TITLE: Bound for difference of quadratic form
QUESTION [1 upvotes]: Suppose $A$ is an $n\times n$ positive definite matrix, and let $f:\mathbb{R}^n \to \mathbb{R}$ be defined as $f(x) = x^TAx$.
Then suppose we know that for two vectors $x$ and $y$, we have $|x_i-y_i| \leq r$, and $0 \leq x_i,y_i \leq 1$ for all $i \in \{1,2,\ldots,n\}$, then can we bound |f(x)-f(y)| in terms of $r$?

REPLY [1 votes]: Some hints:
Let $A=P'DP$ where D is diagonal.
(More work is needed for the case where the case where the decomposition is not possible).
Then $f(x)=x'P'DPx = x'Dx$ where $\bar{x}=Px$
$$f(x)-f(y)=\bar{x}'D\bar{x}-\bar{y}'D\bar{y}=(\bar{x}+\bar{y})'D(\bar{x}-\bar{y})\leftarrow \mbox{here is where the decomposition is  useful}\\
=\overline{(x+y)}'D\overline{(x-y)}=(x+y)'A(x-y)$$
Since $\forall i, |x_i-y_i| \leq r$ and $|x_i+y_i| \leq 2$,
$$|f(x)-f(y)|=|(x+y)'A(x-y)|\leq 2r \sum_{i=1}^n \sum_{j=1}^n |a_{ij}|$$


TITLE: $f_n(x)$ are eigenfunctions for a sturm-liouville eq. Prove $f(x) = \sum_{n=0}^{\infty}a_nf_n(x)$ is uniquely represented
QUESTION [1 upvotes]: Let $f_n(x)$ be eigenfunctions and $f(x)$ be continuous. Show that the
  expression:
$$f(x) = \sum_{n=0}^{\infty}a_nf_n(x)$$
is unique

These eigenvalues are from the sturm-liouville equation
$$\frac{d}{dx}\left[p(x)\frac{d}{dx}f_n(x)\right]+q(x)f_n(x) + \lambda_nw(x)f_n(x) = 0$$
(I guess).
By the sturm-liouville theory, I know that if $f_n$ and $f_k$ are solutions to the sturm-liouville equations:
$$Lf_n+\lambda_nwf_n = 0$$
$$Lf_k+\lambda_kwf_k = 0$$
then doing this multiplication:
$$f_k(Lf_n+\lambda_nwf_n) = 0$$
$$f_n(Lf_k+\lambda_kwf_k) = 0$$
,subtracting and integrating, we have:
$$\int_a^b f_kLf_n \ dx -\int_a^b f_nLf_k \ dx = (\lambda_n-\lambda_k)\int_a^b wf_kf_n \ dx$$
that is, for different eigenvalues $\lambda_n, \lambda_k$, we have $f_k$ and $f_n$ orthogonal.
I think the $f_n$'s are already expected to be there, I must therefore prove that the $a_n$'s are unique, right?
What I tried:
Take the last equation but suppose there are two possibilities:
$$f(x) = \sum_{n=0}^{\infty}a_nf_n(x)$$
$$f(x) = \sum_{n=0}^{\infty}a_mf_n(x)$$
and do:
$$f(x)-f(x) = 0 = \sum_{n=0}^{\infty}(a_n-a_m)f_n(x)$$
Multiply by another eigenfunction $f_m$ and integrate:
$$0 =  \sum_{n=0}^{\infty}\int_a^b(a_n-a_m)f_n(x)f_m(x)\ dx = \sum_{n=0}^{\infty}(a_n-a_m)\int(f_n(x)f_m(x))\ dx$$
I thought that the orthogonality of the eigenfunctions would help but I didn't use an weight function $w$ and also that expression won't make $a_n-a_m = 0$. What should I do?

REPLY [1 votes]: You are on the right track! You just made a slight mistake when writing down the summations. They should say:
$$ f(x) = \sum_{n = 0}^\infty a_n f_n(x), \ \ \ \ \ f(x) = \sum_{n = 0}^\infty b_n f_n(x).$$
The goal is to prove that $a_n = b_n$ for every value of $n$.
Taking the differences, we get
$$ \sum_{n=0}^\infty (a_n - b_n)f_n (x) = 0.$$
Now suppose we want to extract the coefficeint $a_{2017}-b_{2017}$. To do this, we multiply both sides by $f_{2017}(x)w(x)$ and integrate:
$$ \sum_{n = 0}^\infty \int_a^b (a_n - b_n) f_n(x) f_{2017}(x) w(x) dx = 0.$$
[Notice that I multiplied by $f_{2017}(x)w(x)$, not by $f_{2017}(x)$. This resolves your problem about the missing $w(x)$.]
But by the orthogonality relations that you proved in the first half of your post, you know that $$\int_a^b f_n(x) f_{2017}(x) w(x) dx = 0 \ \ \ \ \ {\rm for \ } n \neq 2017$$
Therefore, only the $n = 2017$ term contributes to the sum, and hence
$$ (a_{2017} - b_{2017}) \int f_{2017}(x) f_{2017}(x) w(x) dx = 0,$$
and since $\int f_{2017}(x) f_{2017}(x) w(x) dx > 0$, this implies that
$$a_{2017} - b_{2017} = 0.$$
Of course, there is nothing special about the number $2017$. You can do this for any value of $n$.

While we're discussing this, I strongly suggest that you try to use the same method to show that
$$ a_n = \frac{\int_a^b f(x)f_n(x) w(x) dx}{\int_a^b f_n(x)f_n(x) w(x) dx}.$$
This is a very useful result.
I also suggest you think about how this method is similar to the way you solve the linear algebra problem of extracting the coordinates $c_n$ of a vector $v = \sum_n c_n e_n$ with respect to an orthogonal basis $e_n$ by orthogonal projection.


TITLE: How can I prove this for $a+b+c=0$
QUESTION [2 upvotes]: We wish to prove that
$$\left(\frac{a-b}{c}+\ \frac{b-c}{a}+\frac{c-a}{b}\right)\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)=9,$$ for $a+b+c=0$ and $$abc \neq 0 , a\neq b , b\neq c, c \neq a.$$
I tried working with only $2$ fractions at a time, switching some of the variables (ex.$a=-b-c$) and then adding them up, and multiplying first but could not go further.
Thank you in advance.

REPLY [0 votes]: Comment:As commented user376343 we have:
$$B=-A\times A=-[a+b+c-(\frac ab+\frac bc+\frac ca)]^2=-(\frac ab+\frac bc+\frac ca)^2$$
$$\frac ab+\frac bc+\frac ca=\frac{ca^2+ab^2+bc^2}{abc}\geq3$$
This inequality becomes equality if $a=1, b=2$ and $c=3$. I checked it for $a+b+c=0$ and it is not equal to 3. It is a negative value.I think correct constrain is : a, b and c positive and product must have a negative sign and the sign of inequality must be $\geq$.


TITLE: Solve $ c(n-1)^2 + n \leq cn^2$ for $c$
QUESTION [2 upvotes]: I have been trying to solve the following inequality
$$ c(n-1)^2 + n \leq cn^2 $$
and have been unsuccessful. I typed it into wolfram alpha and it said the solution was c = 1/2. However, it did not show an option to show all of the steps. I'm assuming wolfram is using some sort of approximation technique to solve it. Any help would be appreciated thanks...
I got as far as $$ cn^2 -2cn +c + n <= cn^2 $$
$$2cn - c >= n $$
$$ c>= n\div(2n-1) $$

REPLY [1 votes]: $$ cn^2 -2cn +c + n \leq cn^2 $$
$$2cn - c \geq n $$
$$ c \geq \frac{n}{2n-1} \text{if $n \geq \frac{1}{2}$ or } c \leq \frac{n}{2n-1} \text{if $n < \frac{1}{2}$} $$


TITLE: Is my definition of discontinuity correct?
QUESTION [0 upvotes]: I think I have made a rigorous and intuitive definition of discontinuity.

Definition: For a function $f(x)$ from $\mathbf{R}$ to $\mathbf{R}$, there is a discontinuity of $a$ at $x_0 \in \mathbf{R}$ if for all $\delta > 0$ around $x_0$; whenever $|x-x_0|<\delta$, there exists an $x$ such that $|f(x)-f(x_0)|$ $\geq a$

How much rigor is my definition?

REPLY [0 votes]: Your definition doesn’t quite work; “largest” may not exist (what if there is an infinite sequence approaching a bound, for example). Using maximum won’t work, you’d need to use supremum. You could simply replace “greatest value of” with “there exists an $x$ such that”, and then your definition works.


TITLE: Gamma Integral with constant in exponent
QUESTION [2 upvotes]: If there is a constant, c, in the exponent of the Gamma integral $e^{-cx}$, how is the reflected in the gamma function? Is it something like $\frac{\Gamma(t)}{c^{2}}$ ?
$$\Gamma(t)=\int_0^\infty x^{t-1}e^{-x}\,dx$$

REPLY [3 votes]: \begin{align}
& \overbrace{\int_0^\infty x^{t-1} e^{-cx}\,dx = \frac 1 {c^t} \int_0^\infty (cx)^{t-1} e^{-cx} (c\,dx)}^\text{First, grasp this first step. Then the rest is routine.} \\[12pt]
= {} & \frac 1 {c^t} \int_0^\infty u^{t-1} e^{-u}\,du = \frac 1 {c^t} \Gamma(t).
\end{align}


TITLE: Can't understand an equality between sums
QUESTION [0 upvotes]: During an induction proof I came across an equality that I can't understand.
During the last step of the induction there is:
$$ \sum_{k=1}^{n}{(k+1){n \choose k-1}} = \sum_{k=0}^{n-1}{(k+2){n \choose k}} $$
My question is which sum and combination   identities were used to achieve this transformation. Using simply the index shift of sums I couldn't work out the same result.
Edit: I understand that the equality stands, I can't understand how I can produce the RHS from the LHS. Which identities do I have to use?

REPLY [1 votes]: If you're unfamiliar with index shifts, change the variable first.
\begin{align}
\sum_{k=1}^{n} (k+1)\binom{n}{k-1}
&=\sum_{h=1}^{n} (h+1)\binom{n}{h-1} && \text{(change dummy variable)} \\
&=\sum_{k=0}^{n-1} (k+2)\binom{n}{k} &&
  \begin{gathered}
  h-1=k \\
  h+1=k+2 \\
  \begin{aligned}
  & h=1\implies k=0\\
  & h=n\implies k=n-1
  \end{aligned}
  \end{gathered}
\end{align}


text:
\begin{document}
\author{Nora Ganter}
\email{nganter@unimelb.edu.au}
\address{Department of Mathematics and Statistics\\
The University of Melbourne\\ Parkville VIC 3010\\ Australia}
\thanks{The author was supported by an Australian Research Fellowship
  and by ARC grant DP1095815.}
\title{Categorical tori}
\classification{22E99, 18D99}
\keywords{categorification, Lie group cohomology}
\begin{abstract}
We give explicit and elementary constructions of the categorical
extensions of a torus by the circle and discuss an application to loop
group extensions. Examples include
maximal tori of simple and simply connected compact Lie groups and the
tori associated to the Leech and Niemeyer lattices.
We obtain the extraspecial 2-groups as the isomorphism classes of
categorical fixed points under an involution action.
\end{abstract}
\maketitle

\section{Introduction}
By a {\em categorical group}, or a {\em 2-group},
we mean a monoidal groupoid $(\mG,\bullet,1)$ with weakly invertible
objects. If $\mG$ is a Lie groupoid, the monoidal structure is required
to be locally smooth in an appropriate sense
\cite{Schommer-Pries:Central_extensions}. In this situation, one
speaks of a {\em Lie 2-group}.

Categorical groups play as important a
role in string theory as groups do in particle physics
\cite{Baez:Huerta:An_invitation}, and
a number of prominent groups
seem to be most naturally understood via their categorical refinements.
Most famously, the infinite dimensional groups $String(n)$
come from compact Lie 2-groups
\cite{Schommer-Pries:Central_extensions}.
Also, Weyl groups and some of the sporadic
groups, including the Monster, are known or
conjectured to be the isomorphism classes of categorical groups (see Section
\ref{sec:examples}). In recent years, there is a growing understanding
that the categorical
nature of these groups is worth exploring, and we will see that this perspective
sheds new light on some old and important mathematics.

The purpose of this paper is to give a simple and hands-on description
of a basic class of examples, namely all central extensions of Lie 2-groups
of the form
  \begin{center}
  \begin{tikzpicture}
    \node at (0,0) [name=a] {$1$};
    \node at (2,0) [name=b] {$\pt\mmod U(1)$};
    \node at (4.4,0) [name=c] {$\mT$};
    \node at (6.3,0) [name=d] {$T$};
    \node at (7.75,0) [name=e] {$1$,};

    \draw[->] (a) -- (b);
    \draw[->] (b) -- (c);
    \draw[->] (c) -- (d);
    \draw[->] (d) -- node [midway, above] {} (e);
\end{tikzpicture}
\end{center}
where $T$ is a compact torus, and $\pt\mmod U(1)$ is the one-object groupoid
with $aut(\pt)=U(1)$.
We will refer to such a $\mathcal T$ as a {\em categorical torus}.
Categorical tori are important for a number of reasons:
first, they turn up as maximal tori of Lie
2-groups. So, any character theory of Lie
2-groups should sensibly start here.
Second, categorical tori incorporate many aspects relevant to the construction of
sporadic groups (see Sections \ref{sec:examples} and
\ref{sec:Extraspecial}).

Finally, the theory of Lie 2-groups is closely related to that
of loop groups, a point
we will explore in Section \ref{sec:loops}.
Our approach is to work from 2-groups to loop groups. In particular,
our first construction of $\mT$ does not involve loops or any other infinite
dimensional objects.
The {transgression} machine of
\cite{Waldorf:Transgression_to_loop_spaces_and_its_inverse_II} then
yields a simple description of central extensions of the loop group $\mL T$.
What we hope to
get across is that categorical groups can be fairly simple to construct and
easy to work with and that some of the complicated features of loop
groups are merely the shadow of rather obvious phenomena on the
categorical side.
This is the first in a series of papers describing the representation
theory of categorical tori.
\subsection{Acknowledgments} It is a pleasure to thank David Roberts for
very helpful conversations and correspondence. The idea for
Construction \ref{cons:strict} came from a conversation with him, and
I understand that he will also write about it elsewhere.
Many thanks for helpful and inspiring conversations also go to Matthew
Ando, Konrad Waldorf, Thomas Nikolaus, Geoffrey Mason,
and Arun Ram.
\section{Constructions of categorical tori}\label{sec:cons}
Let $\Lat$ be a free $\ZZ$-module of finite rank, and let $J$ be
an integer-valued bilinear form on $\Lat$. Tensoring $\Lat$ over $\ZZ$
with the short exact sequence
\begin{center}
\begin{tikzpicture}
  \node at (0.2,0) [name=a] {$0$};
  \node at (2,0) [name=b] {$\ZZ$};
  \node at (4,0) [name=c] {$\RR$};
  \node at (6.2,0) [name=d] {$U(1)$};
  \node at (8.2,0) [name=e] {$0$};
  \node at (4,-1) [name=f] {$r$};
  \node at (6.2,-1) [name=g] {$e^{2\pi ir}$,};

  \draw[->] (a) -- (b);
  \draw[->] (b) -- (c);
  \draw[->] (c) -- node [midway, above] {$\exp$} (d);
  \draw[->] (d) --  (e);
  \draw[|->] (f) -- (g);
\end{tikzpicture}
\end{center}
we obtain the short exact sequence
\begin{center}
\begin{tikzpicture}
  \node at (0.2,0) [name=a] {$0$};
  \node at (2,0) [name=b] {$\Lat$};
  \node at (4,0) [name=c] {$\mathfrak t$};
  \node at (6.2,0) [name=d] {$T$};
  \node at (8.2,0) [name=e] {$0$.};

  \draw[->] (a) -- (b);
  \draw[->] (b) -- (c);
  \draw[->] (c) -- node [midway, above] {$\exp$} (d);
  \draw[->] (d) -- (e);
\end{tikzpicture}
\end{center}
Here $T$ is the compact torus with
coweight lattice $\Lat$, and $\mathfrak t$ is its Lie-algebra.
We will give three equivalent constructions of the categorical torus
associated to $(\Lat,J)$.
\begin{Cons}\label{cons:strict}
  Let $\mathfrak t$ act on $\Lat\times U(1)$ by

  \begin{center}
    \begin{tikzpicture}
      \node at (0,0) [name=A, anchor=east] {$\mathfrak t\times\Lat\times U(1)$};
      \node at (1,0) [name=B, anchor=west] {$\Lat\times U(1)$};

      \node at (0,-0.8) [name=C, anchor=east] {$(x,m,z)$};
      \node at (1,-0.8) [name=D, anchor=west] {$(m,z\cdot\exp(J(m,y))$.};

      \draw[->] (A) -- (B);
      \draw[|->] (C) -- (D);
    \end{tikzpicture}
  \end{center}

  Then the strict categorical group $(\mT,\bullet, 0)$ associated
  to the crossed module

  \teqnarray{\Psi}{\Lat\times U(1)}{\mathfrak t}{(m,z)}{m}

  is a categorical torus.
\end{Cons}
Explicitly, the Lie groupoid $\mT$ has objects $\mathfrak t$ and
arrows $\mathfrak t\ltimes(\Lat\times U(1))$, which we write as
\begin{center}
  \begin{tikzpicture}
    \node at (0,0) [name=a] {$x$};
    \node at (1.8,0) [name=b] {$x+m$};
    \draw[->] (a) -- node [midway, above] {$z$} (b);
  \end{tikzpicture}
\end{center}
with $x\in \mathfrak t$, $m\in\Lat$ and $z\in U(1)$, to indicate
source and target. Composition in $\mT$ is
\begin{center}
  \begin{tikzpicture}
    \node at (0,0) [name=a] {$x$};
    \node at (1.8,0) [name=b] {$x+m$};
    \node at (4.3,0) [name=c] {$x+m+n$,};
    \draw[->] (a) -- node [midway, above] {$z$} (b);
    \draw[->] (b) -- node [midway, above] {$w$} (c);
    \draw[->] (a) .. controls (1.5,-1) and (2.8,-1) .. node [midway, below] {$zw$} (c);
  \end{tikzpicture}
\end{center}
and tensor multiplication is
\begin{eqnarray*}
  x\bullet y & = & x+y \\
  \left(x\xrightarrow{\,\,\,\,z\,\,\,\,}x+m\right)\bullet
    \left(y\xrightarrow{\,\,\,\,w\,\,\,\,}y+n\right) & = &
    \left(x+y\xrightarrow{\,\,\,\,zw\exp(J(m,y))\,\,\,\,}x+y+m+n\right).
\end{eqnarray*}
The unit object is $0$, and the
associativity and unit isomorphisms are identities.
This gives $\mT$ the structure of a group object in the
category of Lie groupoids.
  When we want to emphasize the dependence on $J$, we will use
  the notation $\bullet_J$ for $\bullet$.

In the following, we will use the notation $X\mmod G$ for the translation
groupoid associated to the action of a group $G$ on a space $X$.
With this notation, the underlying groupoid of Construction \ref{cons:strict} is
$$
  \mT \cong \left(\mathfrak t\mmod\Lat\right)\times\left(\pt\mmod U(1)\right),
$$
where $\Lat$ acts on $\mathfrak t$ by translations.
\begin{Cons}\label{cons:compact}
  Consider the torus $T$ as a Lie groupoid with only identity
  arrows, and let
  \begin{eqnarray*}
    p\negmedspace : \mathfrak t\mmod\Lat & \xrightarrow{\,\,\sim\,\,} & T
  \end{eqnarray*}
  be the equivalence of Lie-groupoids sending the object $x$ to $\exp(x)$.
  Note that $p$ does not possess a continuous inverse.
  The language developed in \cite{Schommer-Pries:Central_extensions}
  interprets  \begin{eqnarray*}
    \mathfrak T & = & T\times \pt\mmod U(1),
  \end{eqnarray*}
  together with the data inherited from
  Construction \ref{cons:strict}, as a 2-group object with multiplication
  \begin{equation}
    \label{eq:zig-zag}
    \mathfrak T\times\mathfrak T \xleftarrow{\,\,\,\,\sim\,\,\,\,}
    \mT\times\mT \xrightarrow{\,\,\,\,\bullet\,\,\,\,} \mT
    \xrightarrow{\,\,\,\,\sim\,\,\,\,} \mathfrak T
  \end{equation}
  in a suitable
  localization of the bicategory of Lie-groupoids.
  The equivalences in \eqref{eq:zig-zag} are those induced by $p$.
  Different communities have different language for the
  1-morphisms in this localized bicategory
  $\mathbf{Bibun}$. Depending on taste, the reader may
  wish to think of this multiplication
  on $\mathfrak T$ as a {\em zig-zag}, a {\em span}, an {\em orbifold map},
  an {\em anafunctor}, or a {\em bibundle}.
\end{Cons}
Our third construction is as a multiplicative bundle gerbe in the
sense of \cite{Brylinski:Loop_spaces_characteristic_classes} and
\cite{Carey:Johnson:Murray:Stevenson:Wang}.
\begin{Cons}\label{cons:mbg}
  Let $\LE$ be the line bundle over $T\times T$ with multipliers
  \begin{eqnarray*}
    f_{(m,n)}\negmedspace :\mathfrak t\times\mathfrak t & \longrightarrow & U(1)\\
                    (x,y) & \longmapsto & \exp\({J(m,y)}\),
  \end{eqnarray*}
  $(m,n)\in\Lat\times\Lat$.

  We claim that over $T\times T\times T$ we have a canonical isomorphism
  \begin{eqnarray*}
    \alpha\negmedspace : m_{12}^*\LE\tensor pr_{12}^*\LE &\cong&
    m_{23}^*\LE\tensor pr_{23}^*\LE,
  \end{eqnarray*}
  where $pr_{ij}$ is the
  projection onto factors $i$ and $j$ and $m_{ij}$ is
  given by multiplication of these two factors.
  Indeed, source and
  target of $\alpha$ have identical multipliers
  \begin{eqnarray*}
    f_{(k,m,n)}\negmedspace :\mathfrak t\times\mathfrak
    t\times\mathfrak t & \longrightarrow & U(1)\\
                      (x,y,z) & \longmapsto & \exp(J(k,y) + J(k,z) + J(m,z)),
  \end{eqnarray*}
  $(k,m,n)\in\Lat\times\Lat\times\Lat$.
  The pair $(\LE,\alpha)$ equips the trivial bundle
  gerbe over $T$ with a multiplicative structure.
  It is well known
  (e.g. \cite[Thm.3.2.5]{Waldorf:A_construction_of_string_2-group_models}) that
  the data of a multiplicative bundle gerbe over $T$ are equivalent to
  those of a Lie 2-group extension of $T$ by $\pt\mmod U(1)$.
\end{Cons}
Explicitly, the line bundle $\LE$ is constructed as
  \begin{eqnarray*}
    \LE & = & \CC\times \mathfrak t\times\mathfrak
    t\,\,/\,\, \sim
  \end{eqnarray*}
  with
  \begin{eqnarray*}
    \(z,x,y\) &\sim& (z\cdot \exp(J(m,y)),x+m,y+n)
  \end{eqnarray*}
  for $(m,n)\in \Lat\times\Lat$.
The trivial bundle gerbe over $T$ corresponds to the groupoid $\mI$ whose objects are
pairs $(t,L)$ with $t\in T$
and $L$ a hermitian line, and whose arrows $$(t,L)\longrightarrow (t,L')$$
are the unitary isomorphisms from $L$ to $L'$. The line bundle $\LE$
equips $\mI$ with the monoidal structure
\begin{eqnarray*}
  (s,L_1)\circledast(t,L_2) & = & (s\cdot t,\mathbf L_{s,t}\tensor L_1\tensor L_2),
\end{eqnarray*}
whose associativity isomorphisms
\begin{eqnarray*}
  \alpha_{r,s,t}:\mathbf L_{r\cdot s,t}\tensor \mathbf L_{r,s}&\, \cong \:&\mathbf
  L_{r,s\cdot t}\tensor \mathbf L_{s,t}
\end{eqnarray*}
are encoded in $\alpha$.
\begin{Prop}
  The three constructions yield equivalent extensions of $T$ by $\pt\mmod U(1)$.
\end{Prop}
\begin{proof}
  It is clear that Construction \ref{cons:strict} and Construction
  \ref{cons:compact} are equivalent. To see
  their equivalence with Construction \ref{cons:mbg},
  let $F$ be the functor
  $$\begin{array}{rcll}
    F:\mT & \longrightarrow & \mI&\\
    x  & \longmapsto & (exp(x),\CC) &\quad\quad \text{on objects}\\
    (x,m,z) & \longmapsto & z
    &\quad\quad\text{on arrows}.
  \end{array}
  $$
  We want to make $F$ into a monoidal equivalence. For this, we need
  to specify an isomorphism
  $$
    \eps\negmedspace : (1,\CC) \cong F(0)
  $$
  and a natural isomorphism
  $$
  \begin{array}{rcl}
    \phi: F(-)\circledast F(-)  & \Longrightarrow & F(-\bullet-),
  \end{array}
  $$
  satisfying the usual unit and associativity conditions.
  Take $\eps=id$ and note that $\phi$ needs to assign an isomorphism
  $$
    \phi_{x,y}\negmedspace : \LE_{\exp(x),\exp(y)} \longrightarrow \CC
  $$

  \noindent to each pair of objects $(x,y)$ of $\mT$.
  By construction of $\LE$, we have a trivialization of
  $\LE$ over $\mathfrak t\times\mathfrak t$, and this trivialization
  serves as our $\phi$.
  One checks that $(F,\phi,\eps)$ satisfies the axioms of a monoidal
  equivalence from $\(\mT,\bullet,0\)$ to $\(\mI,\circledast,(1,\CC)\)$.
\end{proof}
\begin{Rem}\label{rem:homom}
  Construction \ref{cons:mbg} takes the sum of two bilinear
  forms to the tensor product of the corresponding multiplicative bundle gerbes.
\end{Rem}
\begin{Def}
  We will write $J^t$ for the bilinear form
  \begin{eqnarray*}
    J^t(m,n) & = & J(n,m).
  \end{eqnarray*}
  We say that $J$ is {\em symmetric} if $J=J^t$ and that
  $J$ is {\em skew symmetric} if $J=-J^t$. A symmetric bilinear form
  $I$ is called {\em even} if
  \begin{eqnarray}\label{eq:2Z}
    I(m,m) & \in & 2\,\ZZ \quad\quad \text{for }m\in \Lat.
  \end{eqnarray}
\end{Def}

\begin{Prop}\label{prop:J^t}
  The Lie 2-groups $(\mT,\bullet_J,0)$ and $(\mT,\bullet_{J^t},0)$ are
  equivalent as Lie 2-group extensions of $T$ by $\pt\mmod U(1)$.
\end{Prop}
\begin{proof}
  A monoidal equivalence is given by the functor
  $$
  \begin{array}{rcll}
    F \negmedspace : \mT & \longrightarrow & \mT & \\
    \mathfrak t  & \stackrel\id\longrightarrow &\mathfrak t &\quad\quad \text{on objects,}\\
    (x,m,z) & \longmapsto & (x,m,z\cdot\exp(J(x,m)))
    &\quad\quad\text{on arrows,}
  \end{array}
  $$
  together with the natural transformation
  \begin{eqnarray*}
    \phi\negmedspace : F(-)\bullet_{J^t} F(-) &\: \Longrightarrow \:& F(-\bullet_{J}-)

  \end{eqnarray*}
  with
  $$
    \phi_{x,y}\negmedspace : x+y
    \xrightarrow{\,\,\,\,\,\,\exp(J(x,y))\,\,\,\,\,\,} x+y.
  $$
\end{proof}
\begin{Cor}\label{cor:skew}
  \begin{enumerate}
  \item If $I$ is an even symmetric bilinear form on \upshape{$\Lat$}, then the multiplicative
    bundle gerbe associated to $I$ possesses a square root.
  \item If $B$ is a skew symmetric integral bilinear form on \upshape{$\Lat$},
    then $B$ yields a trivial 2-group
    extension of $T$.
  \end{enumerate}
\end{Cor}
\begin{proof}
  Claim (i) follows from Proposition \ref{prop:J^t}, using Remark \ref{rem:homom} and the fact
    that every even symmetric bilinear form $I$ can be
    written in the form $I=J+J^t$ for an integer-valued bilinear form
    $J$. For instance, fix a basis $(b_1,\dots,b_r)$ of $\Lat$ and set
    \begin{eqnarray*}
      J(b_i,b_j) & = &
      \begin{cases}
        \frac12 I(b_i,b_i) & \text{ if }i=j,\\
        I(b_i,b_j) & \text{ if }i<j,\\
        0 & \text{ else.}
      \end{cases}
    \end{eqnarray*}
  Claim (ii) follows, similarly, from the fact that every skew symmetric bilinear form
    $B$ can be written in the form $B=J-J^t$ for an integer-valued
    bilinear form $J$.
\end{proof}
\begin{Cor}
  Let $J$ be an integer-valued bilinear form on \upshape{$\Lat$}.
  Then, up to equivalence, the categorical extension $(\mT,\bullet_J)$
  of $T$ by $\pt\mmod U(1)$
  only depends on the even bilinear form
  \begin{eqnarray*}
    I(m,n) & = & J(m,n) + J(n,m).
  \end{eqnarray*}
\end{Cor}
\begin{proof}
  Let $J_1$ and $J_2$ be two integer-valued bilinear forms on $\Lat$,
  and assume that
  \begin{eqnarray*}
    J_1 + J_1^t  & = & J_2 + J_2^t.
  \end{eqnarray*}
  Then $J_1-J_2$ is skew symmetric.
  By Corollary \ref{cor:skew}, it follows that the
  multiplicative bundle gerbe obtained from $J_1-J_2$ is
  trivial. Using Remark \ref{rem:homom}, we conclude that the
  multiplicative bundle gerbes obtained from $J_1$ and
  $J_2$ are isomorphic.
\end{proof}
\section{The example of the circle}
\label{sec:circle}
Let $\Lat=\ZZ$ and $$J(m,n) = mn.$$
The {\em basic circle extension} $\mU(1)$ of the circle group
$U(1)$ consists of the following data:
\begin{enumerate}
  \item the trivial bundle gerbe over $U(1)$,
  \item the line bundle $\LE$ on
    \begin{eqnarray*}
      U(1)\times U(1)  & = & \RR^2/\ZZ^2
    \end{eqnarray*}
    defined by the multipliers
    \begin{eqnarray*}
      f_{(m,n)}\negmedspace :\RR^2 & \longrightarrow & U(1)\\
                      (x,y) & \longmapsto & e^{2\pi i my}
    \end{eqnarray*}
    for $(m,n)\in\ZZ^2$,
  \item the canonical isomorphism
    \begin{eqnarray*}
      \alpha\negmedspace : m_{12}^*\LE\tensor pr_{12}^*\LE &\cong&
      m_{23}^*\LE\tensor pr_{23}^*\LE
    \end{eqnarray*}
    over $U(1)\times U(1)\times U(1)$.
\end{enumerate}
For $k\in\ZZ$,
the {\em $k$th circle extension $\mU(1)_k$} of $U(1)$
is obtained by replacing the multipliers with $\exp({kmy})$.
\begin{Rem}\label{rem:Chern-Weil}
  Recall that gerbes on $U(1)$ are classified, up to stable isomorphism, by
  their Dixmier-Douady class in
  \begin{eqnarray*}
    H^3(U(1);\ZZ) & = & 0.
  \end{eqnarray*}
  So, any bundle gerbe over $U(1)$ is trivializable.
  A multiplicative structure on the trivial bundle gerbe over $U(1)$
  consists of a line bundle on $U(1)\times U(1)$ plus extra data, encoding associativity.
  Line bundles on $U(1)\times U(1)$ are classified, up
  to isomorphism, by their first Chern class in
  $$
    H^2(U(1)\times U(1);\ZZ)\:\cong\: Alt^2\(\ZZ^2\),
  $$
  the group of skew symmetric bilinear forms on
  $\ZZ^2$. This group is infinite
  cyclic, generated by the determinant, and we claim
  \begin{eqnarray*}
    c_1(\LE) & = & -\det.
  \end{eqnarray*}
  This fact is proved using Chern-Weil theory, as illustrated in
  Figure 1:
  \begin{center}
  \begin{tikzpicture}
    \matrix (m) [matrix of math nodes, row sep=3em, column sep=3em]
      { {\Omega^2_{U(1)\times U(1)}} & {-dx\wedge dy} &  &  \\
         {\Omega^1_{U(1)\times U(1)}} & -xdy & -mdy &
          \\
        {\ul\TT_{U(1)\times U(1)}} & {} & \exp(-my) &
        1   \\
        {U(1)\times U(1)} & \RR^2 & \RR^2\times\ZZ^2 &
        \RR^2\times\ZZ^2\times\ZZ^2  \\ };
        \draw[|->] (m-2-2) -- (m-1-2);
        \draw[|->] (m-3-3) -- (m-2-3);
        \draw[|->] (m-2-2) -- (m-2-3);
        \draw[|->] (m-3-3) -- (m-3-4);
      { [start chain] \chainin (m-3-1);
        \chainin (m-2-1)
          [join={node[left,labeled] {dlog}}];
        \chainin (m-1-1)
          [join={node[left,labeled] {d}}];
      }
      \draw[->, very thick]  (-3.2,-3.6) -- (-3.2,3.6) ;
      \draw[dashed, very thick, red]  (-6,2) -- (6,2) ;
      \draw[->, very thick]  (-6,-2) -- (6,-2) ;
      \node at (0,3.8) {};
      \node at (0,-4) [anchor=north, black!65, text width=12.5cm]
      {Figure 1. This 2-cocycle in the truncated
      \v Cech-deRham double complex relates the
      multipliers of $\LE$ to the 2-form $-\det$.};
  \end{tikzpicture}
  \end{center}
The entry $\exp(-my)$ is the transition function of the bundle $\LE$
for the cover $\RR^2\to\RR^2/\ZZ^2$. This convention, that the
transition functions in \v Cech cohomology are taken to be the inverse multipliers,
comes from algebraic geometry: with the identifications $\RR^2\cong
\CC$ and $\ZZ^2\cong \langle 1,\tau\rangle$, the line bundle $\LE$ is
the topological line bundle underlying any degree $-1$ line bundle on
the elliptic curve
$\CC/\langle1,\tau\rangle$.

Note also that this argument equips $\LE$ with a
connection, namely th one with 1-form
\begin{eqnarray*}
  \omega &=& -xdy
\end{eqnarray*}
on ${\RR\times\RR}$
and with curvature
\begin{eqnarray*}
  d\omega  & = &  -dx\wedge dy.
\end{eqnarray*}
Similarly, the bundle $\LE^{\tensor k}$ with  multipliers
$\exp(kmy)$ has connection $-kxdy$ and Chern class $-k \cdot det$.
\end{Rem}
\begin{Rem}
  This example is not new.
  A construction of the multiplicative gerbe $\mU(1)$ (but with a
  different connection) already turns up in
  \cite[Sec.3]{Brylinski:McLaughlin:The_geometry_of_degree-4_characteristic_classes_II}.
  I understand that there is also unpublished work of Murray, Roberts,
  Stevenson and Vozzo constructing
  $\mU(1)$.
\end{Rem}
\section{The classification}\label{sec:classification}
Recall from \cite{Schommer-Pries:Central_extensions} that,
up to equivalence, the 2-group extensions of $T$ by $\pt\mmod U(1)$
are classified by degree three Lie
group cohomology classes of $T$ with coefficients in $U(1)$.
There are a number of constructions of Lie group cohomology, and a
unifying axiomatic framework was recently given in \cite{Wagemann:Wockel:A_cocycle_model}.
We choose to work with the \v Cech-simplicial double complex
of \cite{Brylinski:Loop_spaces_characteristic_classes} and
\cite[Prop 5.2]{Carey:Johnson:Murray:Stevenson:Wang}. So, the relevant
cohomology group is
$$
  H^3_{gp}(BT;U(1)) \,=\, \vH^3(BT_\bullet;U(1)).
$$

The goal of this section is to analyze the
composite of isomorphisms
\begin{equation}\label{eq:H^4}
  S^*(\Lambda) \,\, \cong \,\, H^{2*}(BT;\ZZ)
  \,\, \cong \,\, \vH^{2*}(BT_\bullet;\ZZ) \,\, \cong \,\,
  \vH^{2*-1}(BT_\bullet;U(1))
\end{equation}
in degree $*=2$.
Here $S^*(\Lambda)$ is the symmetric algebra of the weight lattice
\begin{eqnarray*}
  \Lambda & = & \Hom(\vL,\ZZ).
\end{eqnarray*}
We may think of elements of
\begin{eqnarray*}
  S^2\Lambda & = & (\Lambda\tensor\Lambda) /S_2
\end{eqnarray*}
as homogeneous polynomials of degree 2
in the weights, and we have the symmetrization map
\begin{eqnarray*}
  S^2(\Lambda) & \longrightarrow & Bil(\vL,\ZZ)\\
  \lambda\mu & \longmapsto & \lambda\tensor\mu + \mu\tensor\lambda,
\end{eqnarray*}
identifying $S^2(\Lambda)$ with the group of even symmetric bilinear forms on $\Lat$.
\begin{Thm}\label{thm:classification}
  Let $I$ be an even symmetric bilinear form on \upshape{$\Lat$}, and let $J$ be
  an integral bilinear form on \upshape{$\Lat$} satisfying $I=J+J^t$. Then $I$
  classifies the multiplicative bundle gerbe obtained from \upshape{$(\Lat,-J)$}
  via Construction \ref{cons:mbg}.
\end{Thm}

\begin{proof}
  Let $ET$ be a contractible space on which $T$ acts freely, and let
  $BT=ET/T$ be our model for the clasifying space of $T$. On $BT$, we
  have the line bundles
  $$
    L_\lambda \: = \: ET\times_T\CC_\lambda \: \cong \: \(ET\times\CC_{-\lambda}\)/T,
  $$
  $\lambda\in\Lambda$, where $\CC_\lambda$ denotes the irreducible complex
  representation of $T$ with weight $\lambda$.
  The first isomorphism in \eqref{eq:H^4} is
  \begin{eqnarray}
  \label{eq:Borel}    S^*(\Lambda) & \stackrel\cong\longrightarrow & H^{2*}(BT;\ZZ)\\
  \notag \lambda & \longmapsto & c_1(L_\lambda).
  \end{eqnarray}
  Its definition goes back to Borel, and it is an isomorphism of graded rings. It is, of
   course, well known that the odd cohomology groups vanish, but
   this fact will not concern us here.

  To define the second isomorphism in \eqref{eq:H^4},
  we use the homeomorphism
  \begin{eqnarray*}
   ET \times T \times  \dots\times T  & \stackrel\cong\longrightarrow &
    ET\times_{BT}\dots\times_{BT} ET\\
   (e,t_1,\dots,t_n) & \longmapsto &    (e,t_1e,\dots,t_ne)
  \end{eqnarray*}
  to interpret
  the maps
  \begin{eqnarray*}
    ET & \longrightarrow & BT \\
    ET \times \mathfrak t & \longrightarrow &
    ET\times_{BT} ET
  \end{eqnarray*}
  as a hypercover of $BT$. The \v Cech double complex of this
  hypercover can be identified with the \v Cech simplicial double complex
  $\vC^*(BT_\bullet;\ZZ)$.
  Under this identification, the cup product becomes
  \begin{eqnarray*}
    (f\cup g) (x_0,\dots,x_{r+s}) &=& (pr_{1}^*f)(x_0,\dots x_r)\cdot
    (pr_{2}^*g)(x_r,\dots x_{r+s}),
  \end{eqnarray*}
  where $r$ and $s$ are the \v Cech degrees of $f$ and $g$, while
  $pr_1$ is the projection onto the first $deg_{simp}(f)$ factors, and
  $pr_2$ is the projection onto the last $deg_{simp}(g)$ factors.

  To determine the image of $\lambda\mu$ in $H^4(BT_\bullet;\ZZ)$,
  we determine the images of $\lambda$ and $\mu$ and then take the cup product.
  The first Chern class of $L_\lambda$ in \v Cech hypercohomology is
  obtained by applying $\delta_{\text{\v Cech}}log$ to the inverse multipliers
  \begin{eqnarray*}
    ET\times \mathfrak t & \longrightarrow & U(1)\\
    (\eta,x) & \longmapsto & \exp(\lambda(x))
  \end{eqnarray*}
  of $L_\lambda$.
  So, $c_1(L_\lambda)$ is represented by the degree $(1,1)$ cocycle
  \begin{eqnarray*}
    \mathfrak t\times\vL & \longrightarrow & \ZZ \\
     (x,m) & \longmapsto & \lambda(m)
  \end{eqnarray*}
  in the \v Cech-simplicial double complex.
  Given two weights, $\lambda$ and $\mu$, their cup product is represented by the
  cocycle
  \begin{eqnarray*}
    \mathfrak t^2\times(\vL)^2\times(\vL)^2 & \longrightarrow & \ZZ \\
    \(\(\!\!\begin{array}{c}
      x\\y
    \end{array}\!\!\),
    \(\!\!\begin{array}{c}
      m\\n
    \end{array}\!\!\),
    \(\!\!\begin{array}{c}
      k\\l
    \end{array}\!\!\)\)
    & \longmapsto & \lambda(k)\cdot\mu(n)
  \end{eqnarray*}
  in $\vC^2(BT_2;\ZZ)$.
  The last isomorphism in \eqref{eq:H^4}
  is the inverse of the connecting homomorphism for the short exact sequence of
  presheaves
  $$
    0 \longrightarrow \ZZ   \longrightarrow \ul\RR      \longrightarrow
    \ul{U(1)}     \longrightarrow 0.
  $$
  Let $J=\mu\tensor\lambda$. As in Remark \ref{rem:Chern-Weil}, the transition
  function for $\LE_{-J}$ is the inverse multiplier. So,
  Construction \ref{cons:mbg} associates to $-J$
  the multiplicative bundle gerbe
  corresponding to the $U(1)$-valued \v Cech-simplicial
  3-cocycle $$(1,\lambda(m)\mu(y),1,1).$$
  The image of this cocycle under the connecting homomorphism is
  the integral
  4-cocycle $$(1,1,\lambda(k)\mu(n),1,1)\:=\:c_1\(L_\lambda\)\cdot c_1\(L_\mu\).$$
  \begin{center}
  \begin{tikzpicture}[scale=.8]
    \matrix (m) [matrix of math nodes, row sep=2.4em, column sep=3em]
      { {T\times T\times T} &  &  &
          \\
        {T\times T} &  & {\lambda(m)\mu(y)} & \lambda(k)\mu(n)
          \\
        {T} & {\lambda(x)} &\lambda(m) &
           \\
        {\pt} & {} & &
          \\
        {} & \mathfrak t^\bullet & (\mathfrak t\times\vL)^\bullet &
        (\mathfrak t \times\vL\times\vL)^\bullet \\ };
        \draw[|->] (m-2-3) -- (m-2-4) node [midway, above] {$\delta$};
        \draw[|->] (m-3-2) -- (m-3-3) node [midway, above] {$\delta$};
      \draw[->, very thick]  (-3,-4) -- (-3,4.5) ;
      \draw[->, very thick]  (-6,-3) -- (7.5,-3) ;
      \node at (0,-5) [anchor=north, black!65]
      {Figure 2. \quad $\delta_{\text{\v Cech}}:\vC^1(BT_2;\ul\RR) \longrightarrow
        \vC^2(T_2;\ul\RR)$
         };
  \end{tikzpicture}
  \end{center}
  The general case now follows from Remark \ref{rem:homom} together
  with the fact that any choice of $J$ can be written as linear
  combination of summands of the form $\mu\tensor\lambda$.
\end{proof}
\begin{Rem}
  In particular, we have seen that the underlying gerbe of
  any multiplicative bundle gerbe on $T$ is trivial. This is consistent with
  work of Waldorf
  \cite[Prop.2.10]{Waldorf:Multiplicative_bundle_gerbes_with_connection}, who
  identifies the forgetful map
  \begin{eqnarray*}
    \{\text{multiplicative bundle gerbes on }G\}/\cong & \longrightarrow &
        \{\text{bundle gerbes on }G\}/\cong
  \end{eqnarray*}
  with the inverse transgression
  \begin{eqnarray*}
    \tau\negmedspace : H^4(BG;\ZZ) & \longrightarrow & H^3(G;\ZZ)
  \end{eqnarray*}
  in the Leray-Serre spectral sequence for the fibration $EG\to BG$.
  For a compact, connected Lie group $G$,
  this inverse transgression map was calculated (in all degrees) by
  Chern and Simons. In the relevant degrees, their result is
  summarized by the commuting diagram\footnote{The formula
    \cite[(3.10)]{Chern:Simons:Characteristic_forms_and_geometric_invariants}
    is often cited in its original form
    \begin{eqnarray*}
      I &\longmapsto& -\frac1{12}\, I(\omega\wedge[\omega,\omega]),
    \end{eqnarray*}
    where $\omega$
    is the right-invariant Maurer-Cartan form on $G$.
    A look at the definitions on page 50 of
    \cite{Chern:Simons:Characteristic_forms_and_geometric_invariants}
    identifies $I(\omega\wedge[\omega,\omega])$ with the bi-invariant
    3-form on $G$ whose restriction to $\mathfrak g=T_1G$ equals
    $12\,\nu$.}
  \begin{center}
    \begin{tikzpicture}
      \node at (0,3) [name=A] {$H^4(BG;\RR)$};
      \node at (4,3) [name=B] {$H^3(G;\RR)$};
      \node at (0,0) [name=C] {$\(S^2\mathfrak g^*\)^{Ad}$};
      \node at (4,0) [name=D] {$\(\Lambda^3\mathfrak g^*\)^{Ad}$};
      \node at (0,-1) [name=E,text  width=1cm] {\quad$I$};
      \node at (4,-1) [name=F, text width=1.5cm] {\quad$-\nu$,};

      \draw[->] (A) -- (B) node[midway,above]{$\tau$};
      \draw[->] (C) -- (A) node[midway,right]{$\cong$}
      node[midway,left, text width= 2.5cm]{Chern-Weil isomorphism};
      \draw[->] (B) -- (D) node[midway,left]{$\cong$}
      node[midway,right, text width= 3.8cm]{\quad Chevalley-Eilenberg\\
        \quad isomorphism};
      \draw[->] (C) -- (D);
      \draw[|->] (E) -- (F);
    \end{tikzpicture}
  \end{center}
  where $\nu$ is the Cartan 3-form
  associated to $I$,
  \begin{eqnarray*}
    \nu(x,y,z) & = &  I([x,y],z).
  \end{eqnarray*}
  So, the multiplicative bundle gerbe on $G$
  classified by $I$ has as its Dixmier-Douady class the left-invariant
  3-form on $G$ whose restriction to $\mathfrak g$ is $-\nu$.
  Since the Lie bracket on a torus is zero, it follows that the
  Dixmier-Douady class of any
  multiplicative bundle gerbe on $T$ vanishes.
\end{Rem}

\subsection{Connections}
Let $\LE$ be the line bundle of Construction \ref{cons:mbg}.
If $J=\lambda\tensor\mu$, then we have
the connection $\nabla$ on $\LE$
whose 1-form on ${\mathfrak t\times\mathfrak t}$ is
\begin{eqnarray*}
  \omega & = & -\lambda\, d\mu.
\end{eqnarray*}
It has curvature 2-form
\begin{eqnarray*}
  \kappa & = & -d\lambda\wedge d\mu.
\end{eqnarray*}
For arbitrary $J$, we introduce the maps
\begin{eqnarray*}
  J^\sharp(x):\mathfrak t & \to & \RR \\
   v & \longmapsto & J(x,v).
\end{eqnarray*}
Then $\nabla$ is defined by the 1-form
\begin{eqnarray*}
  \omega_{(x,y)} & = & -d(J^\sharp(x))_y.
\end{eqnarray*}
The pair $(\LE,\nabla)$ turns the
trivial bundle gerbe $\mI$ on $T$ into
a multiplicative bundle gerbe with
connection (in the sense of
\cite{Waldorf:Multiplicative_bundle_gerbes_with_connection},
with the remaining data trivial).
\section{Examples}\label{sec:examples}
\subsection{Maximal tori}\label{sec:maximal_tori}
Let $G$ be a simple and simply connected compact connected Lie group
with maximal torus $T$ and Weyl group $W$.
Then we have
$$
  \ZZ\,\cong\,  H^3_{gp}(G;U(1)) \, \cong \,
    H^4(BT;\ZZ)^W \, \cong \,  \( Bil_{ev}(\Lat;\ZZ)^{S_2}\)^W.
$$
The elements of this group are multiples of the Killing form, and the
the positive
definite generator $I_{basic}$
classifies the Lie 2-group extension of $G$ denoted
$String(G)$ in \cite{Schommer-Pries:Central_extensions}.
We arrive at a recognition principle for these 2-groups: the
extension $\mG$ is equivalent to $String(G)$ if and only if its
to $T$ is equivalent to the categorical torus
classified by $(\vL,I_{basic})$.
\subsection{The Leech lattice}\label{sec:Leech}
  Another interesting example is given by the Leech lattice
  $\Lat=\Lambda_{Leech}$ inside $\RR^{24}$, together with the standard symmetric
  bilinear form $I$.
  The group of linear isometries $O(\Lat,I)$ of the Leech lattice is the {\em
    Conway group} $Co_0$. In analogy to the previous example, we view
  $I$ as a $Co_0$-invariant cohomology element,
  $$
    I\in H^4(BT,\ZZ)^{Co_0},
  $$
  where $T=\RR^{24}/\Lambda_{Leech}$ is the Leech torus.
  We arrive at a categorical extension $\mT_{Leech}$ of the
  Leech torus, on which $Co_0$ acts by autoequivalences.
\subsection{Niemeyer lattices}
  Similarly, we can choose
  $\Lat$ as one of the Niemeyer lattices,
  and $I$ as the standard
  symmetric bilinear form on $\RR^{24}$, making $\Lat$ an even unimodular
  lattice. If $A_1^{24}$ or $A_2^{12}$,

  then the group of linear isometries of $(\Lat,I)$ is the
  {\em Mathieu group} $M_{24}$ (respectively $M_{12}$), and we have
  $$
    I\in H^4(BT,\ZZ)^{M_{24}} \quad\text{ respectively }\quad
    I\in H^4(BT,\ZZ)^{M_{12}},
  $$
  classifying two categorical tori on which the respective Mathieu
  groups act by autoequivalences.

\section{Extraspecial categorical 2-groups}\label{sec:Extraspecial}
  Assume we are given a strict action of a group $G$ by functors on a category
  $\mC$. In other words, assume that $\mC$ comes equipped with
  endofunctors $\varrho(g)$, one for each $g\in G$, satisfying
  $$
    \varrho(g)\varrho(h)\,=\,\varrho(gh)\quad\quad\text{and}\quad\quad\varrho(1)=id.
  $$
  In this situation, we will allow ourselves to drop $\varrho$ from the notation and
  simply write $g$ for the functor $\varrho(g)$.
\begin{Def}[(Grothendieck)]
  An {\em equivariant object} of $\mC$ is a pair
  $$(x,\mathbf e)=(x,\{e_g\}_{g\in G}),$$ where $x\in ob(\mC)$ and
  the $\isomap{e_g}xgx$ are isomorphisms in $\mC$ satisfying
  \begin{equation}
    \label{eq:equivariant_object}
    (ge_h)\circ e_g = e_{gh},
  \end{equation}
  for all $g$ and $h$ in $G$. An {\em equivariant arrow} from
  $(x,\mathbf e)$ to $(y,\mathbf f)$ is an arrow $\map axy$ in $\mC$
  that is compatible with $\mathbf e$ and $\mathbf f$ in the following
  sense
   $$
    f_g\circ a = (ga)\circ e_g,
  $$
  for all $g\in G$.
\end{Def}
\begin{Def}
  We will refer to the
  category $\mC^G$ of equivariant objects of $\mC$ and equivariant
  arrows between them as the {\em categorical fixed points} $\mC^G$ of
  the action of $G$ on $\mC$.
\end{Def}
The goal of this section is to identify the fixed point category
of the action of $\{\pm1\}$ on $\mT$, where $-1$ acts by
the auto-equivalence associated to the crossed module
automorphism
\begin{center}
  \begin{tikzpicture}
    \node at (-1.5,-2) {$\operatorname{inv}\negmedspace:$};
    \node at (0,0) [name=a] {$(m,z)$};
    \node at (3,0) [name=b] {$(-m,z)$};
    \node at (0,-1) [name=c] {$\Lat\times U(1)$};
    \node at (3,-1) [name=d] {$\Lat\times U(1)$};
    \node at (0,-3) [name=e] {$\mathfrak t$};
    \node at (3,-3) [name=f] {$\mathfrak t$};
    \node at (0,-4) [name=g] {$x$};
    \node at (3,-4) [name=h] {$-x$.};

    \draw[|->] (a) -- (b);
    \draw[->] (c) -- (d);
    \draw[->] (e) -- (f);
    \draw[->] (c) -- (e);
    \draw[->] (d) -- (f);
    \draw[|->] (g) -- (h);
  \end{tikzpicture}
\end{center}
Explicitly, the functor $\varrho(-1)$ sends the object $x$
to $-x$ and the arrow \makebox{$x\xrightarrow{\,\,\,\,z\,\,\,\,}x+m$}
to the arrow
\makebox{$-x\xrightarrow{\,\,\,\,z\,\,\,\,}-x-m$}.
Since this is an action by strictly monoidal functors, the
categorical fixed points $\mT^{\{\pm1\}}$ inherit the
structure of a strict categorical group.

Consider now the points of order two in $T$. These form the elementary
abelian 2-group\footnote{In the context of this Section, the term
  2-group is used in the sense of $p$-group, $p$ a prime, not in the
  sense of categorical group.}
$$
  T[2] \,\cong \,\Lat\tensor\,\FF_2.
$$
We have a central extension $\widetilde{T[2]}$ of $T[2]$,
defined by the $\FF_2$-valued 2-cocycle
$$
  J_{\FF_2}\,:=\,J\tensor \,\FF_2.
$$
Central extensions of this form are know as {\em extraspecial
  2-groups} and classified by their {\em Arf invariant}, i.e., by
the quadratic form
$$
  v \,\longmapsto\,J_{\FF_2}(v,v)
$$
on the $\FF_2$-vector space $\Lat\!\tensor\FF_2$. Writing its center multiplicatively,
our central extension takes the form
\begin{center}
\begin{tikzpicture}
  \node at (0.2,0) [name=a] {$1$};
  \node at (2,0) [name=b] {$\{\pm1\}$};
  \node at (4.3,0) [name=c] {$\widetilde{T[2]}$};
  \node at (6.5,0) [name=d] {$T[2]$};
  \node at (8.4,0) [name=e] {$1$.};

  \draw[->] (a) -- (b);
  \draw[->] (b) -- (c);
  \draw[->] (c) -- (d);
  \draw[->] (d) -- node [midway, above] {} (e);
\end{tikzpicture}
\end{center}
Similarly, we have the (non-canonically trivial) central extension
\begin{center}
\begin{tikzpicture}
  \node at (0.2,0) [name=a] {$1$};
  \node at (2,0) [name=b] {$\{\pm1\}$};
  \node at (4.3,0) [name=c] {$\widetilde{\Lat}$};
  \node at (6.5,0) [name=d] {$\Lat$};
  \node at (8.4,0) [name=e] {$1$};

  \draw[->] (a) -- (b);
  \draw[->] (b) -- (c);
  \draw[->] (c) -- (d);
  \draw[->] (d) -- node [midway, above] {} (e);
\end{tikzpicture}
\end{center}
with 2-cocycle
$$
  (m,n) \longmapsto (-1)^{J(m,n)}.
$$
\begin{Thm}
  The crossed module corresponding to the strict categorical group
  $\mT^{\{\pm1\}}$ is isomorphic to the crossed module\upshape{
  \begin{eqnarray*}
    \Sigma\negmedspace :\Lat\times U(1) & \longrightarrow & \widetilde\Lat\\
     (m,z) & \longmapsto & (2m,1),
  \end{eqnarray*}}
  where $(n,\eps)\in\widetilde{\Lat}$ acts on $\Lat\times U(1)$ by
  $$(m,z)\longmapsto \(m,z\cdot(-1)^{J(m,n)}\).$$
  In particular, $\mT^{\{\pm1\}}$ is part of a categorical central
  extension
  \begin{center}
  \begin{tikzpicture}
    \node at (-0.2,0) [name=a] {$1$};
    \node at (1.8,0) [name=b] {$\pt\mmod U(1)$};
    \node at (4.4,0) [name=c] {$\mT^{\{\pm1\}}$};
    \node at (6.5,0) [name=d] {$\widetilde{T[2]}$};
    \node at (8.2,0) [name=e] {$1$.};

    \draw[->] (a) -- (b);
    \draw[->] (b) -- (c);
    \draw[->] (c) -- (d);
    \draw[->] (d) -- node [midway, above] {} (e);
\end{tikzpicture}
\end{center}
\end{Thm}
\begin{proof}
  For $\mT^{\{\pm1\}}$, Condition \eqref{eq:equivariant_object} reads
  $$e_1=id_x \quad\quad\quad \text{and} \quad\quad\quad {inv} (e_{-1})\circ e_{-1} = \id_x.$$
  So, we have a fully faithful and strictly monoidal embedding
  \begin{center}
    \begin{tikzpicture}
      \node at (0,0) [name=A, anchor=east] {$\mT^{\{\pm1\}}$};
      \node at (1,0) [name=B, anchor=west] {$\mT^I$};

      \node at (0,-0.8) [name=C, anchor=east] {$(x,\boldsymbol e)$};
      \node at (1,-0.8) [name=D, anchor=west] {$e_{-1}$,};

      \draw[->] (A) -- (B);
      \draw[|->] (C) -- (D);
    \end{tikzpicture}
  \end{center}
  where $\mT^I$ denotes the arrow category of $\mT$.
  This embedding identifies the objects of $\mT^{\{\pm1\}}$ with the image of the
  injective group homomorphism
  \begin{center}
    \begin{tikzpicture}
      \node at (0,0) [name=A, anchor=east] {$\widetilde\Lat$};
      \node at (1,0) [name=B, anchor=west] {$ob\(\mT^I\)$};

      \node at (0,-0.8) [name=C, anchor=east] {$(m,\eps)$};
      \node at (1,-0.8) [name=D, anchor=west]
      {$\(-\frac
        m2\xrightarrow{\,\,\,\,\eps\,\,\,\,}\frac m2\)$.};

      \draw[->] (A) -- (B);
      \draw[|->] (C) -- (D);
    \end{tikzpicture}
  \end{center}
  The source kernel of $\mT^{\{\pm1\}}$ is identified with the image
  of the group homomorphism
  \begin{center}
    \begin{tikzpicture}
      \node at (0,0) [name=A, anchor=east] {$\Lat\times U(1)$};
      \node at (1,0) [name=B, anchor=west] {$arrows\(\mT^I\)$};

      \node at (0,-1.5) [name=C, anchor=east] {$(m,\eps)$};
      \node at (1,-1.5) [name=D, anchor=west]
      {\begin{tikzpicture}
          \node (X) {$0$};
          \node [below=of X] (Y) {$0$};
          \node [right=of X] (Z) {$-m$};
          \node [below=of Z] (W) {$m$,};
          \draw[double equal sign distance] (X) -- (Y);
          \draw[->] (X) -- node [midway, above] {$z$} (Z);
          \draw[->] (Y) -- node [midway, above] {$z$} (W);
          \draw[->] (Z) -- node [midway, right] {$1$} (W);
        \end{tikzpicture}
        };

      \draw[->] (A) -- (B);
      \draw[|->] (C) -- (D);
    \end{tikzpicture}
  \end{center}
  and under these identifications, the target map becomes the
  homomorphism $\Sigma$ of the Theorem. It
  remains to identify the conjugation action of the objects on the source kernel.

We have
\begin{center}
  \begin{tikzpicture}[scale=1.1]
    \node at (0.6,2) [name=a] {$0$};
    \node at (2.6,2) [name=g] {$-m$};
    \node at (4,2) [name=c] {$-\frac n2$};
    \node at (6,2) [name=i] {$-\frac n2$};
    \node at (8,2) [name=e] {$-\frac n2$};
    \node at (11.5,2) [name=k] {$-\frac{2m+n}2$};
    \node at (0.6,0) [name=b] {$0$};
    \node at (2.6,0) [name=h] {$m$};
    \node at (4,0) [name=d] {$\frac n2$};
    \node at (6,0) [name=j] {$\frac n2$};
    \node at (8,0) [name=f] {$\frac n2$};
    \node at (11.5,0) [name=l] {$\frac{2m+n}2$};
    \node at (3.3,1) {$\bullet$};
    \node at (7,1) {$=$};

    \draw[double equal sign distance] (a) -- (b);
    \draw[->] (c) -- node[midway,right] {$\eps$}  (d);
    \draw[->] (e) -- node[midway,right] {$\eps$}  (f);
    \draw[->] (a) -- node[midway,above] {$z$}  (g);
    \draw[->] (b) -- node[midway,above] {$z$}  (h);
    \draw[double equal sign distance] (c) -- (i);
    \draw[double equal sign distance] (d) -- (j);
    \draw[->] (e) -- node[midway,above] {\!\small{$z(-1)^{J(m,n)}$}}  (k);
    \draw[->] (f) -- node[midway,above] {\!\small{$z(-1)^{J(m,n)}$}}  (l);
    \draw[->] (g) -- node[midway,right] {$1$}  (h);
    \draw[->] (i) -- node[midway,right] {$\eps$}  (j);
    \draw[->] (k) -- node[midway,right] {$\eps$}  (l);
  \end{tikzpicture}
\end{center}

and

\begin{center}
  \begin{tikzpicture}[scale=1.1]
    \node at (4,2) [name=a] {$0$};
    \node at (6,2) [name=g] {$-m$};
    \node at (.6,2) [name=c] {$-\frac n2$};
    \node at (2.6,2) [name=i] {$-\frac n2$};
    \node at (8,2) [name=e] {$-\frac n2$};
    \node at (10.5,2) [name=k] {$-\frac{2m+n}2$};
    \node at (4,0) [name=b] {$0$};
    \node at (6,0) [name=h] {$m$};
    \node at (.6,0) [name=d] {$\frac n2$};
    \node at (2.6,0) [name=j] {$\frac n2$};
    \node at (8,0) [name=f] {$\frac n2$};
    \node at (10.5,0) [name=l] {$\frac{2m+n}2$.};
    \node at (3.3,1) {$\bullet$};
    \node at (7,1) {$=$};

    \draw[double equal sign distance] (a) -- (b);
    \draw[->] (c) -- node[midway,right] {$\eps$}  (d);
    \draw[->] (e) -- node[midway,right] {$\eps$}  (f);
    \draw[->] (a) -- node[midway,above] {$z$}  (g);
    \draw[->] (b) -- node[midway,above] {$z$}  (h);
    \draw[double equal sign distance] (c) -- (i);
    \draw[double equal sign distance] (d) -- (j);
    \draw[->] (e) -- node[midway,above] {{$z$}}  (k);
    \draw[->] (f) -- node[midway,above] {$z$}  (l);
    \draw[->] (g) -- node[midway,right] {$1$}  (h);
    \draw[->] (i) -- node[midway,right] {$\eps$}  (j);
    \draw[->] (k) -- node[midway,right] {$\eps$}  (l);
  \end{tikzpicture}
\end{center}
So, the conjugation action is as claimed.
\end{proof}

\begin{Rem}
 Let $I=J+J^t$, and consider the
  integer-valued quadratic form
  \begin{eqnarray*}
    \phi(m) & = & \frac12 I(m,m)
  \end{eqnarray*}
  on $\Lat$. Its reduction mod 2 is the Arf invariant of $\widetilde{T[2]}$.
  The form $I$ can be recovered from $\phi$ by the identity
  \begin{eqnarray*}
    I & = & \delta_{simp}(\phi),
  \end{eqnarray*}
  i.e.,
  \begin{eqnarray*}
    I(m,n) & = & \phi(m+n)-\phi(m)-\phi(n).
  \end{eqnarray*}
\end{Rem}

\begin{Exa}

There is a prominent
subgroup $C$ of the Monster, sitting in a non-split extension
\begin{center}
\begin{tikzpicture}
  \node at (0.2,0) [name=a] {$1$};
  \node at (2,0) [name=b] {$\widetilde{T[2]}$};
  \node at (4.3,0) [name=c] {$C$};
  \node at (6.5,0) [name=d] {$Co_1$};
  \node at (8.4,0) [name=e] {$1$,};

  \draw[->] (a) -- (b);
  \draw[->] (b) -- (c);
  \draw[->] (c) -- (d);
  \draw[->] (d) -- node [midway, above] {} (e);
\end{tikzpicture}
\end{center}
where $Co_1$ is the
Conway group $Co_1=Co_0/{\pm1}$. This subgroup is typically the first
step in the construction of the Monster, see for instance
\cite{Tits:Le_Monstre} or \cite{Conway:Sloane:Sphere_packings}.
\end{Exa}

\section{From categorical groups to loop groups}\label{sec:loops}
The relationship between loop groups and categorical Lie groups has
been the subject of extensive study.
In \cite{Baez:Stevenson:Crans:Schreiber}, the authors use central
extensions of loop groups to define 2-groups.
We will mostly be interested in the other direction and work with the transgression-regression
machine of \cite{Waldorf:Transgression_to_loop_spaces_and_its_inverse_II} and
\cite{Waldorf:A_construction_of_string_2-group_models}. In this context,
the term {\em `transgression'} refers
to a recipe for turning multiplicative bundle
  gerbes with connection into central extensions of loop
  groups. Let $\mL T$ be the group of piecewise smooth loops in $T$.
  Applied to the trivial bundle gerbe $\mI$ on $T$ equiped with
  the trivial connection, transgression yields the trivial principal
  $U(1)$-bundle over $\mL T$. If we further apply
  transgression to our multiplicative structure with
  connection $((\LE,\nabla),\alpha)$, we obtain
  the central extension of $\mL T$ whose
  2-cocycle $c$ is given by the holonomy of $(\LE,\nabla)$. So, if
  $\phi$ and $\gamma$ are loops in $T$, and $(f,g)$ is any choice of lift of
  $(\varphi,\gamma)$ to $\mathfrak t\times\mathfrak t$, then we have
  \begin{eqnarray}
    \notag
    c(\varphi,\gamma) & = & Hol_{\(\LE,\nabla\)}((\varphi,\gamma))\\[+4pt]
    \label{eq:cocycle}
    & = & \exp\(\int_0^1 J(f(t),\dot{g}(t))\, dt\).
  \end{eqnarray}

(The underlying principal bundle of this central extension is trivial.)
\begin{Lem}
  The central extension $\widetilde{\Omega T}$ defined
  by the restriction of $c$ to the based loops $\Omega T$ is
  isomorphic to that in \cite[4.8]{Pressley:Segal}.
\end{Lem}
\begin{proof}
  Pressley and Segal denote our $I$ by $\langle-,-\rangle$, their $b$
  can be taken to be our $J$, and they write $\Lambda$ for the
  cocharacter lattice $$\vT=\Hom(U(1),T).$$ We first note that $c$
  describes the correct extension of this lattice
  \begin{eqnarray*}
    \Lat\cong \vT \subset \mL T,
  \end{eqnarray*}
  namely
  \begin{eqnarray*}
    c_\Lat(m,n) & = & (-1)^{J(m,n)}.
  \end{eqnarray*}
  The commutators in $\widetilde{\mL T}$ are
  \begin{eqnarray}
    \notag
    \widetilde\varphi\cdot\widetilde\gamma\cdot\widetilde\varphi\inv\cdot
    \widetilde\gamma\inv
    &=&\frac{c(\varphi,\gamma)}{c(\gamma,\varphi)}\\
    \notag
    & = & \exp\(\int_0^1 J(f(t),\dot g(t))\, dt -\int_0^1J(g(t),\dot f(t))\,dt\)\\
    \label{eq:commutator1}
    & = & \exp\(\int_0^1 I(f(t),\dot g(t))\,dt -
    \left[J(g(t),f(t))\right]^1_0\)\\
    \label{eq:commutator2}
    & = & \exp\(-\int_0^1 I(\dot f(t),g(t))\,dt + \left[J(f(t),g(t))\right]^1_0\)
  \end{eqnarray}
  (integration by parts).
  Over the identity component of $\mL T$, our extension is completely
  described by its Lie-algebra cocycle,
  \begin{eqnarray*}
    \omega & = & \int_0^1 I(f(t),\dot g(t)) \,dt
  \end{eqnarray*}
  (from \eqref{eq:commutator1} and \cite[2.1]{Pressley:Segal}), which
  agrees with the expression for $\omega$ in \cite[(4.2.2)]{Pressley:Segal}. The
  adjoint action of $\varphi\in\mL T$ on $\widetilde{L\mathfrak t}$
  can be read off from \eqref{eq:commutator2}, it sends
  $(g,r)\in\widetilde{L\mathfrak t}$ to
  $$
    \(g,r - \int_0^1I(\dot f(t),g(t))dt + J\(\Delta_f,g(0)\)\),
  $$
  where $\Delta_f=f(1)-f(0)$. If $g\in\Omega\mathfrak t$ is a based
  loop, this agrees with the expression in
  \cite[(4.3.2)]{Pressley:Segal}.
\end{proof}
\begin{Rem}
  This definition of $\widetilde{\mL T}$, as transgression of a
  categorical torus, leads to a considerable simplification of the
  picture in the standard literature on loop groups
  \cite[4.8]{Pressley:Segal}. The expression
  \eqref{eq:cocycle} is less complicated than
  \cite[(4.8.3)]{Pressley:Segal}. It is, by construction, invariant
  under the action of $Diff^+(\SS^1)$. There is no need to restrict ourselves to
  maximal tori of simply laced groups. Further, the cocycle $c$
  satisfies the {\em fusion rule}
  \begin{eqnarray}\label{eq:fusion}
    c(\varphi_1*\overline\varphi_2,\gamma_1*\overline\gamma_2) \cdot
    c(\varphi_2*\overline\varphi_3,\gamma_2*\overline\gamma_3) & = &
    c(\varphi_1*\overline\varphi_3,\gamma_1*\overline\gamma_3).
  \end{eqnarray}
  Here $*$ stands for concatenation of paths, and $(\varphi_1,\varphi_2,\varphi_3)$ and
  $(\gamma_1,\gamma_2,\gamma_3)$ are triples of paths with identical
  start and end points, i.e.,
  $$\varphi_1(0)=\varphi_2(0)=\varphi_3(0)\quad\quad\text{and}\quad\quad
  \varphi_1(1)=\varphi_2(1)=\varphi_3(1),$$ and likewise for the
  $\gamma_i$.
\end{Rem}

Our fourth construction of $\mT$ applies Waldorf's
regression machine to reconstruct our categorical torus from the cocycle $c$.
\begin{Cons}\label{cons:loops}
  Given $(\Lat,J)$, we let $T$ be the torus $\Lat\tensor U(1)$ and
  $\widetilde{\mL T}$ the central extension of
  the loop group $\mL T$ defined in \eqref{eq:cocycle}. We write
  $\mP_1T$ for the space of smooth paths based at $1$ in $T$ and
  $$
    \Omega T\,\cong\, \mP_1T\times_T\mP_1T
  $$
  for the group of smooth loops based at $1$.
  Then we have the (infinite dimensional) Lie groupoid
  \begin{center}
    \begin{tikzpicture}
    \node at (1.2,0) [name=a, anchor=west] {$\widetilde{\Omega T}\Large)$,};
    \node at (0,0) [name=b, anchor=east] {$\mG\,\, :=\,\, \Large(\mP_1T$};
    \draw[->] (1.2,.1) -- (0,.1);
    \draw[->] (1.2,-.1) -- (0,-.1);
  \end{tikzpicture}
  \end{center}
  where the composition of arrows is
  $$
    (z,\gamma_2,\gamma_3)\circ (w,\gamma_1,\gamma_2) \,=\,
    (zw,\gamma_1,\gamma_3).
  $$
  The fusion rule \eqref{eq:fusion} ensures that this composition is a
  group homomorphism, making $\mG$ a groupoid in Lie groups.
\end{Cons}

By a general result of Waldorf's this construction recovers our
categorical torus. We give an explicit equivalence in three steps.
\begin{Lem}
  The strict categorical Lie group $\mG$ of Construction
  \ref{cons:loops} corresponds to the crossed module
  \teqnarray{\Pi}{\widetilde{\Omega T}}{\mP_1T}{(z,\gamma)}\gamma
  where $\widetilde{\Omega T}$ is the central extension with cocycle
  $c\inv$, equipped
  with the $\mP_1T$-action
  $$
    (z,\gamma)^\varphi\,=\,(ez,\gamma),
  $$
  with
  $$
    e \,=\, \exp\(\int_0^1I(f(t),\dot g(t)) dt\).
  $$
  Here $(f,g)$ is the lift of $(\varphi,\gamma)$ to $\mathfrak
  t\times\mathfrak t$ that starts at $(0,0)$.
\end{Lem}
\begin{proof}
  The source kernel of $\mG$ consists of triples $(z,1,\gamma)$ where
  $\gamma$ is a closed loop and $1$ denotes the constant loop. Since
  $c$ is invariant under reparametrization, we have
  $$
    c(1*\overline\gamma,1*\overline\beta) \,=\,
    c(\overline\gamma,\overline\beta) \,=\, c(\gamma,\beta)\inv.
  $$
  Similarly, the conjugation action of the loop $\varphi *\overline\varphi$ on
  $(z,1,\gamma)$ is calculated by \eqref{eq:commutator1}.
\end{proof}
In the following, we deviate from the notation in Section
\ref{sec:Extraspecial} and write $\widetilde\Lat$ for the central
extension of $\Lat$ by $U(1)$ with cocycle $(-1)^J$. Recall that a
(weak) map of crossed modules, in the sense of
  \cite[Def.8.4]{Noohi:Notes_on_2-groupoids}, is an equivalence if it
  induces a monoidal equivalence of the corresponding categorical groups.
\begin{Lem}
  The crossed module of the previous lemma is equivalent to the
  crossed module
  \upshape{
    \teqnarray{\Xi}{\widetilde\Lat}{\mathfrak t}{(z,m)}m
  }
  with action
  $$
    (z,m)^x\,=\,\(z\cdot\exp\(\frac12I(x,m)\),m\).
  $$
\end{Lem}
\begin{proof}
  For $x\in\mathfrak t$, define the path
  \teqnarray{\gamma_x}{[0,1]}{T}t{\exp(tx)}
  from $1$ to $\exp(x)$ in $T$. The resulting group homomorphism from
  $\mathfrak t$ to $\mP_1T$

  identifies $\Lat$ with the subgroup $\vT\sub\Omega T$, and makes $\Xi$ into an
  equivalent sub-crossed module of $\Pi$.
\end{proof}
\begin{Lem}
  We have an equivalence
  $$\Psi\,\simeq \,\Xi$$
  between the crossed module of
  Construction \ref{cons:strict} and that of the previous lemma.
\end{Lem}
\begin{proof}
  Let $\longmap a\Lat\{\pm1\}$ be any 1-cochain on $\Lat $ with
  boundary $(-1)^J$. Then we have the group homomorphism
  \teqnarray{p_2}{\Lat\times U(1)}{\widetilde\Lat}{(m,z)}{(z\cdot a(m),m).}
  Together with the maps $p_1=id_{\mathfrak t}$ and
  \teqnarray{\eps}{\mathfrak t\times\mathfrak
    t}{\widetilde\Lat}{(x,y)}{\(\exp\(\frac12J(x,y)\),0\),}
  this defines a weak map of crossed modules inducing an equivalence
  of the corresponding categorical groups.
\end{proof}
This concludes our argument that the four constructions of our
categorical torus are equivalent.

\bibliographystyle{amsalpha}
\bibliography{weyl.bib}

\end{document}

config: arxiv


TITLE: Use tableau to convert formula to DNF/CNF form
QUESTION [0 upvotes]: Is there any method that can be used to convert any formula do a DNF/CNF form using only the truth table?
For example if I have the following formula

p → ¬(q∨r)

How can I convert it into DNF?
p q r  (q∨r)   ¬(q∨r)   p → ¬(q∨r)
0 0 0    0         1         1
0 0 1    1         0         1
0 1 0    1         0         1
0 1 1    1         0         1
1 0 0    0         1         1
1 0 1    1         0         0
1 1 0    1         0         0
1 1 1    1         0         0

REPLY [2 votes]: If you write out the truth table for your formula...
p q r  p → ¬(q∨r)
0 0 0    1
0 0 1    1
0 1 0    1
0 1 1    1
1 0 0    1
1 0 1    0
1 1 0    0
1 1 1    0

To get the DNF, look at every result that is True; those lines directly correspond to the following clauses in DNF: $$(\lnot p\land\lnot q\land\lnot r)\lor(\lnot p\land\lnot q\land r)\lor(\lnot p\land q\land \lnot r)\lor(\lnot p\land q\land r)\lor(p\land \lnot q\land \lnot r)$$
This is a complete description of the formula in DNF - look at how the zeros in the table inputs result in a negation of a literal, and ones result in plain literals.
To get the CNF, look at every result that is False; those lines directly correspond to the following clauses in CNF: $$(p\lor \lnot q\lor r)\land(p\lor q\lor \lnot r)\land(p\lor q\lor r)$$
Again, this is a complete description of the formula in CNF.
It follows then, that if you also wish to convert a DNF to a CNF then what you need to do is reverse engineer this process. Say for instance you had a DNF:

map each clause in the DNF to a line in the truth table that results in True
map all other combinations of inputs to False
then follow the process above to get the CNF from the resulting truth table

This process can be followed to convert CNF to DNF as well if we swap True and False, DNF and CNF etc..


TITLE: Problem with the definition of sample function (realization).
QUESTION [1 upvotes]: Suppose our sample space is $\Omega = \{H,T\}$ and we have two random variables $X_1 , X_2$ such that $X_1(H) = 1,X_1(T) = 2$ and $X_2(H) = 3,X_2(T) = 4$. If we define stochastic process $S = \{X_1,X_2\}$ is $\{1, 4\}$ a realization of $S$ or only sets $\{1, 3\},\{2, 4\}$ are possible realization of process $S$? I though a realization means a sequence of sampling from each random variable in a process but reading this sample function definition (also page 9)  seems i was totally wrong and $\{1, 4\}$ cant be realization of $S$ because realization is  defined for a fixed $\omega \in \Omega$ and in the case of $\{1, 4\}$ the first outcome is $H$ and the second outcome is $T$.

REPLY [1 votes]: If $\omega=H$ then $X_1=1$ and $X_2=3$, and similarly if $\omega=T$ then $X_1=2$ and $X_2=4$. So indeed the only realizations of $S$ are $\{1,3\}$ and $\{2,4\}$.


TITLE: Collections (families of sets) and powersets
QUESTION [1 upvotes]: I've recently been able to prove that $P(E) \cap P(F) = P(E \cap F)$ and a few similar dealing with powersets. However the notation of collections is rather confusing, and I'd like to have a clear understanding.
Ive been asked to show E is always equal to $E = \bigcup P(E)$. At best I believe I can interpret this as $\{X: X \in P(E)\ or\ X \in X\}$. Note that the question may also be written E = $\bigcup_{X \in P(E)} X$. Which makes even less sense. Could someone go into detail into the notation of collections of intersections/unions and its use with this question?

REPLY [2 votes]: The notation $\bigcup P(E)$ does not mean what you said.
In general, $$\def\S{\mathscr S}\bigcup_{S\in \S} S$$ means the union of all the sets in $\S$, which is a family of sets. The notation $$\bigcup \S$$ (with no subscript) is a shorthand for this. Some entity $x$ is an element of  this union if it is an element of one of the sets in the union.
This is the way unions always work.  For example, $x$ is an element of the union $A\cup B\cup C$ if it is an element of one of the sets in the union; that is if it is a element of $A$ or an element of $B$ or an element of $C$.
So:  $$x\in \bigcup_{S\in\S} S\\\text{means}\\ x \in S \text{ for some $S\in \S$.}$$
Or using the shorthand:
$$x\in \bigcup \color{darkblue}{\S}\\\text{means}\\ x \in S \text{ for some $S\in \S$.}$$
(That was the crucial point, which you should make sure you understand before continuing.)
Applying this when $\S = P(E)$ we have $$x\in \bigcup \color{darkblue}{P(E)}\\\text{means}\\ x \in S \text{ for some $S\in \color{darkblue}{P(E)}$.}$$
But $S\in P(E)$ is true exactly when $S\subset E$; this is the definition of $P(E)$.  So we can rewrite the formula above as:
$$x\in \bigcup P(E)\\\text{means}\\ x \in S \text{ for some $\color{darkblue}{S\subset E}$.}$$
If at this point you could show that $$x\in S\text{ for some } S\subset E$$ and $$x\in E$$ were equivalent conditions, you would be done, because then you would have
$$x\in \bigcup P(E)\\\text{is equivalent to}\\ x \in E.$$
which is what you were asked to show.
I hope this is what you were looking for.


TITLE: Why is regularity a problem in cubical type theory?
QUESTION [11 upvotes]: In my current understanding, regularity in cubical type theory is the following definitional equality (I'm using $A~\textbf{type}$ to emphasize the fact that $i \notin FV(A)$):
$$
\cfrac{A~\textbf{type} \quad a:A}
{\text{transport}(\langle i \rangle A, a) \mapsto a}
$$
I am not sure how is transport specified here (does it have a standard formulation?). I just assume it to be a special case of transp in de Morgan cubical type theory and a special case of coe in Cartesian cubical type theory.
In Cubical Agda, this definitional equality doesn't hold. It's proved as transportRefl in the cubical library. I'm not sure how it works in Cartesian cubical type theory, but I guess it is also a long-standing problem (Jon Sterling said he solved it a while ago, but it's on Twitter, which is not a reliable source of academic information).
I don't quite get the difficulty -- it seems like a problem in the model. In Arend's type theory, regularity is a part of the definition of coe, and it's implemented as "normalizing the path, reduce to $a$ if $i \notin FV(A)$". Why can't we do this in cubical type theory? We already have eta laws for functions where we need FV checks, so this is not a problem implementation-wise (or is it? Please tell me!).

REPLY [13 votes]: The difficulty is in making such a reduction compatible with all the other reductions involving transport/coe. From one perspective it is a “confluence” problem. It is unfortunate that in the community, so much was said about "decidability of degeneracies" that we began to give the impression that this was the hard part of the algorithm (since as you say, in syntax you can decide degeneracy)...
The real problem is competing reductions. If you have both the type-based reductions and the regularity rule, then the term Transport(P, a) can reduce in potentially many different ways. It is necessary to ensure that the type-based reductions satisfy the regularity law, then. The problem in cubical type theory was that the type-based reductions do not satisfy the regularity law (actually, most of them do satisfy it, but ones involving the glue type do not). So the difficulty with regularity was to find new versions of these type-based reductions that preserve regularity, and I personally believe this cannot be done without extending the language of cofibrations beyond what appears in the CCHM or ABCFHL papers. (If you add further cofibrations, it is trivial to correct the reductions for CCHM & ABCFHL --- but my ideas in this direction do not seem possible to implement.)
The reason that Arend doesn't seem to have this problem is that Arend doesn't support the other reductions... So in principle it is possible to have a sound type theory that supports the regularity reduction for paths by simply deleting the other reductions that might conflict with it (but I do not want to make a claim about Arend's soundness).


TITLE: Do these 2 mathematical expressions both refer to the same thing?
QUESTION [0 upvotes]: (1)$ (\forall a \in S)(\exists e \in S) $
(2)$(\exists e \in S)(\forall a \in S)$
Are  $(1)\equiv(2) $ or the order in which I write them has some different meaning ?
Basically, I have been told that (1) and (2) are not the same mathematical expressions ... Could someone please explain why aren't they the same ?

REPLY [3 votes]: Everyone is someone's child, but nobody is everyone's parent.
In symbols: taking "$C(a, b)$" to mean "$a$ is $b$'s child," we have $\forall x\exists y C(x, y)$ but $\neg \exists y\forall x C(x, y)$.
"$\exists x\forall y\varphi(x, y)$" says that not only does every $y$ have an $x$ such that $\varphi(x, y)$ holds, but there is one single $x$ which works for every $y$. This is in general a much stronger claim.
Another example worth considering is that the statement "There is no largest number" can be written in two natural ways: "For all $x$ there is some $y$ such that $x<y$" and "NOT(there is some $y$ such that for all $x$, $x<y$)." In this example, not only are "$\forall x\exists y$" and "$\exists y\forall x$" not equivalent, they're actually direct opposites!

REPLY [1 votes]: $\forall$ means "for all" but sometimes more cleanly translates as "for any" or "for every."
$\exists$ means "there exists" but might be interpreted as "we can find."
For any point, $a,$ on the number line, there is a point $e$ on the number line that is greater than $a.$
Clearly true.
There exists a point on the number line $e,$ for every point $a$ on the number line $a$ is greater than $e.$
This implies that $e$ is the smallest point.  Or that there is a least element which might be true in $\mathbb N$ but is not true in $\mathbb Z$ or $\mathbb R$


TITLE: Transforming a QFT identity (quantisation of the Dirac field)
QUESTION [0 upvotes]: When (falsely) quantizing the Dirac-field, Peskin/Schroeder (Introduction to Quantum Field Theory) get with
$$\psi(\vec{x})=\int\dfrac{d^3p}{(2\pi)^3}\dfrac{1}{\sqrt{2E_{\vec{p}}}}e^{i\vec{p}\cdot\vec{x}}\sum\limits_{s=1,2}\left(a_{\vec{p}}^su^s(\vec{p})+b_{-\vec{p}}^sv^s(-\vec{p})\right)$$
and
$$ \left[a_{\vec{p}}^r,~a_{\vec{q}}^{s\dagger}\right] = \left[b_{\vec{p}}^r,~b_{\vec{q}}^{s\dagger}\right] = (2\pi)^3\delta^{(3)}(\vec{p}-\vec{q})\delta^{rs}$$
the relationship
$$\left[\psi(\vec{x}),~\psi^{\dagger}(\vec{y})\right] = \int\dfrac{d^3p~d^3q}{(2\pi)^6}\dfrac{1}{\sqrt{2E_{\vec{p}}2E_{\vec{q}}}}e^{i(\vec{p}\cdot\vec{x}-\vec{q}\cdot\vec{y})}\times\sum\limits_{r,s}\left(\left[a_{\vec{p}}^r,~a_{\vec{q}}^{s\dagger}\right]u^r(\vec{p})\bar{u}^s(\vec{q})+\left[b_{-\vec{p}}^r,~b_{-\vec{q}}^{s\dagger}\right]v^r(-\vec{p})\bar{v}^s(-\vec{q})\right)\gamma^0$$
This is all easy to see but I am wondering where the $\gamma^0$ at the end comes from?

REPLY [3 votes]: $$\psi(\vec{x})=\int\dfrac{d^3p}{(2\pi)^3}\dfrac{1}{\sqrt{2E_{\vec{p}}}}e^{i\vec{p}\cdot\vec{x}}\sum\limits_{s=1,2}\left(a_{\vec{p}}^su^s(\vec{p})+b_{-\vec{p}}^sv^s(-\vec{p})\right)$$ so $$\psi^{\dagger}(\vec{y})=\int\dfrac{d^3q}{(2\pi)^3}\dfrac{1}{\sqrt{2E_{\vec{q}}}}e^{-i\vec{q}\cdot\vec{y}}\sum\limits_{s=1,2}\left(a_{\vec{q}}^{\dagger s}u^{\dagger s}(\vec{q})+b_{-\vec{q}}^{\dagger s}v^{\dagger s}(-\vec{q})\right)$$
This gives the relationship $$\left[\psi(\vec{x}),~\psi^{\dagger}(\vec{y})\right] = \int\dfrac{d^3p~d^3q}{(2\pi)^6}\dfrac{1}{\sqrt{2E_{\vec{p}}2E_{\vec{q}}}}e^{i(\vec{p}\cdot\vec{x}-\vec{q}\cdot\vec{y})}\times\sum\limits_{r,s}\left(\left[a_{\vec{p}}^r,~a_{\vec{q}}^{s\dagger}\right]u^r(\vec{p})u^{\dagger s}(\vec{q})+\left[b_{-\vec{p}}^r,~b_{-\vec{q}}^{s\dagger}\right]v^r(-\vec{p})v^{\dagger s}(-\vec{q})\right)$$
Now, $$\gamma^0 = \begin{pmatrix} I & 0 \\ 0 & -I \end{pmatrix}$$ and so we can define the Dirac adjoint of $u$, $\bar{u} = u^{\dagger}\gamma^0$ which also gives $u^{\dagger} = \bar{u}\gamma^0$ as $\left(\gamma^0\right)^{-1} = \gamma^0$ and we obtain the required result


TITLE: If vectors in a set is linearly independent to each other, does it make the set an linearly independent set? Why or why not?
QUESTION [0 upvotes]: A set S has vectors $a_1$,$a_2$,$a_3$,.....,$a_n$.
$a_1$ is linearly independent to $a_2$,$a_3$,......,$a_n$.
$a_2$ is linearly independent to $a_1$, $a_3$,......,$a_n$.
$a_3$ is linearly independent to $a_1$, $a_2$,......,$a_n$.
......
$a_n$ is linearly independent to $a_1$, $a_2$,$a_3$,......,$a_(n-1)$.
does this make the set S a linearly independent set?
Why or why not?

REPLY [0 votes]: Let $S=\{v_1,\dots,v_n\}$ be the set you describe in the question. Let $a_1,\dots,a_n$ be scalars such that $$a_1v_1+\dots+a_nv_n=0$$
This implies that the $$a_1=\dots=a_n=0$$ as if that were not the case, then there would exist a vector that could be written as a linear combination of the other vectors in $S$, contradicting the hypothesis that each vector is linearly independent to all the other vectors.
Trivial point but worth mentioning: None of the vectors can be the zero vector as then that vector would not be linearly independent to the other vectors.