text:
\section{Opetopes} \label{sec:opetopes}

\subsection{Definition}
\label{sec:opetopes:definition}

\begin{definition}
    [The $\optPolyFun^n$ monad]
    \label{def:zn}
    Let $\optPolyFun^0$ be the identity polynomial monad on $\Set$, as depicted
    on the left below, and let $\optPolyFun^n \eqdef
    (\optPolyFun^{n-1})^+$\index{$\optPolyFun^n$}. Write $\optPolyFun^n$ as on
    right:
    \begin{equation}
        \label{eq:zn}
        \polynomialfunctor
            {\{* \}}{\{* \}}{\{* \}}{\{* \},}
            {}{}{}
        \qqquad
        \polynomialfunctor
            {\bbOO_n}{E_{n+1}}{\bbOO_{n+1}}{\bbOO_n .}
            {\src}{p}{\tgt}
    \end{equation}
\end{definition}

\begin{definition}
    [Opetope]
    \label{def:opetope}
    An \emph{$n$-dimensional opetope} (or simply \emph{$n$-opetope}) $\omega$
    is by definition an element of $\bbOO_n$, and we write $\dim \omega = n$.
    If $n \geq 2$, then opetopes are exactly $\optPolyFun^{n-2}$-trees. In this
    case, an opetope $\omega \in \bbOO_n$ is called
    \emph{degenerate}\index{degenerate opetope} if its underlying tree has no
    nodes (thus consists of a unique edge); it is \emph{non
    degenerate}\index{non degenerate opetope} otherwise.

    Following \cref{eq:polynomial-monad-structure-map}, for $\omega \in
    \bbOO_{n+2}$, the structure of polynomial monad $(\optPolyFun^n)^\star
    \longrightarrow \optPolyFun^n$ gives a bijection $\readdress_\omega :
    \omega^\leafsymbol \longrightarrow (\tgt \omega)^\nodesymbol$ between the
    leaves of $\omega$ and the nodes of $\tgt \omega$, preserving the
    decoration by $n$-opetopes.
\end{definition}

\begin{example}
    \begin{enumerate}
        \item The unique $0$-opetope is denoted $\optZero$\index{$\optZero$|see
        {point (opetope)}} and called the \emph{point}\index{point (opetope)}.
        \item The unique $1$-opetope is denoted $\optOne$\index{$\optOne$|see
        {arrow (opetope)}} and called the \emph{arrow}\index{arrow (opetope)}.
        \item If $n \geq 2$, then $\omega\in \bbOO_n$ is a
        $\optPolyFun^{n-2}$-tree, i.e. a tree whose nodes are labeled in
        $(n-1)$-opetopes, and edges are labeled in $(n-2)$-opetopes. In
        particular, $2$-opetopes are $\optPolyFun^0$-trees, i.e. linear trees,
        and thus in bijection with $\bbNN$. We will refer to them as
        \emph{opetopic integers}\index{opetopic integers}, and write
        $\optInt{n}$\index{$\optInt{n}$|see {opetopic integer}} for the
        $2$-opetope having exactly $n$ nodes.
    \end{enumerate}
\end{example}

\subsection{The category of opetopes}

Akin to the work of Cheng \cite{Cheng2003}, we define a category of opetopes by
means of generators and relations. The difference with the aforementioned
reference is our use of polynomial opetopes (also equivalent to Leinster's
definition \cite{Leinster2004, Kock2010}), while Cheng uses an approach by
multicategorical slicing, yielding ``symmetric'' opetopes.

\begin{theorem}
    [Opetopic identities]
    \label{th:opetopic-identities}
    Let $\omega \in \bbOO_n$ with $n \geq 2$.
    \begin{enumerate}
        \item (Inner edge) For $[p[q]] \in \omega^\nodesymbol$ (forcing
        $\omega$ to be non degenerate), we have $\tgt \src_{[p[q]]} \omega =
        \src_{[q]} \src_{[p]} \omega$.
		\item (Globularity 1) If $\omega$ is non degenerate, we have $\tgt
		\src_{[]} \omega = \tgt \tgt \omega$.
        \item (Globularity 2) If $\omega$ is non degenerate, and $[p[q]] \in
        \omega^\leafsymbol$, we have $\src_{[q]} \src_{[p]} \omega =
        \src_{\readdress_\omega [p[q]]} \tgt \omega$.
        \item (Degeneracy) If $\omega$ is degenerate, we have $\src_{[]} \tgt
        \omega = \tgt \tgt \omega$.
    \end{enumerate}
\end{theorem}
\begin{proof}
	\begin{enumerate}
		\item (Inner edge) By definition of a $\optPolyFun^{n-2}$-tree.

		\item (Globularity 1 and 2) By
		\cref{th:polynomial-monads-star-algebras}, the monad structure on
		$\optPolyFun^{n-2}$ amounts to a structure map
		$(\optPolyFun^{n-2})^\star \longrightarrow \optPolyFun^{n-2}$, which,
		taking the notations of \cref{def:readdressing}, is written as
		\[
			\begin{tikzcd}
				\bbOO_{n-2}
					\ar[d, equal] &
				\treewithleaf \optPolyFun^{n-2}
					\pullbackcorner
					\ar[r, "p"]
					\ar[d, "\readdress" left]
					\ar[l, "\src" above] &
				\tree \optPolyFun^{n-2}
					\ar[r, "\edg_{[]}"]
					\ar[d, "\tgt" left] &
				\bbOO_{n-2}
					\ar[d, equal] \\
				\bbOO_{n-2} &
				\bbOO^\nodesymbol_{n-1}
					\ar[r, "p"]
					\ar[l, "\src" above] &
				\bbOO_{n-1}
					\ar[r, "\tgt"] &
				\bbOO_{n-2} .
			\end{tikzcd}
		\]
		The claims follow from the commutativity of the right and left square
		respectively.

		\item (Degeneracy) Let $\omega = \itree{\phi}$, for $\phi \in
		\bbOO_{n-2}$. Then $\tgt \omega = \ytree{\phi}$, and $\src_{[]}
		\ytree{\phi} = \src_{[]} \mu \ytree{\phi} = \tgt \ytree{\phi}$ where
		$\mu$ is the monad law of $\optPolyFun^{n-2}$.
		\qedhere
	\end{enumerate}
\end{proof}

\begin{definition}
    [The category $\bbOO$ of opetopes]
    \label{def:o}
    With the identities of \cref{th:opetopic-identities}, we define the
    category $\bbOO$ of opetopes by generators and relations as follows.
    \begin{enumerate}
        \item (Objects) We set $\ob \bbOO = \sum_{n \in \bbNN} \bbOO_n$.
        \item (Generating morphisms) Let $\omega \in \bbOO_n$ with $n \geq 1$.
        We introduce a generator $\tgt : \tgt \omega \longrightarrow
        \omega$\index{$\tgt$|see {target embedding}}, called the \emph{target
        embedding}\index{target embedding}. If $[p] \in \omega^\nodesymbol$,
        then we introduce a generator $\src_{[p]} : \src_{[p]} \omega
        \longrightarrow \omega$\index{$\src_{[p]}$|see {source embedding}},
        called a \emph{source embedding}\index{source embedding}. A \emph{face
        embedding}\index{face embedding} is either a source or the target
        embedding.
        \item (Relations) We impose 4 relations described by the following
        commutative squares, that are well defined thanks to
        \cref{th:opetopic-identities}. Let $\omega \in \bbOO_n$ with $n \geq 2$
        \begin{enumerate}
            \item \condition{Inner}\index{inner@\condition{Inner}} for $[p[q]]
            \in \omega^\nodesymbol$ (forcing $\omega$ to be non degenerate),
            the following square must commute:
            \[
                \squarediagram
                    {\src_{[q]} \src_{[p]} \omega}{\src_{[p]} \omega}
                        {\src_{[p[q]]}\omega}{\omega}
                    {\src_{[q]}}{\tgt}{\src_{[p]}}{\src_{[p[q]]}}
            \]
            \item \condition{Glob1}\index{glob1@\condition{Glob1}} if $\omega$
            is non degenerate, the following square must commute:
            \[
                \squarediagram
                    {\tgt \tgt \omega}{\tgt \omega}{\src_{[]} \omega}{\omega .}
                    {\tgt}{\tgt}{\tgt}{\src_{[]}}
            \]
            \item \condition{Glob2}\index{glob2@\condition{Glob2}} if $\omega$
            is non degenerate, and for $[p [q]] \in \omega^\leafsymbol$, the
            following square must commute:
            \[
                \diagramsize{2}{4}
                \squarediagram
                    {\src_{\readdress_\omega [p [q]]} \tgt \omega}{\tgt \omega}
                        {\src_{[p]} \omega}{\omega .}
                    {\src_{\readdress_\omega [p[q]]}}{\src_{[q]}}{\tgt}
                        {\src_{[p]}}
                \diagramsize{2}{3}
            \]
            \item \condition{Degen}\index{degen@\condition{Degen}} if $\omega$
            is degenerate, the following square must commute:
            \[
                \squarediagram
                    {\tgt \tgt \omega}{\tgt \omega}{\tgt \omega}{\omega .}
                    {\tgt}{\src_{[]}}{\tgt}{\tgt}
            \]
        \end{enumerate}
    \end{enumerate}
\end{definition}

\begin{remark}
	\label{rem:o}
	Let us explain this definition a little more. Opetopes are trees whose
	nodes (and edges) are decorated by opetopes. The decoration is now
	interpreted as a geometrical feature, namely as an embedding of a lower
	dimensional opetope. Further, the target of an opetope, while not an
	intrinsic data, is also represented as an embedding. The relations can be
	understood as follows.
	\begin{enumerate}
		\item \condition{Inner} The inner edge at $[p[q]] \in
		\omega^\nodesymbol$ is decorated by the target of the decoration of the
		node ``above'' it (here $\src_{[p[q]]} \omega$), and in the
		$[q]$-source of the node ``below'' it (here $\src_{[p]} \omega$). By
		construction, those two decorations match, and this relation makes the
		two corresponding embeddings $\src_{[q]} \src_{[p]} \omega
		\longrightarrow \omega$ match as well. On the left is an informal
		diagram about $\omega$ as a tree (reversed gray triangle), and on the
		right is an example of pasting diagram represented by an opetope, with
		the relevant features of the \condition{Inner} relation colored or
		thickened.
		\[
			\tikzinput{optids-informal}{inner}
			\qqquad
			\tikzinput{optids-informal}{inner.ps}
		\]
		\item \condition{Glob1-2} If we consider the underlying tree of
		$\omega$ (which really is $\omega$ itself) as its ``geometrical
		source'', and the corolla $\ytree{\tgt \omega}$ as its ``geometrical
		target'', then they should be parallel. The relation \condition{Glob1}
		expresses this idea by ``gluing'' the root edges of $\omega$ and
		$\ytree{\tgt \omega}$ together, while \condition{Glob2} glues the
		leaves according to $\readdress_\omega$.
		\[
			\tikzinput{optids-informal}{glob1}
			\qqquad
			\tikzinput{optids-informal}{glob1.ps}
		\]
		\[
			\tikzinput{optids-informal}{glob2}
			\qqquad
			\tikzinput{optids-informal}{glob2.ps}
		\]
		\item \condition{Degen} If $\omega$ is a degenerate opetope, depicted
		as on the right, then its target should be a ``loop'', i.e. its only
		source and target should be glued together.
		\[
			\tikzinput{optids-informal}{degen}
			\qqquad\qqquad
			\tikzinput{optids-informal}{degen.ps}
		\]
	\end{enumerate}
\end{remark}

\begin{notation}
    \label{not:subcategories-of-o}
    For $n \in \bbNN$, let $\bbOO_{\leq n}$ be the full subcategory of $\bbOO$
    spanned by opetopes of dimension at most $n$. We define $\bbOO_{< n}$,
    $\bbOO_{\geq n}$, and $\bbOO_{> n}$ similarly. We also consider the set
    $\bbOO_n$ as a discrete category.
\end{notation}


TITLE: Reconciling the 'column' and the 'row' pictures for matrices
QUESTION [3 upvotes]: I understand the column picture (namely considering a matrix equation $Ax=b$ as a vector equation with each column of the matrix $A$ scalarly multiplied by each coefficient of the vector $x$ to give the vector $b$) and the row picture (intersection of the planes, or hyperplanes/lines according to the dimensions).
But since these are equivalent, there has to be some way to show this equivalency? What I mean is, the vectors in the column picture must somehow directly correspond to the planes in the row picture, and the procedures we follow in either of the pictures must be equivalent. Is there an easy way to visualise this equivalency?
I do not mean equivalence, in terms of determining the solutions, that is, to determine if the planes are parallel based on the column vectors. What I mean is to get to a system/visualisation, which can connect the two pictures.
The row and the column pictures are as mentioned in the Introduction to Linear Algebra by Gilbert Strang.

REPLY [6 votes]: You must realise that a $n\times m$ matrix $A$ encodes a linear map $\Bbb R^m\to\Bbb R^n$ (note the order of $m,n$) and that the spaces at departure and arrival are otherwise unrelated. Moreover, matrices are usually just a tool to numerically represent a linear map $f:V\to W$ between two more abstract vector spaces, after choosing a basis in both; that actual matrix entries one gets depend heavily on the basis used. As a consequence there cannot be a direct geometrical relation between columns, which represent vectors of $W$ in coordinates with respect to the basis chosen there, and rows, which represent linear combinations of coordinate functions (for the chosen basis) on$~V$.
Nonetheless there is a way to understand what is going on from two points of view.
Solving a system of equation means trying to find the pre-image$~x$ (which must live in$~V$) under $f$ of a given vector $y\in W$.
The column picture says that the columns of $A$ describe particular vectors in$~W$ that are $f$-images of the chosen basis in$~V$; the question then is to find a linear combination of these that equals the given$~y$, and the corresponding linear combination of the basis vectors of $V$ will be our solution $x$ (of course there might be no solution, or infinitely many of them).
The row picture uses the coordinates in$~W$ with respect to the chosen basis, one at a time. In order for an $f$-image to match$~y$, each of its coordinates must be right. Computing one coordinate of the $f$-image of an unknown $x$ amounts to computing a linear combination of the coordinates of$~x$ with respect to the basis chosen in $V$, and the coefficients of such a linear combination are given by a row of $A$.
Requiring that one coordinate of the image matches the one of $y$ means that this linear combination of the coordinates of$~x$ must have a specific value, and that (usually) defines a plane in $V$. Combining these requirements means intersecting the planes in $V$ to find the set of possibilities for$~x$.


TITLE: How do you find the area?
QUESTION [0 upvotes]: I tried to use trigonometry to find the area but I found no formula.How can I can find the area of this triangle?[![enter image description here][2]][2]

REPLY [0 votes]: From figure you can see,
$$AD/CD=\cot \angle A$$ & $$BD/CD=\cot \angle B$$
So
$$AB/CD=(AD+BD)/CD=\cot \angle A+\cot \angle B$$
Applying this you get,
$$4.7/CD=\cot 55^° + \cot 33^° \implies CD=4.7/(\cot 55^° + \cot 33^°)$$
Hence area of the given triangle be:
$$\frac 12 . AB.CD=\frac 12 .(4.7).\frac{4.7}{(\cot 55^° + \cot 33^°)}$$