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\def\myurl{http://www.math.toronto.edu/~drorbn}
\def\thistalk{USC-240205}
\def\title{Shifted Partial Quadratics, their Pushforwards, and Signature Invariants for Tangles}

\def\navigator{{
  \href{\myurl}{Dror Bar-Natan}:
  \href{\myurl/Talks}{Talks}:
  \href{\myurl/Talks/\thistalk/}{\thistalk}:
}}
\def\thanks{{Thanks for allowing me at USC!}}
\def\webdef{{\href{http://drorbn.net/usc24}{http://drorbn.net/usc24}}}
\def\web#1{{\href{\myurl/Talks/\thistalk/#1}{{\greektext web}/#1}}}
\def\titleA{{\title}}
\def\titleB{{\title}}
\def\titleC{{\title}}

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\def\red{\color{red}}
\def\yellowm#1{{\setlength{\fboxsep}{0pt}\colorbox{yellow}{$#1$}}}
\def\myellowm#1{{\setlength{\fboxsep}{0pt}\colorbox{myellow}{$#1$}}}
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\def\cbox#1#2{{\setlength{\fboxsep}{0pt}\colorbox{#1}{#2}}}

\def\arXiv#1{{\href{http://arxiv.org/abs/#1}{{\tiny arXiv:}\linebreak[0]{#1}}}}

\def\qed{{\linebreak[1]\null\hfill\text{$\Box$}}}
\def\qedqed{{\linebreak[1]\null\hfill\text{$\Box\ \Box$}}}

\def\act{{\hspace{-1pt}\sslash\hspace{-0.75pt}}}
\def\ad{\operatorname{ad}}
\def\Ad{\operatorname{Ad}}
\def\aft{$\overrightarrow{\text{4T}}$}
\def\AS{\mathit{AS}}
\def\bbZZ{{\mathbb Z\mathbb Z}}
\def\CW{\text{\it CW}}
\def\diag{\operatorname{diag}}
\def\eps{\epsilon}
\def\FL{\text{\it FL}}
\def\Hom{\operatorname{Hom}}
\def\IHX{\mathit{IHX}}
\def\mor{\operatorname{mor}}
\def\PvT{{\mathit P\!v\!T}}
\def\remove{\!\setminus\!}
\def\STU{\mathit{STU}}
\def\SW{\text{\it SW}}
\def\TC{\mathit{TC}}
\def\tr{\operatorname{tr}}
\def\vT{{\mathit v\!T}}

\def\bara{{\bar a}}
\def\barb{{\bar b}}
\def\barF{{\bar F}}
\def\bart{{\bar t}}
\def\barT{{\bar T}}
\def\barX{{\bar X}}
\def\bbC{{\mathbb C}}
\def\bbE{{\mathbb E}}
\def\bbe{\mathbbm{e}}
\def\bbH{{\mathbb H}}
\def\bbN{{\mathbb N}}
\def\bbO{{\mathbb O}}
\def\bbQ{{\mathbb Q}}
\def\bbR{{\mathbb R}}
\def\bbZ{{\mathbb Z}}
\def\bcA{{\bar{\mathcal A}}}
\def\calA{{\mathcal A}}
\def\calC{{\mathcal C}}
\def\calD{{\mathcal D}}
\def\calF{{\mathcal F}}
\def\calG{{\mathcal G}}
\def\calH{{\mathcal H}}
\def\calI{{\mathcal I}}
\def\calK{{\mathcal K}}
\def\calL{{\mathcal L}}
\def\calM{{\mathcal M}}
\def\calO{{\mathcal O}}
\def\calP{{\mathcal P}}
\def\calR{{\mathcal R}}
\def\calS{{\mathcal S}}
\def\calT{{\mathcal T}}
\def\calU{{\mathcal U}}
\def\fraka{{\mathfrak a}}
\def\frakb{{\mathfrak b}}
\def\frakg{{\mathfrak g}}
\def\frakh{{\mathfrak h}}
\def\tilE{\tilde{E}}
\def\tilq{\tilde{q}}

\def\tDelta{\tilde{\Delta}}
\def\tf{\tilde{f}}
\def\tF{\tilde{F}}
\def\tg{\tilde{g}}
\def\tI{\tilde{I}}
\def\tm{\tilde{m}}
\def\tR{\tilde{R}}
\def\tsigma{\tilde{\sigma}}
\def\tS{\tilde{S}}
\def\tSW{\widetilde{\SW}}

\def\car{\reflectbox{\usym{1F697}}}
\def\rac{\usym{1F697}}

\newcommand{\ann}{\operatorname{ann}}
\newcommand{\gen}{\operatorname{span}}
\newcommand{\im}{\operatorname{im}}
\newcommand{\minuseq}{\mathrel{{-}{=}}}
\newcommand{\pluseq}{\mathrel{{+}{=}}}
\newcommand{\rad}{\operatorname{rad}}
\newcommand{\sign}{\operatorname{sign}}

% From http://tex.stackexchange.com/questions/154672/how-to-get-a-medium-sized-otimes
\DeclareMathOperator*{\midotimes}{\text{\raisebox{0.25ex}{\scalebox{0.8}{$\bigotimes$}}}}

%%%

\def\Abstract{{\raisebox{2mm}{\parbox[t]{3.33in}{
{\red\bf Abstract.} Following a general discussion of the computation of
zombians of unfinished columbaria (with examples), I will tell you about
my recent joint work w/ Jessica Liu on what we feel is the ``textbook''
extension of knot signatures to tangles, which for \text{unknown} reasons,
is not in any of the textbooks that we know.
}}}}

\def\freepik{{Image: \href{http://freepik.com}{Freepik.com}}}

\def\PriorArt{{\raisebox{2mm}{\parbox[t]{3.96in}{
{\red\bf Prior Art} on signatures for tangles / braids.
Gambaudo and Ghys~\cite{GambaudoGhys:BraidsAndSignatures},
Cimasoni and Conway~\cite{CimasoniConway:ColoredTanglesAndSignatures},
Conway~\cite{Conway:SignaturesSurvey},
Merz~\cite{Merz:CimasoniConway}. All define signatures of tangles / braids by first closing them to
links and then work hard to derive composition properties.
}}}}

\def\WhyTangles{{\raisebox{2mm}{\parbox[t]{3.96in}{
{\red\bf Why Tangles?} $\bullet$ Faster!
\newline$\bullet$ Conceptually clearer proofs of invariance
\newline\null\quad (and of skein relations).
\newline$\bullet$ Often fun and consequential:
\newline\null\ $\circ$ The Jones Polynomial $\leadsto$ The Temperley-Lieb Algebra.
\newline\null\ $\circ$ Khovanov Homology $\leadsto$ ``Unfinished complexes'', complexes
\newline\null\qquad in a category.
\newline\null\ $\circ$ The Kontsevich Integral
\newline\null\qquad $\leadsto$ Associators.
\newline\null\ $\circ$ HFK $\leadsto$ OMG, type $D$,
\newline\null\qquad type $A$, $\calA_\infty$, \ldots
}}}}

\def\savings{{$2^{n/2}+2^{n/2}+2^{\sqrt{n}} \ll 2^n$}}

\def\Computing{{\raisebox{2mm}{\parbox[t]{2.87in}{
\parshape 1 0in 4in
{\red\bf Computing Zombians of Unfinished Columbaria.}
\begin{itemize}
\item Must be no slower than for finished ones.
\item Future zombies must be able to complete the computation.
\item Future zombies must not even know the size of the task that today's zombies were facing.
\item We must be able to extend to ZPUCs, Zombie Processed Unfinished Columbaria!
\end{itemize}

\parshape 1 0in 2.87in
{\red\bf Example / Exercise.} Compute the determinant of a $1,000\times 1,000$ matrix in which $50$
entries are not yet given.

\parshape 1 0in 3.96in
{\red\bf Homework / Research Projects.} $\bullet$ What with
ZPUCs? $\bullet$ Use this to get an Alexander tangle invariant.
}}}}

\def\zfreepik{{Zombies: \href{http://freepik.com}{Freepik.com}}}

\def\Reminders{{\raisebox{4mm}{\parbox[t]{3.96in}{
{\red\bf Reminders.} $\{\text{links}\} \rightrightarrows \{\text{matrices / quadratic
forms}\} \xrightarrow[\sigma]{\text{signature}} \bbZ$:
}}}}

\def\ConjThm{{\raisebox{0mm}{\parbox[t]{3.96in}{
\sout{{\bf\red Kashaev's Conjecture}} \cite{Kashaev:SymmetricMatrices}
\par{\bf\red Liu's Theorem} \cite{Liu:Kashaev}.
}}}}
\def\LiuThm{For links, {$\sigma_{\text{\it Kas}}=2\sigma_{\text{\it TL}}$.}}

\def\sss{$\sigma_+ \!-\! \sigma_- \!+\! s$}

\def\X{$X_{-i,j,k,-l}$}
\def\Xb{$\barX_{-i,j,k,-l}$}

\def\TLp{$A\pluseq
  \begin{pmatrix}
	-r	&	-t	& 2t	& \bart	\\
	-\bart	&	0	& \bart	& 0	\\
	2\bart	&	t	& -r	& -\bart\\
	t	&	0	& -t	& 0
  \end{pmatrix}
  \begin{matrix}i\\j\\k\\l\end{matrix}
$}
\def\TLm{$A\pluseq
  \begin{pmatrix}
	r	&	-t	& -2\bart& \bart\\
	-\bart	&	0	& \bart	& 0	\\
	-2t	&	t	& r	& -\bart\\
	t	&	0	& -t	& 0
  \end{pmatrix}
  \begin{matrix}i\\j\\k\\l\end{matrix}
$}

\def\Kasp{$A\pluseq
  \begin{pmatrix}
    v & u & 1 & u \\
    u & 1 & u & 1 \\
    1 & u & v & u \\
    u & 1 & u & 1
  \end{pmatrix}
  \begin{matrix}i\\j\\k\\l\end{matrix}
$}
\def\Kasm{$A\minuseq
  \begin{pmatrix}
    v & u & 1 & u \\
    u & 1 & u & 1 \\
    1 & u & v & u \\
    u & 1 & u & 1
  \end{pmatrix}
  \begin{matrix}i\\j\\k\\l\end{matrix}
$}

\def\sp{$s\minuseq 1$}
\def\sm{$s\pluseq 1$}

\def\wtrvu{{With $|\omega|=1$, $t=1-\omega$, $r=t+\bart$, $v=\operatorname{Re}(\omega)$, and
$u=\operatorname{Re}(\omega^{1/2})$:}}

\def\PQ{{\raisebox{0mm}{\parbox[t]{3.96in}{
A {\red\em Partial Quadratic (PQ)} on $V$ is a quadratic $Q$
defined only on a subspace $\calD_Q\subset V$. We add PQs with
$\calD_{Q_1+Q_2}\coloneqq\calD_{Q_1}\cap\calD_{Q_2}$. Given a linear $\psi\colon V\to W$ and a PQ $Q$
on $W$, there is an obvious {\em pullback} $\psi^*Q$, a PQ on $V$.

{\bf\red Theorem 1.} Given a linear $\phi\colon V\to W$ and a PQ $Q$ on $V$, there is a unique {\em
pushforward} PQ $\phi_*Q$ on $W$ such that for every $PQ$ $U$ on $W$,
$\yellowm{\sigma_V(Q+\phi^*U) = \sigma_{\ker\phi}(Q|_{\ker\phi}) + \sigma_W(U+\phi_*Q)}$.
\par {\small(If you must,
$\calD(\phi_*Q)=\phi(\operatorname{ann}_Q(\calD(Q)\cap\ker\phi))$
and $(\phi_*Q)(w)=Q(v)$, where $v$ is s.t.\ $\phi(v)=w$ and
$Q(v,\operatorname{rad} Q|_{\ker\phi})=0$).}
}}}}

\def\Gist{{\raisebox{2mm}{\parbox[t]{3.96in}{
{\red\bf Gist of the Proof.}
}}}}

\def\BT{$\bar{B}^T$}
\def\CT{$\bar{C}^T$}
\def\sigker{$\sigma(Q|_{\ker\phi})$}

\def\NowD{{\raisebox{-1mm}{\parbox[t]{3.96in}{
\ldots and the quadratic $F\eqqcolon\phi_*Q$ is well-defined only on $D\coloneqq\ker C$.
}}}}

\def\Exactly{{\raisebox{2mm}{\parbox[t]{3.96in}{
{\red\bf Exactly} what we want, if the Zombian is the signature!
\par $V$: The full space of {\em faces}.
\par $W$: The boundary, made of {\em gaps}.
\par $Q$: The known parts.
\par $U$: The part yet unknown.
\par $\sigma_V(Q+\phi^*(U))$: The overall Zombian.
\par $\sigma(Q|_{\ker\phi})$: An internal bit.  $U+\phi_*Q$: A boundary bit.
\par And so our ZPUC is the pair $S=(\sigma(Q|_{\ker\phi}), \phi_*Q)$.
}}}}

\def\SPQ{{\raisebox{0mm}{\parbox[t]{3.96in}{
A {\red\em Shifted Partial Quadratic (SPQ)} on $V$ is a pair $S=(s\in\bbZ, Q\text{ a PQ on }V)$. addition also adds the
shifts, pullbacks keep the shifts, yet $\phi_*S\coloneqq(s+\sigma_{\ker\phi}(Q|_{\ker\phi}),\phi_*Q)$ and
$\sigma(S)\coloneqq s+\sigma(Q)$.

{\bf\red Theorem 1' (\em Reciprocity).} Given $\phi\colon V\to W$, for SPQs $S$ on $V$ and $U$ on $W$ we have
$\yellowm{\sigma_V(S+\phi^*U) = \sigma_W(U+\phi_*S)}$ (and this characterizes $\phi_*S$).
\hfill
{\red Note.} $\psi^*$ is additive but $\phi_*$ is not.

\parshape 1 0in 3.33in
{\red\bf Theorem 2.} $\psi^*$ and $\phi_*$ are functorial.

\parshape 3 0in 3.33in 0in 3.33in 0in 3.96in
{\red\bf Theorem 3.} ``The pullback of a pushforward scene is a pushforward scene'':
If, on the right, $\beta$ and $\delta$ are arbitrary, $Y = \operatorname{EQ}(\beta,\gamma) =
V\oplus_ZW = \{(v,w)\colon\beta v=\gamma w\}$ and $\mu$ and $\nu$ are the obvious projections, 
then $\gamma^*\beta_* = \nu_*\mu^*$.
}}}}

\def\pushpull{$\xymatrix@C=3mm@R=3mm{
  Y \ar[r]^\nu \ar[d]_\mu & W \ar[d]^\gamma \\
  V \ar[r]_\beta \ar@{.>}[ru] & Z
}$}

\def\PlanarAlgebra{{\raisebox{2.5mm}{\parbox[t]{3.96in}{
{\red\bf Definition.} $\calS\begin{pmatrix}\import{../ICERM-2305}{figs/Tgaps.pdf_t}\end{pmatrix} \coloneqq
  \left\{\parbox{0.44in}{
    SPQ $S$ \\ on $\langle g_i\rangle$
  }\right\}$.

\parshape 6 0in 2.4in 0in 2.4in 0in 2.4in 0in 2.4in 0in 2.4in 0in 3.96in 
{\red\bf Theorem 4.} $\{\calS(\text{cyclic sets})\}$ is a planar
algebra, with compositions $\calS(D)((S_i))\coloneqq\phi^D_*(\psi_D^*(\bigoplus_i S_i))$,
where $\psi_D\colon\langle f_i\rangle\to\langle g_{\alpha i}\rangle$
maps every face of $D$ to the sum of the input gaps adjacent to it and
$\phi^D\colon\langle f_i\rangle\to\langle g_i\rangle$ maps every face
to the sum of the output gaps adjacent to it. So for our $D$,
$\psi_D\colon$ {\footnotesize
  $f_1\mapsto g_{34}$, $f_2\mapsto
  g_{31}\!+\!g_{14}\!+\!g_{24}\!+\!g_{33}$, $f_3\mapsto g_{32}$, $f_4\mapsto
  g_{11}$, $f_5\mapsto g_{13}\!+\!g_{21}$, $f_6\mapsto g_{23}$, $f_7\mapsto
  g_{12}\!+\!g_{22}$
} and $\phi^D\colon$ {\footnotesize
  $f_1\mapsto g_1$,
  $f_2\mapsto g_2\!+\!g_6$, $f_3\mapsto 0$, $f_4\mapsto g_3$, $f_5\mapsto 0$,
  $f_6\mapsto g_5$, $f_7\mapsto g_4$
}.

\vskip 1mm
\parshape 4 0in 2.2in 0in 2.2in 0in 2.2in 0in 3.96in
{\red\bf Theorem 5.} {\it TL} and {\it Kas}, defined on $X$ and $\barX$
as before, extend to planar algebra morphisms $\{\text{tangles}\}\to
\{\calS\}$. Restricted to links, $\text{\it TL} = \sigma_{\text{\it TL}}$
and $\text{\it Kas} = \sigma_{\text{\it Kas}}$.
}}}}

\def\f#1{$f_{#1}$}
\def\g#1{$g_{#1}$}

\def\refs{{\raisebox{5mm}{\parbox[t]{3.95in}{
%{\red\bf References.}
{\footnotesize
%\par\vspace{-3mm}
\renewcommand{\section}[2]{}%
\begin{thebibliography}{}
\setlength{\parskip}{0pt}
\setlength{\itemsep}{0pt plus 0.3ex}
\def\nl{} \import{../Geneva-231201}{refs.tex}
\end{thebibliography}}
}}}}

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\def\nbpdfEcho#1{\vskip 1mm\par\noindent\includegraphics{#1}}
\def\nbpdfPrint#1{\vskip 1mm\par\noindent\includegraphics{#1}}
\def\nbpdfText#1{\vskip 1mm\par\noindent\includegraphics{#1}}
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\def\nbpdfOutput#1{\vskip 1mm\par\noindent\includegraphics[max width=\linewidth]{#1}}
\def\nbpdfSubsection#1{\vskip 1mm\par\noindent\includegraphics{#1}}
\def\nbpdfSubsubsection#1{\vskip 1mm\par\noindent\includegraphics{#1}}
\def\nbpdfgraphInput#1{\vskip 1mm\par\noindent\includegraphics{#1}}
\def\nbpdfgraphOutput#1{\vskip 1mm\par\noindent\includegraphics[width=1.5in]{#1}}

\pagestyle{empty}

\begin{document} \latintext
%\setlength{\jot}{0ex}
\setlength{\abovedisplayskip}{0.5ex}
\setlength{\belowdisplayskip}{0.5ex}
\setlength{\abovedisplayshortskip}{0ex}
\setlength{\belowdisplayshortskip}{0ex}

\begin{center}
\null\vfill\input{PQ1.pdftex_t}\vfill\null
\end{center}

\newgeometry{textwidth=8in,textheight=10.5in}

\def\Cordon{{
\includegraphics[width=0.2\linewidth]{figs/Cordon.pdf}
\hfill\includegraphics[width=0.75\linewidth]{Cordon.png}
\[
  \arraycolsep=0pt \renewcommand{\arraystretch}{1}
  \begin{array}{c} s \\ i \\ \null \end{array}
  \left(\begin{array}{c|cc}
    0 & \phi & C_{\text{rest}} \\ \hline
      \bar{\phi}^T & \lambda & \theta \\ \bar{C}_{\text{rest}}^T & \bar{\theta}^T & A_{\text{rest}}
  \end{array}\right)
  \to
  \arraycolsep=0pt \renewcommand{\arraystretch}{0}
  \begin{cases}
    \exists p\,\phi_p\neq 0 & \parbox{1.9in}{use $\phi_p$ to kill its row and \text{column}, drop a
      $\left(\begin{array}{cc}0&1\\[-5pt]1&0\end{array}\right)$ summand} \\
    \phi\!=\!0,\lambda\!\neq\!0 & \parbox{1.9in}{use $\lambda$ to kill $\theta$, let $s\pluseq\operatorname{sign}(\lambda)$} \\
    \phi\!=\!0,\lambda\!=\!0 & \parbox{1.9in}{append $\theta$ to $C_{\text{rest}}$.}
  \end{cases}
\]
}}

\begin{multicols*}{2}
\import{../ICERM-2305}{Signatures.tex}

\vskip -3mm\rule{\linewidth}{1pt}\vspace{0mm}

\vskip 1mm
\resizebox{\linewidth}{!}{
  \parbox{3.5in}{
    $\begin{CD}
      \begin{pmatrix} A & B \\ C & U \end{pmatrix}
      @>\det(A)>>
      \begin{pmatrix} I & A^{-1}B \\ C & U \end{pmatrix}
      @>1>>
    \begin{pmatrix} I & A^{-1}B \\ 0 & U-CA^{-1}B \end{pmatrix}
    \end{CD}$,
    \newline so $\det\begin{pmatrix} A & B \\ C & U \end{pmatrix} = \det(A)\det(U-CA^{-1}B)$.
  }
    \parbox{2.4in}{
    Roughly, $\det(A)$ is ``$\det$ on $\ker$'',
    \newline $-CA^{-1}B$ is ``a pushforward of $\begin{pmatrix} A & B \\ C & U \end{pmatrix}$''.
    \newline (what if $\not\exists A^{-1}$?)
  }
}

\vskip 1mm\rule{\linewidth}{1pt}\vspace{-1mm}

%\setlist[enumerate]{left=0pt .. 12pt}

\needspace{20mm}
{\bf\red Questions.}
%\begin{enumerate}[left=5pt .. 17pt]
\begin{enumerate*}

\item Does this have a topological meaning?

\item Is there a version of the Kashaev Conjecture for tangles?

\item Find all solutions of R123 in our ``algebra''.

\item Braids and the Burau representation.

\item Recover the work in ``Prior Art''.

\item Are there any concordance properties?

\item What is the ``SPQ group''?

\item The jumping points of signatures are the roots of the Alexander polynomial. Does this
generalize to tangles?

\item Which of the three Cordon cases is the most common?

\item Are there interesting examples of tangles for which rels is non-trivial?

\item Is the $pq$ part determined by $\Gamma$-calculus?

\item Is the $pq$ part determined by finite type invariants?

\item Does it work with closed components / links?

\item Strand-doubling formulas?

\item A multivariable version?

\item Mutation invariance?

\item Ribbon knots?

\item Are there ``face-virtual knots''?

\item Does the pushforward story extend to ranks? To formal Gaussian measures? To super Gaussian measures?

\end{enumerate*}

\vskip -3mm\rule{\linewidth}{1pt}\vspace{0mm}

\par{\red\bf References.}
\par\vspace{-3mm}
\renewcommand{\section}[2]{}%
\begin{thebibliography}{}
\setlength{\parskip}{0pt}
\setlength{\itemsep}{0pt plus 0.3ex}
\def\nl{}
\import{../Geneva-231201}{refs.tex}
\end{thebibliography}

\vskip -3mm\rule{\linewidth}{1pt}\vspace{0mm}

{\bf\red Acknowledgement.} This work was partially supported by NSERC
grant RGPIN-2018-04350 and by the Chu Family Foundation (NYC).

\newpage
{\bf\red Some Rigor.}\hfill{\footnotesize (Exercises hints and partial solutions at end)}

\Opensolutionfile{Hints}

\begin{ex} \label{ex:SameD} Show that if two SPQ's $S_1$ and $S_2$ on $V$
satisfy $\sigma(S_1+U) = \sigma(S_2+U)$ for every quadratic $U$ on $V$,
then they have the same shifts and the same domains.
\end{ex}

\begin{Filesave}{Hints}\hint{\ref{ex:SameD}}{
On a vector in the domain of one but not the other, take an outrageous
value for $U$, that will raise or lower the signature.
}\end{Filesave}

\begin{ex} \label{ex:SameQ} Show that if two full quadratics $Q_1$
and $Q_2$ satisfy $\sigma(Q_1+U) = \sigma(Q_2+U)$ for every $U$, then
$Q_1=Q_2$.
\end{ex}

\begin{Filesave}{Hints}\hint{\ref{ex:SameQ}}{
WLOG, $Q_1$ is diagonal and $Q_1=0$.
}\end{Filesave}

{\bf\red Proof of Theorem 1'.} Fix $W$ and consider triples
$(V_1,S_1,\phi_1\colon V\to W)$ where $S_1=(s_1,D_1,Q_1)$ is an
SPQ on $V_1$. Say that two such triples are ``push-equivalent'',
$(V_1,S_1,\phi_1)\sim(V_2,S_2,\phi_2)$ if for every quadratic $U$ on $W$,
\[ \sigma_{V_1}(S_1+\phi_1^*U) = \sigma_{V_2}(S_2+\phi_2^*U). \]

Given our $(V,S,\phi)$, we need to show:
\begin{enumerate}
\item There is an SPQ $S'$ on $W$ such that $(V,S,\phi)\sim(W,S',I)$.
\item If $(W,S',I)\sim(W,S'',I)$ then $S'=S''$.
\end{enumerate}

Property 2 is easy (Exercises~\ref{ex:SameD},~\ref{ex:SameQ}). Property
1 follows from the following three claims, each of which is easy.

{\red Claim 1.} If $v\in\ker\phi\cap D(S)$, and $\lambda\coloneqq Q(v,v)\neq 0$, then $(V,S,\phi)\sim$
\[ \left(
  V/\langle v\rangle,
  \left(
    s \!+\! \sign(\lambda), D(S)/\langle v\rangle, Q \!-\! \lambda^{-1}Q(-,v)\otimes Q(v,-)
  \right),
  \phi/\langle v\rangle
\right). \]
So wlog $Q|_{\ker\phi}=0$ (meaning, $Q|_{\ker\phi\otimes\ker\phi}=0$).\qed

{\red Claim 2.} If $Q|_{\ker\phi}=0$ and $v\in\ker\phi\cap D(S)$, let $V'=\ker Q(v,-)$ and then
$(V,S,\phi)\sim(V',S|_{V'},\phi|_{V'})$ so wlog $Q|_{V\otimes\ker\phi+\ker\phi\otimes V}=0$.\qed

{\red Claim 3.} If $Q|_{V\otimes\ker\phi+\ker\phi\otimes V}=0$ then $S=\phi^*S'$ for some SPQ $S'$ on
$\operatorname{im}\phi$ and then $(V,S,\phi)\sim(W,S',I)$.\qedqed

{\bf\red Proof of Theorem 2.} The functoriality of pullbacks
needs no proof. Now assume $V_0 \overset{\alpha}{\longrightarrow}
V_1 \overset{\beta}{\longrightarrow} V_2$ and that $S$ is an SPQ on
$V_0$. Then for every SPQ $U$ on $V_2$ we have, using reciprocity three
times, that $\sigma(\beta_*\alpha_*S+U) = \sigma(\alpha_*S+\beta^*U)
= \sigma(S+\alpha^*\beta^*U) = \sigma(S+(\beta\alpha)^*U)
= \sigma((\beta\alpha)_*S+U)$. Hence $\beta_*\alpha_*S =
(\beta\alpha)_*S$.\qed

{\bf\red Proof of Theorem 3.}

\newpage

To prove Theorem~4, given three\footnote{Truly, we need the same
for any number of input SPQ's that are divided into two groups, ``multiply
in the first step'' and ``multiply in the second step''.  But there's no
added difficulty here, only an added notational complexity.} SPQ's $S_1$,
$S_2$, and $S_3$, we need to show that planar-multiplying them in two
steps, first using a planar connection diagram $D_I$ ($I$ for Inner) to
yield $S_6=\calS(D_I)(S_2,S_3)$ and then using a second planar connection
diagram $D_O$ ($O$ for Outer) to yield $\calS(D_O)(S_1,S_6)$, gives
the same answer as multiplying them all at once using the composition
planar connection diagram $D_B=D_O\circ_6D_I$ ($B$ for Big) to yield
$\calS(D_B)(S_1,S_2,S_3)$.\footnote{Aren't we sassy? We picked ``6''
for the name of the product of ``2'' and ``3''.} An example should help:

\[ \import{../Geneva-231201}{figs/DB.pdf_t} \]

In this example, if you ignore the dotted green line (marked ``6''), you see
the planar connection diagram $D_B$, which has three inputs (1,2,3)
and a single output, the cycle 0. If you only look inside the green line, you
see $D_I$, with inputs 2 and 3 and an output cycle 6. If you ignore the
inside of 6 you see $D_O$, with inputs 1 and 6 and output cycle 0.

\parpic[r]{$\xymatrix@C=4mm@R=4mm{
  (MD) & & G_0 \\
  & F_B \ar[r]^\nu \ar[d]^\mu \ar[ur]^\phi \ar[dl]_\psi & F_O \ar[u]_\delta \ar[d]^\gamma \\
  G_1\oplus G_2\oplus G_3 & G_1\oplus F_I \ar[l]^<>(0.5){\alpha_+} \ar[r]_{\beta^+} & G_1\oplus G_6
}$}
Let $F_B$ (Big Faces) denote the vector space whose basis are the faces
of $D_B$, let $F_I$ (Inner Faces) be the space of faces of $D_I$, and let $F_O$
(Outer Faces) be the space of faces of $D_O$. Let $G_1$, $G_2$, $G_3$,
$G_6$, and $G_0$ be the spaces of gaps (edges) along the cycles
1,2,3,6, and 0, respectively.  Let $\psi\coloneqq\psi_{D_B}$ and
$\phi\coloneqq\phi^{D_B}$ be the maps defining $\calS(D_B)$ and let
$\gamma\coloneqq\psi_{D_O}$ and $\delta\coloneqq\phi^{D_O}$ be the maps
defining $\calS(D_O)$.  Further, let $\alpha\coloneqq\psi_{D_I}\colon
F_I\to G_2\oplus G_3$ and $\beta\coloneqq\phi^{D_I}\colon F_I\to G_6$ be the
maps defining $\calS(D_I)$, and let $\alpha_+\coloneqq I\oplus\alpha$ and
$\beta^+\coloneqq I\oplus\beta$ be the extensions of $\alpha$ and $\beta$
by an identity on an extra factor of $G_1$, so that $\beta^+_*\alpha_+^* =
I_{G_1}\oplus\calS(D_I)$. Let $\mu$ map any big face to the sum of $G_1$
gaps around it, plus the sum of the inner faces it contains. Let $\nu$
map any big face to the sum of the outer faces it contains. It is easy
to see that the master diagram $(MD)$ shown on the right, made of all
of these spaces and maps, is commutative.

\parpic[r]{$\xymatrix@C=3mm@R=2.5mm{
  F_B \ar[r]^\nu \ar[d]^\mu & F_O \ar[d]^\gamma \\
  G_1\oplus F_I \ar[r]_{\beta^+} & G_1\oplus G_6
}$}
\picskip{3}
{\red Claim.} The bottom right square of $(MD)$ is an equalizer square, namely $F_B\simeq
EQ(\beta^+,\gamma)$. Hence $\nu_*\mu^*=\gamma^*\beta^+_*$.

{\red Proof.} A big face (an element of $F_B$) is a sum of outer faces
$f_o$ and a sum of inner faces $f_i$, and it has a boundary $g_1$ on input
cycle~1, such that the boundary of the outer pieces $f_o$ is equal to the
boundary of the inner pieces $f_i$ plus $g_1$. That matches perfectly with
the definition of the equalizer: $EQ(\beta^+,\gamma) = \{(g_1,f_i,f_o)
\colon \beta^+(g_1,f_i)=\gamma(f_o)\} = \{(g_1,f_i,f_o)\colon\gamma(f_o) =
(g_1,\beta(f_i))\}$.\qed

{\bf\red Proof of Theorem 4.} With notation as above, with the
example above (which is general enough), and with the claim above,
and also using functoriality, we have $\calS(D_B) = \phi_*\psi^* =
\delta_*\nu_*\mu^*\alpha_+^* = \delta_*\gamma^*\beta^+_*\alpha_+^* =
\calS(D_O)\circ(I_{G_1}\oplus\calS(D_I)) $, as required.\qed

{\bf\red Proof of Theorem 5.} We need to verify the Reidemeister moves
and that was done in the computational section, and the statement about
the restriction to links, which is easy: simply assemble an $n$-crossing
knot using an $n$-input planar connection diagram, and the formulas
clearly match.\qed

\vskip 1mm
{\bf\red Further Homework.}\small

\begin{ex} \label{ex:Norad} Show that if two SPQ's $S_1$ and $S_2$
on $V\oplus A$ satisfy $A\subset\rad S_i$ and $\sigma(S_1+\pi^*U) =
\sigma(S_2+\pi^*U)$ for every quadratic $U$ on $V$, where $\pi\colon
V\oplus A\to V$ is the projection, then $S_1=S_2$.
\end{ex}

\begin{ex} \label{ex:SurjectivePullbacks}
Show that if $\phi\colon V\to W$ is surjective and $Q$ is a quadratic on $W$, then
$\sigma(Q)=\sigma(\phi^*Q)$.
\end{ex}

\begin{ex} \label{ex:PushOfPull} Show that always, $\phi_*\phi^*S = S|_{\im\phi}$.
\end{ex}

\begin{Filesave}{Hints}\hint{\ref{ex:PushOfPull}}{
It's enough to test that against $U$ with $\calD(U)=\im\phi$.
}\end{Filesave}

\begin{ex} \label{ex:Uzero} By taking $U=0$ in the reciprocity
statement, prove that always $\sigma(\phi_*S) = \sigma(S)$. But that
seems wrong, if $\phi=0$. What saves the day?
\end{ex}

\begin{Filesave}{Hints}\hint{\ref{ex:Uzero}}{
The ``shift'' part of $0_*S$ is $\sigma(S)$.
}\end{Filesave}

\begin{ex} \label{ex:Szero} By taking $S=0$ in the reciprocity statement, frove that
always $\sigma(\phi^*U)=\sigma(U)$. But wait, this is nonsense! What went wrong?
\end{ex}

\begin{Filesave}{Hints}\hint{\ref{ex:Szero}}{
$\phi_*S$ isn't $0$, it's the {\em partial} quadratic ``$0$ on
$\im\phi$'' (and indeed, $\sigma(\phi^*U)=\sigma(U)$ if $\phi$ is surjective).
}\end{Filesave}

\begin{ex} Given $\phi\colon V\to W$ and a subspace $D\subset V$, show
that there is a unique subspace $\phi_*D\subset W$ such that for every
quadratic $Q$ on $W$, $\sigma(\phi^*Q|_D) = \sigma(Q|_{\phi_*D})$.
\end{ex}

\parpic[r]{\def\side{2.3mm}
  $\xymatrix@C=\side@R=\side{Y \ar[r] \ar[d] & 0 \ar[d] \\ V \ar[r] & Z}$
  \quad
  $\xymatrix@C=\side@R=\side{Y \ar[r] \ar[d] & W \ar[d] \\ V \ar[r] & 0}$
}
\begin{ex} \label{ex:exact} When are diagrams as on the right equalizer
diagrams? What then do \text{we} learn from Theorem~3?
\end{ex}

\begin{ex} \label{ex:Eleven} There are 11 types or irreducible commutative squares:
\def\side{2.3mm}
$\xymatrix@C=\side@R=\side{1 \ar[r] \ar[d] & 0 \ar[d] \\ 0 \ar[r] & 0}$,
$\xymatrix@C=\side@R=\side{0 \ar[r] \ar[d] & 1 \ar[d] \\ 0 \ar[r] & 0}$,
$\xymatrix@C=\side@R=\side{0 \ar[r] \ar[d] & 0 \ar[d] \\ 1 \ar[r] & 0}$,
$\xymatrix@C=\side@R=\side{0 \ar[r] \ar[d] & 0 \ar[d] \\ 0 \ar[r] & 1}$,
$\xymatrix@C=\side@R=\side{1 \ar[r]_1 \ar[d] & 1 \ar[d] \\ 0 \ar[r] & 0}$,
$\xymatrix@C=\side@R=\side{0 \ar[r] \ar[d] & 1 \ar[d]_1 \\ 0 \ar[r] & 1}$,
$\xymatrix@C=\side@R=\side{0 \ar[r] \ar[d] & 1 \ar[d]_1 \\ 0 \ar[r] & 1}$,
$\xymatrix@C=\side@R=\side{0 \ar[r] \ar[d] & 0 \ar[d] \\ 1 \ar[r]^1 & 1}$,
$\xymatrix@C=\side@R=\side{0 \ar[r] \ar[d] & 1 \ar[d]_1 \\ 1 \ar[r]^1 & 1}$,
$\xymatrix@C=\side@R=\side{1 \ar[r]_1 \ar[d]^1 & 1 \ar[d] \\ 1 \ar[r] & 0}$,
and $\xymatrix@C=\side@R=\side{1 \ar[r]_1 \ar[d]^1 & 1 \ar[d]_1 \\ 1 \ar[r]^1 & 1}$.
Show that pushing commutes with pulling for all but four of them. Compare with the statement of
Theorem~3.
\end{ex}

\begin{Filesave}{Hints}\hint{\ref{ex:Eleven}}{
The exceptions are ${01}\atop{00}$, ${00}\atop{10}$, ${01}\atop{11}$, and ${11}\atop{10}$.
}\end{Filesave}

\begin{ex} \label{ex:iff} Prove that a square is admissible iff it is an
equalizer square, with an additional direct summand $A$ added to the $Y$
term, and with the maps $\mu$ and $\nu$ extended by 0 on $A$.
\end{ex}

\begin{ex} \label{ex:sums} Prove that the direct sum of two admissible
squares is admissible. {\em Warning:} Harder than it seems! Not all
quadratics on $V_1\oplus V_2$ are direct sums of quadratics on
$V_1$ and on $V_2$.
\end{ex}

\begin{Filesave}{Hints}\hint{\ref{ex:sums}}{
Use Exercise~\ref{ex:iff}.
}\end{Filesave}

\begin{ex} \label{ex:modrad} Given a quadratic $Q$ on a space $V$,
let $\pi$ be the projection $V\to V/\rad(Q)$ and show that $\pi_*Q =
Q/\rad(Q)$, with the obvious definition for the latter.
\end{ex}

\begin{ex} Show that for any partial quadratic $Q$ on a space $W$ there
exists a space $A$ and a fully-defined quadratic $F$ on $W\oplus A$
such that $\pi_*F=Q$, where $\pi\colon W\oplus A\to W$ is the projection
(these are not unique). Furthermore, if $\phi\colon V\to W$, then $\phi^*Q
= \pi_*\phi_+^*F$, where $\phi_+=\phi\oplus I\colon V\oplus A\to W\oplus
A$ and $\pi$ also denotes the projection $V\oplus A\to V$.
\end{ex}

\Closesolutionfile{Hints}

{\bf\red Solutions / Hints.}
\def\hint#1#2{{\par\scalebox{1}[1]{\parbox{\linewidth}{
  \vskip 1mm{\red Hint~for~#1.} #2
}}}}
\input Hints.tex

\end{multicols*}

\end{document}

\endinput
