TITLE: Show that the permutation symbol, $\epsilon_{ijk}$ can be expressed as $\det\{\delta_{k,l}\}$
QUESTION [1 upvotes]: This question relates to this continuum mechanics online course. So far it has excellent lectures and problem sets but no solutions.

Problem 3.
Show that the permutation symbol, $\epsilon_{ijk}$ can be expressed as:
$$ \epsilon_{ijk} =  \begin{vmatrix}
                            \delta_{i1} & \delta_{i2} & \delta_{i3} \\
                            \delta_{j1} & \delta_{j2} & \delta_{j3} \\
                            \delta_{k1} & \delta_{k2} & \delta_{k3} \\
                       \end{vmatrix}$$

I am also posting my answer. I am working on my rigorous mathematical writing as well as the continuum mechanics, so I would appreciate feedback of my proof-writing style as well.

REPLY [2 votes]: Starting with the conjecture:
$$ \epsilon_{ijk} =  \begin{vmatrix}
                            \delta_{i1} & \delta_{i2} & \delta_{i3} \\
                            \delta_{j1} & \delta_{j2} & \delta_{j3} \\
                            \delta_{k1} & \delta_{k2} & \delta_{k3} \\
                       \end{vmatrix}$$
It should be true that:
$$ \epsilon_{ijk}A_iB_jC_k =  \begin{vmatrix}
                                  \delta_{i1} & \delta_{i2} & \delta_{i3} \\
                                  \delta_{j1} & \delta_{j2} & \delta_{j3} \\
                                  \delta_{k1} & \delta_{k2} & \delta_{k3} \\
                              \end{vmatrix}A_iB_jC_k$$
We can expand the left hand side of the statement to be the following, keeping in mind that even though there are 27 permutations of $ijk$, only the following six of them are non-zero.
$$\epsilon_{ijk}A_iB_jC_k = A_1B_2C_3 + A_2B_3C_1 + A_3B_1C_2 - A_3B_2C_1 - A_2B_1C_3 - A_1B_3C_2$$
Likewise expanding the right hand side of statement:
$$\begin{vmatrix}
       \delta_{i1} & \delta_{i2} & \delta_{i3} \\
       \delta_{j1} & \delta_{j2} & \delta_{j3} \\
       \delta_{k1} & \delta_{k2} & \delta_{k3} \\
  \end{vmatrix}A_iB_jC_k =
  (\delta_{i1}\delta_{j2}\delta_{k3} +
  \delta_{i2}\delta_{j3}\delta_{k1} +
  \delta_{i3}\delta_{j1}\delta_{k2} -
  \delta_{i3}\delta_{j2}\delta_{k1} -
  \delta_{i2}\delta_{j1}\delta_{k3} -
  \delta_{i1}\delta_{j3}\delta_{k2})A_iB_jC_k$$
$$\begin{vmatrix}
       \delta_{i1} & \delta_{i2} & \delta_{i3} \\
       \delta_{j1} & \delta_{j2} & \delta_{j3} \\
       \delta_{k1} & \delta_{k2} & \delta_{k3} \\
  \end{vmatrix}A_iB_jC_k =
  A_1B_2C_3 + A_2B_3C_1 + A_3B_1C_2 - A_3B_2C_1 - A_2B_1C_3 - A_1B_3C_2$$
$$\therefore$$
$$\epsilon_{ijk}A_iB_jC_k = \begin{vmatrix}
       \delta_{i1} & \delta_{i2} & \delta_{i3} \\
       \delta_{j1} & \delta_{j2} & \delta_{j3} \\
       \delta_{k1} & \delta_{k2} & \delta_{k3} \\
  \end{vmatrix}A_iB_jC_k$$
$$\epsilon_{ijk} = \begin{vmatrix}
       \delta_{i1} & \delta_{i2} & \delta_{i3} \\
       \delta_{j1} & \delta_{j2} & \delta_{j3} \\
       \delta_{k1} & \delta_{k2} & \delta_{k3} \\
  \end{vmatrix}$$
QED


TITLE: (Vectors) Finding the scalar of a perpendicular vector
QUESTION [0 upvotes]: Let u = (u1, u2, u3) v = (v1, v2, v3) and w = (w1, w2, w3) be non collinear vectors.
Find the scalar $\alpha$ so that u + $\alpha$ v is perpendicular to w.
I know that for a vector to be perpendicular to another, their dot product needs to be 0.
I get something like:
(u1 + $\alpha$ v1)(w1) = 0
(u2 + $\alpha$ v2)(w2) = 0
(u3 + $\alpha$ v3)(w3) = 0
My problem is that I don't know how to isolate the scalar in the formula.
In this exercise, the real values are u = (2,3,1) v = (0,2,3) and z = (4,3,3)
Any tips on how to solve this?

REPLY [0 votes]: You shouldn’t be getting a system of linear equations. Remember that the dot product produces a scalar, not a vector, so $$\begin{align}(\mathbf u+\alpha\mathbf v)\cdot\mathbf w &= \mathbf u\cdot\mathbf w+\alpha(\mathbf v\cdot\mathbf w) \\ &= (u_1w_1+u_2w_2+u_3w_3)+\alpha(v_1w_1+v_2w_2+v_3w_3).\end{align}$$


TITLE: $Y^3$ congruent to $1 \pmod {p}$
QUESTION [3 upvotes]: How to get the condition on $p$ for which $y^3$ congruent to $1$ modulo $p$ has $3$ solutions ( $1$ solution $x= 1$ is always possible, right ?).

REPLY [0 votes]: The group $\mathbb{Z}/p\mathbb{Z}^*$ is cyclic of order $p-1$ (look up primitive element/root). The question is about the existence of elements of order 3 (the element of order 1 is always there). The basic theory of cyclic groups tells that these exist, iff $3\mid p-1$ or, equivalently, $p\equiv1\pmod3$.
If $p$ does not satisfy this congruence, then the primitive cubic roots of unity reside in the (unique up to isomorphism) quadratic extension field $GF(p^2)$.


TITLE: How would I explain Ohm's Law in terms of Electrical Fields and Force?
QUESTION [1 upvotes]: In terms of current, resistance, and voltage, it's easy: Ohm's Law is the relationship between current, voltage, and resistance of a circuit. Boom, simple as that. How could I put this in terms of $E$ and $F$? I can sort of see a way to do it by relating the formulas $E=F/q$ and $I=q/t$ to Ohm's Law, $V=IR$, but I'm not entirely sure how I could explain this in words.

REPLY [1 votes]: Imagine at the face of the resistor that $N$ electrons each with charge $q$ are collected and move along it with a constant average drift velocity $v$. So in time $dt$ a charge $dQ$ of $N q vdt$ moves past any point, that is, the current $I = \frac {dQ}{dt} = N q v$.
Start with Peltio’s first equation above assuming that $v’ = 0$ (a reasonable approximation), so that
$v = - q E / (m*gamma)$, where m is the mass of the electron.
So $I$ and $v$ are shown to be proportional. Now we finish the argument by showing that also the voltage $V$ across the resistor and v are proportional as follows:
The force $F$ on a single electron $F$ is $q E$, and the work $W$ to move it along a distance $s$, the length of the resistor, is $F \cdot s$. So $W = q E \cdot s$ = $- m*gamma*v*s$. So the voltage $V = \frac{W}{q}$$ = - m*gamma*v*s$$ = -\frac{m*gamma*s*I} {(N q)}$.
Finally,
$V = I [\frac{m*gamma*s}{N (-q)^2}] = I R$, where R is the bracketed expression.


TITLE: Derivatives of polar coordiantes?
QUESTION [1 upvotes]: I'm a undergraduate and I was reading about the polar coordinate system specifically this paper. I don't understand the term: $$\frac{de_r}{d\theta} =  e_\theta \text{,  and } \frac{de_\theta}{d\theta} = -e_r$$
I don't see how you can have a the derivative of $de_r$ over $d\theta$ since they are not the same variable, if that makes any sence.

REPLY [2 votes]: In reference to the diagram, ask yourself what happens to the unit vector $\hat{e}_r$ as you increase the angle $\varphi$ (as measured counterclockwise from the rightward pointing horizontal axis) by an infinitesimal amount $d\varphi$. Note that they use $\varphi$ in this diagram instead of $\theta$.
In words, the tail of the vector $\hat{e}_r$ moves counterclockwise along the circumference of the circle, with its tip always pointing radially outwards from the centre.
The difference between the transformed $\hat{e}_r$ (call it $\hat{e}_{r1}$), and the original $\hat{e}_r$ (call it $\hat{e}_{r0}$), is another vector (call it $d\hat{e}_r$), which points from the tip of $\hat{e}_{r0}$ to the tip of $\hat{e}_{r1}.$ That is, $d\hat{e}_r=\hat{e}_{r1}-\hat{e}_{r0}.$
In the limit as $d\varphi\rightarrow 0,$ that is, as $\hat{e}_{r1}$ and $\hat{e}_{r0}$ become closer and closer together, $d\hat{e}_r$ becomes closer and closer to a vector pointing perpendicular to $\hat{e}_{r0}$, i.e. it becomes close and closer to the vector $\hat{e}_\varphi$. Draw some pictures to convince yourself of this
In other words, as $d\varphi\rightarrow 0,$ $d\hat{e}_r\rightarrow \hat{e}_\varphi,$ which is to say that $$\frac{d\hat{e}_r}{d\varphi} = \hat{e}_\varphi.$$

Reference for the diagram
Edit:
To address @R.W.Bird's comment. I'll try to show that the magnitude of $\frac{d\hat{e}_r}{d\varphi} = \hat{e}_\varphi$ indeed is 1.
The magnitude of $d\hat{e}_r$ is given by
$$ \left|{d\hat{e}_r}\right|=\left|\hat{e}_{r1}-\hat{e}_{r0}\right|=\sqrt{\hat{e}_{r1}^2-2\hat{e}_{r1}\cdot\hat{e}_{r0}+\hat{e}_{r0}^2}.$$
If we use the fact that by definition $\left|\hat{e}_{r}\right|=1$, then
$$ \left|{d\hat{e}_r}\right|=\sqrt{1-2\cos(d\varphi)+1}=\sqrt{2-2\cos(d\varphi)},$$
since the angle between $\hat{e}_{r0}$ and $\hat{e}_{r1}$ is $d\varphi$. Now,
$$ \left|\frac{{\left|d\hat{e}_r\right|}}{d\varphi}\right|=\left|\frac{d}{d\varphi}\sqrt{2-2\cos(d\varphi)}\right|=\left|\frac{\sin(d\varphi)}{\sqrt{2-2\cos(d\varphi)}}\right|.$$
As $d\varphi\rightarrow 0,$
$$\lim_{d\varphi\rightarrow 0}\left|\frac{{\left|d\hat{e}_r\right|}}{d\varphi}\right|=\lim_{d\varphi\rightarrow 0}\left|\frac{\sin(d\varphi)}{\sqrt{2-2\cos(d\varphi)}}\right|$$
$$=\lim_{d\varphi\rightarrow 0}\left|\frac{d}{d\varphi}\frac{\sin(d\varphi)}{\sqrt{2-2\cos(d\varphi)}}\right|$$
$$=\lim_{d\varphi\rightarrow 0}\left|\frac{\cos (d\varphi)}{\sqrt{2-2 \cos (d\varphi)}}-\frac{\sin ^{2}(d\varphi)}{(2-2 \cos (d\varphi))^{3 / 2}}\right|$$
$$=1$$
where from the first to second line we have used l'Hopital's rule. This argument is a little hand-wavy, but it gets us the right answer... sorry about that.

REPLY [0 votes]: $\hat{e}_r$ is a radial unit vector. It obviously points in different directions for different values of $\theta$. So it has a derivative expressing that rate of change.


text:
\section{The equivalence}
\label{sec:equialence}

We now aim to prove that the category of opetopic sets, i.e. $\Set$-presheaves
over the category $\bbOO$ defined previously, is equivalent to the category of
many-to-one polygraphs $\PolMTO$. We achieve this by first constructing the
\emph{polygraphic realization} functor $|-| : \bbOO \longrightarrow \PolMTO$ in
\cref{sec:polygraphic-realization}. This functor ``realizes'' an opetope as a
polygraph in that it freely implements all its tree structure by means of
adequately chosen generators in each dimension. Secondly, writing $\PshO =
\Set^{\bbOO^\op}$ we consider the left Kan extension $| - | : \PshO
\longrightarrow \PolMTO$ along the Yoneda embedding. This functor has a right
adjoint, the ``opetopic nerve'' $N : \PolMTO \longrightarrow \PshO$, and we
prove this adjunction to be an adjoint equivalence. This is done using the
shape function, defined in \cref{sec:shape}, which to any generator
$x$ of a many-to-one polygraph $P$ associates an opetope $\shape{x}$ along with
a canonical morphism $\tildX : |\shape{x}| \longrightarrow P$.

\subsection{Polygraphic realization}
\label{sec:polygraphic-realization}

An opetope $\omega \in \bbOO_n$, with $n \geq 1$, has one target $\tgt \omega$,
and sources $\src_{[p]} \omega$ laid out in a tree. If the sources $\src_{[p]}
\omega$ happened to be generators in some polygraph, then that tree would
describe a way to compose them. With this in mind, we define a many-to-one
polygraph $|\omega|$, whose generators are essentially iterated faces (i.e.
sources or targets) of $\omega$ (hypothesis \condition{PR1} below). Moreover,
$|\omega|$ will be ``maximally unfolded'' (or ``free''), in that two (iterated)
faces that are the same opetope, but located at different addresses, will
correspond to distinct generators.

The rest of this section is devoted to inductively define the realization
functor $|-|$ together with its \emph{boundary} $\partial |-|$. We bootstrap
the process with \cref{def:polygraphic-realization-low-dimension}, state our
induction hypothesis in \cref{axiom:polygraphic-realization}.

\begin{definition}
	[Low dimensional cases]
	\label{def:polygraphic-realization-low-dimension}
	For $\optZero$ the unique $0$-opetope, let $\partial |\optZero|$ be the
	empty polygraph, and $|\optZero|$ be the polygraph with a unique generator
	in dimension $0$, which we denote by $\optZero$. For $\optOne$ the unique
	$1$-opetope, let $\partial |\optOne| \eqdef |\optZero| + |\optZero|$, and
	let $|\optOne|$ be the cellular extension
	\[
		\partial |\optOne| \xleftarrow{\src, \tgt} \{ \optOne \} ,
	\]
	where $\src$ and $\tgt$ map $\optOne$ to distinct $0$-generators. There are
	obvious functors $|\src_{[]}|, |\tgt| : |\optZero| \longrightarrow
	|\optOne|$, mapping $\optZero$ to $\src \optOne$ and $\tgt \optOne$,
	respectively.
\end{definition}

At this stage, we defined a functor $|-| : \bbOO_{\leq 1} \longrightarrow
\PolMTO$ (\cref{not:subcategories-of-o}). Let $N \geq 2$ and assume by
induction that $\partial |-|$ and $|-|$ are defined on $\bbOO_{< n}$. Assume
further that the following induction hypothesis hold (they are easily verified
for $N = 2$).

\begin{axiom}
	\label{axiom:polygraphic-realization}
	For all $\psi \in \bbOO_k$ with $k < N$, the following hold
	\begin{enumerate}
		\item \condition{PR0}\index{\condition{PR0}} the polygraphs $\partial
		|\psi|$ and $|\psi|$ are many-to-one, and $\partial |\psi|$ is the
		sub-polygraph of $|\psi|$ spanned by its $k$-generators;
		\item \condition{PR1}\index{\condition{PR1}} for all $j \in \bbNN$, the
		set $|\psi|_j$ of $j$-generators of $|\psi|$ is the slice $\bbOO_j /
		\psi$, i.e. of the form $\left( \phi \xrightarrow{\sfA} \psi \right)$
		for $\phi \in \bbOO_j$ and $f : \phi \longrightarrow \psi$ a morphism
		in $\bbOO$;
		\item \condition{PR2}\index{\condition{PR2}} for $\left( \phi
		\xrightarrow{\sfA} \psi \right)$ a generator of $|\psi|$, its target is
		$\tgt \left( \phi \xrightarrow{\sfA} \psi \right) = \left( \tgt \phi
		\xrightarrow{\tgt} \phi \xrightarrow{\sfA} \psi \right)$;
		\item \condition{PR3}\index{\condition{PR3}} for $l \leq k$, and for $\left( \phi \xrightarrow{\sfA}
		\psi \right)$ a $l$-generator of $|\psi|$, the composition tree of its
		source $\ct \src \left(\phi \xrightarrow{\sfA} \psi \right) \in \tree
		\nabla_{l - 1} |\psi|$ is
		\begin{align*}
			\underlyingtree{\phi} &\longrightarrow \nabla_{l - 1} |\psi| \\
			[p] &\longmapsto \left(
				\src_{[p]} \phi \xrightarrow{\src_{[p]}} \phi \xrightarrow{\sfA} \psi
			\right) .
		\end{align*}
		Recall that by \cref{prop:composition-tree-duality}, this completely
		determines $\src \left(\phi \xrightarrow{\sfA} \psi \right) \in
		|\psi|_{l-1}^*$.
	\end{enumerate}
\end{axiom}

We now define $\partial |\omega|$ and $|\omega|$ when $\omega \in \bbOO_N$.
Defining the former is easy, and done in
\cref{def:polygraphic-realization:inductive-boundary}. The latter is defined in
\cref{def:polygraphic-realization:inductive} as a cellular extension
\[
	\partial |\omega| \xleftarrow{\src, \tgt} \left\{ \omega \right\}
\]
of $\partial |\omega|$, where the target and source of the new generator are
given by \condition{PR2} and \condition{PR3}. Lastly, we check the inductive
hypothesis in \cref{prop:polygraphic-realization:inductive}.

\begin{definition}
	[Inductive step for $\partial |-|$]
	\label{def:polygraphic-realization:inductive-boundary}
	For $\omega \in \bbOO_N$, let
	\[
		\partial |\omega| \eqdef \colim_{\psi \in \bbOO_{< N} / \omega} |\psi| .
	\]
	For $\sfA : \psi \longrightarrow \omega$ in $\bbOO_{< N} / \omega$, this
	colimit comes with a corresponding coprojection $|\sfA| : |\psi|
	\longhookrightarrow \partial |\omega|$.
\end{definition}

\begin{lemma}
	\label{lemma:polygraphic-realization:generators-of-boundary}
	For $\omega \in \bbOO_N$, and $j < N$, the set $\partial |\omega|_j$ of
	$j$-generators of $\partial |\omega|$ is the slice $\bbOO_j / \omega$.
\end{lemma}
\begin{proof}
	Follows from the induction hypothesis \condition{PR1} and
	\cref{th:structure-of-mto}.

\end{proof}

\begin{corollary}
	\label{lemma:polygraphic-realization:nabla-boundary}
	For $\omega \in \bbOO_N$ and $1 \leq k < N$, the polynomial functor
	$\nabla_k \partial |\omega|$ is described as follows:
	\[
		\polynomialfunctor
			{\bbOO_{k-1} / \omega}{E}{\bbOO_k / \omega}
				{\bbOO_{k-1} / \omega}
			{\src}{p}{\tgt}
	\]
	where for $\left( \psi \xrightarrow{\sfA} \omega \right) \in \bbOO_k
	\omega$,
	\begin{enumerate}
		\item the fiber $E \left( \psi \xrightarrow{\sfA} \omega \right)$ is
		simply $\psi^\nodesymbol$;
		\item for $[p] \in E \left( \psi \xrightarrow{\sfA} \omega \right)
		\cong \psi^\nodesymbol$, we have $\src [p] = \left( \src_{[p]} \psi
		\xrightarrow{\src_{[p]}} \psi \xrightarrow{\sfA} \omega \right)$;
		\item $\tgt \left( \psi \xrightarrow{\sfA} \omega \right) = \left( \tgt
		\psi \xrightarrow{\tgt} \psi \xrightarrow{\sfA} \omega \right)$.
	\end{enumerate}
\end{corollary}
\begin{proof}
	Direct consequence of
	\cref{lemma:polygraphic-realization:generators-of-boundary} and
	\condition{PR1}, \condition{PR2}, and \condition{PR3}.
\end{proof}

\begin{definition}
	\label{def:polygraphic-realization:nabla-boundary:forgetful}
	For $\omega \in \bbOO_N$ and $1 \leq k < N$, we have a morphism $u :
	\nabla_k \partial |\omega| \longrightarrow \optPolyFun^{k-1}$
	\[
		\begin{tikzcd}
			\bbOO_{k-1} / \omega
				\ar[d, "u_0"'] &
			E
				\ar[r, "p"]
				\ar[d, "u_2" left]
				\ar[l, "\src" above] &
			\bbOO_k / \omega
				\ar[r, "\tgt"]
				\ar[d, "u_1" left] &
			\bbOO_{k-1} / \omega
				\ar[d, "u_0" left] \\
			\bbOO_{k-1} &
			\bbOO_k^\nodesymbol
				\ar[r, "p"]
				\ar[l, "\src" above] &
			\bbOO_k
				\ar[r, "\tgt"] &
			\bbOO_{k-1}
		\end{tikzcd}
	\]
	induced by the forgetful maps $\bbOO_{k-1} / \omega \longrightarrow
	\bbOO_{k-1}$ and $\bbOO_k / \omega \longrightarrow \bbOO_k$.
\end{definition}

\begin{lemma}
	\label{lemma:polygraphic-realization:lift}
	Let $\omega \in \bbOO_N = \tree \optPolyFun^{N-2}$. The map $\omega :
	\underlyingtree{\omega} \longrightarrow \optPolyFun^{N-2}$ factors through
	$u : \nabla_{N-1} \partial |\omega| \longrightarrow \optPolyFun^{N-2}$
	(\cref{def:polygraphic-realization:nabla-boundary:forgetful}):
	\[
		\triangleDRdiagram
			{\nabla_{N-1} \partial |\omega|}{\underlyingtree{\omega}}
				{\optPolyFun^{N-2} .}
			{\baromega}{u}{\omega}
	\]
\end{lemma}
\begin{proof}
	Let $\baromega$ map $[p] \in \omega^\nodesymbol$ to $\left( \src_{[p]}
	\omega \xrightarrow{\src_{[p]}} \omega \right) \in \bbOO_{N-1} / \omega$,
	and map an edge $[l]$ to $\left( \edg_{[l]} \omega \xrightarrow{\edg_{[l]}}
	\omega \right) \in \bbOO_{N-2} / \omega$ (\cref{not:edg}).
\end{proof}

\begin{proposition}
	\label{prop:polygraphic-realization:parallel}
	On the one hand, consider the tree $\nabla_{N-1} \partial |\omega|$-tree
	$\baromega$ of \cref{lemma:polygraphic-realization:lift}, and on the other
	hand, recall that there is a $(N-1)$-generator $\left( \tgt \omega
	\xrightarrow{\tgt} \omega \right)$ of $\partial |\omega|$ corresponding to
	the target embedding of $\omega$. Then, in $\partial |\omega|$, the
	composite $\baromega^\circ$ (\cref{def:composition}) and the generator
	$\left( \tgt \omega \xrightarrow{\tgt} \omega \right)$ are parallel.
\end{proposition}
\begin{proof}
	If $\omega$ is degenerate, say $\omega = \itree{\phi}$ for some $\phi \in
	\bbOO_{N-2}$, then $\composition{\baromega} = \id_{( \phi \xrightarrow{\tgt
	\tgt} \omega )}$, while $\left( \tgt \omega
	\xrightarrow{\tgt} \omega \right) = \left( \ytree{\phi}
	\xrightarrow{\tgt} \omega \right)$. By \condition{Degen}, those two
	cells are parallel.

	For the rest of the proof, we assume that $\omega$ is not degenerate.
	First, we have
	\[
		\tgt \composition{\baromega}
		= \tgt \src_{[]} \baromega
		= \tgt \left( \tgt \src_{[]} \omega
			\xrightarrow{\tgt \src_{[]}} \omega \right)
		= \left( \tgt \tgt \omega \xrightarrow{\tgt \tgt} \omega \right)
		= \tgt \left( \tgt \omega \xrightarrow{\tgt} \omega \right) .
	\]
	Then, in order to show that $\src \composition{\baromega} = \src (\tgt
	\omega \xrightarrow{\tgt} \omega)$, we show that the $(N-2)$-generators
	occurring both sides are the same, and that the way to compose them is
	unique.
	\begin{enumerate}
		\item Generators in $\src \composition{\baromega}$ are of the form
		$\left( \phi \xrightarrow{[q]} \psi \xrightarrow{[p]} \omega \right)$,
		for $[p [q]] \in \omega^\leafsymbol$. By \condition{Glob2}, those are
		equal to $\left( \phi \xrightarrow{\readdress_\omega [p[q]]} \tgt
		\omega \xrightarrow{\tgt} \omega \right)$, which are exactly the
		generators in $\src \left( \tgt \omega \xrightarrow{\tgt} \omega
		\right)$.

		\item To show that there is a unique way to compose all the
		$(N-2)$-generators of the form $(\phi \xrightarrow{\src_{[q]}} \psi
		\xrightarrow{\src_{[p]}} \omega)$, where $[p[q]]$ ranges over
		$\omega^\leafsymbol$, it is enough to show that no two have the same
		target. Assume $(\phi_i \xrightarrow{\src_{[q_i]}} \psi_i
		\xrightarrow{\src_{[p_i]}} \omega)$, with $i = 1, 2$, are
		$(N-2)$-generators occuring in $\src \composition{\baromega}$ with the
		same target. Consider the following diagram:
		\[
			\begin{tikzcd} [column sep = large]
				&
				\phi_1
					\ar[r, "\src_{[q_1]}"]
					\ar[dr, "\src_{[r_1]}"'] &
				\psi_1
					\ar[dr, "\src_{[p_1]}"] \\
				\rho
					\ar[ur, "\tgt"]
					\ar[dr, "\tgt"'] &
				&
				\tgt \omega
					\ar[r, "\tgt"] &
				\omega \\
				&
				\phi_2
					\ar[r, "\src_{[q_2]}"]
					\ar[ur, "\src_{[r_2]}"] &
				\psi_2
					\ar[ur, "\src_{[p_2]}"']
			\end{tikzcd}
		\]
		where $[r_i] \eqdef \readdress_\omega [p_i [q_i]] \in \tgt
		\omega^\nodesymbol$. The outer hexagon commutes by assumption, the two
		squares on the right are instances of \condition{Glob2}, and the left
		square commutes as $\tgt : \tgt \omega \longrightarrow \omega$ is a
		mono, since $\omega$ is non degenerate. By inspection of the opetopic
		identities, the only way for the left square to commute is the trivial
		way, i.e. $[r_1] = [r_2]$. Since $\readdress_\omega$ is a bijection, we
		have $[p_1 [q_1]] = [p_2 [q_2]]$, thus $[p_1] = [p_2]$ and $[q_1] =
		[q_2]$.
		\qedhere
	\end{enumerate}
\end{proof}

\begin{definition}
	[Inductive step for $|-|$]
	\label{def:polygraphic-realization:inductive}
	For $\omega \in \bbOO_N$, let $|\omega|$ be the cellular extension
	\[
		\partial |\omega| \xleftarrow{\src, \tgt} \left\{ \omega \right\} ,
	\]
	where $\tgt$ maps $\omega$ to the $(N-1)$-generator $\left( \tgt \omega
	\xrightarrow{\tgt} \omega \right)$, and where the composition tree of $\src
	\omega$ is $\baromega$ (\cref{lemma:polygraphic-realization:lift}). For
	consistency, we also write $\left( \omega \xrightarrow{\id} \omega \right)$
	for the unique $N$-generator of $|\omega|$. This is well defined by
	\cref{prop:polygraphic-realization:parallel}, and gives a functor $|-| :
	\bbOO_{\leq N} \longrightarrow \Pol$.
\end{definition}

\begin{proposition}
	\label{prop:polygraphic-realization:inductive}
	For $\omega \in \bbOO_N$, the polygraphs $\partial |\omega|$
	(\cref{def:polygraphic-realization:inductive-boundary}) and $|\omega|$
	(\cref{def:polygraphic-realization:inductive}) satisfy the inductive
	hypothesis of \cref{axiom:polygraphic-realization}.
\end{proposition}
\begin{proof}
	\begin{enumerate}
		\item \condition{PR0} By definition.
		\item \condition{PR1} For $j < N$, by
		\cref{lemma:polygraphic-realization:generators-of-boundary}, we already
		have $|\omega|_j = \partial |\omega|_j = \bbOO_j / \omega$. In
		dimension $N$, the only element of $\bbOO_N / \omega$ is $\id : \omega
		\longrightarrow \omega$, which corresponds to the unique $N$-generator
		of $|\omega|$. If $j > N$, then both $\bbOO_j / \omega$ and
		$|\omega|_j$ are empty.
		\item \condition{PR2} and \condition{PR3}. By definition, those
		hypothesis hold for the unique $N$-generator $\left( \omega
		\xrightarrow{\id} \omega \right)$ of $|\omega|$. By induction, they
		also hold on the other generators.
		\qedhere
	\end{enumerate}
\end{proof}

To conclude, we defined a functor $|-| : \bbOO \longrightarrow \PolMTO$ which
satisfies \cref{axiom:polygraphic-realization} for all $N \in \bbNN$.

\subsection{The shape function}
\label{sec:shape}

This subsection is devoted to define the \emph{shape function} $\shape{(-)}$.
We first sketch the idea. Take $P \in \PolMTO$ and define $\shape{(-)} : P_n
\longrightarrow \bbOO_n$ by induction. The cases $n = 0, 1$ are trivial, since
there is a unique $0$-opetope and a unique $1$-opetope. Assume $n \geq 2$, and
take $x \in P_n$. Then the composition tree of $\src x$ is a coherent tree
whose nodes are $(n-1)$-generators, and edges are $(n-2)$-generators. Replacing
those $(n-1)$ and $(n-2)$-generators by their respective shape, we obtain a
coherent tree whose nodes are $(n-1)$-opetopes, and edges are $(n-2)$-opetopes,
in other words, we obtain an $n$-opetope, which we shall denote by $\shape{x}$.
\begin{center}
	\tikzinput{shape}{}
\end{center}
The fact that $\shape{x}$ corresponds to the intuitive notion of ``shape'' of
$x$ is justified by \cref{th:shape}. The rest of this subsection makes this
sketch formal.

We first define a many-to-one polygraph $\termMTO$, that will turn out in
\cref{prop:terminal-mtop} to be terminal in $\PolMTO$. We then proceed to
define the shape function for $\termMTO$, before stating the general case.

\begin{definition}
	[The $\termMTO$ polygraph]
	\label{def:terminal-mtop}
	We define a polygraph $\termMTO \in \PolMTO$\index{$\termMTO$} as follows:
	we first set $\termMTO_0 \eqdef \{ \optZero \}$, and inductively,
	$\termMTO_{n+1} \eqdef \left\{(u, v) \in \termMTO_n^\MTO \times \termMTO_n
	\mid u \parallel v \right\}$
	(\cref{def:many-to-one-polygraph,def:parallel-cells}), with $\src (u, v)
	\eqdef u$ and $\tgt (u, v) \eqdef v$.
\end{definition}

\begin{lemma}
	\label{lemma:terminal-mtop:parallel-generators}
	If $x, y \in \termMTO_n$ are two parallel generators, then they are equal.
\end{lemma}
\begin{proof}
	We have $x = (\src x, \tgt x) = (\src y, \tgt y) = y$.
\end{proof}

\begin{proposition}
	\label{prop:terminal-mtop}
	The polygraph $\termMTO$ is terminal in $\PolMTO$.
\end{proposition}
\begin{proof}
	For $P \in \PolMTO$, we show that there exists a unique $! : P
	\longrightarrow \termMTO$.
	\begin{enumerate}
		\item (Existence) For $x \in P_0$ let $!_0 x \eqdef \optZero$, and for
		$x \in P_n$ with $n \geq 1$, let $!_n x = (!_{n-1} \src x, !_{n-1} \tgt
		x)$. The source and target compatibility is trivial.
		\item (Uniqueness) Let $f : P \longrightarrow \termMTO$ be a morphism
		different from $!$ defined above. Necessarily $f_0 = !_0$ as
		$\termMTO_0$ is a singleton. Let $x \in P_n$ be such that $f_n x \neq
		!_n x$, with $n$ minimal. Then $n \geq 1$, and we have $!_n x =
		(!_{n-1} \src x, !_{n-1} \tgt x) = (f_{n-1} \src x, f_{n-1} \tgt x)
		\parallel f_n x$. By \cref{lemma:terminal-mtop:parallel-generators},
		$!_n x = f_n x$, a contradiction.
		\qedhere
	\end{enumerate}
\end{proof}

\begin{proposition}
	\label{prop:terminal-mtop:opetopes}
	For $x \in \termMTO_n$ there exists a unique $\shape{x} \in \bbOO_n$ such
	that the terminal morphism $!^{\shape{x}} : |\shape{x}| \longrightarrow
	\termMTO$ maps $\shape{x}$ (the unique $n$-generator of $|\shape{x}|$) to
	$x$.
\end{proposition}
\begin{proof}
	\begin{enumerate}
		\item (Uniqueness) Assume that there exists two distinct opetopes
		$\phi, \phi' \in \bbOO_k$ such that $!^\phi_k (\phi) = !^{\phi'}_k
		(\phi')$, with $k$ minimal for this property. Then necessarily, $k \geq
		2$. On the one hand, we have $\underlyingtree{\phi} =
		\underlyingtree{\ct \src \phi} = \underlyingtree{\ct \src \phi'} =
		\underlyingtree{\phi'}$. On the other hand, for $[p] \in
		\phi^\nodesymbol = (\phi')^\nodesymbol$, we have
		\begin{align*}
			!^{\src_{[p]} \phi}_{k-1} \src_{[p]} \phi
			&= !^\phi_{k-1} \src_{[p]} \phi
				& \text{since $!^{\src_{[p]} \phi}$ is also }
				|\src_{[p]} \phi| \hookrightarrow |\phi| \rightarrow \termMTO \\
			&= \src_{[p]} !^\phi_k \phi
				& \text{since $!^\phi$ is a mor. of polygraphs} \\
			&= \src_{[p]} !^{\phi'}_k \phi'
				& \text{by assumption} \\
			&= !^{\phi'}_{k-1} \src_{[p]} \phi'
				& \text{since $!^{\phi'}$ is a mor. of polygraphs} \\
			&= !^{\src_{[p]} \phi'}_{k-1} \src_{[p]} \phi'
				& \text{since $!^{\src_{[p]} \phi'}$ is also }
				|\src_{[p]} \phi'| \hookrightarrow |\phi'| \rightarrow \termMTO,
		\end{align*}
		and by minimality of $k$, we have $\src_{[p]} \phi = \src_{[p]} \phi'$,
		for all address $[p]$. Consequently, $\phi = \phi'$, a contradiction.

		\item (Existence) The cases $n = 0, 1$ are trivial, so assume $n \geq
		2$, and that by induction, the result holds for all $k < n$, i.e. that
		for $g \in P_k$, there is a unique opetope $\shape{g} \in \bbOO_k$ such
		that $!^{\shape{g}}_k \shape{g} = g$. In particular the following two
		triangles commute:
		\[
			\triangleURdiagram
				{|\src_{[p]} \shape{g}|}{|\shape{g}|}{P,}{|\src_{[p]}|}
				{!^{\src_{[p]} \shape{g}}}{!^{\shape{g}}}
			\qqquad
			\triangleURdiagram
				{|\tgt \shape{g}|}{|\shape{g}|}{P ,}
				{|\tgt|}{!^{\tgt \shape{g}}}{!^{\shape{g}}}
		\]
		where $[p] \in (\shape{g})^\nodesymbol$. Consequently,
		$\shape{(\src_{[p]} g)} = \src_{[p]} (\shape{g})$ and $\shape{(\tgt g)}
		= \tgt (\shape{g})$, and the following displays an isomorphism
		$\nabla_{n-1} \termMTO \longrightarrow \optPolyFun^{n-2}$:
		\[
			\begin{tikzcd}
				\termMTO_{n-2}
					\ar[d, "\shape{(-)}"'] &
				\termMTO_{n-1}^\nodesymbol
					\ar[r, "p"]
					\ar[d]
					\ar[l, "\src" above] &
				\termMTO_{n-1}
					\ar[r, "\tgt"]
					\ar[d, "\shape{(-)}" left] &
				\termMTO_{n-2}
					\ar[d, "\shape{(-)}" left] \\
				\bbOO_{n-2} &
				\bbOO_{n-1}^\nodesymbol
					\ar[r, "p"]
					\ar[l, "\src" above] &
				\bbOO_{n-1}
					\ar[r, "\tgt"] &
				\bbOO_{n-2} .
			\end{tikzcd}
		\]
		Hence, the composite $\underlyingtree{\ct \src x} \xrightarrow{\ct \src
		x} \nabla_{n-1} \termMTO \xrightarrow{\shape{(-)}} \optPolyFun^{n-2}$
		defines an $n$-opetope $\shape{x}$ with $\underlyingtree{\shape{x}} =
		\underlyingtree{\ct \src x}$.

		We claim that $!^{\shape{x}}_n \shape{x} = x$. We first show that
		$!^{\shape{x}}_{n-1} \src \shape{x} = \src x$. We have
		\begin{align*}
			\underlyingtree{\ct \src x}
			&= \underlyingtree{\shape{x}}
				& \text{by definition} \\
			&= \underlyingtree{\ct \src \shape{x}}
				& \text{by \condition{PR3}} \\
			&= \underlyingtree{\ct !^{\shape{x}}_{n-1} \src \shape{x}}
				& \text{since $!^{\shape{x}}$ is a mor. of polygraphs.}
		\end{align*}
		Then, for any address $[p]$ in $\underlyingtree{\ct \src x}$, we have
		\begin{align*}
			\src_{[p]} x
			&= !^{\shape{x}}_{n-1} \shape{(\src_{[p]} x)}
				& \text{by induction} \\
			&= !^{\shape{x}}_{n-1} \src_{[p]} \shape{x}
				& \text{by definition of $\shape{x}$} \\
			&= \src_{[p]} !^{\shape{x}}_n \shape{x}
				& \text{since $!^{\shape{x}}$ is a mor. of polygraphs,}
		\end{align*}
		and therefore, by \cref{prop:composition-tree-duality}, $\src x = \src
		(!^{\shape{x}}_n \shape{x})$. Next,
		\begin{align*}
			\tgt (!^{\shape{x}}_n \shape{x})
			&= !^{\shape{x}}_{n-1} \tgt \shape{x}
				& \text{by induction} \\
			&\parallel !^{\shape{x}}_{n-1} \src \shape{x}
				& \text{since } \src \shape{x} \parallel \tgt \shape{x} \\
			&= \src x
				& \text{showed above} \\
			&\parallel \tgt x,
		\end{align*}
		and therefore, $\tgt !^{\shape{x}}_n \shape{x} = \tgt x$. Finally,
		$!^{\shape{x}}_n \shape{x} \parallel x$, and by
		\cref{lemma:terminal-mtop:parallel-generators}, $!^{\shape{x}}_n
		\shape{x} = x$.
		\qedhere
	\end{enumerate}
\end{proof}

\begin{notation}
	In the light of \cref{prop:terminal-mtop:opetopes}, we identify
	$\termMTO_n$ with $\bbOO_n$. This identification is compatible with faces,
	i.e. $\src_{[p]}$ and $\tgt$. Then, $!^\omega : |\omega| \longrightarrow
	\termMTO$ maps a generator $(\phi \rightarrow \omega)$ to $\phi$.
\end{notation}

\begin{theorem}
	\label{th:shape}
	For $P \in \PolMTO$ and $x \in P_n$, there exists a unique pair\footnote{In
	\cite[proposition 2.2.3]{Henry2017}, $\shape{x}$ is written $[x]$.}
	\[
		\left( \shape{x}, |\shape{x}| \xrightarrow{\tildX} P \right)
		\in |-| / P
	\]
	such that $\tildX_n (\shape{x}) = x$. Moreover, $\shape{x} = !^P_n x$,
	where $!^P$ is the terminal map $P \longrightarrow \termMTO$. In
	particular, the map
	\[
		\widetilde{(-)} : P_n \longrightarrow
			\sum_{\omega \in \bbOO_n} \PolMTO (|\omega|, P)
	\]
	is an isomorphism.
\end{theorem}
\begin{proof}
	\begin{enumerate}
		\item (Uniqueness) Assume $|\omega| \xrightarrow{f} P \xleftarrow{f'}
		|\omega'|$ are different morphisms such that $f_n (\omega) = x = f'_n
		(\omega')$. Then $!^\omega_n (\omega) = !^P f_n (\omega) = !^P f'_n
		(\omega') = !^{\omega'} (\omega')$, and by
		\cref{prop:terminal-mtop:opetopes}, $\omega = \omega'$. Let $\left(\phi
		\xrightarrow{\sfA} \omega \right) \in |\omega|_k$ be such that $f_k
		\left( \phi \xrightarrow{\sfA} \omega \right) \neq f'_k \left(\phi
		\xrightarrow{\sfA} \omega \right)$, with $k$ minimal for this property.
		Then $k < n$ (since by assumption $f_n (\omega) = x = f'_n (\omega')$),
		and $\sfA$ factorizes as $\left(\phi \xrightarrow{\sfJ} \psi
		\xrightarrow{\sfB} \omega \right)$, where $\sfJ$ is a face embedding,
		i.e. either $\tgt$ or $\src_{[p]}$ for some $[p] \in
		\omega^\nodesymbol$. Then by assumption,
		\begin{align*}
			f_k \left( \phi \xrightarrow{\sfA} \omega \right)
			&= \sfJ \left( f_{k+1} \left(
				\psi \xrightarrow{\sfB} \omega \right) \right) \\
			&= \sfJ \left( f'_{k+1} \left(
				\psi \xrightarrow{\sfB} \omega \right) \right)
				& \text{by minimality of $k$} \\
			&= f'_k \left( \phi \xrightarrow{\sfA} \omega \right) ,
		\end{align*}
		a contradiction.

		\item (Existence) The cases $n = 0, 1$ are trivial, so assume $n \geq
		2$, and that by induction, the result holds for all $k < n$. Let
		$\shape{x} = !^P_n x \in \bbOO_n$. We wish to construct a morphism $O
		[\shape{x}] \xrightarrow{\tildX} P$ having $x$ in its image. For
		$\left( \psi \xrightarrow{\sfJ} \shape{x} \right)$ a face of
		$\shape{x}$ (i.e. $\tgt$ or $\src_{[p]}$ for some $[p] \in
		(\shape{x})^\nodesymbol$), we have $\shape{(\operatorname{\sfJ} x)} =
		\psi$, so that by induction, there exists a morphism $|\psi|
		\xrightarrow{\widetilde{\operatorname{\sfJ} x}} P$ having
		$\operatorname{\sfJ} x$ in its image, providing a commutative square
		\[
			\squarediagram
				{|\psi|}{P}{|\shape{x}|}{\termMTO .}
				{\widetilde{\sfJ x}}{|\sfJ|}{!^P}{!^{\shape{x}}}
		\]
		To alleviate upcoming notations, write $\bar{\sfJ} \eqdef
		\widetilde{\operatorname{\sfJ} x} : |\psi| \longrightarrow P$. Let
		$\left( \phi \xrightarrow{\sfA} \shape{x} \right) \in \bbOO_{< n} /
		\shape{x}$. If $\sfA$ is a face embedding, define $\bar{\sfA}$ as
		before. If not, then it factors through a face embedding as $\sfA =
		\left( \phi \xrightarrow{\sfJ} \psi \xrightarrow{\sfB} \omega \right)$,
		and let $\bar{\sfA} \eqdef \bar{\sfB} \cdot |\sfJ|$. Then the left
		square commutes, and passing to the colimit over $\bbOO_{< n} /
		\shape{x}$, we obtain the right square:
		\[
			\squarediagram
				{|\phi|}{P}{|\shape{x}|}{\termMTO ,}
				{\bar{\sfA}}{|a|}{!^P}{!^{\shape{x}}}
			\qqquad
			\squarediagram
				{\partial |\shape{x}|}{P}{|\shape{x}|}{\termMTO .}
				{f}{}{!^P}{!^{\shape{x}}}
		\]
		We want a diagonal filler of the right square. Since $|\shape{x}|$ is a
		one-generator cellular extension of $\partial |\shape{x}|$
		(\cref{def:polygraphic-realization:inductive}), it is enough to check
		that $f_{n-1} \src \shape{x} = \src x$, and $f_{n-1} \tgt \shape{x} =
		\tgt x$. The latter is clear, as $f$ extends $\bar{\tgt} : |\tgt
		\shape{x}| \longrightarrow P$, and $f_{n-1} \tgt \shape{x} =
		\bar{\tgt}_{n-1} \tgt \shape{x} = \tgt x$ by definition. We now proceed
		to prove the former. First, $\underlyingtree{\ct \src \shape{x}} =
		\underlyingtree{\ct \src x}$ since both are mapped to the same element
		of $\termMTO_n$. Then, for $[p]$ a node address of $\ct \src
		\shape{x}$, we have $f_{n-1} \src_{[p]} \shape{x} =
		\overline{\src_{[p]}}_{n-1} \src_{[p]} \shape{x} = \src_{[p]} x$. Hence
		$f_{n-1} \src \shape{x} = \src x$.
		\qedhere
	\end{enumerate}
\end{proof}

\subsection{The adjoint equivalence}

\begin{definition}
	[Polygraphic realization-nerve adjunction]
	\label{def:polygraphic-realization-nerve}
	The polygraphic realization functor $|-| : \bbOO \longrightarrow \PolMTO$
	of \cref{sec:polygraphic-realization} extends to a left adjoint
	\[
		| - | : \PshO \adjunction \PolMTO : N ,
	\]
	by left Kan extension of along the Yoneda embedding $\yoneda : \bbOO
	\longrightarrow \PshO$. Explicitly, the polygraphic realization of an
	opetopic set $X \in \PshO$ can be computed with the coend on the left,
	while the and the \emph{polygraphc nerve}\index{polygraphic nerve} $N P$ of
	a polygraph $P \in \PolMTO$ is given on the right:
	\[
		|X| = \int^{\omega \in \bbOO} X_\omega \times |\omega| ,
		\qqquad
		N P = \PolMTO (|-|, P) : \bbOO^\op \longrightarrow \Set .
	\]
	We note $\eta : \id_{\PshO} \longrightarrow N |-|$ the unit, $\epsilon :
	|N| \longrightarrow \id_{\PolMTO}$ the counit, and $\Phi : \PshO (-, N)
	\longrightarrow \PolMTO (| - |, -)$ the natural $\hom$-set isomorphism.
\end{definition}

\begin{notation}
	Unfolding the coend formula of \cref{def:polygraphic-realization-nerve},
	for $X \in \PshO$, its realization can be written as the quotient
	\[
		|X| = \frac{
			\sum_{\omega \in \bbOO} X_\omega \times |\omega|
		}{
			\binom{f^* x}{\phi \xrightarrow{h} \psi}
			\sim \binom{x}{\phi \xrightarrow{h} \psi \xrightarrow{f} \omega}
			\text{ with } x \in X_\omega, h : \phi \rightarrow \psi,
			f : \psi \rightarrow \omega
		} .
	\]
	For $\binom{x}{\psi} \in \sum_{\omega \in \bbOO} X_\omega \times |\omega|$,
	let $\bbinom{x}{\psi}$ be its equivalence class in $|X|$. Note that all
	classes have a representative of the form $\bbinom{y}{\id_\phi}$, for some
	$\phi \in \bbOO$ and $y \in X_\phi$.
\end{notation}

\begin{proposition}
	Take $X \in \PshO$, $P \in \PolMTO$, and $f : X \longrightarrow NP$.
	The unit $\eta$ at $X$, the transpose $\Phi f$ of $f$, and the counit
	$\epsilon$ at $P$ are respectively given by:
	\begin{align*}
		\eta : X_\omega
			&\longrightarrow N |X|_\omega
		&
		\Phi f : |X|_\omega
			&\longrightarrow P_\omega
		&
		\epsilon : |NP|_\omega
			&\longrightarrow P_\omega \\
		x
			&\longmapsto \widetilde{\bbinom{x}{\id_\omega}} ,
		&
		\bbinom{x}{\id_\omega}
			&\longmapsto f (x) (\omega) ,
		&
		\bbinom{\tildX}{\id_\omega}
			&\longmapsto x .
	\end{align*}
\end{proposition}
\begin{proof}
	\begin{enumerate}
		\item (Unit and transpose) We have to check that the following diagram
		commutes
		\[
			\diagramarrows{}{}{<-}{}
			\triangleULdiagram{X}{N|X|}{NP ,}{\eta}{f}{N \Phi f}
		\]
		and that $f$ is unique for that property. For $x \in X_\omega$ we have
		\[
			(N \Phi f) (\eta (x))
			= (N \Phi f) \left( \widetilde{\bbinom{x}{\id_\omega}} \right)
			= (\Phi f) \widetilde{\bbinom{x}{\id_\omega}} ,
		\]
		which maps $\omega$ to $f (x) (\omega)$. Since a map $|\omega|
		\longrightarrow P$ is uniquely determined by the image of $\omega$, we
		have $(N \Phi f) \eta = f$. Let $g : X \longrightarrow NP$ be another
		morphism such that $(N \Phi g) \cdot \eta = f$. Then for $x \in
		X_\omega$ we have
        \[
			f (x) (\omega)
			= \left( (N \Phi g) \cdot \eta \right) (x) (\omega)
			= (N \Phi g) \widetilde{\bbinom{x}{\id_\omega}} (\omega)
			= g (x) (\omega) ,
		\]
		whence $f = g$.
		\item (Counit) The counit is given by $\epsilon = \Phi (\id_{NP})$, so
		that
		\[
			\epsilon \bbinom{\tildX}{\id_\omega}
			= (\Phi \id_{NP}) \bbinom{\tildX}{\id_\omega}
			= \tildX (\omega)
			= x .
		\]
	\end{enumerate}
\end{proof}

\begin{theorem}
	The unit and counit are natural isomorphisms. Consequently, the polygrahic
	realization-nerve adjunction of \cref{def:polygraphic-realization-nerve} is
	an adjoint equivalence between $\PshO$ and $\PolMTO$.
\end{theorem}
\begin{proof}
	\begin{enumerate}
		\item (Unit) Remark that for $x, y \in X_\omega$, if
		$\bbinom{x}{\id_\omega} = \bbinom{y}{\id_\omega}$, then $x = y$, which
		shows that $\eta$ is injective. Take $f \in N|X|_\omega$. Then $f
		(\omega)$ is of the form $\bbinom{x}{\id_\omega}$, hence $f = \eta
		(x)$, and $\eta$ is surjective.
		\item (Counit) The following triangle identity shows that $N \epsilon$
		is a natural isomorphism:
		\[
			\diagramlines{=}{}{}{} \diagramarrows{-}{}{<-}{}
			\triangleDdiagram{N}{N}{N | N - | .}{}{\eta N}{N \epsilon}
		\]
		It is easy to check that the following square commutes, and since
		$\widetilde{(-)}$ is a bijection by \cref{th:shape}, $\epsilon$ is a
		natural isomorphism:
		\[
            \squarediagram
                {|NP|}{P}{N|NP|}{NP .}
                {\epsilon}{\widetilde{(-)}}{\widetilde{(-)}}{N \epsilon}
        \]
	\end{enumerate}
\end{proof}

Many-to-one polygraphs have been the subject of other work \cite{Harnik2002,
Harnik2008}, and proved to be equivalent to the notion of \emph{multitopic
sets}. This, together with our present contribution, proves the following:

\begin{corollary}
	The category $\PshO$ of opetopic sets is equivalent to the category
	of multitopic sets.
\end{corollary}

In \cite[corollary 2.4.6 and paragraph 2.4.7]{Henry2017}, Henry shows that
$\PolMTO$ is a presheaf category over some category $\bbOO\mathrm{plex}$ of
\emph{opetopic plexes}, and asks wether they are the same as opetope. We now
answer positively to this question.

\begin{corollary}
	The category $\bbOO\mathrm{plex}$ of opetopic plexes is equivalent to
	$\bbOO$.
\end{corollary}
\begin{proof}
	Opetopic plexes are proved to be generators of the terminal many-to-one
	polygraph $\mathfrak{1}$ in \cite[Proposition 2.2.3]{Henry2017}, and so
	together with \cref{prop:terminal-mtop:opetopes}, we have that opetopic
	plexes are exactly opetopes. On the other hand, morphisms of opetopic
	plexes are by definition morphisms of polygraphs between the representables
	they induce, which by the Yoneda lemma are exactly morphisms of opetopes.
\end{proof}

config: arxiv


TITLE: Volume of tetrahedron when equation of planes are given
QUESTION [0 upvotes]: Find the volume of the tetrahedron formed by the planes whose equations are $x+y=0$, $y+z=0$, $z+x=0$ and $x+y+z-1=0$.
I am trying to use the formula $\frac{1}{3}$ Area of base $\times$ Height. I achieved vertex as $(0,0,0)$ and $a=\sqrt2$ as side of base but I am achieving volume as $\frac{1}{6}$ which is wrong as per the given answer (which is $\frac{2}{3}$).
Could someone help me with this?

REPLY [1 votes]: Find the vertices, which are the planes intersections, i.e.,
$$\left(\pi_{1}\right)x+y=0\;;\left(\pi_{2}\right)y+z=0\;;\left(\pi_{3}\right)z+x=0\;\rightarrow A\left(0,0,0\right)$$
$$\left(\pi_{1}\right)x+y=0\;;\left(\pi_{2}\right)y+z=0\;;\left(\pi_{4}\right)x+y+z-1=0\;\rightarrow B\left(1,-1,1\right)$$
$$\left(\pi_{1}\right)x+y=0\;;\left(\pi_{3}\right)z+x=0\;;\left(\pi_{4}\right)x+y+z-1=0\;\rightarrow C\left(-1,1,1\right)$$
$$\left(\pi_{2}\right)y+z=0\;;\left(\pi_{3}\right)z+x=0\;;\left(\pi_{4}\right)x+y+z-1=0\;\rightarrow D\left(1,1,-1\right)$$
Thus, you have the tetrahedron edge vectors $$\overrightarrow{AB}=\left(1,-1,1\right)\;;\;\overrightarrow{AC}=\left(-1,1,1\right)\;;\;\overrightarrow{AD}=\left(1,1,-1\right)$$
The volume is given by the triple product
$$V=\frac{1}{6}|\overrightarrow{AB}\cdot\left(\overrightarrow{AC}\times\overrightarrow{AD}\right)|=\frac{1}{6}\begin{vmatrix}1 & -1 & 1 \\ -1 & 1 & 1 \\ 1 & 1 & -1\end{vmatrix}=\frac{|-4|}{6}=\frac{2}{3}$$


TITLE: How to find the probability of a state from non-degenerate spectral decomposition
QUESTION [0 upvotes]: I have just begun learning the topics of time evolution in quantum mechanics, and I'm having trouble understanding how to calculate the probabilities of certain eigenvalues of an operator at a later time.
In my textbook, there is this example: For a quantum system with a Hamiltonian $H$ and operator $A$ with spectral decomposition given by
$$A = \sum_n a_n \lvert \psi_n \rangle \langle \psi_n \rvert \, .$$
The system is initially in the eigenstate $|\psi\rangle$ of $A$, with eigenvalue $a_n$. $H$ and $A$ do not commute. After time θ, the probability of obtaining the eigenvalue $a_n$ is given by
$$w_{nn}(\theta) \approx 1 - (\Delta E)_n^2 \, \theta^2 \, .$$
I am completely clueless as to how this equation is derived. My attempt at deriving this would be to begin with $|\langle\psi_n|\psi_n\rangle|^2$ as this gives the probability of the system being at the same initial state, but I'm unsure if this approach is correct for time-evolving systems. I also noticed the θ in the equation, which leads me to think the time evolution operator $U$ is involved
$$U = \exp(-i H t / \hbar) \, .$$
May someone provide an explanation or clue as to how measurements are calculated in this case?

REPLY [0 votes]: Well, how we would find the probability of being in the same initial state (say $|\psi_n\rangle$) is as follows:
$$P(|\psi_n\rangle, t) = |\langle\psi_n| \Psi \rangle|^2$$
where $|\Psi\rangle = e^{-i H t}|\psi_n \rangle$. Physically, how you should think about this expression is that I am finding the projection of my quantum state at time $t$ on my initial state which yields the probability of measuring my initial state. Now, we can do a simple calculation:
$$P(|\psi_n\rangle, t) = |\langle\psi_n|e^{-i H t}|\psi_n \rangle|^2 \approx |\langle\psi_n| 1- i H t - \frac{1}{2} H^2 t^2|\psi_n \rangle|^2 \approx | 1 - i \langle E \rangle - \frac{1}{2} \langle E^2\rangle t^2|^2 $$
$$\approx 1 - \langle E^2 \rangle t^2 - \langle E\rangle ^2 t^2 \approx 1 - \langle \Delta E^2\rangle t^2  $$
Note that this is a short time approximation where "short" is set by the characteristic energy scale.


TITLE: Homework: calculation about differential form
QUESTION [4 upvotes]: Here is the question: Let $\omega = A dy\wedge dz + B dz \wedge dx + C dx \wedge dy$ in $\mathbf{R}^3$, and $d\omega = 0$. Denote
\begin{eqnarray}
\alpha = \int_0^1 tA(tx,ty,tz)dt\cdot(ydz-zdy)\\
+\int_0^1 tB(tx,ty,tz)dt\cdot(zdx-xdz)\\
+\int_0^1 tC(tx,ty,tz)dt\cdot(xdy-ydx).
\end{eqnarray}
Then show that $d\alpha = \omega$.
Can you teach me how to do this calculation?

REPLY [1 votes]: Just follow the definitions. Write $\alpha$ as $adx+bdy+cdz$, then show that $b_x-a_y=C$ etc. On the way, you will need to use that $d\omega=0$, ie $A_x+B_y+C_z=0$.


TITLE: Bracket satisfies Jacobi-identity?
QUESTION [1 upvotes]: For all $F,G \in C^{\infty}(\mathbb{R}^2)$, define the following bracket:
$$\lbrace F, G \rbrace= y\Big(\frac{\partial F}{\partial x}\frac{\partial G}{\partial y}-\frac{\partial F}{\partial y}\frac{\partial G}{\partial x}\Big)$$
I have to verify that this defines a Poisson-bracket. I've already checked that it is anti-symmetric, linear in both variables and that it satisfies the Leibniz-rule. So I still have to check that the Jacobi-identity is satisfied. I started calculating all the brackets but I got stuck since the terms don't cancel out. Is there a direct way to see that the Jacobi-identity holds?

REPLY [0 votes]: You definitely made a mistake. There are several ways to see this. One is to only consider the terms in the triple bracket linear in y since the combinatorics of the ones bilinear in y are identical to the flat metric one. The ones linear in y,
$$
y\Bigl(H_x(F_xG_y-F_yG_x)+ F_x(G_xH_y-G_yH_x)+G_x(H_xF_y-H_yF_x)     \Bigr) =0,
$$
have the proper pairwise cancellations.
But it is even more trivial than that. You may absorb the loose y into the $\partial y$ denominators, converting them to $\partial \ln y$, or, equivalently, changing variables from y to $z\equiv \ln y$. The PB is then the conventional form in x and z, whose Jacobi compliance you have already checked.