Lambda$ can be written as a direct sum
\begin{equation}\label{Eq:LambdaDecomp}
\Lambda = \bigoplus_{k=0}^\infty \Lambda_k\quad\text{with}\quad
\Lambda_k = \bigoplus_{\substack{r, m \ge 0 \\ r + m = k}}\Lambda_r(w)\otimes \Lambda_m(z),
\end{equation}
where $\Lambda_r(w)$ and $\Lambda_m(z)$ are the graded vector spaces generated by monomials $w_I = w_{i_1}\dotsc w_{i_r}$ and $z_J = z_{j_1}\dotsc z_{j_m}$ for all multiindices $I=\{i_1, \dotsc, i_r\}$ and $J=\{j_1,\dotsc,j_m\}$, respectively.
The direct sum decompositions~\eqref{Eq:VDecomp} and~\eqref{Eq:LambdaDecomp} induce a direct sum decomposition of $\hat{V} = V \otimes \Lambda$ via the distributivity of $\otimes$ and $\oplus$.
We denote
\[
\hat{V}_k \coloneqq V \otimes \Lambda_k\quad\text{for }k\ge 0.
\]
Let $\hat{\Or}: \hat{V} \to \R$ be the linear map satisfying
\begin{align}
	\hat{\Or}(v) &\coloneqq \Or(v) && \text{for all }v\in V,\\
	\hat{\Or}(\xi^i \wedge z_j) &\coloneqq \Or(\xi^i\wedge \xi_j) && \text{and} \\
	\hat{\Or}(\Dd \xi^i \wedge w_j) &\coloneqq (-1)^{\deg \xi^i + 1}\Or(\xi^i \wedge \xi_j) && \text{for all }i, j = 1, \dotsc, m,
\end{align}
and which is zero on $(\Harm\oplus\Dd V\oplus C^\perp \oplus \bigoplus_{i\ge 0, i \neq n-l}E^i)\otimes \Lambda_1(z)$, on $(\Harm\oplus C \oplus \Dd C^\perp \oplus \bigoplus_{i\ge 0, i\neq n-l} \Dd E^i)\otimes\Lambda_1(w)$ and on $\hat{V}_k$ for $k \ge 2$.

In order to show that $\hat{\Or}$ is an orientation, we must check that $\hat{\Or}\neq 0$ and $\hat{\Or}\circ \Dd = 0$.
The first condition is clear from $\Restr{\hat{\Or}}{V} = \Or \neq 0$.
As for the second condition, $\hat{V}$ is generated by elements $v \wedge w_I \wedge z_J$ for $v\in V$ and multiindices $I$, $J$.
It holds $\Dd \hat{V}_k \subset \hat{V}_k$ for all~$k\ge 0$, and hence $\Dd \hat{V} = \bigoplus_{k=0}^\infty \Dd \hat{V}_k$.
From the definition of $\hat{\Or}$, we have immediately $\Dd \hat{V}_0 = \Dd V \subset \ker \Or \subset \ker \hat{\Or}$ and $\bigoplus_{k=2}^\infty \Dd \hat{V}_k \subset \ker\hat{\Or}$.
As for $\Dd\hat{V}_1$, we write $\hat{V}_{1} = \Span \{v\wedge w_j, v\wedge z_j \mid v\in V, j=1,\dotsc, m\}$ as a graded vector space and compute
\begin{equation}\label{Eq:DVI}
\begin{aligned}
	\Dd \hat{V}_1 &= \Span\{\Dd(v\wedge w_j), \Dd(v\wedge z_j) \mid v\in V, j=1, \dotsc, m \}\\
	&=\Span\{\Dd v \wedge w_j + (-1)^{\Deg v} v \wedge z_j \mid v\in V, j=1, \dotsc, m\}.
\end{aligned}
\end{equation}
Write $v\in V^{n-l}$ as $v = h  + \Dd c + c^\perp + \sum_{i=1}^m \alpha_i \xi^i$ for $h\in \Harm^{n-l}$, $c\in C^{n-l-1}$, $c^\perp\in C^{\perp n-l}$ and $\alpha_i\in \R$, and compute for every $j=1$, $\dotsc$, $m$ the following:
\begin{align*}
\hat{\Or}(\Dd v\wedge w_j) & = \hat{\Or}\Bigl(\Dd c^\perp \wedge w_j + \sum_{i=1}^m \alpha_i \Dd \xi^i \wedge w_j\Bigr)\\
&= \sum_{i=1}^m \alpha_i \hat{\Or}(\Dd \xi^i \wedge w_j)\\
&= \sum_{i=1}^m (-1)^{\deg \xi^i + 1}\alpha_i \hat{\Or}(\xi^i \wedge \xi_j) \\
&= (-1)^{n-l+1}\sum_{i=1}^m \alpha_i \hat{\Or}(\xi^i \wedge z_j)\\
&= (-1)^{n-l+1}\hat{\Or}\Bigl((h + \Dd c + c^\perp)\wedge z_j +\sum_{i=1}^m \alpha_i \xi^i \wedge z_j \Bigr) \\
&= (-1)^{\deg v + 1}\hat{\Or}(v\wedge z_j).
\end{align*}
Consequently, $\Dd \hat{V}_{1}\subset \ker \hat{\Or}$.
This shows $\hat{\Or}\circ\Dd = 0$.

The inclusion $V\hookrightarrow\hat{V}$ is clearly orientation preserving.

\item[Construction of $\hat{\Harm}$:]
It holds $\bar{\H}(\Lambda) = 0$ for the reduced homology, and hence $\H(\hat{V}) \simeq \H(V)\otimes\H(\Lambda) = \H(V)$ by K\"unneth's formula. Because $\Harm\subset \ker \Dd$, $\Harm\cap\im\Dd = 0$ and $\dim(\Harm) = \dim \H(\hat{V})$, $\Harm$ is a complement of $\Dd\hat{V}$ in $\ker \Dd$. Therefore,
\[
\hat{\Harm} \coloneqq \Harm
\]
is a harmonic subspace of $\hat{V}$.
Also note that $\H(\hat{V})=\bigoplus_{k=0}^\infty \H(\hat{V}_k)$ and $\dim \H(\hat{V}) = \dim \H(\hat{V}_0)$, and hence $\H(\hat{V}_k) = 0$ for all $k\ge 1$.

\item[Construction of $\hat{C}$:]
We construct $\hat{C}$ as a direct sum $\hat{C} = \bigoplus_{k=0}^\infty \hat{C}_k$.
Set
\[
\hat{C}_0 \coloneqq C.
\]
For $k=1$, define
\begin{equation}\label{Eq:CIDef}
\begin{aligned}
 \tilde{C}_1 &\coloneqq \Span\{v \wedge w_i \mid v\in V, i=1, \dotsc,m\} \subset \hat{V}_1, \\
  \hat{C}_1 &\coloneqq \{\tilde{c} - \pi(\tilde{c}) \mid \tilde{c}\in \tilde{C}_1\}\subset\hat{V}_0\oplus\hat{V}_1,
\end{aligned}
\end{equation}
where $\pi: \hat{V}\rightarrow\hat{\Harm}$ is the orthogonal projection.
For all $k\ge 2$, let $\hat{C}_k \subset \hat{V}_k$ be an arbitrary complement of $\ker\Dd$ in $\hat{V}_k$ as a graded vector space.

We show first that $\hat{C}_i$ for $i\ge 0$ are disjoint.
Clearly, $\hat{C}_j \cap \sum_{i=0, i\neq j}^\infty \hat{C}_i = 0$ for all $j\ge 2$.
Because $(\Harm + C)\cap\tilde{C}_1 = 0$ and $\Harm \cap C = 0$, it holds $\hat{C}_1 \cap \hat{C}_0 = 0$, and $(\hat{C}_0 + \hat{C}_1) \cap \sum_{i\ge 2} \hat{C}_i = 0$ implies that $\hat{C}_j \cap \sum_{i=0, i\neq j}^\infty \hat{C}_i = 0$ holds also for $j=0$, $1$.

We show that $\hat{C}=\bigoplus_{k\ge 0}\hat{C}_k$ is a complement of $\ker \Dd$ in~$\hat{V}$.
For $k\ge 2$, $\hat{C}_k$ are complements of $\ker \Dd$ in $\hat{V}_k$ by construction.
For $k=0$, $\hat{C}_0 \oplus \ker \Dd \cap \hat{V}_0 = \hat{V}_0$ follows from \eqref{Eq:DecompOfV}.
For $k=1$, we compare \eqref{Eq:DVI} and \eqref{Eq:CIDef} to see that $\tilde{C}_1 \oplus \Dd \hat{V}_1 = \hat{V}_1$.
Because $\Harm\subset \ker \Dd$, it is easy to see that $\hat{C}_0\oplus\hat{C}_1$ is a complement of $\ker \Dd$ in $\hat{V}_0\oplus\hat{V}_1$.

Finally, $\hat{C}_0 \perp \hat{\Harm}$ holds by \eqref{Eq:DecompOfV}, $\hat{C}_1\perp\Harm$ holds by the construction of $\hat{C}_1$ from $\tilde{C}_1$ and $\hat{C}_k \perp \Harm$ follows from the definition of $\hat{\Or}$.

\item[Degreewise description of $\hat{C}$ and $\hat{C}^\perp$:]
We are interested in $\hat{C}^i$ for $0\le i\le l$.
For all $k\ge 2$, the graded vector space $\hat{C}_k$ is concentrated in degrees $i\ge 2(l-1)$.
But $2(l-1)>l$ due to $n\ge 5$.
Therefore, $\hat{C}_k^i = 0$ for $k\ge 2$ and $0\le i \le l$.
We denote $\bar{V} \coloneqq \bigoplus_{i=1}^\infty V^i$ and write $\tilde{C}_1 = \Lambda_1(w) \oplus (\bar{V}\otimes \Lambda_1(w))$.
Because $V^1=0$, the graded vector space $\bar{V}\otimes\Lambda_1(w)$ is concentrated in degrees $i\ge 2 + (l-1) = l + 1$. We obtain
\begin{equation}\label{Eq:CDegreewise}
\hat{C}^i = (\hat{C}_0 \oplus \hat{C}_1)^i =
 \begin{cases}
 	C^i & \text{for }0 \le i \le l-1, \\
	C^i + \Span\{ w_j \mid j = 1, \dotsc, m\} & \text{for }i = l - 1, \\
	C^i & \text{for }i = l.
 \end{cases}
\end{equation}
Here we used that $\hat{C}_1^{l-1} = \tilde{C}_1^{l-1} = \Span\{ w_i \mid i = 1, \dotsc, m\}$, which is true because $\Lambda_1(w)\perp\Harm$ from the definition of $\hat{\Or}$.

We are now interested in $\hat{C}^{\perp i}$ for $n-l\le i \le l$.
Note that $n-l \le i \le l$ is equivalent to $n-l \le n-i \le l$.
By definition, $\hat{\Or}$ vanishes on $C\wedge\Lambda_1(w)=C\otimes\Lambda_1(w)$ and $\Lambda_1(w)\wedge\Lambda_1(w) \subset \Lambda_2(w)$.
Looking at \eqref{Eq:CDegreewise}, we see the following:
\begin{equation}\label{Eq:CPerpDegreewise}
	\hat{C}^{\perp i} =
		\begin{cases}
			C^{\perp i} &  \text{for }n-l \le i \le l - 2, \\
			C^{\perp i} + \Span\{w_i\mid i=1,\dotsc,m\} & \text{for }i = l-1, \\
			C^{\perp i} & \text{for }i = l.
		\end{cases}
\end{equation}
Because $\hat{V}$ is non-negatively graded, it holds $\hat{C}^{\perp i} = \hat{C}^i$ for $i>n$.
Notice that $\hat{C}^{\perp i}$ might be smaller than $C^i$ for $0\le i \le n-l-1$.
The reason for this is a possible existence of $v_1$, $v_2\in V$ such that $v_1 \wedge v_2$ has a non-trivial $\Dd E$-component and $\langle v_1, v_2\wedge w_i \rangle = \hat{\Or}((v_1\wedge v_2)\wedge w_i) \neq 0$.

\item[Construction of $\hat{E}$:]
Because $\hat{V}_i \wedge \hat{V}_j \subset \hat{V}_{i+j}$ and $\hat{V}_k \subset \ker \hat{\Or}$ for $k\ge 2$, we have $\hat{C}_k \subset \hat{C}^\perp$ for all $k\ge 2$.
It follows that
\[
\hat{C}^\perp = (\hat{C}_0\oplus\hat{C}_1)\cap\hat{C}^\perp \oplus \bigoplus_{k\ge 2}\hat{C}_k,
\]
and hence it is enough to construct $\hat{E} \subset \hat{C}_0\oplus\hat{C}_1$ such that $\hat{E}\oplus (\hat{C}_0\oplus\hat{C}_1)\cap\hat{C}^\perp = \hat{C}_0\oplus\hat{C}_1$.
Because $E\cap \hat{C}^\perp \subset E \cap C^\perp = 0$, we can get $\hat{E}$ by extending $E$.
For $0\le i \le n$, we define
%For $n-l \le i \le l$, we set $\hat{E}^i\coloneqq E^i$.
%For $i=0$, $\dotsc$, $n-l-1$, we obtain $\hat{E}^i$ from $E^i$ by adding elements from $C^i$.
%For $i=l+1$, $\dotsc$, $n$, we obtain $\hat{E}^i$ from $E^i$ by adding elements from $\hat{C}_0^i \oplus \hat{C}_1^i$.
\begin{equation}\label{Eq:EDegreewise}
\hat{E}^i \coloneqq
	\begin{cases}
		E^i\oplus \Span\{c_1^i,\dotsc, c_{k_i}^i\}\text{ for }c_1^i, \dotsc, c_{k_i}^i\in C^i & \text{for }0\le i \le n-l-1,\\
		E^i & \text{for }n-l\le i \le l,\\
		E^i\oplus\Span\{\hat{v}_1,\dotsc,\hat{v}_k\}\text{ for }\hat{v}_1^i,\dotsc,\hat{v}_{k_i}^i\in(\hat{C}_0\oplus\hat{C}_1)^i & \text{for }l+1\le i \le n,
	\end{cases}
\end{equation}
where the existence of $c^i_j$ and $\hat{v}^i_j$ and the fact that $\hat{C} = \hat{E}\oplus\hat{C}^\perp$ are justified by \eqref{Eq:CDegreewise}, \eqref{Eq:CPerpDegreewise} and \eqref{Eq:ConditionTempII}.

\item[Property $(P_l)$:] We define $\hat{\rho}: \hat{E}^{\lceil \frac{n}{2}\rceil}\oplus\dotsb\oplus\hat{E}^l \rightarrow \Dd\hat{V}$ by $\hat{\rho} \coloneqq \rho$ on $\hat{E}^{\lceil \frac{n}{2}\rceil}\oplus\dotsb\oplus\hat{E}^{l-1}=E^{\lceil \frac{n}{2}\rceil}\oplus\dotsb\oplus E^{l-1}$ and by
\[
\hat{\rho}(\xi_i) \coloneqq z_i\quad\text{for all }i=1, \dotsc, m.
\]
Conditions \eqref{Eq:ConditionTemp} and \eqref{Eq:ConditionTempII} are checked easily using $(P_{l-1})$, \eqref{Eq:CDegreewise}, \eqref{Eq:CPerpDegreewise}, \eqref{Eq:EDegreewise} and the definition of $\hat{\Or}$.\qedhere
%For $\lceil\frac{n}{2}\rceil \le i \le l - 1$,
%\eqref{Eq:ConditionTemp} \eqref{Eq:ConditionTempI}
%
%Given $\lceil \frac{n}{2} \rceil \le i \le l - 1$ and $e\in \hat{E}^i = E^i$, there is $z\in(\im\Dd)^i$ such that $\langle e,f\rangle = \langle z,f\rangle$ and $\langle z,c^\perp\rangle=0$ for all $f\in E^{n-i}$ and $c^\perp\in (C^\perp)^{n-i}$.
%Due to $j\le l$ and $l\le $
%$n - j \ge n - l$
%
%
% we can write an $\hat{e}\in\hat{E}^i$ as $\hat{e} = e + c^\perp + v \wedge w + w$ for $e\in E$, $c^\perp\in C^\perp$, $v\in V^{\ge 2}$ and $w\in \Lambda_1(w)$.
% It holds $\deg(v\wedge w) = l+1$, and hence $v=0$ for $k+1\le i \le l$; this summand is thus always zero.
% It holds $\deg w = l - 1$, and hence $w=0$ for $i\neq l-1$.
% In this case, $c^\perp + w = \hat{e}-e = \hat{E}^{l-1}$ can pair non-trivially only with an element in degree $n - (l-1) \le n - k = k+1 \le l$.
% However, since $\hat{V}^l = V^l \oplus \Lambda_1(z)$, $\hat{V}^{l-1} = V^{l-1}\oplus\Lambda_1(w)$ and $\hat{V}^i = V^i$ for $i<l-1$, we see that $(c^\perp + w)\perp \hat{C}$.
% This implies $\hat{e} = e$.
% In the case $i = l$, we have $w=0$ and $v=0$.
% Now, $\hat{e}-e = c^\perp \in \hat{E}^l$ can pair non-trivially only with an element in degree $n - l \le n - k - 1 = k \le l - 1$.
% However, since $\hat{C}^{l-1} = C^{l-1}\oplus \Lambda_1(w)$, we see that $c^\perp \hat{C}$.
% This implies again that $\hat{e} = e$.
%
% Let us check $(P_{l+1})$ for the decomposition consisting of $\Harm$, $\hat{C}$ and $\hat{E}$. Let $k+1 \le i \le l$, and let $e\in \hat{E}^i = E^i$ (the equality has been previously proven).
%
%
% We also need that $\hat{E}^i = E^i$ for all $n-l+1 \le i\le k$. However, this is clear.
\end{description}
\end{proof}

\begin{Questions}\phantomsection\label{Q:QuestOnPoinc}
\begin{RemarkList}
\item Given a cochain complex $(V,\Dd)$ with a symmetric pairing $\langle\cdot,\cdot\rangle$, which conditions have to be satisfied by the maps $\pi_\Harm$, $\StdHtp: V\rightarrow V$ so that $V=\Im \pi_\Harm \oplus \Im \Dd \oplus \Im \StdHtp$ is a Hodge decomposition with Hodge pair $(\Im \pi_\Harm, \StdHtp)$? This would characterize Hodge decompositions in terms of Hodge pairs.
\item It should be possible to prove Lemma~\ref{Lemma:Exte} also for $n\le 4$ by hand. Check that!
\qedhere
\end{RemarkList}
\end{Questions}
\end{document}


\documentclass[a4paper]{article}
\usepackage{amsmath}
\usepackage[colorlinks]{hyperref}
\usepackage{CJKutf8}
\title{Stokes' Theorem}
\author{Naitree Zhu}
\date{Last modified: \today}
\begin{document}
\maketitle
In differential geometry, Stokes' theorem (also called the generalized Stokes' theorem) is a statement about the integration of differential forms on manifolds, which both simplifies and generalizes several theorems from vector calculus.
 Stokes' theorem says that the integral of a differential form $\omega$ over the boundary of some orientable manifold $\Omega$ is equal to the integral of its exterior derivative $\mathrm{d}\omega$ over the whole of $\Omega$, i.e.
\begin{equation}
\int_{\partial\Omega} \omega=\int_\Omega \mathrm{d}\omega
\end{equation}

This modern form\footnote{For more deep informations about the \emph{generalized} stokes' theorem:\url{http://en.wikipedia.org/wiki/Stokes'_theorem}} of Stokes' theorem is a vast generalization of a classical result first discovered by Lord Kelvin, who communicated it to George Stokes in a letter dated July 2, 1850. Stokes set the theorem as a question on the 1854 Smith's Prize exam, which led to the result bearing his name.
This classical Kelvin--Stokes theorem relates the surface integral of the curl of a vector field $\boldsymbol{F}$ over a surface $\Sigma$ in Euclidean three-space to the line integral of the vector field over its boundary $\partial\Sigma$:
\begin{equation}
\iint_S \nabla\times \boldsymbol{F}\cdot \mathrm{d}\boldsymbol{S}=\oint_{\partial S}\boldsymbol{F}\cdot\mathrm{d}\boldsymbol{l}
\end{equation}

This classical statement, as well as the classical Divergence theorem, fundamental theorem of calculus, and Green's Theorem are simply special cases of the general formulation stated above.
\part{Special cases}
The general form of the Stokes theorem using differential forms is more powerful and easier to use than the special cases. The traditional versions can be formulated using Cartesian coordinates without the machinery of differential geometry, and thus are more accessible. Further, they are older and their names are more familiar as a result.
 The traditional forms are often considered more convenient by practicing scientists and engineers but the non-naturalness of the traditional formulation becomes apparent when using other coordinate systems, even familiar ones like spherical or cylindrical coordinates. There is potential for confusion in the way names are applied, and the use of dual formulations.
\section{Kelvin-Stokes theorem}
This is a (dualized) 1+1 dimensional case, for a 1-form (dualized because it is a statement about vector fields). This special case is often just referred to as the Stokes' theorem in many introductory university vector calculus courses and as used in physics and engineering. It is also sometimes known as the curl theorem.

The classical Kelvin–-Stokes theorem:
\begin{equation}
\iint_S \nabla\times \boldsymbol{F}\cdot \mathrm{d}\boldsymbol{S}=\oint_{\partial S}\boldsymbol{F}\cdot\mathrm{d}\boldsymbol{l}
\end{equation}

which relates the surface integral of the curl of a vector field over a surface $\Sigma$ in Euclidean three-space to the line integral of the vector field over its boundary, is a special case of the general Stokes theorem (with n = 2) once we identify a vector field with a 1-form using the metric on Euclidean three-space.
The curve of the line integral, $\partial\Sigma$, must have positive orientation, meaning that $d\boldsymbol{l}$ points counterclockwise when the surface normal, $d\boldsymbol{S}$, points toward the viewer, following the right-hand rule.

One consequence of the formula is that the field lines of a vector field with zero curl cannot be closed contours.
\begin{CJK}{UTF8}{gbsn}
在式(3)中， 考虑场区内的任一曲面，由于场的旋度为0， 左边为0， 则右面说明在场区内场矢量沿任何一条闭合曲线的积分为0. 这就说明了该矢量场的场线是绝不闭合的，因为在任一场线上矢量场的方向沿着该场线，沿闭合场线的积分非0.
\end{CJK}

In Cartesian coordinates,the formula can be rewritten as:
\begin{equation}
\begin{split}
\iint_\Sigma \left\lbrace\left(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\right) dydz + \left(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\right) dzdx + \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) dxdy\right\rbrace\\
= \oint_{\partial\Sigma} \left\lbrace P dx + Q dy + R dz\right\rbrace
\end{split}
\end{equation}
\section{2-dimensional: Green's theorem}
Green's theorem is immediately recognizable as the third integrand of both sides in the integral in terms of P, Q, and R cited above.
\section{Divergence theorem}
Likewise, the Divergence theorem
\begin{equation}
\int_V \nabla\cdot\boldsymbol{F} \mathrm{d}V=\oint_{\partial V} \boldsymbol{F}\cdot \mathrm{d}\boldsymbol{S}
\end{equation}
is a special case if we identify a vector field with the n-1 form obtained by contracting the vector field with the Euclidean volume form.
\end{document}


\section{Theory}

Our dataset consists of images.  One could use each individual pixel as an individual input feature, but that loses any relative spacial locality information between pixels, and is therefore prone to errors due to location in the image of the object to be identified \cite{lecun1998gradient}.  For example, if we want to identify if an image has a circle or not, looking at a set of pixels without knowing where they are located relative to eachother cannot be done in general.

For handwriting recognition, there are some features that some people have tried to extract to use in a learning algorithm.  These include aspect ratio, percent of thresholded pixels above the horizontal half plane, percent of thresholded pixels to the right of the vertical half plane, and so forth \cite{web:wikihandwritfeat}.  It is unknown if these features will be sufficient to classify the characters.

\subsection{Convolutional Neural Networks}

Early on in the project, we received a recommendation to use Convolutional Neural Networks (CNNs).  CNNs have been shown to be very effective in image recognition \cite{korekado2003convolutional} \cite{ciresan2012multi}.  One significant advantage of CNNs is the ability to make the representation \emph{invariant} to small translations of the input, such as for image recognition tasks \cite{Bengio-et-al-2015-Book}.  Another useful aspect of CNNs are their ability to automate feature identification and extraction \cite{Bengio-et-al-2015-Book}.

\begin{figure}[ht]
  \centering
  \includegraphics[width=\textwidth]{images/mylenet.png}
  \caption{
    Convolution and pooling layers that make up a CNN followed by a
    fully-connected Multi-layered Perceptron (MLP).
    (from \cite{Bengio-et-al-2015-Book})
    }
  \label{fig:convnet}
\end{figure}

CNNs are an example of multi-layered neural networks.  As can be seen in Figure~\ref{fig:convnet}, CNNs have convolution layers and sub-sampling layers, also known as pooling layers.  The convolution layer uses multiple weight matrices to generate other images.  These are then pooled in order to reduce the dimensionality while at the same time introducing non-linearities.  The type of non-linearity depends on the function used in the pooling layer, such as a max function over a $4 \times 4$ region.  We have a loss function at the end of the neural pipeline, and train the previous layers by performing back-propagation.  This back-propagation is simply gradient descent over all of the weight vectors in the whole network (by doing chain rule).

We studied some example code of CNN on \url{deeplearning.net}.   By reading their code, we made two interesting observations.
\begin{description}
  \item[Kernels]
    Multiple weight filters are generated at each convolution layer.  How do they not converge to the same weight vector?  We believe this has to do with the space not being convex and having multiple local minima, because of the introduced non-linearities at each layer.  We initialize the weight vectors with random values and hope that they find different local minima.  We treated this value as a hyper-parameter.  We suspect that having this value too high will both cause a significant increase in running time and cause the resultant classifier to overfit.
  \item[Weight bounds]
    The initialization of the weight bounds seemed to be determined
\end{description}

\subsection{Theano}

We were pointed by Dustin Webb to the Theano python module for implementation of CNNs.  Theano takes mathematical expressions and compiles them into C++ to make for a robust and efficient framework \cite{bergstra+al:2010-scipy}.  Theano was something completely new to us.  It uses symbolic expressions that can be compounded together.  You can define a loss function and have Theano symbolically generate the gradient for use in stochastic gradient descent.

\subsection{Linear Classifiers}

Even though we chose to use CNNs, we decided to first employ the linear classifiers we learned in class for comparison and to learn how to use Theano.  We chose to use Perceptron, Averaged Perceptron, SVM, and Logistic Regression.  These all can be implemented using stochastic gradient descent even though Perceptron is usually not directly implemented in that way.


\section{Introduction}\label{sec:introduction}

Single-particle cryo-electron microscopy (cryo-EM) has revolutionized the field of structural biology over the last decades~\cite{dubochet1988cryo, frank2006three, chap0-nat2015MethodYear}.
The use of electron beams to image ice-embedded samples has permitted the recovery of 3D bio-structures at unprecedented resolution.
This ``resolution revolution'' has had a tremendous impact in biomedical research, providing invaluable insights into the biological processes that underlie many current diseases.

In single-particle cryo-EM, every 3D particle adopts a random orientation $\bth_i$ in the ice layer before being imaged.
Hence, the projection geometry associated to each acquired 2D projection (\figref{imaging-geometry}) is unknown.
Yet, this knowledge is essential for the tomographic reconstruction of bio-structures~\cite{Natterer2001mathematics}.
We consider that a cryo-EM measurement (\ie, a projection) $\p_i \in \R^{n_p}$ is acquired through
\begin{equation}
    \label{eqn:imaging-model}
    \p_i = \mathbf{C}_{\boldsymbol\varphi} \mathbf{S}_{\mathbf{t}_i} \mathbf{P}_{\bth_i} \x + \mathbf{n},
\end{equation}
where $\x \in \R^{n_x}$ is the unknown 3D density map~\cite{dimaio_creating_2007} (Coulomb potential).
%\mdeff{Alternative notation: $\p_i = \mathbf{C}(\boldsymbol\varphi) \mathbf{S}(\mathbf{t}) \mathbf{P}(\bth_i) \x + \mathbf{n}$ if we want to write $\mathbf{R}(\bth_i) = \mathbf{R}(q_i) = \mathbf{R}(-q_i)$ later.}
%\mdeff{Alternative notation: $\x: \R^3 \rightarrow \R$ and $\p_i: \R^2 \rightarrow \R$ as functions, so $\x$ can be rotated by $\mathbf{R} \in \mathbf{SO}(3)$ as $\mathbf{R}\x$. But then $\x$ is not a vector for the SiameseNN.}
The operator $\mathbf{P}_{\bth_i}: \R^{n_x} \to \R^{n_p}$ is the projection along the orientation $\bth_i$ (\ie, the x-ray transform).
% consistency: 3D pose = orientation
The operator $\mathbf{S}_{\mathbf{t}_i}: \R^{n_p} \to \R^{n_p}$ is a shift of the projection by $\mathbf{t}_i = (t_{i_1}, t_{i_2})$.
The convolution operator $\mathbf{C}_{\boldsymbol\varphi}: \R^{n_p} \to \R^{n_p}$ models the microscope point-spread function (PSF) with parameters $\boldsymbol\varphi = (d_1, d_2, \alpha_\mathrm{ast})$, where $d_1$ is the defocus-major, $d_2$ is the defocus-minor, and $\alpha_\mathrm{ast}$ is the angle of astigmatism~\cite{vulovic_image_2013,rullgard_simulation_2011}.
Finally, $\mathbf{n} \in \R^{n_p}$ represents additive noise.
\figref{different-projections} illustrates the effect of projection, shift, and noise.
The challenge is then to reconstruct $\x$ from a set of projections $\{\p_i\}_{i=1}^P$ acquired along unknown orientations.
%\mdeff{We shall decompose $\mathbf{P}_{\bth_i}$ into a rotation $R(\bth_i)$ and an integration.} \lau{The most straightforward way is to simply rewrite $P_theta$ as a rotation operator, followed by a summation operator along lines. That being said, now that I look at it, I would suggest not to rewrite it like this; this is really not so standard in the cryo-EM community. Plus everyone now that the rotation operator is implicitly contained in the projection one, so it's not really needed either.}

% \tdplotsetmaincoords{60}{110}
% \pgfmathsetmacro{\rvec}{.8}
% \pgfmathsetmacro{\thetavec}{30}
% \pgfmathsetmacro{\phivec}{60}
% \begin{figure}
% \centering
% \begin{tikzpicture}[scale=4,tdplot_main_coords]
%     \coordinate (O) at (0,0,0);
%     \draw[thick,->] (0,0,0) -- (1,0,0) node[anchor=north east]{$\bsx_1$};
%     \draw[thick,->] (0,0,0) -- (0,1,0) node[anchor=north west]{$\bsx_2$};
%     \draw[thick,->] (0,0,0) -- (0,0,1) node[anchor=south]{$\bsx_3$};
%     \tdplotsetcoord{P}{\rvec}{\thetavec}{\phivec}
%     \draw[-stealth,very thick,color=red] (O) -- (P);
%     \draw[dashed, color=red] (O) -- (Pxy);
%     \draw[dashed, color=red] (P) -- (Pxy);
%     \tdplotdrawarc{(O)}{0.2}{0}{\phivec}{anchor=north}{$\theta_1$}
%     \tdplotsetthetaplanecoords{\phivec}
%     \tdplotdrawarc[tdplot_rotated_coords]{(0,0,0)}{0.5}{0}%
%         {\thetavec}{anchor=south west}{$\theta_2$}
%     \draw[dashed,tdplot_rotated_coords] (\rvec,0,0) arc (0:90:\rvec);
%     \draw[dashed] (\rvec,0,0) arc (0:90:\rvec);
%     \tdplotsetrotatedcoords{\phivec}{\thetavec}{0}
%     \tdplotsetrotatedcoordsorigin{(P)}
%     \draw[dashed,blue,tdplot_rotated_coords,-] (-.4,0,0)
%         -- (.4,0,0) node[anchor=north west]{};
%     \draw[dashed,blue,tdplot_rotated_coords,-] (0,-.4,0)
%         -- (0,.4,0) node[anchor=west]{};
%     \draw[blue,tdplot_rotated_coords,-]  (-.4,.4,0) -- (.4,.4,0)  -- (.4,-.4,0) -- (-.4,-.4,0) -- (-.4,.4,0)   node[anchor=north]{};
%     \tdplotdrawarc[tdplot_rotated_coords]{(0,0,0)}{0.2}{0}%
%         {30}{anchor=north west,color=black}{$\theta_3$}
%     \tdplotsetrotatedcoords{\phivec}{\thetavec}{30}
%     \draw[thick,tdplot_rotated_coords,->] (0,0,0)
%         -- (.3,0,0) node[anchor=north west]{$\bsy_1$};
%     \draw[thick,tdplot_rotated_coords,->] (0,0,0)
%         -- (0,.3,0) node[anchor=west]{$\bsy_2$};
%     \node[blue] at (0.4,0.45,1.2) {$\Omega_{\mathrm{2D}}$};
%     \node[red] at (1.5,0.75,1.2) {$\bvth_{\bth}$};
%     \tdplotsetrotatedthetaplanecoords{45}
% \end{tikzpicture}
% \caption*{\mdeff{Kept as reference for \figref{imaging-geometry} as this tikz figure is beautiful. Will be commented out in the end.}}
% \end{figure}

\begin{figure}
    \begin{minipage}[t]{0.48\linewidth}
        \centering
        \includegraphics[height=0.7\linewidth]{geomProj3D}
        \caption{%
            % * (i) what we mean by a projection and an orientation (the two most important objects of our paper), and
            % * (ii) how a projection (=integration through z in the new coordinate system) is made from a 3D volume.
            % $\p_i = \mathbf{P} \Rot(\bth) \x$ \todo{($\mathbf{P}$ is a projection/integration)}.
            Geometry of the imaging model defined in \eqnref{imaging-model}.
            The 3D density $\x$ in the coordinate system $(x_1, x_2, x_3)$ is imaged along the \textit{orientation} $\bth$ to produce the 2D \textit{projection} $\p$ in the coordinate system $(y_1, y_2)$ of the microscope's detector plane.
            The orientation $\bth = (\theta_3, \theta_2, \theta_1)$ is decomposed as the direction $(\theta_2, \theta_1) \in [0,\pi] \times [0,2\pi[$ (parameterizing the sphere $\mathbb{S}^2$) and the in-plane rotation $\theta_3 \in [0,2\pi[$ (parameterizing the circle $\mathbb{S}^1$).
            In our work, we represent the orientation $\bth$ as a unit quaternion $q$.
            %The 3D rotation $\Rot(\bth) = \Rot(q) \in \SO(3)$ maps the object coordinate system to the projection coordinate system.
            %\mdeff{$\Rot$, $\bth$, and $q$ all represent orientations. That's the problem with notation that separates the representation not the semantic (though we might need that separation later).}
        }\label{fig:imaging-geometry}
    \end{minipage}
    \hfill
    \begin{minipage}[t]{0.48\linewidth}
        \centering
        \includegraphics[height=0.7\linewidth]{intuition-method}
        \caption{%
            Single-particle cryo-EM produces $P$ projections (with $P$ in the order of $10^5$) from unknown orientations: $\{(\p_i, q_i)\}_{i=1}^P$.
            Observing that distances between orientations constrain the latter, we aim to \textit{recover the orientations} $\{q_i\}$ from $\{d_q(q_i, q_j)\}$, where $d_q(q_i, q_j)$ is the distance (angle) between orientations $q_i$ and $q_j$.
            Observing that the similarity between projections depends on their relative orientation, we aim to \textit{estimate the distance} $d_q(q_i, q_j)$ from the projections $(\p_i, \p_j)$.
            % \todo{The projections don't look to be on the surface of the sphere: They should be on tangent planes.}
            % \todo{The gray line to the center is the projection direction to be labeled $(\theta_1, \theta_2)$ with a color consistent with \figref{imaging-geometry}.}
            % \todo{The two green arcs seems too much like a single one.}
        }\label{fig:intuition-method}
    \end{minipage}
\end{figure}

% RELATED WORKS

A popular approach is to alternatively refine the 3D structure and estimated orientations~\cite{penczek1994ribosome,Baker1996,Dempster1977,sigworth1998maximum,scheres2012bayesian,zehni2020joint}.
Yet, the outcome of these iterative-refinement procedures is often predicated on the quality of the initial reconstruction, or, equivalently, on the initial estimation of the orientations~\cite{sorzano2006optimization,henderson2012outcome}.

Several methods have been designed to produce a first rough \textit{ab initio} structure for the refinement procedure~\cite{singer2020computational}. \textit{Moment-matching} techniques~\cite{kam1980reconstruction, salzman1990method,goncharov1988integral,sharon2019method} reconstruct an initial structure such that the first few moments of the distribution of its theoretical measurements match the ones of its experimental projections; however, they typically remain sensitive to error in data and can require relatively high computational complexity. Based on the central-slice theorem, \textit{common-lines} methods~\cite{van1987angular, penczek1994ribosome,mallick2006structure,singer2010detecting,wang2013orientation,greenberg2017common,pragier2019common} aim at uniquely determining the orientations of each projection by identifying the common-lines between triplets of projections---a real challenge given the massive amount of noise. Alternatively, the marginalized maximum likelihood (ML) formulation of the reconstruction problem~\cite{sigworth1998maximum}---classically used for the iterative-refinement procedures themselves---can be minimized using stochastic gradient descent~\cite{punjani2017cryosparc}.
This permits to avoid the need for an initial volume estimate, at the possible cost of greater convergence instability.

More recently, the recovery of geometrical information from unknown view tomography of 2D point sources has been proposed~\cite{zehni2019distance}, but the extension to 3D cryo-EM tomography is not straightforward.
Finally,~\cite{miolane2019estimation} proposed to recover the in-plane rotations by learning to embed projections in an appropriate latent space, but only after directions had been estimated through three rounds of 2D classification in RELION\@.

%Several methods have been designed to produce a first rough \textit{ab initio} structure for the refinement procedure~\cite{singer2020computational}.
%An early approach~\cite{kam1980reconstruction} proposed to reconstruct an initial structure such that the first few moments of the distribution of its theoretical measurements match the ones of its experimental projections.
%Since then, \textit{moment-matching} techniques have been refined and extended~\cite{salzman1990method,goncharov1988integral,sharon2019method}, \eg, to accommodate for non-uniform orientation configurations.
%However, they typically remain sensitive to error in data and can require relatively high computational complexity.

%Another popular approach relies on the central-slice theorem, which relates the Fourier transform of a projection to a plane (orthogonal to the projection direction) in the Fourier transform of the 3D object~\cite{Natterer2001mathematics}.
%Hence, every two projections \textit{de facto} share a common 1D intersection in the 3D Fourier domain, and three projections theoretically suffice to define a coordinate system from which their orientations can be deduced~\cite{van1987angular}.
%Exploiting this principle, \textit{common-lines} methods aim at uniquely determining the orientations of each projection by identifying the common-lines between triplets of projections~\cite{penczek1994ribosome,mallick2006structure,singer2010detecting,wang2013orientation,greenberg2017common,pragier2019common}---a real challenge given the massive amount of noise.
%If needs be for downside: sensitivity to high-noise levels, small particles, etc.

Despite the aforementioned advances, providing a robust initial volume remains a challenge due to the high-dimensionality and ill-posedness of the underlying optimization problem.
%\mdeff{How is our method different from previous work? What's its singularity? Why do we expect it to solve the aforementioned issues? This isn't a survey: The related work is here to build a stage for our work to stand on.}
On the other hand, the remarkable ability of convolutional neural networks to capture relevant representations of images has had a profound influence in imaging~\cite{lecun2015deep}.
% and our ability to model the imaging / forward model / simulate training data.
%%%
In this work, we present a learning-based approach to recover the unknown orientations directly from the acquired set of projections---without the need for an intermediate reconstruction procedure or an initial volume estimate.
%, hence bypassing some of the aforementioned limitations.
% \mdeff{That's method:} The novelty consists in first estimating the relative distances between pairs of projections (by using a Siamese neural network previously trained on a synthetic dataset), and then recovering the absolute orientation of each projections from these relative distances.
% By doing so, orientations are recovered without the need for an intermediate reconstruction procedure or an initial volume estimate, which has useful implications for SPA\@.
%\mdeff{``bypassing some limitations'' and ``has useful implications'' are too vague IMO\@. Could we be more specific? Not on the how (that's the method) but on the why. Think about an ML audience.}
%\mdeff{Idea: Our method is direct (as opposed to iterative refinement schemes) and does not require any prior model of the reconstructed protein.}
%Our two-step method  learns to estimate the unknown orientation $\bth_i$ (represented by a unit quaternion $q_i$) associated to each projection $\p_i$ in a single-particle cryo-EM dataset without relying on any intermediate reconstruction procedure.

%\mdeff{Content. (p1) Why is the general problem of protein reconstruction important and difficult. (p2) Short background on single-particle cryo-EM and how reconstruction is done. (p3) Previous work on orientation estimation (or initial structure estimation). If there's none (because people have researched other routes), state it and write why it's an interesting route to explore. How does it compare to initial rough structure estimation? Or whatever the other routes are. (p4) Our contribution on top of previous work.}