text:
\section{Symmetric self-adjoint Hopf categories}
\label{sec:SSHDef}
\subsection{The notion of positive self-adjoint Hopf algebra and its categorification}
In the book \cite{Zelbook} Zelevinsly defines positive self-adjoint Hopf (PSH) algebra as follows:
\begin{Definition}
\label{def:PSH}
A positive self-adjoint Hopf (PSH) algebra is a graded connected Hopf algebra over $\ZZ$ with an inner product and a distinguished finite orthogonal $\ZZ$ basis in each grade s.t. multiplication and comultiplication are adjoint and take elements with positive coefficients in the basis to elements with positive coefficients in the basis (that is to say, they are \emph{positive} maps).
\end{Definition}
Let us analyze this definition and restate it in a form which can be adapted to the category setting.

Given a PSH algebra $A$ we have an adjoint pair of multiplication and comultiplication maps which we will denote $m$ and $\Delta$:
\begin{align*}
A\otimes A \xrightarrow{m} A & {} & A\xrightarrow{\Delta}A\otimes A
\end{align*}
The Hopf axiom is the requirement that \[
\forall x,y\in A: \Delta(xy)=\Delta(x)\Delta(y)
\]
This can be restated as the commutativity of the following square:
\[
\stik{1}{
\AAS{4} \arrow[]{r}{\overline m} \arrow[leftarrow]{d}[left]{\Delta^{\otimes 2}}\& \AAS{2} \arrow[leftarrow]{d}{\Delta}\\
\AAS{2} \arrow[]{r}{m} \& A}
\]
where
\begin{align*}
\overline{m}(x\otimes y\otimes z \otimes w)&=m(x\otimes z)\otimes m(y \otimes w)\\
\Delta^{\otimes 2}(x \otimes y)&=\Delta(x)\otimes \Delta(y)
\end{align*}
Using the adjointness of $m$ and $\Delta$ this square can be obtained from the commutative square of multiplications
\begin{equation}
\label{squareHopf}
\stik{1}{} \stik{1}{
\AAS{4} \arrow[]{r}{\overline m} \arrow[]{d}[left]{m^{\otimes 2}}\& \AAS{2} \arrow[]{d}{m}\\
\AAS{2} \arrow[]{r}{m} \& A
}
\end{equation}
by replacing the verticals with their adjoints. This operation is called "taking the mate" of the square. We will say that a square satisfies the (1-categorical) \emph{Beck-Chevalley condition} if its mate commutes.

Observe that the square \eqref{squareHopf} corresponds to the Cartesian square of finite sets
\begin{align*}
\stik{1}{
\AAS{4} \arrow[]{r}{\overline m} \arrow[]{d}[left]{m^{\otimes 2}} \& \AAS{2} \arrow[]{d}{m}\\
\AAS{2} \arrow[]{r}{m} \& A
}
& {} & \leftarrow & {} &
\stik{1}{
\fset{4} \arrow[]{r}{} \arrow[]{d}[left]{} \& \fset{2} \arrow[]{d}{} \ar[draw=none]{dl}[above,yshift=-0.5pc,xshift=-0.8pc,scale=2]{\ulcorner}\\
\fset{2} \arrow[]{r}{} \& \fset{1}
}
\end{align*}
where we denote the finite set $\{1,2,\ldots,n\}$ by $\fset{n}$.
More generally, given the multiplication map $m:A\otimes A \rightarrow A$, we have for every map of finite sets $a:S\rightarrow T$ an obvious extension $m_{a}:\AAS{S} \rightarrow \AAS{T}$.
The Hopf axiom implies that all squares formed using the above extension that correspond to Cartesian squares of finite sets, satisfy the Beck-Chevalley condition.

These observations allow us to reformulate the \autoref{def:PSH} as follows: Denote by $\ZAd$ the category of graded free $\ZZ$ modules with chosen finite basis in each grade and positive graded maps which admit positive graded adjoints (positivity and adjointness is considered with respect to the inner product induced by the chosen basis). Then the following holds:
\begin{Proposition}
\label{th:PSHfunctor}
A PSH algebra is the same as a functor \[
\mathcal{A}:\FinSet\xrightarrow{}\ZAd
\] which takes disjoint union to tensor product (i.e. a symmetric monoidal functor), sends morphisms in $\FinSet$ to degree zero morphisms and sends Cartesian squares to squares satisfying the Beck-Chevalley condition.
\end{Proposition}

The proof of \autoref{th:PSHfunctor} is straightforward in the view of the discussion above. Note in addition that the unit and counit maps are given by the image of the map $\emptyset \rightarrow \fset{1}$ and the connectedness follows from the fact that the image of the Cartesian square
\[
\stik{1}{
\emptyset \arrow[equal]{d} \arrow[equal]{r} \& \emptyset \ar{d}\\
\emptyset \ar{r} \& \fset{1}
}
\]
satisfies the Beck-Chevalley condition.

Given in this form, the definition of a PSH algebra can be categorified to define a class of objects which we call the \emph{symmetric self-adjoint Hopf categories}. In place of the category $\ZAd$ we will consider the bicategory $\CatAd$.

\begin{Definition} The objects of $\CatAd$ are non-negatively indexed sequences of 2-vector spaces, where a 2-vector space, as defined in \cite{2Vect}, is a semisimple $\kk$-category equivalent to a finite sum of copies of $\Vect$. We think of objects of $\CatAd$ as graded 2-vector spaces. The 1-morphisms are finite sums of exact morphisms of bounded degree and the 2-morphism are the natural transformations between those. \end{Definition}

\begin{Remark}
Note that 1-morphisms in $\CatAd$ admit left and right adjoints.
\end{Remark}

The natural analog  of the tensor of $\ZZ$-modules in $\CatAd$ is the Deligne tensor of categories from \cite{deligne}. To categorify the definition of PSH algebra we would like to replace the functor $\mathcal{A}:\FinSet\xrightarrow{}\ZAd$ with a pseudo functor of bicategories $\Hop:\FinSet \rightarrow \CatAd$ that takes the disjoint union of finite sets into the Deligne tensor of categories and sends Cartesian squares of finite sets to squares satisfying the Beck-Chevalley condition (see \autoref{sec:BC} for the general definition). Principally speaking, this definition replaces the commutative squares corresponding to the Hopf relation on the level of algebras with squares that commute up to isomorphisms forming a coherent system encoded in the functor from the category of finite sets.

In the next section we will formulate the requirement that $\Hop$ preserves tensor product in terms of it being a coCartesian functor of coCartesian fibrations of bicategories in the sense of \cite{Buckley}. This approach is motivated by the approach by J.Lurie in \cite{luriealgebra} to symmetric monoidal $(\infty,1)$-categories. Thinking about $\Hop$ as functor of fibrations leads to construction of morphisms categorifying the Heiseberg double relations in \autoref{sec:heisenbergaction}.

\subsection{CoCartesian fibrations over the category of pointed finite sets}
Denote by $\FinSet_*$ the category of pointed finite sets. Since many objects of interest in this article are fibrations over $\FinSet_*$, we recall here some basic definitions relating to it.

Let $I$ be a pointed set. We denote $\nobase{I}$ the complement of the basepoint. A map $I\xrightarrow{\varphi} J$ in $\FinSet_*$ can be thought of as a partially defined map $\nobase{I}\rightarrow\nobase{J}$. i.e. a map $\nobase{I}\supseteq T\rightarrow \nobase{J}$.

\begin{Definition}
A map $I\rightarrow J$ is called \emph{active} if the preimage of the basepoint is the basepoint. In other words, a map is active if it is just a regular map of sets $\nobase{I}\rightarrow\nobase{J}$.
\end{Definition}

Denote by $(\FinSet_*)_{ac}$ the subcategory spanned by the active maps. There is an obvious equivalence $\FinSet\cong (\FinSet_*)_{ac}$. It is easy to see that calculating pullbacks in $(\FinSet_*)_{ac}$ is the same as calculating them in $\FinSet$. Note however that the imbedding $\FinSet\hookrightarrow \FinSet_*$ does \emph{not} preserve all limits. In particular, the final object of $\FinSet$ is not a final object in $\FinSet_*$.

\begin{Definition}[cf \cite{Buckley}]
Let $\CCC$ be a bicategory, $B$ a 1-category, and $p:\CCC\rightarrow B$ a functor.
\begin{enumerate}
\item Let $f:x\rightarrow y$ be an arrow in $B$ and $F:X\rightarrow Y$ an arrow in $\CCC$ with $p(F)=f$. Suppose that for any $Z\in\CCC$ we have that the square
\[
\stik{1}{
\CCC(Y,Z) \ar{r}{-\circ F} \ar{d} \& \CCC(X,Z) \ar{d} \\
B(p(Y),p(Z)) \ar{r}{-\circ f} \& B(p(X),p(Z))
}
\]
is a pseudo pullback of categories, then we say that $F$ is a coCartesian lift of $f$ starting at $X$.
\item We say that $p:\CCC\rightarrow B$ is a \emph{coCartesian fibration} if for any $f:x\rightarrow y$ and any $X$ such that $p(X)=x$ we have a coCartesian lift of $f$ starting at $X$.
\end{enumerate}
\end{Definition}

\begin{Remark}
In \cite{Buckley} there are further requirements on 2-cells which become trivial when the base is a 1-category. Also, the definition there is the dual one, giving Cartesian instead of coCartesian fibrations.
\end{Remark}
when working with a coCartesian fibration, we have the following slightly simpler description for coCartesian lifts following \cite{lurietopos}  Proposition 2.4.2.8:

\begin{Lemma} \label{lem:coCartInitial} Suppose that $p:\CCC\rightarrow B$ is a coCartesian fibration, then an arrow $F:X\rightarrow Y$ is a coCartesian lift over $f:x\rightarrow y$ iff it is initial in the category of arrows over $f$ with source $X$, i.e. that we can assume in the definition that $p(Z)\cong p(Y)$.
\end{Lemma}

\begin{proof}
The relevant part of the proof of \cite{lurietopos} Proposition 2.4.2.8 carries over word for word, using that coCartesian fibrations of bicategories are closed on iso-comma pullback, which is proven in \cite{Buckley} Proposition 4.3.7.
\end{proof}

The three coCartesian fibrations of interest to us are:

\begin{Definition}[$\FinDisj\xrightarrow{p}\FinSet_*$] $ $\\
\label{def:finsettensor}
\begin{itemize}
\item An object of $\FinDisj$ over $I\in\FinSet_*$ is a collection $(S_i)_{i\in I}$ of finite sets, with $\base{S}$ being a terminal object in $\FinSet$, i.e. a one point set.
\item A map $(S_i)\rightarrow(T_j)$ is a map $I\xrightarrow{\varphi}J$ in $\FinSet_*$ and a collection of functions of finite sets $S_i\rightarrow T_{\varphi(i)}$ for all $i\in I$.
\end{itemize}
\end{Definition}

\begin{Remark}
\label{rem:findisjalt}
Another way to see $\FinDisj$ is as the subcategory of arrows in $\FinSet_*$ spanned by the active maps, with the projection given by taking the target.
\end{Remark}

\begin{Definition}[$\CatAdTen\xrightarrow{p}\FinSet_*$] $ $\\
\label{SquaresInCatAd}
\label{def:CatAdTen}
\begin{itemize}
\item An object of $\CatAdTen$ over $I\in\FinSet_*$ is a collection $(\CCC_i)_{i\in I}$ of elements of $\CatAd$, with $\base{\CCC}$ being the final object in $\CatAd$, i.e. the zero category.
\item A map $(\CCC_i)\rightarrow(\DDD_j)$ over a map $I\xrightarrow{\varphi}J$ in $\FinSet_*$ is a collection of multi-exact \emph{bounded} functors ( i.e. sums of multi-exact functors of non-negative bounded degree).
\[
f_j:\prod_{\varphi(i)=j}\CCC_i\rightarrow \DDD_j
\]
\begin{Note}If $\varphi^{-1}(j)=\emptyset$, then we should give a map from the empty product, which is defined as $\Vect$, to $\DDD_j$. This is the same as specifying an object in $\DDD_j$.
\end{Note}
\item a 2-cell in $\CatAdTen$ between two maps $F,G:(\CCC_i)\rightarrow (\DDD_j)$ over $\varphi:I\rightarrow J$ is a collection of 2-morphisms in $\CatAd$ between the functors defining $F,G$, i.e. natural transformations $\varphi_j:F_j\rightarrow G_j$.
\begin{Note}
When $\varphi^{-1}(j)=\emptyset$, then giving $\alpha_j:F_j\rightarrow G_j$ amounts to specifying a morphism between the specified objects.
\end{Note}
\end{itemize}
\end{Definition}

\begin{Definition}[$\ZAdTen\xrightarrow{p}\FinSet_*$] $ $\\
\label{def:ZAdTen}
\begin{itemize}
\item An object of $\ZAdTen$ over $I\in\FinSet_*$ is a collection $(V_i)_{i\in I}$ of free graded $\ZZ$ modules with a chosen finite basis in each degree, with $\base{V}$ being the 0 module.
\item A map $(V_i)\rightarrow(U_j)$ over a map $I\xrightarrow{\varphi}J$ in $\FinSet_*$ is a collection of multilinear \emph{bounded} (as above) maps
\[
f_j:\prod_{\varphi(i)=j}V_i\rightarrow U_j
\]
\begin{Note}If $\varphi^{-1}(j)=\emptyset$, this means specifying a map from $\ZZ$ to $U_j$, or just an element in $U_j$.
\end{Note}
\end{itemize}
\end{Definition}

\subsubsection{coCartesian lifts}
For simplicity let us consider what are coCartesian lifts over the unique active map $I\rightarrow\pset{1}$.

Using \autoref{lem:coCartInitial}, in $\FinDisj$ this is a collection of maps $S_i\rightarrow T$ which is initial among collections of maps $S_i\rightarrow T'$, i.e. a presentation of $T$ as the disjoint union (or coproduct) of the $S_i$. In the point of view of \autoref{rem:findisjalt} this is just a map $S\rightarrow T$ initial among all maps from $S$, so just an isomorphism. Clearly a lift exists for any map in $\FinSet_*$.

In $\CatAdTen$ this is a multi-exact graded functor $\prod\CCC_i\rightarrow \DDD$ with the property that
\[
\stik{1}{
\CatAdTen(D,E) \ar{r} \ar{d} \& \CatAdTen((C_i),E) \ar{d} \\
\FinSet_*(\pset{1},\pset{1}) \ar{r} \& \FinSet_*(I,\pset{1})
}
\]
is a pseudo pullback, which in this case just means the upper arrow is an equivalence.

This is precisely what it means to present $\DDD$ as the (graded) Deligne tensor of the $\{\CCC_i\}$ (see \cite{deligne}). From \cite{lopez} it follows that in $\CatAd$ the Deligne tensor always exists and so $\CatAdTen$ is a coCartesian fibration. Note also that in $\ZAdTen$ this is a presentation of a module as a tensor of several modules.

In all of our fibrations all coCartesian lifts can be constructed from the ones above. The only non-obvious case is the lift of the map $\pset{0}\rightarrow\pset{1}$ starting at the empty list. In $\FinDisj$ this is the map $()\rightarrow(\emptyset)$. In $\CatAdTen$ by definition a map from the empty list to a category is a map from $\Vect$ to that category, and so the lift is $()\rightarrow(\Vect)$ given by the identity map of $\Vect$. Similarly, note that in $\ZAdTen$ this is $()\rightarrow(\ZZ)$ given by the identity of $\ZZ$.

We will sometimes write $\boxtimes$ to denote coCartesian arrows in $\CatAdTen$ or $\ZAdTen$, and $+$ to denote coCartesian arrows in $\FinDisj$.

\subsection{Hopf categories}

We can now formulate our main definition

\begin{Definition}
\label{maindef}
A \emph{symmetric self-adjoint Hopf }(SSH) category is a morphism of fibrations $\FinDisj\xrightarrow{\Hop}\CatAdTen$, which
\begin{enumerate}
\item Takes maps in $\FinDisj$ to degree zero maps in $\CatAdTen$.
\item Is coCartesian, i.e. takes coCartesian arrows to coCartesian arrows.
\item Takes Cartesian squares in the fiber over $\pset{1}$ to squares satisfying the Beck-Chevalley condition.
\end{enumerate}
\end{Definition}

We will show in \autoref{sec:2Cartesian} that the Beck-Chevalley squares in $\CatAd$ can be viewed as a certain 2-categorical generalization of Cartesian squares that we call \comma squares. With this generalization in mind we can think of the functor $\Hop$ as preserving \comma squares and we will refer to this property as "the continuity" of $\Hop$.

We will say that $\Hop$ gives a structure of SSH category on $\Hop(\fset{1})$ and that $\Hop(\fset{1})$ is an SSH category.

Explicitly, Specifying the SSH category structure amounts to the following data:
\begin{itemize}
\item For every finite set $S$, a category $\Hop(S)$.
\item For every finite collection of maps of finite sets $(S_i\xrightarrow{\varphi_i}T)_{i\in I}$, a functor $\Hop(\varphi_i):\prod_i \Hop(S_i)\rightarrow \Hop(T)$.
\item For every commutative square
\[
\stik{1}{
(S_i)_{i\in I} \arrow{d}\arrow{r} \& (Q_k)_{k\in K} \ar{d}\\
(T_j)_{j\in J} \arrow{r} \& U
}
\]
(where commutativity is understood separately for each $i$), a 2-commutative square
\[
\stik{1}{
\Hop((S_i)_{i\in I}) \arrow{d}\arrow{r} \& \Hop((Q_k)_{k\in K}) \ar{d} \ar[Rightarrow,shorten <=0.7em,shorten >=0.7em]{dl}[above,sloped]{\sim}\\
\Hop((T_j)_{j\in J}) \arrow{r} \& \Hop(U)
}
\]

\end{itemize}

And it has to satify the following properties: \begin{itemize}
\item $3$-cells in $\FinDisj$ map to $3$-cells in $\CatAdTen$, e.g. commutative cubes go to commutative cubes.
\item coCartesian arrows map to coCartesian arrows.
\item Cartesian squares of sets in the fiber over $\pset{1}$ map to squares of categories satisfying the Beck-Chevalley condition.
\end{itemize}
\begin{Notation}
\label{fiberover1}
The restriction of the functor $\Hop$ to the fiber over $\pset{1}$ gives a functor that we will denote $\Hop^1:\FinSet \rightarrow \CatAd$ which takes Cartesian squares to Beck-Chevalley squares. One can think of the images of the maps in $\FinSet$ under $\Hop^1$ and their adjoints as the categorical analogs of the maps of multiplication and comultiplication in the Hopf algebra (for any number of variables). Hence we will denote the image of the arrows $S \xrightarrow{a} T$ by $m_a$ and the image of the arrow $\fset{2}\rightarrow \fset{1}$ will be denoted just by $m$.

Let $\CCC=\Hop(\fset{1})$ (in other words $\Hop$ gives an SSH structure on the category $\CCC$). Then elsewhere in the article we will denote the image of a finite set $U$ under $\Hop$ by $\CS{U}$. This notation is motivated by the fact that, as we noted earlier, the image of $U$ should satisfy the universal property of being a Deligne tensor.
\end{Notation}

\subsection{Basic properties}
\label{ssec:SSHbasicproperties}

Throughout, let $\Hop$ be a SSH structure on $\CCC$.

\begin{Proposition}
\label{frenkelcranehopfcat}
$\Hop$ induces on $\CCC$ the structure of a Hopf category in the sense of Frenkel and Crane from \cite{FrenkelCrane} (we recall the definition in the proof).
\end{Proposition}

\begin{proof}
The data of a Hopf category is the following:
\begin{itemize}
\item Functors $m:\CCC\otimes\CCC\rightarrow \CCC$ and $\Delta:\CCC\rightarrow\CCC\otimes\CCC$ which are (co)associative (i.e. monoidal and comonoidal structures on $\CCC$).
\item A 2-isomorphism
\[
\widestik{1}{
\CCC^{\otimes\fset{4}} \arrow[shorten >=0.4cm,shorten <=0.4cm,Rightarrow]{dr}[above,sloped]{\sim} \ar{r}{m_{13}\otimes m_{24}} \ar[leftarrow]{d}[left]{\Delta\otimes \Delta}  \& \CCC^{\otimes\fset{2}} \ar[leftarrow]{d}{\Delta} \\
\CCC^{\otimes\fset{2}} \ar{r}[below]{m} \& \CCC
}
\]

satisfying the relations imposed by requiring that the following cubes commute (abbreviating $n$ for $\CCC^{\otimes\fset{n}}$ and e.g $1,(23)$ for $\Id_1\otimes m_{23}$ and $\overline{1,(23)}$ for $\Id_1\otimes \Delta_{23}$):
\begin{align*}
\widestik{0.8}{
{} \& 3 \ar[leftarrow]{rr}{\overline{(12),3}} \ar[leftarrow]{dd}[sloped,xshift=0.5pc,yshift=0.5pc]{\overline{1,(23)}} \& {} \& 2 \ar[leftarrow]{dd}{\Delta}\\
6 \ar[crossing over,leftarrow]{rr}{\overline{(13),5,(24),6}} \ar[leftarrow]{dd}[left,sloped,yshift=-0.5pc,xshift=2pc]{\overline{1,2,(35),(46)}} \ar{ur}[sloped,xshift=2pc]{(12),(34),(56)} \& {} \& 4 \ar[leftarrow]{dd}[sloped,yshift=0.7pc,xshift=-1.5pc]{\overline{(12),(34)}} \ar{ur}[sloped,xshift=1pc]{(13),(24)} \& {}\\
{} \& 2 \ar[leftarrow]{rr}[xshift=-1pc]{\Delta} \& {} \& 1\\
4 \ar[leftarrow]{rr}{\overline{(13),(24)}} \ar{ur}[sloped,xshift=1pc]{(12),(34)} \& {} \& 2 \ar{ur}[below,sloped]{m}
\latearrow{commutative diagrams/crossing over}{4-3}{2-3}{}
} & {} & \widestik{0.8}{
{} \& 3 \ar{rr}{(12),3} \ar{dd}[sloped,xshift=0.5pc,yshift=0.5pc]{1,(23)} \& {} \& 2 \ar{dd}{m}\\
6 \ar[crossing over]{rr}{(13),5,(24),6} \ar{dd}[left,sloped,yshift=-0.5pc,xshift=2pc]{1,2,(35),(46)} \ar[leftarrow]{ur}[sloped,xshift=2pc]{\overline{(12),(34),(56)}} \& {} \& 4 \ar{dd}[sloped,yshift=0.7pc,xshift=-1.5pc]{(12),(34)} \ar[leftarrow]{ur}[sloped,xshift=1pc]{\overline{(13),(24)}} \& {}\\
{} \& 2 \ar{rr}[xshift=-1pc]{m} \& {} \& 1\\
4 \ar{rr}{(13),(24)} \ar[leftarrow]{ur}[sloped,xshift=1pc]{\overline{(12),(34)}} \& {} \& 2 \ar[leftarrow]{ur}[below,sloped]{\Delta}
\latearrow{commutative diagrams/crossing over}{2-3}{4-3}{}
}
\end{align*}
\end{itemize}

The 2-isomorphism is obtained as the left mate of $\Hop$ of the cartesian square of finite sets
\[
\stik{1}{
\fset{4} \ar{r}{} \ar{d}{} \& \fset{2} \ar{d}{}\\
\fset{2} \ar{r}{} \& \fset{1}
}
\]
The relation cubes commute since they are both mates of $\Hop$ of the cube of sets (see \autoref{sec:BC} for details regarding mates of cubes and squares)
\[
\widestik{1}{
{} \& \fset{3} \ar{rr}{(12),3} \ar{dd}[sloped,xshift=0.5pc,yshift=0.5pc]{1,(23)} \& {} \& \fset{2} \ar{dd}{}\\
\fset{6} \ar[crossing over]{rr}{(13),5,(24),6} \ar{dd}[left,sloped,yshift=-0.5pc,xshift=2pc]{1,2,(35),(46)} \ar{ur}[sloped,xshift=2pc]{(12),(34),(56)} \& {} \& \fset{4} \ar{dd}[sloped,yshift=0.7pc,xshift=-1.5pc]{(12),(34)} \ar{ur}[sloped,xshift=1pc]{(13),(24)} \& {}\\
{} \& \fset{2} \ar{rr}[xshift=-1pc]{} \& {} \& \fset{1}\\
\fset{4} \ar{rr}{(13),(24)} \ar{ur}[sloped,xshift=1pc]{(12),(34)} \& {} \& \fset{2} \ar{ur}[below,sloped]{}
\latearrow{commutative diagrams/crossing over}{2-3}{4-3}{}
}
\]
where e.g. by $1,2,(35),(46)$ we mean the map $1\mapsto 1;2\mapsto 2;3,5\mapsto 3;4,6\mapsto 4$.

The left one is obtained by taking the left mate twice, and the right one by taking the right mate. All faces end up invertible because they are all either mates of $\Hop$ of cartesian squares or $\Hop$ of squares, or double mates of $\Hop$ of squares.
\end{proof}

\begin{Remark}
The cube in the above proof can be recovered as follows:
\begin{enumerate}
\item Consider the associator square:
\[
\stik{1}{
\fset{3} \ar{d}[left,sloped,yshift=-0.5pc,xshift=1pc]{1,(23)} \ar{r}{(12),3} \& \fset{2} \ar{d} \\
\fset{2} \ar{r} \& \fset{1}
}
\]
\item Add the map $\fset{2}\rightarrow\fset{1}$ in a transversal direction:
\[
\stik{1}{
{} \& \fset{3} \ar{rr}{(12),3} \ar{dd}[sloped,xshift=-1.5pc,yshift=0.5pc]{1,(23)} \& {} \& \fset{2} \ar{dd}{}\\
{} \& {} \& {} \& {}\\
{} \& \fset{2} \ar{rr}[xshift=-1pc]{} \& {} \& \fset{1}\\
{} \& {} \& \fset{2} \ar{ur}[below,sloped]{}
}
\]
\item Form the pullbacks on the right and bottom:
\[
\widestik{1}{
{} \& \fset{3} \ar{rr}{(12),3} \ar{dd}[sloped,xshift=-1.5pc,yshift=0.5pc]{1,(23)} \& {} \& \fset{2} \ar{dd}{}\\
{} \& {} \& \fset{4} \ar{dd}[sloped,yshift=0.7pc,xshift=-1.5pc]{(12),(34)} \ar{ur}[sloped,xshift=1pc]{(13),(24)} \& {}\\
{} \& \fset{2} \ar{rr}[xshift=-1pc]{} \& {} \& \fset{1}\\
\fset{4} \ar{rr}{(13),(24)} \ar{ur}[sloped,xshift=1pc]{(12),(34)} \& {} \& \fset{2} \ar{ur}[below,sloped]{}
\latearrow{commutative diagrams/crossing over}{2-3}{4-3}{}
}
\]
\item Form the pullback cube.
\end{enumerate}
\end{Remark}

\begin{Proposition}
$\Hop(\emptyset)$ is canonically equivalent to $\Vect$.
\end{Proposition}

\begin{proof}
Let $\alpha:\pset{0}\rightarrow\pset{1}$ be the unique map. In $\CatAdTen$, over $\pset{0}$, there is only the list $(\base{C})$ consisting of the trivial category. Therefore, a map in $\CatAdTen$ over $\alpha$ is a map $(\base{C})\rightarrow (\base{C},D)$, which amounts to a functor from the empty product to $D$, i.e. the specification of an object in $D$.

We have:
\begin{enumerate}
\item The arrow $(\base{S})\rightarrow (\base{S},\emptyset)$ in $\FinDisj$ is a coCartesian arrow over $\alpha$.
\item The arrow $(\base{C})\rightarrow (\base{C},\Vect)$ in $\CatAdTen$ which sends the empty product to  $\kk$, is a coCartesian arrow over $\alpha$.
\end{enumerate}

The first is obvious, and second is just saying that a choice of an element in $D$ is the same as a functor $\Vect\rightarrow D$.

As a result, the first arrow must go to an arrow which is canonically equivalent to the second arrow and in particular $\Hop(\emptyset)$ is canonically equivalent to $\Vect$.
\end{proof}

\begin{Proposition}
$\Hop$ defines a symmetric monoidal structure on $\CCC$.
\label{prop:symmonstruct}
\end{Proposition}

\begin{proof}
Consider the section $s$ of $\FinDisj\rightarrow\FinSet_*$ given by $\pset{n}\mapsto(\fset{1},\ldots,\fset{1})$ which sends a map $\alpha:\pset{m}\rightarrow\pset{n}$ to the appropriate collection of identity maps. Composing this section with the map $\FinDisj\rightarrow\CatAdTen$ we get a section $\bar{s}$ of $\CatAdTen\rightarrow\FinSet_*$, which gives a symmetric monoidal structure on $\bar{s}(\fset{1})=H(\fset{1})=\CCC$.
\end{proof}

Explicitly, this monoidal structure is given by applying $\Hop$ to the map $(\fset{1},\fset{1})\rightarrow (\fset{1})$ given by the two copies of $\Id: \fset{1} \rightarrow \fset{1}$ which lies over the active map $\pset{2}\rightarrow\pset{1}$.

Let us denote this map by $(F,G)\mapsto F\otimes G$, where $F,G$ are objects in $\CCC$
\begin{Remark}
The map $(F,G)\mapsto F\otimes G$ factors essentially uniquely as $\Hop$ applied to the composition
\[
\nnstik{1}{
\fset{1} \ar{dr} \\
\& \fset{2} \ar{r} \& \fset{1} \\
\fset{1} \ar{ur}
}
\]
The factorization is given on $F,G \in \CCC$ as $(F,G)\mapsto F\boxtimes G \mapsto m(F\boxtimes G)$ i.e. a coCartesian arrow followed by an arrow over $\Id_{\pset{1}}$.
\end{Remark}

\begin{Corollary}
\label{SSH-kgroup}
The functor $\Hop$ canonically defines the structure of a PSH structure on the Grothendieck group $K(\CCC)$.
\end{Corollary}

\begin{proof}
Let $A:=K(\CCC)$.

Denote by $(\CatAdTen)^0$ the same bicategory where all non-invertible 2-morphisms have been discarded.

Then we have a functor $(\CatAdTen)^0\xrightarrow{K(-)}\ZAdTen$ which obviously sends Deligne tensor to tensor of $\ZZ$-modules.

The monoidal structure on $\CCC$ yields a structure of algebra on $A$. The Hopf axiom comes from the continuity of the original functor, as follows:

Consider the Cartesian square in $\FinDisj$, and its image under $\Hop$
\[
\stik{1}{
\fset{4} \ar{r}{a} \ar{d}{b} \& \fset{2} \ar{d}{c} \\
\fset{2} \ar{r}{d} \& \fset{1}
}\mapsto
\stik{1}{
\Hop(\fset{4}) \ar{r}{\Hop(a)} \ar{d}[left]{\Hop(b)} \& \Hop(\fset{2}) \ar{d}{\Hop(c)} \ar[Rightarrow,shorten <=1pc,shorten >=1pc]{dl}[above,sloped]{\alpha} \\
\Hop(\fset{2}) \ar{r}{\Hop(d)} \& \Hop(\fset{1})
}
\]
Since the square is Cartesian, its image satisfies the Beck-Chevalley condition. Taking $K$ groups, we get the commutative square of algebras
\[
\widestik{1}{A^{\otimes 4} \ar{r}{m_{12}\otimes m_{34}} \ar{d}[left]{m_{13}\otimes m_{24}} \& A^{\otimes 2} \ar{d}{m}\\
A^{\otimes 2} \ar{r}{m} \& A
}
\]
The Hopf axiom amounts to showing that this square with the top and bottom arrows replaced by adjoints should commute. But replacing top and bottom with adjoints in the square of modules is the same as taking the left mate of the square $\alpha$ and then taking $K$ groups. The square $\alpha$ satisfies the Beck-Chevalley condition, hence its left mate has an isomorphism in the middle, so we conclude that the square of modules commutes.

Regarding connectedness of $K(\CCC)$ see the discussion in \autoref{rem:connected}.
\end{proof}

\begin{Proposition}
\label{prop:shonpower}
For any finite set $T_0$ we have a natural SSH structure on $\Hop(T_0)$.
\end{Proposition}

\begin{proof}
Define a functor by $S\mapsto \Hop(S\times T_0)$, then it is obviously an SSH functor, and it sends $\fset{1}$ to $\Hop(T_0)$.
\end{proof}

\subsection{The categorical analogs of Hopf algebra structures}

\subsubsection{Connectedness}
\label{rem:connected}
\label{connectedness}

Recall that a PSH algebra $A$ is defined to be a \emph{connected} Hopf algebra, namely it has unit and counit morphisms $\ZZ\rightarrow A_0$ and $A\rightarrow\ZZ$ such that they give an isomorphism of $A_0$ with $\ZZ$.\\
This is categorified as follows:
The unit and counit in $K(\CCC)$ are the image of the map $\emptyset\rightarrow 1$ in $\FinDisj$ and its left adjoint. Denote the image of this map under $\Hop$ by $m_\emptyset$ and its adjoint by $\Delta_\emptyset$.

Consider the diagrams
\begin{align*}
\stik{1}{
\emptyset \arrow[equal]{d} \arrow[equal]{r} \& \emptyset \ar{d}\\
\emptyset \ar{r} \& 1
}
& {\stik{1}{}} &
\stik{1}{
\emptyset \arrow{d} \arrow{r} \& 1 \ar[equal]{d}\\
1 \ar[equal]{r} \& 1
}
\end{align*}

They go to squares in $\CatAd$

\begin{align*}
\stik{1}{
\Vect \arrow[equal]{d} \arrow[equal]{r} \& \Vect \ar{d}{m_\emptyset}\ar[Rightarrow,shorten <=0.7pc,shorten >=0.7pc]{dl}[above, sloped]{\Id}\\
\Vect \ar{r}[below]{m_\emptyset} \& \CCC
}
& {\stik{1}{}} &
\stik{1}{
\Vect \arrow{d}[left]{m_\emptyset} \arrow{r}{m_\emptyset} \& \CCC \ar[equal]{d} \ar[Rightarrow,shorten <=0.7pc,shorten >=0.7pc]{dl}[above, sloped]{\Id}\\
\CCC \ar[equal]{r} \& \CCC
}
\end{align*}
their left mates are
\begin{align*}
\stik{1}{
\Vect \arrow[equal]{d} \arrow[equal]{r} \& \Vect \ar{d}{m_\emptyset}\\
\Vect \ar[leftarrow]{r}[below]{\Delta_\emptyset} \& \CCC \ar[Rightarrow,shorten <=0.7pc,shorten >=0.7pc]{ul}[above, sloped]{\alpha}
}
& {\stik{1}{}} &
\stik{1}{
\Vect \arrow{d}[left]{m_\emptyset} \arrow[leftarrow]{r}{\Delta_\emptyset} \& \CCC \ar[equal]{d}\\
\CCC \ar[equal]{r} \& \CCC \ar[Rightarrow,shorten <=0.7pc,shorten >=0.7pc]{ul}[above, sloped]{\beta}
}
\end{align*}
The left square came from an \comma square, so contains an isomorphism, i.e. $\alpha$ is invertible, but $\beta$ need not be, and in a non trivial situation will not be.

Define $\CCC_00$ to be the full subcategory of objects $X\in\CCC$ for which $\beta:X\rightarrow m_\emptyset\Delta_\emptyset X$ is invertible. Note that all objects of $\CCC_00$ are necessarily of degree 0 because the map $m_\emptyset$ lands in degree 0.

\begin{Definition}
We say that an SSH category is \emph{connected} if $\CCC_00=\CCC_0$.
\end{Definition}

\begin{Proposition}
$m_\emptyset,\Delta_\emptyset$ give an adjoint equivalence between $\Vect$ and $\CCC_00$.
\end{Proposition}
\begin{proof}
All we need to show is that $m_\emptyset$ always lands in $\CCC_0$. This holds because $\alpha,\beta$ are just the counit and unit of the adjunction $(m_\emptyset\dashv \Delta_\emptyset)$, so the composition\[
m_\emptyset V\xrightarrow{\beta}m_\emptyset\Delta_\emptyset m_\emptyset V\xrightarrow{\alpha}m_\emptyset V
\]
is the identity of $m_\emptyset V$. In particular $\beta$ is invertible for $m_\emptyset V$.
\end{proof}

\begin{Corollary}
The $K$-group of a connected SSH category is a (connected) PSH algebra.
\end{Corollary}

\subsubsection{The categorical Hopf axiom}
In the notation of \autoref{fiberover1}, for any \comma square of maps of sets we have the corresponding diagram in $\CatAd$
\[
\tik
S\ar{r}{a} \ar{d}{c} &T \ar{d}{b} \\
R \ar{r}{d} & U
\tak
\rightsquigarrow
\tik
\CS{S}\arrow[]{d}{m_c} \arrow[]{r}{m_a} & \CS{T} \arrow[]{d}{m_b} \arrow[shorten >=0.4cm,shorten <=0.4cm,Rightarrow]{dl}[above,sloped]{\sim} \\
\CS{R} \arrow[]{r}{m_d} & \CS{U}
\tak
\]
(Here $S=T\times_U R$). Denoting the left adjoint to $m$ by $\Delta^l$ we can consider the left mate of this square
\[
\tik
\CS{S} \arrow[shorten >=0.4cm,shorten <=0.4cm,Leftarrow]{dr}[above,sloped]{} \arrow[leftarrow]{r}{\Delta^l_a} & \CS{T} \\
\CS{R} \arrow[leftarrow]{u}{m_c} \arrow[leftarrow]{r}{\Delta^l_d} & \CS{U} \arrow[leftarrow]{u}{m_b}
\tak
\]

And the Beck-Chevalley condition tells us that the $2$-morphism in this square is invertible. This can be viewed as a natural categorification of the Hopf axiom in the algebra "for any number of variables".

This system of isomorphisms includes a compatibility with the monoidal structure as we saw in the proof of \autoref{frenkelcranehopfcat}.

Note that taking the right adjoint of $m$ gives an additional system of "Hopf isomorphisms". This phenomenon appears in the article \cite{savageyacobi} where the authors construct a dual pair of Hopf algebras from a monoidal category. We suggest that the underlying structure of their construction should be a non-semisimple selfadjoint Hopf category. See also \autoref{rem:leftright}.

\subsection{\comma squares}
\label{sec:2Cartesian}

A \comma square in a $2$-category can be thought of as a square \smallttik{
A \ar{r} \ar{d} \& B \ar{d} \ar[Rightarrow,shorten <=0.7em,shorten >=0.7em]{dl} \\
C \ar{r} \& D
}
which is final in the category of squares which share its right and bottom sides. This is a slight weakening of the notion of comma square - the precise definition is given in \autoref{app:WeightedLimits}.

Morally, it means that for any square \smallttik{
E \ar{r} \ar{d} \& B \ar{d} \ar[Rightarrow,shorten <=0.7em,shorten >=0.7em]{dl}\\
C \ar{r} \& D
}
there exists an "essentially unique" oriented commutative cube
\begin{equation}
\label{commacube}
\stik{0.7}{
{} \& C \ar{rr}[name=topback,below]{} \ar[equal]{dd}[name=backleft,sloped,xshift=-1.5pc]{} \& {} \& D \ar[equal]{dd}\\
E \ar[crossing over]{rr}[name=topfront,above]{}\ar{dd}[name=frontleft]{} \ar{ur}[name=topleft,below]{} \& {} \& B \ar[equal]{dd}[name=frontright,sloped,xshift=1pc]{} \ar{ur}[name=topright,below]{} \& {}\\
{} \& C \ar{rr} \& {} \& D\\
A \ar{rr}[name=bottomfront,below]{} \ar{ur}[name=bottomleft,sloped,yshift=-0.2pc]{} \& {} \& B \ar{ur}
\arrow[Rightarrow,to path={(topright) to[bend left] (topback)}]{}{}
\arrow[Rightarrow,to path={(frontright) to[bend right] (bottomfront)}]{}{}
\arrow[equal,crossing over,to path={(frontright) to[bend right] (topright)}]{}{}
\arrow[Rightarrow,to path={(bottomfront) to[bend right] (bottomleft)}]{}{}
\arrow[Rightarrow,to path={(bottomleft) to[bend left] (backleft)}]{}{}
\arrow[equal,to path={(topback) to[bend left] (backleft)}]{}{}
\latearrow{/tikz/commutative diagrams/crossing over,/tikz/commutative diagrams/equal}{2-3}{4-3}{}
\latearrow{/tikz/commutative diagrams/crossing over}{2-1}{2-3}{}
}
\end{equation}
where vertices are objects, edges are $1$-morphisms and faces are $2$-morphisms composed as indicated by the arrows in the diagram (more on the subject of commutative cubes in \autoref{app:commcube}).

In $\FinSet$, or any 1-category, an \comma square is the same as a usual Cartesian square.

\begin{Proposition}
\label{prop:bccart}
\label{proof:cartesianBC}
In $\CatAd$ a square is \comma iff it satisfies the Beck-Chevalley condition.
\end{Proposition}

\begin{proof}
For the definitions and details regarding the Beck-Chevalley condition see \autoref{sec:BC}.

Let \[
\stik{1}{
A \ar{r}{h} \ar{d}{i} \& B \ar{d}{g} \ar[Rightarrow,shorten <=0.7em,shorten >=0.7em]{dl}[above,sloped]{\alpha}\\
C \ar{r}{f} \& D
}
\]
be a square, and suppose it is \comma. Denote the $2$-morphism in its left mate by $\alpha_L$. We want to show that $\alpha_L$ is invertible. For any functor $x\in\CatAd$, denote by $x_L,x_R$ the left and right adjoints of $x$, and consider the square
\[
\stik{1}{
D \ar[leftarrow]{r}{g} \ar[equal,shorten <=0.7em,shorten >=0.7em]{dr}[above,sloped]{} \ar{d}{f_L} \& B \ar{d}{g} \\
C \ar[leftarrow]{r}{f_L} \& D
}
\]
Take its right mate to get the square
\[
\stik{1}{
D \ar{r}{g_R}  \ar{d}{f_L} \& B \ar[Rightarrow,shorten <=0.7em,shorten >=0.7em]{dl}[above,sloped]{\beta} \ar{d}{g} \\
C \ar{r}{f} \& D
}
\]
which has the same right and bottom sides as our original square. As a consequence we have a commutative cube
\[
\stik{0.7}{
{} \& C \ar{rr}[name=topback,below]{} \ar[equal]{dd}[name=backleft,sloped,xshift=-1.5pc]{} \& {} \& D \ar[equal]{dd}\\
D \ar[crossing over]{rr}[name=topfront,above]{}\ar{dd}[name=frontleft]{} \ar{ur}[name=topleft,below]{} \& {} \& B \ar[equal]{dd}[name=frontright,sloped,xshift=1pc]{} \ar{ur}[name=topright,below]{} \& {}\\
{} \& C \ar{rr} \& {} \& D\\
A \ar{rr}[name=bottomfront,below]{} \ar{ur}[name=bottomleft,sloped,yshift=-0.2pc]{} \& {} \& B \ar{ur}
\arrow[Rightarrow,to path={(topright) to[bend left] (topback)}]{}{}
\arrow[Rightarrow,to path={(frontright) to[bend right] (bottomfront)}]{}{}
\arrow[Rightarrow,crossing over,to path={(frontright) to[bend right] (topright)}]{}{}
\arrow[Rightarrow,to path={(bottomfront) to[bend right] (bottomleft)}]{}{}
\arrow[Rightarrow,to path={(bottomleft) to[bend left] (backleft)}]{}{}
\arrow[Rightarrow,to path={(topback) to[bend left] (backleft)}]{}{}
\latearrow{/tikz/commutative diagrams/crossing over,/tikz/commutative diagrams/equal}{2-3}{4-3}{}
\latearrow{/tikz/commutative diagrams/crossing over}{2-1}{2-3}{}
}
\]

Taking the left mate of this cube as described in \autoref{lem:cubemate}, we get a commutative cube that on one side has the composition of three identity morphisms and on the other side has a composition of the form $\gamma_1\circ \gamma_2\circ\alpha_L$. In particular, $\alpha_L$ is invertible, as required.

In the other direction, suppose that our originial square $\alpha$ satisfies the Beck-Chevalley condition. We want to show it is \comma using the definitions in \autoref{app:PBcont}. Let $X$ be the category of squares with bottom right corner \stik{0.7}{\& B \ar{d}{g}\\ C \ar{r}{f} \& D} as described in \autoref{app:WeightedLimits}. We need to show that the map $X\rightarrow\point$ has a lax right adjoint given by our square.

Consider a square
\[
\stik{1}{
M \ar{r}{b} \ar{d}{a} \& B \ar{d}{g} \ar[Rightarrow,shorten <=0.7em,shorten >=0.7em]{dl}[above,sloped]{\beta}\\
C \ar{r}{f} \& D
}
\]
(for simplicity we have taken a square with precisely the same corner and not the more general case described in \autoref{app:WeightedLimits}, but this does not affect the proof).

Denote $S=\Hom_X(\beta,\alpha)$. We will show that $S$ has a final object (as discussed in \autoref{app:PBcont}).

Let $C_R=i_R\circ a$ and consider the following object in $S$ which we also denote $C_R$:
\[
\stik{1.2}{
{} \& C \ar{rr}[name=topback,below]{}[above]{f} \ar[equal]{dd}[name=backleft,sloped,xshift=-1.5pc]{} \& {} \& D \ar[equal]{dd}\\
M \ar[crossing over]{rr}[name=topfront,above]{}[above,near end,yshift=1pt]{b} \ar{dd}[name=frontleft]{}[left]{C_R} \ar{ur}[name=topleft,below]{}[above]{a} \& {} \& B \ar[equal]{dd}[name=frontright,sloped,xshift=1pc]{} \ar{ur}[name=topright,below]{}[below,near end]{g} \& {}\\
{} \& C \ar{rr}[above,near start]{f} \& {} \& D\\
A \ar{rr}[name=bottomfront,below]{}[below]{h} \ar{ur}[name=bottomleft,sloped,yshift=-0.2pc]{}[above,near start]{i} \& {} \& B \ar{ur}[below]{g}
\arrow[Rightarrow,to path={(topright) to[bend left] (topback)}]{}{}
\arrow[Rightarrow,to path={(frontright) to[bend right] (bottomfront)}]{}{}
\arrow[equal,crossing over,to path={(frontright) to[bend right] (topright)}]{}{}
\arrow[Rightarrow,to path={(bottomfront) to[bend right] (bottomleft)}]{}{}
\arrow[Rightarrow,to path={(bottomleft) to[bend left] (backleft)}]{}{}
\arrow[equal,to path={(topback) to[bend left] (backleft)}]{}{}
\latearrow{/tikz/commutative diagrams/crossing over,/tikz/commutative diagrams/equal}{2-3}{4-3}{}
\latearrow{/tikz/commutative diagrams/crossing over}{2-1}{2-3}{}
}
\]
where the left face is given by the counit $\epsilon$ of the adjunction $(i\dashv i_R)$ and the front face is given by the composition (reading from left to right)
\[
g_R:=\nstik{1}{
{} \ar{d}{b} \& \ar{d}{b} \ar[yshift=-1em,shorten <=1em,shorten >=0.2em]{dr}[xshift=0em]{\beta} \& \ar{d}{a} \& \ar{d}{a}\\
{} \ar[equal]{d} \ar[yshift=-1em,shorten <=1em,shorten >=0.2em]{dr}[xshift=-0.5em]{\eta} \& \ar{d}{g} \ar[yshift=-1.5em,shorten <=1.2em]{ur} \& \ar{d}{f} \ar[yshift=-1em,shorten <=1em,shorten >=0.2em]{dr}[xshift=0em]{\alpha_R^{-1}} \& \ar{d}{i_R}\\
{} \ar[equal]{d} \ar[yshift=-1.5em,shorten <=1.2em]{ur} \& \ar{d}{g_R} \& \ar{d}{g_R} \ar[yshift=-1.5em,shorten <=1.2em]{ur} \& \ar{d}{h}\\
{} \& {} \& {} \& {}
}
\]
It can be easily checked that this cube commutes.

\begin{Claim}
The cube $C_R$ is final in $S$.
\end{Claim}

\begin{Note}
In addition to the cube we constructed above, we should also give a morphism between the compositions in the corner as described in the end of \autoref{app:WeightedLimits} which in this case can be taken to be the identity. As a result this part of the 1-morphism does not affect the finality and so for the sake of readability we disregard it in the proof.
\end{Note}

Suppose that we have another element $Y\in S$, i.e. a cube
\[
\stik{1.2}{
{} \& C \ar{rr}[name=topback,below]{}[above]{f} \ar[equal]{dd}[name=backleft,sloped,xshift=-1.5pc]{} \& {} \& D \ar[equal]{dd}\\
M \ar[crossing over]{rr}[name=topfront,above]{}[above,near end,yshift=1pt]{b} \ar{dd}[name=frontleft]{}[left]{Y} \ar{ur}[name=topleft,below]{}[above]{a} \& {} \& B \ar[equal]{dd}[name=frontright,sloped,xshift=1pc]{} \ar{ur}[name=topright,below]{}[below,near end]{g} \& {}\\
{} \& C \ar{rr}[above,near start]{f} \& {} \& D\\
A \ar{rr}[name=bottomfront,below]{}[below]{h} \ar{ur}[name=bottomleft,sloped,yshift=-0.2pc]{}[above,near start]{i} \& {} \& B \ar{ur}[below]{g}
\arrow[Rightarrow,to path={(topright) to[bend left] (topback)}]{}{}
\arrow[Rightarrow,to path={(frontright) to[bend right] (bottomfront)}]{}{}
\arrow[equal,crossing over,to path={(frontright) to[bend right] (topright)}]{}{}
\arrow[Rightarrow,to path={(bottomfront) to[bend right] (bottomleft)}]{}{}
\arrow[Rightarrow,to path={(bottomleft) to[bend left] (backleft)}]{}{}
\arrow[equal,to path={(topback) to[bend left] (backleft)}]{}{}
\latearrow{/tikz/commutative diagrams/crossing over,/tikz/commutative diagrams/equal}{2-3}{4-3}{}
\latearrow{/tikz/commutative diagrams/crossing over}{2-1}{2-3}{}
}
\]

A morphism $Y\rightarrow C_R$ can be given by a commutative 4-cube
\[
\begin{tikzpicture}[commutative diagrams/every diagram,scale=2]
	 \tikzstyle{over}=[-,line width=4pt,draw=white];
	 \tikzstyle{CD}=[commutative diagrams/.cd, every arrow, every label]
	 \tikzstyle{equal}=[-,line width=1pt,double,double distance=3 pt]
	 \node (IBFL) at (0,0) {$A$};
	 \node (ITFL) at (0,1) {$M$};
	 \node (IBFR) at (1,0) {$B$};
	 \node (ITFR) at (1,1) {$B$};
	 \node (IBKL) at (0.23, 0.4) {$C$};
	 \node (ITKL) at (0.23,1.4) {$C$};
	 \node (IBKR) at (1.23,0.4) {$D$};
	 \node (ITKR) at (1.23,1.4) {$D$};
	 \node (OBFL) at (-1,-1) {$A$};
	 \node (OTFL) at (-1,2) {$M$};
	 \node (OTKL) at (-0.66,2.7) {$C$};
	 \node (OBKL) at (-0.66,-0.3) {$C$};
	 \node (OBFR) at (2,-1) {$B$};
	 \node (OBKR) at (2.34,-0.3) {$D$};
	 \node (OTFR) at (2,2) {$B$};
	 \node (OTKR) at (2.34,2.7) {$D$};
	 \path[CD]

        (OTKL) edge (OTKR)
        (OTKL) edge[equal] (OBKL)
        (OBKL) edge (OBKR)
        (OTKR) edge[equal] (OBKR)

        (OTKL) edge[equal] (ITKL)
        (OBKL) edge[equal] (IBKL)
        (OTKR) edge[equal] (ITKR)
        (OBKR) edge[equal] (IBKR)

        (ITKL) edge (ITKR)
        (ITKL) edge[equal] (IBKL)
        (IBKL) edge (IBKR)
        (ITKR) edge[equal] (IBKR)

        (ITFL) edge (ITKL)
        (IBFL) edge (IBKL)
        (ITFR) edge (ITKR)
        (IBFR) edge (IBKR)

        (ITFL) edge[over] (ITFR) edge (ITFR)
        (ITFL) edge[over] (IBFL) edge node[scale=1.5,left] {$C_R$} (IBFL)
        (IBFL) edge[over] (IBFR) edge (IBFR)
        (ITFR) edge[over] (IBFR) edge[equal] (IBFR)

        (OTFL) edge (OTKL)
        (OBFL) edge (OBKL)
        (OTFR) edge (OTKR)
        (OBFR) edge (OBKR)

        (ITFL) edge[over] (OTFL) edge[equal] (OTFL)
        (IBFL) edge[over] (OBFL) edge[equal] (OBFL)
        (ITFR) edge[over] (OTFR) edge[equal] (OTFR)
        (IBFR) edge[over] (OBFR) edge[equal] (OBFR)

        (OTFL) edge[over] (OTFR) edge (OTFR)
        (OTFL) edge[over] (OBFL) edge node[scale=1.5,left] {$Y$} (OBFL)
        (OBFL) edge[over] (OBFR) edge (OBFR)
        (OTFR) edge[over] (OBFR) edge[equal] (OBFR)
        ;
\end{tikzpicture}
\]
where the dimensions are ordered as $[\rightarrow,\downarrow$,radial towards center,$\nearrow$]. This defines the orientation of 2-morphisms and the order of their composition - see \autoref{app:commcube} for details.
The diagram of 2-morphisms associated to the oriented 4-cube commutes iff each sub 3-cube commutes (see  \cite{graycoherence}). Hence, in all, the data we should provide is only one 2-face, namely the one shared by $Y$ and $C_R$, and this just amounts to a morphism $Y\xrightarrow{\psi} C_R$. We need to show that such a morphism which makes the 4-cube commute exists and is unique.

Consider the left 3-cube
\[
\stik{1}{
{} \& C \ar[equal]{rr}[name=topback,below]{}[above]{} \ar[equal]{dd}[name=backleft,sloped,xshift=-1.5pc]{}[right,yshift=-1.5pc]{} \& {} \& C \ar[equal]{dd}\\
M \ar[crossing over]{rr}[name=topfront,above]{}[below, xshift=1pc]{Y}\ar[equal]{dd}[name=frontleft]{}[left]{} \ar{ur}[name=topleft,below]{} \& {} \& A \ar[equal]{dd}[name=frontright,sloped,xshift=1pc]{} \ar{ur}[name=topright,below]{} \& {}\\
{} \& C \ar[equal]{rr} \& {} \&C\\
M \ar{rr}[name=bottomfront,below]{C_R} \ar{ur}[name=bottomleft,sloped,yshift=-0.2pc]{} \& {} \& A \ar{ur}
\arrow[Rightarrow,to path={(topright) to[bend left] (topback)}]{}{}
\arrow[Rightarrow,to path={(frontright) to[bend right] (bottomfront)},"\psi"']{}{}
\arrow[equal,crossing over,to path={(frontright) to[bend right] (topright)}]{}{}
\arrow[Rightarrow,to path={(bottomfront) to[bend right] (bottomleft)}]{}{}
\arrow[equal,to path={(bottomleft) to[bend left] (backleft)}]{}{}
\arrow[equal,to path={(topback) to[bend left] (backleft)}]{}{}
\latearrow{commutative diagrams/crossing over,commutative diagrams/equal}{2-3}{4-3}{}
\latearrow{commutative diagrams/crossing over}{2-1}{2-3}{}
\latearrow{commutative diagrams/crossing over}{2-3}{1-4}{}
}
\]
By assumption it commutes. Its bottom face is the left face of $C_R$; its top face is the left face of $Y$, denote it by $f_Y$; its front face is $\psi$; the rest are degenerate. So we get the equation
\[
\nstik{1}{
{} \ar{dd}{Y}  \ar[yshift=-1em,shorten <=1em,shorten >=0.2em]{dr}[xshift=-0em]{\psi}\& \ar{d}{a}  \& \ar{d}{a}\\
{} \ar[yshift=-1.5em,shorten <=1.2em]{ur} \& \ar{d}{i_R}  \ar[yshift=-1em,shorten <=1em,shorten >=0.2em]{dr}[xshift=0em]{\epsilon} \& \ar[equal]{d}{}\\
{} \ar{d}{i}  \& \ar{d}{i} \ar[yshift=-1.5em,shorten <=1.2em]{ur} \& \ar[equal]{d}{}\\
{} \& {} \& {}
}
=
f_Y
\]
So since $\epsilon$ is part of an adjunction, we have that
\begin{align*}
\psi & =
\nstik{1}{
{} \ar{dd}{Y} \ar[yshift=-1em,shorten <=1em,shorten >=0.2em]{dr}[xshift=-0em]{\psi} \& \ar{d}{a}  \& \ar{d}{a}  \& {} \ar{d}{a}\\
{} \ar[yshift=-1.5em,shorten <=1.2em]{ur} \& \ar{d}{i_R}  \&  {} \ar{d}{i_R} \ar[yshift=-1em,shorten <=1em,shorten >=0.2em]{dr}[xshift=0em]{\epsilon} \& {} \ar[equal]{dd}\\
{} \ar[equal]{dd}{}  \& \ar[yshift=-1em,shorten <=1em,shorten >=0.2em]{dr}[xshift=-0em]{\eta} \ar[equal]{dd}{} \&  \ar{d}{i}  \ar[yshift=-1.5em,shorten <=1.2em]{ur}\& {} \\
{}  \& {} \ar[yshift=-1.5em,shorten <=1.2em]{ur} \& {}  \ar{d}{i_R}\& {} \ar{d}{i_R}\\
{}  \& {} \& {} \& {}
}
\\
& =
\nstik{1}{
{} \ar{dd}{Y}  \& {} \ar[yshift=-1em,shorten <=1em,shorten >=0.2em]{dr}[xshift=-0em]{\psi} \ar{dd}{Y}  \& \ar{d}{a}  \& {} \ar{d}{a}\\
{} \& {} \ar[yshift=-1.5em,shorten <=1.2em]{ur}  \&  {} \ar{d}{i_R} \ar[yshift=-1em,shorten <=1em,shorten >=0.2em]{dr}[xshift=0em]{\epsilon} \& {} \ar[equal]{dd}\\
{} \ar[equal]{dd}{}\ar[yshift=-1em,shorten <=1em,shorten >=0.2em]{dr}[xshift=-0em]{\eta}  \&  \ar{d}{i} \&  \ar{d}{i}  \ar[yshift=-1.5em,shorten <=1.2em]{ur}\& {} \\
{} \ar[yshift=-1.5em,shorten <=1.2em]{ur} \& {}\ar{d}{i_R}  \& {}  \ar{d}{i_R}\& {} \ar{d}{i_R}\\
{}  \& {} \& {} \& {}
}
=
 f_Y \circ \eta
\end{align*}
where the second step is justified by the four-interchange law in a bicategory.

So $\psi$ is uniquely defined, and all we need to check is that it makes the front cube in the 4-cube commute.

The front cube is
\[
\stik{1}{
{} \& M \ar{rr}[name=topback,below]{}[above]{} \ar{dd}[name=backleft,sloped,xshift=-1.5pc]{}[right,yshift=-1.5pc]{C_R} \& {} \& B \ar[equal]{dd}\\
M \ar[crossing over]{rr}[name=topfront,above]{}[below, xshift=1pc]{} \ar{dd}[name=frontleft]{}[left]{Y} \ar[equal]{ur}[name=topleft,below]{} \& {} \& B \ar[equal]{dd}[name=frontright,sloped,xshift=1pc]{} \ar[equal]{ur}[name=topright,below]{} \& {}\\
{} \& A \ar{rr} \& {} \&B\\
A \ar{rr}[name=bottomfront,below]{} \ar[equal]{ur}[name=bottomleft,sloped,yshift=-0.2pc]{} \& {} \& B \ar[equal]{ur}
\arrow[equal,to path={(topright) to[bend left] (topback)}]{}{}
\arrow[Rightarrow,to path={(frontright) to[bend right] (bottomfront)},"g_Y"]{}{}
\arrow[equal,crossing over,to path={(frontright) to[bend right] (topright)}]{}{}
\arrow[equal,to path={(bottomfront) to[bend right] (bottomleft)}]{}{}
\arrow[Rightarrow,to path={(bottomleft) to[bend left] (backleft)},"\psi"]{}{}
\arrow[Rightarrow,to path={(topback) to[bend left] (backleft)},"g_R"']{}{}
\latearrow{commutative diagrams/crossing over,commutative diagrams/equal}{2-3}{4-3}{}
\latearrow{commutative diagrams/crossing over}{2-1}{2-3}{}
\latearrow{commutative diagrams/crossing over,commutative diagrams/equal}{2-3}{1-4}{}
}
\]
So we need to show that $\psi\circ g_Y=g_R$.

Recall that
\[
g_R=\nstik{1}{
{} \ar{d}{b} \& \ar{d}{b} \ar[yshift=-1em,shorten <=1em,shorten >=0.2em]{dr}[xshift=0em]{\beta} \& \ar{d}{a} \& \ar{d}{a}\\
{} \ar[equal]{d} \ar[yshift=-1em,shorten <=1em,shorten >=0.2em]{dr}[xshift=-0.5em]{\eta} \& \ar{d}{g} \ar[yshift=-1.5em,shorten <=1.2em]{ur} \& \ar{d}{f} \ar[yshift=-1em,shorten <=1em,shorten >=0.2em]{dr}[xshift=0em]{\alpha_R^{-1}} \& \ar{d}{i_R}\\
{} \ar[equal]{d} \ar[yshift=-1.5em,shorten <=1.2em]{ur} \& \ar{d}{g_R} \& \ar{d}{g_R} \ar[yshift=-1.5em,shorten <=1.2em]{ur} \& \ar{d}{h}\\
{} \& {} \& {} \& {}
}
\]
Denoting the front face of the outer cube (the "$Y$"-cube ) by $g_Y$, we have $\beta=f_Y\circ\alpha\circ g_Y$.

Substituting we get
\[g_R=
\nstik{1}{
{} \ar{dd}{b} \& \ar{dd}{b} \ar[yshift=-1em,shorten <=1em,shorten >=0.2em]{dr}[xshift=0em]{g_Y}\& \ar{d}{Y} \& \ar{d}{Y} \ar[yshift=-1em,shorten <=1em,shorten >=0.2em]{dr}[xshift=0em]{f_Y}\& \ar{dd}{a} \& \ar{dd}{a}\\
{} \& {} \ar[yshift=-1.5em,shorten <=1.2em]{ur} \& \ar{d}{h} \ar[yshift=-1em,shorten <=1em,shorten >=0.2em]{dr}[xshift=-0.5em]{\alpha}\& \ar{d}{i} \ar[yshift=-1.5em,shorten <=1.2em]{ur}\& {}\& {}\\
{} \ar[equal]{d} \ar[yshift=-1em,shorten <=1em,shorten >=0.2em]{dr}[xshift=-0.5em]{\eta} \& \ar{d}{g}  \& \ar{d}{g} \ar[yshift=-1.5em,shorten <=1.2em]{ur}  \& \ar{d}{f} \& {} \ar{d}{f} \ar[yshift=-1em,shorten <=1em,shorten >=0.2em]{dr}[xshift=0em]{\alpha_R^{-1}} \& {} \ar{d}{i_R}\\
{} \ar[equal]{d} \ar[yshift=-1.5em,shorten <=1.2em]{ur} \& \ar{d}{g_R} \& \ar{d}{g_R}  \& {} \ar{d}{g_r} \& {} \ar{d}{g_R} \ar[yshift=-1.5em,shorten <=1.2em]{ur} \& {} \ar{d}{h}\\
{} \& {} \& {} \& {} \& {} \& {}
}
\]

Now we plug in the identity of $i$ in the fourth column as a composition of unit/counit maps, and interchange $g_Y$ and $\eta$:
\[g_R=
\nstik{0.95}{
{} \& {} \& {} \& {}\ar{d}{Y} \& {} \ar{d}{Y} \& {} \ar{d}{Y} \ar[yshift=-1em,shorten <=1em,shorten >=0.2em]{dr}[xshift=0em]{f_Y}\& {}\ar{dd}{a} \& {}\ar{dd}{a}\\
{} \& {} \& {} \& {}\ar[yshift=-1em,shorten <=1em,shorten >=0.2em]{dr}[xshift=0em]{\eta'}\ar[equal]{dd} \& {} \ar{d}{i} \& {} \ar{d}{i} \ar[yshift=-1.5em,shorten <=1.2em]{ur} \& {} \& {}\\
{} \ar{dd}{b} \ar[yshift=-1em,shorten <=1em,shorten >=0.2em]{dr}[xshift=0em]{g_Y} \& \ar{d}{Y} \& \ar{d}{Y} \& {} \ar[yshift=-1.5em,shorten <=1.2em]{ur} \& {} \ar{d}{i_R} \ar[yshift=-1em,shorten <=1em,shorten >=0.2em]{dr}[xshift=0em]{\epsilon'}\& {} \ar[equal]{dd} \& {}  \ar[equal]{dd} \& \ar[equal]{dd}\\
{} \ar[yshift=-1.5em,shorten <=1.2em]{ur} \& {}\ar{d}{h} \& {}\ar{d}{h} \ar[yshift=-1em,shorten <=1em,shorten >=0.2em]{dr}[xshift=-0.5em]{\alpha}\& {}\ar{d}{i} \& {} \ar{d}{i} \ar[yshift=-1.5em,shorten <=1.2em]{ur} \& {}\& {}\& {}\\
{} \ar[equal]{dd}  \& {} \ar[equal]{dd} \ar[yshift=-1em,shorten <=1em,shorten >=0.2em]{dr}[xshift=-0.5em]{\eta} \& \ar{d}{g} \ar[yshift=-1.5em,shorten <=1.2em]{ur}  \& \ar{d}{f} \& {} \ar{d}{f}  \& {} \ar{d}{f} \& {}\ar{d}{f}  \ar[yshift=-1em,shorten <=1em,shorten >=0.2em]{dr}[xshift=0em]{\alpha_R^{-1}}\& {}\ar{d}{i_R} \\
{}  \& \ar[yshift=-1.5em,shorten <=1.2em]{ur} \& \ar{d}{g_R}  \& {} \ar{d}{g_r} \& {} \ar{d}{g_R}  \& {} \ar{d}{g_R}\& {}\ar{d}{g_R} \ar[yshift=-1.5em,shorten <=1.2em]{ur}\& {} \ar{d}{h}\\
{} \& {} \& {} \& {} \& {} \& {}\& {} \& {}
}
\]

using some interchanging we get to
\[g_R=
\nstik{0.95}{
{} \& {}\ar{d}{Y} \& {}\ar{d}{Y} \& {}\ar{d}{Y} \& {} \ar{d}{Y} \& {} \ar{d}{Y} \& {}\ar{d}{Y} \ar[yshift=-1em,shorten <=1em,shorten >=0.2em]{dr}[xshift=0em]{f_Y} \& {}\ar{dd}{a}\\
{} \& {}\ar[equal]{dd} \ar[yshift=-1em,shorten <=1em,shorten >=0.2em]{dr}[xshift=0em]{\eta'} \& {} \ar{d}{i} \& {} \ar{d}{i} \& {} \ar{d}{i} \& {} \ar{d}{i}  \& {}\ar{d}{i}\ar[yshift=-1.5em,shorten <=1.2em]{ur} \& {}\\
{} \ar{dd}{b} \ar[yshift=-1em,shorten <=0em,shorten >=0.2em]{dr}[xshift=0em]{g_Y} \& {} \ar[yshift=-1.5em,shorten <=1.2em]{ur} \& {} \ar{d}{i_R} \& {}  \ar{d}{i_R} \& {} \ar{d}{i_R} \ar[yshift=-1em,shorten <=1em,shorten >=0.2em]{dr}[xshift=0em]{\epsilon'}\& {} \ar[equal]{dd} \& {}  \ar[equal]{dd} \& \ar[equal]{dd}\\
{} \ar[yshift=-1.5em,shorten <=1.2em]{uuur} \& {}\ar{d}{h} \& {}\ar{d}{h} \& {} \ar[yshift=-1em,shorten <=1em,shorten >=0.2em]{dr}[xshift=-0.5em]{\alpha} \ar{d}{h} \& {} \ar{d}{i} \ar[yshift=-1.5em,shorten <=1.2em]{ur} \& {}\& {}\& {}\\
{} \ar[equal]{dd}  \& {} \ar[equal]{dd}  \& {} \ar[equal]{dd} \ar[yshift=-1em,shorten <=1em,shorten >=0.2em]{dr}[xshift=-0.5em]{\eta}   \& \ar{d}{g} \ar[yshift=-1.5em,shorten <=1.2em]{ur} \& {} \ar{d}{f}  \& {} \ar{d}{f}\ar[yshift=-1em,shorten <=1em,shorten >=0.2em]{dr}[xshift=0em]{\alpha_R^{-1}} \& {}\ar{d}{i_R}  \& {}\ar{d}{i_R} \\
{}   \& {} \& {} \ar[yshift=-1.5em,shorten <=1.2em]{ur} \& {} \ar{d}{g_R} \& {} \ar{d}{g_R}  \& {} \ar{d}{g_R} \ar[yshift=-1.5em,shorten <=1.2em]{ur} \& {}\ar{d}{h} \& {} \ar{d}{h}\\
{} \& {} \& {} \& {} \& {} \& {}\& {} \& {}
}
\]

Finally, we can cancel the middle part, which is just $\alpha_R^{-1}\circ\alpha_R$ and get \[
g_R=f_Y\circ \eta'\circ g_Y=\psi\circ g_Y
\]
as required. This completes the proof.
\end{proof}

\begin{Remark}
A completely dual argument yields an initial object in the Hom-category, which is not isomorphic to the final object, so it is not the case that the Hom-category is contractible.
\end{Remark}


TITLE: If $f(0)=0$ and $f''\ge 0$, then $f(a+b)\ge f(a)+f(b)$
QUESTION [0 upvotes]: Given $\ f$ so $\ f''(x) \ge 0$ for every $\ x \ge 0$, also $\ f(0)=0$.
Trying to show that if $\ a,b \ge 0 \Rightarrow f(a+b) \ge f(a) + f(b)$

Using Taylor I used $\ f(0)=0$ and got
$$f(x)=f'(0)x + \frac {f''(0)}2x^2$$
Plugging in$\ a, b$ and $\ a+b$ I got to this:
$$f''(c3)(a+b)^2\ge f''(c1)a^2+f''(c2)b^2$$
What am I missing here? Is there a better way to solve this than Taylor?

REPLY [7 votes]: Wlog. $a\le b$. Then $f(a)-f(0)=af'(\xi)$ with $\xi\in(0,a)$ and $f(a+b)-f(b)=bf'(\eta)$ with $\eta\in(b,a+b)$ by the Mean Value Theorem. As $\xi<a\le b<\eta$ we have $f'(\eta)-f'(\xi)=(\eta-\xi)f''(\zeta)\ge0$ for some $\zeta\in (\xi,\eta)$ again by MVT. Therefore
$$ f(a+b)-f(b)=bf'(\eta)\ge af'(\xi)=f(a)-f(0),$$
whence the claim.


TITLE: Which $A\subseteq\mathbb{N}\cup\{0\}$ satisfy $\bigg(\sum\limits_{a\in A}x^a\bigg)\prod\limits_{a\in A} (1+x^{2^a})=\dfrac{1}{1-x}$ for all $|x|<1$?
QUESTION [2 upvotes]: Determine the subsets  $A \subseteq \mathbb{N} \cup\{ 0 \}$ that  satisfy :
$$\left (\sum_{a \in A}^{ } x^a  \right )\left ( \prod_{a \in A}^{ } (1+x^{2^a}) \right ) = \frac{1}{1-x}$$
for all $|x|<1$.

REPLY [4 votes]: Using the identity
$$
\frac 1 {1-x}=\prod_{i\ge 0} (1+x^{2^i}),
$$
this equation can be rewritten as
$$
\sum_{a\in A} x^a = \prod_{b\ge 0, \ b\notin A} (1+x^{2^b}).
$$
Each side then defines a function which is analytic in $\{x\in {\Bbb C}\mid |x|<1\}$, so the values will match for all $x$ with $|x|<1$ iff the coefficients of $x^n$ on each side are the same for all $n\ge 0$.  So, the equation is equivalent to the set equation
$$
A=\Sigma({\Bbb Z}_{\ge 0}\setminus A),\qquad \qquad (+)
$$
where, for a subset $B$ of ${\Bbb Z}_{\ge 0}$,
$$
\Sigma(B)=\{2^{b_1}+2^{b_2}+\cdots+2^{b_r}\mid r\ge 0, b_1<b_2<\cdots<b_r\in B\}.
$$
Intersecting both sides of $(+)$ with $\{0,\ldots,2^n-1\}$ gives
$$
A\cap\{0,\ldots,2^n-1\}=\Sigma(\{0,1,\ldots,n-1\}\setminus A). \qquad (*)
$$
Setting $n:=0$ in $(*)$ then immediately gives $$A\cap\{0\}=\Sigma(\emptyset)=\{0\}.$$  Then we know that $\{0\}\setminus A=\emptyset$, so we can plug $n:=1$ into $(*)$ to give $$A\cap\{0,1\}=\Sigma(\{0\}\setminus A)=\Sigma(\emptyset)=\{0\}.$$
Now that we know that $\{0,1\}\setminus A=\{1\}$, we can plug $n:=2$ into $(*)$ to get
$$
A\cap\{0,1,2,3\}=\Sigma(\{0,1\}\setminus A)=\Sigma(\{1\})=\{0,2\}.$$
Continue in this way for $n=3$, $n=4$, $\dots$  Solving $(*)$ for $n=m$ determines $A\cap \{0,\ldots,2^m-1\}$, and then, since $2^m-1\ge m$ for all $m\ge 0$, the right-hand side of $(*)$ for the next value of $n$, $n=m+1$, is determined.  This then gives a unique $A$ which satisfies $(*)$ for each $n$, and hence also satisfies $(+)$.  Therefore, the equation has a unique solution
$$
A=\{0, 2, 8, 10, 16, 18, 24, 26, 32, 34, 40, 42, 48, 50, 56, 58, 64, 66, 72, 74, 80, 82, 88, 90, 96, 98, 104, 106, 112, 114, 120, 122, 128, 130, 136, 138, 144, 146, 152, 154, 160, 162, 168, 170, 176, 178, 184, 186, 192, 194, 200, 202, 208, 210, 216, 218, 224, 226, 232,\ldots\}
$$
This sequence is OEIS A079599.


TITLE: How to derive the process noise co-variance matrix Q in this Kalman Filter example?
QUESTION [2 upvotes]: How to understand the process co-variance matrix Q in the example below ( I extracted it from Wikipedia http://en.wikipedia.org/wiki/Kalman_filter  )

Consider a truck on perfectly frictionless, infinitely long straight rails. Initially the truck is stationary at position 0, but it is buffeted this way and that by random acceleration. We measure the position of the truck every Δt seconds, but these measurements are imprecise; we want to maintain a model of where the truck is and what its velocity is. We show here how we derive the model from which we create our Kalman filter.
  Since F, H, R and Q are constant, their time indices are dropped.
The position and velocity of the truck are described by the linear state space
$\textbf{x}_{k} = \begin{bmatrix} x \\ \dot{x} \end{bmatrix}$
where $\dot{x}$ is the velocity, that is, the derivative of position with respect to time.
We assume that between the $(k−1)$ and $k$ timestep the truck undergoes a constant acceleration of $a_k$ that is normally distributed, with mean $0$ and standard deviation $\sigma_a$. From Newton's laws of motion we conclude that
$\textbf{x}_{k} = \textbf{F} \textbf{x}_{k-1} + \textbf{G}a_{k}$
(note that there is no $\textbf{B}u$ term since we have no known control inputs)
  where
$\textbf{F} = \begin{bmatrix} 1 & \Delta t \\ 0 & 1 \end{bmatrix}$
and
$\textbf{G} = \begin{bmatrix} \frac{\Delta t^{2}}{2} \\ \Delta t \end{bmatrix}$
so that
$\textbf{x}_{k} = \textbf{F} \textbf{x}_{k-1} +  \textbf{w}_{k}$
where $\textbf{w}_{k} \sim N(0,  \textbf{Q})$ and
$\textbf{Q}=\textbf{G}\textbf{G}^{\text{T}}\sigma_a^2 =\begin{bmatrix} \frac{\Delta t^4}{4} & \frac{\Delta t^3}{2} \\ \frac{\Delta t^3}{2} & \Delta t^2 \end{bmatrix}\sigma_a^2.$
[...]

I am not sure how the process noise co-variance matrix  $\textbf{Q}=\textbf{G}\textbf{G}^{\text{T}}\sigma_a^2 =\begin{bmatrix} \frac{\Delta t^4}{4} & \frac{\Delta t^3}{2} \\ \frac{\Delta t^3}{2} & \Delta t^2 \end{bmatrix}\sigma_a^2$ is derived here, could someone shed light on this, please?

REPLY [0 votes]: I know that this is a very late response but I think the Q matrix should be following:
$Q=\begin{bmatrix}
T^3/3 & T^2/2 \\
 T^2/2 & T
\end{bmatrix}$


TITLE: Use Least Upper Bound to show that $\mathbb{R}$ is completee
QUESTION [0 upvotes]: Use Least Upper Bound to show that $\mathbb{R}$ is complete.

The following is the proof I did, and it's slightly different from what I see from the book. Can someone check if I'm missing something? Thank you.
Proof: Let $\{x_n\}$ be a Cauchy sequence, and it suffices to show that $\{x_n\}$ is convergent in $\mathbb{R}$. Since sequence is Cauchy, then for any $\epsilon>0$, there is $N$ such that $m,n\ge N$ implies $|x_m-x_n|<\epsilon$. In particular, this means that for all $m\ge N$, $|x_m - x_N|<\epsilon$, which implies that $\{x_m:m\ge N\}$ is bounded above. Thus, it has a supremum $S$.
By definition of supremum, $S-\epsilon<x_m<S$. This shows that $\{x_n\} \to S.$
Am I missing something? I feel like I'm missing something :(

The following is in the proof from the book:

“Let$\{s_k\}$ be a Cauchy sequence in $\mathbb{R}$. Let $S$ be the set of $y$   such that $s_n < y$ for only finitely many integers $n$.”

I don't see how this comes into work?

REPLY [0 votes]: All is good up to here:
"By definition of supremum $S-\epsilon<x_m<S$ " (It should say there exists an $m$ satisfying this, since it is not necessarily true for all $m$).
All that we can know from the above is that there is an $x_m$ that is as close enough to $S$ as required, but this doesn't mean the sequence converges. For example lets consider the following sequence:
$x_n=\frac{1}{n}$ if $n$ is odd and $1$ when $n$ is even. For every $\epsilon$ we can find $a_n$ so that $|a_n-0|<\epsilon$ yet $x_n$ does not converge to $0$.

You have taken some steps in the right direction however, perhaps this can get you going. You proved the sequence $x_n$ is bounded, and we know that every bounded sequence has a converging subsequence (hopefully this has been taught).
So the idea for the proof is to take a converging subsequence that converges to $l$ and use this converging subsequence and the fact that $x_n$ is Cauchy to prove $x_n$ converges to $l$ also.


TITLE: Why is Ohm's Law $V=IR$?
QUESTION [0 upvotes]: More specifically, why is R=V/I. I understand what current is; the amount of charge that passes through a point in one second or Q/t. Voltage I'm not as sure, but from my knowledge its the amount of Electrical Potential Energy required to move a single unit of charge from one location to another.  What I don't understand is how the resistance, a measurement of an opposing force to electrons, is equal to voltage/current. So in simple terms, why is this case?

REPLY [1 votes]: Think of this problem as a real life problem.
Suppose there is a job opening with a salary of one million dollars (The salary is analogue to the Voltage V ). Just like higher the voltage the higher the current we expect to flow, higher the salary higher the number of applicants for the job. If there are absolutely no qualifications required for the job, then every single person would apply for the job. Just like that if the wires do not have any resistance, we expect the current to be very large (i.e. infinite. okay may be not infinite but it will be as large as it can be). Whereas the specific qualifications required for the job will greatly decrease the number of applicants. In the same way the resistance in the connecting wires will hinder the electron flow and limit the current. So the relation between resistance(R) and current (I) is as:
                 I ~ 1/R

Also as we expect higher current with higher voltage
                 I ~ V

Combining the above two relations we get
                I ~ V/R

So
                R ~ V/I