text:
\chapter{Higher structures}\label{chap:higher}

One of the two main conceptual questions in this thesis is what happens to mathematical -- mostly geometric -- structures under the process of quantisation. The other question could be stated as follows: \emph{what are the most suitable structures to begin with?} In this chapter we explore a handful of ideas to this end. In particular, we introduce concepts that involve a \emph{higher} or \emph{categorical} perspective. There is a growing interest in mathematical physics to study such objects in hopes that they would help make the known phenomena conceptually simpler, and perhaps lead to new physics as well. Ultimately one hopes to get a better insight into the mysteries of quantisation, so that at the end of the day these two conceptual questions combine to just one. While even attempting to answer this Great Question properly is well beyond the scope of this thesis, we do try to provide a few glimpses to this direction.

Thus we do not strive to give an in-depth treatment of category theory or higher geometry. Rather, we are motivated by two examples with nontrivial connections to quantum anomalies: namely, categorical groups and bundle gerbes. The first of these gives a context to the earlier results published in \cite{MN2019}, and the second provides a neat geometric handle to the group cohomology aspect of the Hamiltonian anomaly. These two examples are not entirely unrelated, since in both a prominent feature is the Mickelsson-Faddeev extension of current groups on three-dimensional manifolds -- and indeed one of our main motivations in this chapter is to look for a categorical perspective to current groups.

\section{Categorification and higher objects}
\idxalku{categorification}

The classical gauge field theory as introduced in the previous chapter is geometrically a theory of connections on principal bundles over topological spaces, describing the (gauge) dynamics of point particles. However, in theories such as string theory or loop quantum gravity one would need a dynamic description for objects that are not point particles but rather $1$-dimensional paths or loops in a space -- or, going further, $k$-dimensional surfaces (or branes, as they are often called). For the gauge theory formalism to extend to such applications it is necessary to extend the basic geometric concepts beyond the point-space foundations. This is one of the motivations for defining what could be called \emph{higher geometry} through the process of \emph{categorification}. Historically, the study of higher categorical objects has also been closely linked to topological quantum field theory~\cite{BAEZ1997}. More recently, higher symmetry groups have been a subject of growing interest not only among gauge theorists, but also in condensed matter physics. In particular, higher groups appear to be relevant in the study of topological phases of matter.~\cite{GKSW2015,SHARPE2015}

One can roughly divide categorification into two flavours: \term{vertical} and \term{horizontal categorification}.\footnote{Horizontal categorification could also be called \term{oidification} since it often leads to objects like groupoids, algebroids, ringoids, and such.} The following diagram illustrates the general idea: in vertical categorification, one extends objects and functions -- algebras, groups and the like -- to categories and functors of such, and further to higher categories; on the horizontal side, one can realise the same objects with a category of a specific type, and then study the general structure behind this instance.
\[
\begin{tikzcd}
  & \text{$n$-categories} & \\
  & \mlnode{categories\\{\itshape{\small morphisms, functors}}}  \arrow[u] & \\
  \mlnode{sets\\{\itshape{\small equations, functions}}} \arrow[rr] \arrow[ur] & & \text{single-object categories} \arrow[ul]
\end{tikzcd}
\]
Put together, one goes not only higher but also \emph{wider}. A good example is the categorification of groups. The corresponding diagram could look like this:
\[
\begin{tikzcd}
  \text{(categories of) $n$-groups} &  \text{(categories of) $n$-groupoids} \\
  \text{category of groups} \arrow[u] & \text{groupoids} \arrow[u] \\
  \text{group} \arrow[r] \arrow[u] & \text{single-object groupoid} \arrow[u]
\end{tikzcd}
\]

  \idx{groupoid}
  \idx{$2$-group}

While groups, groupoids, and their $n$-categorified ancestors are not entirely alien concepts to a working mathematician, this process is not always straightforward nor is it easy to see as to what exactly would the appropriate definitions for the extended entities be like. Already in the categorification process for Lie algebras one runs into rather complicated questions, see for instance \cite{BC2004} and \cite[Sec.~6.5.2]{SCHREIBER2017}.
  \footnote{Besides specific examples, there seems to be little literature available on the categorification as such. Here we have mostly followed \cite{BD1998} and the more informal ideas found in \cite{NLAB:CAT}.}

Since the horizon of categorification is rather vast, we do not dwell on the general ideas too long. A bit more space will be dedicated to categorical groups from which one may gain a higher perspective to symmetry groups and related anomalies.
However, before the discussion on the categorical groups, we argue that there is also some sense in speaking of higher objects and structures without explicitly diving into category theory as such.

\subsection{What is it like to be a higher object?}
\idx{higher object}

Categorification described above is not necessarily the only way to catch a glimpse of topology or geometry that is in some sense higher.

One characteristics of the generalisation of the point spaces is that of a more complex homotopy on the base. From this perspective it makes sense to call higher objects those reflecting a higher homotopy type of the space. Recall that a space $X$ of a homotopy type $n$ is characterised by the groups $\pi_k(X)$ with $k\leq n$, in addition to the demand that all higher homotopy groups are trivial. For instance, the $2$-group encountered below is a group structure on a homotopy $1$-type objects -- thus the generalised group structure equivalences must preserve isomorphisms of the homotopy groups $\pi_0$ and $\pi_1$, which is not a property of the standard definition of groups. Similarly, the bundle gerbe is an instance of a Lie groupoid coming either from a principal $2$-bundle or a groupoid extension -- in the former case, this is a principal $U(1)$-bundle structure over a $2$-group (homotopy $1$-type).

In Chapter~\ref{chap:gauge} we implicitly introduced a categorical object: the space of gauge connections and their transformations. This is an example of a groupoid -- and since the gauge transformations themselves are homotopies of other geometric objects, it is natural to consider them in the categorical framework.

We will later see in our examples that higher objects can be well defined without explicitly speaking about categories or such. This indeed is one of the aims: to \emph{concretely} build objects that fulfill the necessary properties without making the process itself unduly abstract. It is the application of category-theoretical concepts that drives us, not so much the machinery itself.

\idxloppu{categorification}

\section{Categorical groups and symmetries}\label{sec:cat_groups}

\idxalku{categorical groups}

We introduce two main components in this section: groupoids and $2$-groups as categorification of groups. Groupoids have already appeared in various flavours in mathematics -- also in gauge theory, as argued above -- while $2$-groups have gained more recent interest in mathematical physics. We discuss the necessity of the $2$-group structure when it comes to string theory, and in the next section construct two examples. We refer to \cite{SHARPE2015} for a more general exposition to categorical groups in quantum field theory.

\subsection{Categorical groups}

We begin the categorification of the group structure in the horizontal direction by introducing groupoids.
\begin{definition}
    \idx{groupoid}

  A \termd{groupoid} is a small category in which all morphisms are isomorphisms.
\end{definition}
Recall that in a \term{small category}
  \idx{category!small}

both the objects and the morphisms form proper sets. A single-object groupoid would then take us back to the usual definition of a group, with morphisms serving as the group elements. The definition naturally extends to include smooth structures, and so we get \term{Lie groupoids}
  \idx{groupoid!Lie (smooth)}

(and Lie algebroids, see \cite{MACKENZIE2005}).

The canonical example of a groupoid is the fundamental groupoid of a topological space: the fundamental group $\pi_1(X,p)$ of a space $X$ based at $p\in X$ is the automorphism group of $p$ in the fundamental groupoid $\Pi_1(X)$ -- thus the fundamental groupoid can encode the information provided by fundamental groups without a singled-out base point. The objects of the fundamental groupoid are points in $X$, and its morphisms are homotopy classes of paths between two points. The groupoid axioms are fulfilled by defining composition as concatenation of paths; the rest follows from the properties of the homotopy equivalence.

We have already mentioned that the space of gauge connections forms a groupoid. What we mean is the following.
\begin{example}[Gauge groupoid]\label{ex:gauge_groupoid}
    \idx{groupoid!gauge}

  Let $\A$ be the space of connections on a gauge bundle over a manifold $M$, and let $\G$ be the group of gauge transformations. The transformations $A \mapsto A^g$ given by a $g\in \G$ are morphisms in the Lie groupoid of local connections $A \in \A$. In other words, for an open set $\U \subset M$, the objects of the \term{gauge groupoid} are the smooth Lie algebra valued $1$-forms $A \in \map{\U,\lie{g}}$, and the morphisms are derived from the smooth $G$-valued functions $g \in \map{\U,G}$ by imposing the gauge principle
    \idx{gauge transformation}

  \[
    A^g \defeq \ad_g(A) - g^*(\theta_G) ,
  \]
  where $\theta_G$ is the Maurer-Cartan form on $G$. Often in the context of matrix groups the pullback is written as $g^*(\theta_G) = \inv{g}\ext g \in \map{\U,\lie{g}}$.
\end{example}

Another good example, which we will also use later on, is that of \term{action groupoid}.
  \idx{groupoid!action}

Let $G$ be a group acting on a set $X$. The objects in the action groupoid $\actg{X}{G}$ are the elements of $X$, and the morphisms between two given elements $a$ and $b$ in $X$ are the group elements $g\in G$ such that $g.a = b$; the composition of morphisms is given by the multiplication in $G$. The notation $\actg{X}{G}$ follows from the idea that the action groupoid is in some sense a \emph{weak quotient}: in the proper quotient $X/G$ the elements in the same orbit subset of $X$ are considered equal, whereas in the action groupoid they are considered merely isomorphic.

\idxalku{$2$-group}

In the vertical direction the group structure is generalised by introducing $2$-groups as categories with specified functors analogous to the group operations. One of the many ways to think about $2$-groups is to view them as a group object in the category of groupoids.
  \footnote{See \cite{FORRESTERBARKER2002} for a review on various equivalent formulations of strict $2$-groups.}
\begin{definition}
  A \termd{strict $2$-group} is a monoidal category in which all morphisms are isomorphisms, and every object has an inverse.
\end{definition}
A \term{monoidal category}
  \idx{category!monoidal}
is a category $\Ca$ with a \term{product functor} $\otimes:\Ca\times \Ca \to \Ca$ such that it contains
\begin{enumerate}
  \item a unit object $1 \in \Ca$,
  \item an associator isomorphism $\alpha: (x\otimes y) \otimes z \to x \otimes (y\otimes z)$,
  \item left/right unitor isomorphisms $l : 1 \otimes x \to x$ and $r: x \otimes 1 \to x$,
\end{enumerate}
together with the \term{triangle} and \term{pentagon identities} given by the following commuting diagrams:~\cite{BL2004}

\[
   \begin{tikzcd}[row sep=large]
     (x \otimes 1 ) \otimes y \arrow{dr}[swap]{r\otimes 1} \arrow{rr}{\alpha} & & x \otimes (1  \otimes y ) \arrow{dl}{1\otimes l} \\
        & x \otimes y &
   \end{tikzcd}
\]

and

\[
   \begin{tikzcd}[column sep=-2em, row sep=huge, inner sep=0pt]
     & & (w\otimes x)\otimes(y\otimes z) \arrow{drr}{\alpha} & & \\
     ((w \otimes x ) \otimes y ) \otimes z \arrow{urr}{\alpha} \arrow{dr}[swap]{\alpha \otimes 1} & & & &  w \otimes (x \otimes (y \otimes z)) \\
     &  (w \otimes (x \otimes y)) \otimes z \arrow{rr}{\alpha} & &  w \otimes ((x \otimes y) \otimes z) \arrow{ur}[swap]{1 \otimes \alpha}&
   \end{tikzcd}
\]

Hence a strict $2$-group is a groupoid equipped with a product respecting conditions similar to those of an ordinary group multiplication -- the product is associative, there is a unit object, and objects have inverses under the product. For a method to construct a $2$-group from ordinary groups we refer to the next section in which we discuss the equivalent formulation as crossed modules and give concrete examples of strict $2$-groups.

\idx{$2$-group!Lie (smooth)}

Again, introducing further smoothness conditions gives Lie $2$-groups: we then require that both the objects and the morphisms in the $2$-group carry the Lie group structure. There is a categorical counterpart to Lie algebras as well, and these are called Lie $2$-algebras in the case of Lie $2$-groups. In contrast to ordinary Lie groups and algebras, the relationship between the two is not that straightforward; in particular, given a Lie group $G$ there is a whole family of Lie $2$-algebras arising from the categorification of the Lie algebra $\lie{g}$ of $G$, and only one of them has a natural Lie $2$-group counterpart~\cite{BSCS2007}.
We will not discuss Lie $2$-algebras further, since we do not have any immediate use for them in what follows.

\begin{remark}
    \idx{$2$-group!coherent}
    \idx{$2$-group!weak}

  For the strict $2$-group, the inverse operations hold as equalities of functors, and the group laws hold strictly as equations. Since the construction is categorical, the axioms can be defined also up to coherent isomorphisms. Thus the definition is \emph{strict} as opposed to \emph{weak} -- in the latter case, given an object $g$ there is a weak inverse $h$ such that $g\otimes h \isom 1 \isom h\otimes g$. Such categories are called weak $2$-groups. Another interesting category is formed by \emph{coherent $2$-groups}, which are weak $2$-groups in which all objects have a special weak inverse subject to certain coherence laws. These concepts are naturally more general, and may well turn out be the right choice of group categorification for physical applications as well.~\cite{BL2004}
\end{remark}

\begin{remark}
  These definitions generalise to $n$-groups and $n$-groupoids. If a $2$-group is a category object in the category of groups, then a $3$-group would be a $2$-category object in the category of $2$-groups, and so on. Analogously, an $n$-groupoid is an $n$-category with all morphisms equivalent. Note that this \emph{equivalence} can mean different things depending on the level of $n$ -- with $1$-groupoids as above it is simply an isomorphism.
\end{remark}

\idxloppu{$2$-group}

The usefulness of categorical groups may be illustrated by considering a basic concept from theoretical physics: the string group.

\subsection{String group and the need for categorical groups}
  \idx{string group}
  \idx{spin group}
  \idx{spin structure}

Let $M$ be a manifold. The general structure group for the frame bundle of the manifold $M$ is $O(n)$, where $n = \dim M$. We can refine this structure as follows:
\begin{enumerate}
  \item If $M$ is orientable -- the first Stieffel-Whitney class vanishes -- the structure group is $SO(n)$;
  \item If the second Stieffel-Whitney class vanishes, $M$ has a spin structure and the structure group is $\spin(n)$;
  \item If the first fractional Pontryagin class vanishes,
   $M$ has a string structure and the structure group is $\strng(n)$.
\end{enumerate}

The spin group $\spin(n)$ is well-known as the double cover of the rotation group $SO(n)$, defined via the short exact sequence
\[
	0 \to \Z_2 \to \spin(n) \to SO(n) \to 0 .
\]
The string group comes by a similar construction. Consider the short exact sequence
\[
	0 \to K(\Z,2) \to \strng(n) \to \spin(n) \to 0 ,
\]
where $K(\Z,2)$ is an Eilenberg-MacLane space.\footnote{See \cite[Ch.~9]{HJJS2007} for details on the Eilenberg-Maclane spaces.}
The string group builds from the orthogonal group $O(n)$ via a sequence given by the relevant Whitehead tower:
\[
	\dots \to \strng(n) \to \spin(n) \to \spin(n) \to SO(n) \to O(n)  .
\]
If we take the orthogonal group and kill its zero homotopy, we get the special orthogonal group. Likewise, killing the first homotopy of the special orthogonal group we gain the spin group. If we extend this to the third homotopy group (the second being trivial already), the result is the string group.
  \marginnote{Higher up in the tower is the \term{five-brane group} obtained by eliminating the seventh homotopy group; this structure is significant in the M-theory.}

Hence as the spin group is the $1$-connected cover of the special orthogonal group, so is the string group the $3$-connected cover of the spin group.~\cite{SP2011}

As the Whitehead tower illustrates,
the string group as an abstract (homotopy) object cannot be realised as a finite-dimensional Lie group since by definition it should have a closed subgroup with non-trivial second homotopy, which by Cartan's theorem is not possible for finite-dimensional Lie groups. Various approaches to model the string group more concretely seem to orbit around the idea that whatever the realisation, it is a specific representation of a higher object. Hence a $2$-group object makes a good candidate for being a concrete model. An example of this will be seen in the next section.

\idx{loop space}

The string structure is closely connected to \emph{loop spaces}, which are topological spaces formed from maps $S^1 \to M$ (we will explore the related loop groups below). While loop spaces play a central role in the study of string theories, there are definite shortcomings in trying to describe (string) gauge theory with principal bundles over loop spaces -- for instance, coupling of strings with gauge fields cannot be fully realised without introducing extra structure. This, too, points towards higher geometry as a more appropriate context.~\cite{WALDORF2015}

\idxloppu{categorical groups}

\section[Strict $2$-groups and $n$-loop group extensions]{Strict \mots{{2}}-groups and \mots{{n}}-loop group extensions}
\idx{$2$-group}

Having motivated the introduction of categorical groups, let us now concretely build two examples. The aim is to show how $2$-group objects emerge from extensions of loop groups and justify why this is a useful approach. This section is partly based on the article~\cite{MN2019}.

Before jumping into categorical groups proper, we discuss loop groups to some detail -- these groups will be the basis for the presented examples and play an important role in many gauge theory applications.

\subsection[Loop groups and $n$-loop groups]{Loop groups and \mots{\boldsymbol{n}}-loop groups}
\idxalku{loop group}
\idxalku{current group}
\idx{loop group!$n$-loop}
\idx{loop group!based}

Let $G$ be a compact Lie group. The \term{free loop group $\loopg{G}$ of $G$} is the group of smooth maps
\[
	\gamma : S^1 \to G ,
\]
in which the group multiplication is defined pointwise. A \term{based loop group $\bloopg{G}$} is a loop group with a fixed base point: $\gamma(1) = e \in G$ for all $\gamma \in \bloopg{G}$. Geometrically the free loop group $\loopg{G}$ is a principal $\bloopg{G}$-bundle over the space $G$. An important property of the free loop group (and inherited by the based group $\bloopg{G}$) is the existence of nontrivial central extensions by the circle group $S^1 \isom U(1)$:
\[
  \exact{S^1}{\gext{\loopg{G}}}{\loopg{G}} .
\]
As a topological space the based loop group is dual to the suspension $\Sigma G$ of the space $G$, when assuming compact-open topology. The standard result in homotopy theory then states that in the connected components
\[
  \pi_{k+1}(G) = [S^{k+1},G] \isom [\Sigma S^k,G] \isom [S^k,\bloopg{G}] = \pi_k(\bloopg{G}) ,
\]
following from the sphere suspension map $S^{k+1} \isom \Sigma S^k$. Furthermore, if the group $G$ is simply connected, so is $\loopg{G}$.~\cite[Ch.~4]{PS1986}

The loop groups generalise to groups of smooth maps $S^n \to G$ of arbitrary dimensions $n\in \N$. We call the analogous groups formed this way the \term{free $n$-loop group $S^{n}{G}$ of $G$} and the \term{based $n$-loop group $\bnloopg{n}{G}$}. There is furthermore an analogous $n$-fold suspension map such that the following holds as a homotopy classification~\cite[Ch.~5]{MAY1972}:
\[
  [\Sigma^nS^{k},G] \isom [S^k,\bnloopg{n}{G}] .
\]
From this it follows that $\pi_{k+n}(G) \isom \pi_{k}(\bnloopg{n}{G})$ in the connected components.

\begin{remark}
  As noted in Chapter~\ref{chap:gauge}, the mapping groups $\map{M,G}$ are meaningful as gauge transformation groups (under the assumption of a trivial gauge bundle) or current groups,
  and one often considers spherical spaces $M \defeq S^n$ as the base manifold. The $1$-loop group in particular is important in the study of low-dimensional quantum field theories as well as in string theory, and there exists a sizeable body of mathematical theory for the $1$-loop group -- see \cite{PS1986} for the canonical reference. However, not much is known about the mapping groups for general $M$.
\end{remark}

\idxloppu{loop group}
\idxloppu{current group}

\subsection{Crossed modules}

\idxalku{crossed module}

The central point in this section is to construct practical examples of $2$-groups. We do this by considering objects called \emph{crossed modules}, which are algebraic counterparts to strict $2$-groups. The usefulness of this lies in the fact that crossed modules can be built rather concretely from suitable ordinary groups and homomorphisms between them.

We recall the basic definition:
\begin{definition}
  Let $G$ and $H$ be groups, and consider morphisms
  \[
  	\delta : H \to G \quad \textrm{and} \quad \alpha : G \to \aut(H) .
  \]
  We say that  $[\delta:H \to G]$ is a \termd{crossed module} if the following two diagrams commute.

  \[
  \begin{tikzcd}
  H \times H \arrow[rd,"\ad"] \arrow[r,"\delta\times\id"] & G \times H \arrow[d,"\alpha"]\\
  	& H
  \end{tikzcd}
  \qquad
  \begin{tikzcd}
  G \times H \arrow[d,"\id\times\delta"] \arrow[r,"\alpha"] & H \arrow[d,"\delta"] \\
  G \times G \arrow[r,"\ad"] & G
  \end{tikzcd}
  \]

  Equivalently, if we denote by $h^g$ the element-wise action of $G$ on $H$, the diagrams correspond to the equations
  \[
  	h^{\delta(h')} = \inv{h'} h h'
  \]
  and
  \[
  	\delta(h^g) = \inv{g} \delta(h) g
  \]
  for all $h,h' \in H$ and $g \in G$.
\end{definition}
The following theorem then gives a way to associate strict $2$-groups with crossed modules~\cite{BS1976}:
\begin{theorem}
  The categories of crossed modules and of strict $2$-groups are equivalent. This extends to the level of homotopy, that is, also the $2$-categories of crossed modules and of strict $2$-groups are equivalent.
\end{theorem}
\begin{remark}
  In \cite{BS1976} strict $2$-groups are treated as $G$-groupoids, that is, group objects in the category of groupoids. This is equivalent to our earlier definition.
\end{remark}

Note that by the definition any central extension of groups
\[
  \exact{A}{H}{G}
\]
with action $\alpha:G \to \aut(H)$ gives a crossed module through the epimorphism $\delta:H \to G$. We will use this property below in the examples of constructing $2$-groups from loop groups.

Since we consider mostly Lie groups, it is also necessary to verify that crossed modules can retain the smooth structure.
\begin{definition}
    \idx{crossed module!Lie (smooth)}

  If the groups $G$ and $H$ in a crossed module $[\delta:H\to G]$ are Lie groups and the action defined by the morphism $\alpha$ is smooth, the crossed module is called a \termd{Lie crossed module}, or a \termd{smooth crossed module}.
\end{definition}
In what we follows we work with \emph{smooth group} and \emph{continuous algebra} cohomology, see Appendix~\ref{app:cohomology}.

\idxloppu{crossed module}

\begin{example}[Loop group extension and quasi-periodic automorphisms~\cite{MRW2017}]

Let $G$ be simple, compact and simply connected. Let $\bloopg{G}$ be the group of based loops, and $\flatloopg{G}$ a \term{flattened subgroup} defined by imposing the condition of vanishing derivatives at the base point. There are well-known non-trivial central extensions~\cite[Ch.~4]{PS1986}
  \idx{Kac-Moody!group}

  \marginnote{The group $\gext{\bloopg{G}}$ is often called the \term{affine Kac-Moody group}.}

\[
  \exact{S^1}{\gext{\bloopg{G}}}{\bloopg{G}}
\]
which naturally restrict to central extensions of the flattened subgroup $\flatloopg{G}$. This restriction can be done on the Lie algebra level as well, and the corresponding $2$-cocycle $\kappa_{\flat} : \flatloopg{\lie{g}} \times \flatloopg{\lie{g}} \to \R$ is simply the restriction of the Kac-Moody cocycle
  \idx{Kac-Moody!cocycle}

\[
  \kappa(u,v) \defeq \frac{1}{2\pi}\int_0^{2\pi} \ip{u(t), v'(t)} \ext t .
\]
Here $\ip{\cdot,\cdot}$ is an invariant bilinear symmetric form on $\lie{g}$,
  \footnote{A form $c$ on a Lie algebra is \emph{invariant} if it fulfills the condition $c(\comm{u}{v},w) = c(u,\comm{v}{w})$.}

which can be taken as a scalar multiple of the Killing form if $\lie{g}$ is simple. In a matrix algebra we can write $\ip{u,v} = k \tr (u v)$, and hence
\[
  \kappa(u,v) = k \int_{S^1} \tr u \ext v ,
\]
where a suitable normalisation by $k \in \C$ is assumed.

While the flattened loop group $\flatloopg{G}$ is a subgroup of the based loop group and the free loop group, all of them are \emph{subspaces} of a larger space of smooth maps, that of \term{quasi-periodic paths}:
\[
  \qloopg{G} \defeq \{ \gamma \in \map{\R,G} :~ \gamma(t+1) \cdot \inv{\gamma(t)} \text{ is constant} \}  .
\]
Note that $\qloopg{G}$ is not a group with respect to the point-wise multiplication, since the quasi-periodicity is not a property that is preserved under such an operation; moreover, there is no Lie group structure that would be compatible with the standard structure of the free loop group, and it cannot be realised as a Lie group extension of $G$ with the fibre $\loopg{G}$.

However, we can impose the same flattening condition to the quasi-periodic paths and thus gain an honest group:
\[
  \flatqloopg{G} \defeq \{ \gamma \in \qloopg{G} :~ \gamma^{(k)}(0) = \gamma^{(k)}(1) = 0 \quad \forall k \in \N \} .
\]
Since any quasi-periodic path is uniquely determined by considering the interval $t \in [0,1]$, this flattening at the end points is sufficient to ensure the group structure by point-wise multiplication. Then it holds that $\flatloopg{G} = \flatqloopg{G} \cap \bloopg{G}$. Finally, there is a subgroup of based paths:
\[
  \flatqbloopg{G} = \{ \gamma \in \flatqloopg{G} :~ \gamma(0) = e \} .
\]

We can now construct a crossed module based on the groups $\gext{\flatloopg{G}}$ and $\flatqbloopg{G}$. The group of quasi-periodic flattened paths $\flatqbloopg{G}$ acts on $\flatloopg{G}$ pointwise -- this is the right adjoint action $u.\gamma$ for $\gamma \in \flatqbloopg{G}$ and $u \in \flatloopg{G}$. The effect of this action on the Lie algebra cocycle can be written as
\[
  \kappa_{\flat}(u.\gamma,v.\gamma) = \kappa_{\flat}(u,v) + \int_{S^1} \tr \comm{u}{v} \ext \gamma \inv{\gamma} .
\]
Then by the lifting theorem~\cite[Theorem~V.9]{MN2003}
the right conjugation action of $\flatqbloopg{G}$ on $\flatloopg{G}$ lifts to a unique action on the extension $\gext{\flatloopg{G}}$ which is trivial on the centre $S^1$. Since there is a natural inclusion $\flatloopg{G} \hookrightarrow \flatqbloopg{G}$, we also gain a morphism $\delta: \gext{\flatloopg{G}} \to \flatqbloopg{G}$. This together with the action
$\alpha: \flatqbloopg{G} \to \aut(\gext{\flatloopg{G}})$ gives a smooth crossed module.
\end{example}

\begin{remark}
  \idx{loop space}

Note that the inclusion $\flatloopg{G} \hookrightarrow \flatqbloopg{G}$ together with the corresponding conjugation action would already yield a smooth crossed module. The interest in the central extensions comes from the idea that the spin group given by the extension
\[
  \exact{\Z_2}{\spin(n)}{SO(n)}
\]
should have a string-theoretic analogue. A crucial component in this is the Kac-Moody extension for the free loop group: one begins with a principal $\loopg{G}$-bundle over the loop space $\map{S^1,M}$ of the manifold $M$ and wishes to lift this to a principal $\gext{\loopg{G}}$-bundle. The group $\gext{\loopg{G}}$ can be here interpreted as the group $\spin(n)$, and this lifting can be used as the definition of the string structure.~\cite{KILLINGBACK1987}
\end{remark}

\idx{string group}

Much of the discussion in \cite{MRW2017} is devoted to how this approach gives a coherent (as opposed to strict) $2$-group model for the string group. The value of this particular construction is that -- since it is built upon the smooth free loop group -- it naturally admits the action of the circle group $S^1 \isom SO(2)$. This is important for the definition of $S^1$-equivariant operators on the loop space (which itself is a difficult problem, see \cite{WITTEN1988} for a discussion on the Dirac operator). However, there is an obstruction to using the free loop group as the basis for a strict $2$-group model; one way out of this is to use the group of flattened paths as in the example above. This gives a strict $2$-group which has an underlying action groupoid \emph{equivalent} to the one constructed from the free loop group. This action groupoid is given by the weak quotient $\actg{\qloopg{G}}{\gext{\loopg{G}}}$.
From this equivalence it follows that there is an essentially unique $2$-group structure that fulfills the wanted properties of the string group.

\subsection[$3$-loop group extensions]{\mots{\boldsymbol{3}}-loop group extensions}
\idxalku{loop group!$3$-loop}

The $2$-group construction from the previous example raises the question whether similar construction would work on general $n$-loop groups. A generalisation to the $3$-loop group $\threeloop$ can be built using a similar method as for the $1$-loop case. This subsection is mostly a summary of the article~\cite{MN2019}, although in places we have expanded the treatment (most importantly in Proposition~\ref{prop:3loop-homotopy}).

The central extension of the loop group
\[
	\exact{S^1}{\gext{\bloopg{G}}}{\bloopg{G}}
\]
can be initially replaced with an Abelian extension
\[
	\exact{\map{\threeloop, S^1}}{\gext{\threeloop}}{\threeloop} .
\]
On the level of Lie algebra this extension is characterised by the \term{Mickelsson-Faddeev cocycle}:
  \idx{Mickelsson-Faddeev cocycle}

\[
	\theta(A;x,y) = c_2 \int_{S^3} \tr A ~[\ext x , \ext y ] ,
\]
where $c_2 \in \C$ is the normalisation coefficient.
In the gauge-theoretic framework as in \cite{MICKELSSON1987}, the function $A$ is a gauge connection on the gauge bundle of the symmetry group $G$, a $\lie{g}$-valued $1$-form on the domain $S^3$. As a useful identification we can set $A \defeq \inv{f}\ext f$ for a given $f\in \threeloop$, that is, $A$ is the (pullback of the) left-invariant Maurer-Cartan form.
This is based on the homotopy equivalence of smooth paths in $S^2G$ and the moduli space of connections over $S^3$, see \cite{SINGER1981}; we use the same approach later in Section~\ref{sec:general_comm} when discussing the Hamiltonian anomaly.

Now the idea is essentially the same as in \cite{MRW2017}: a strict $2$-group can be obtained by forming a smooth crossed module of relevant Lie groups. However, the Mickelsson-Faddeev extension is not central and this complicates the process. Consequently the definition of the crossed module will be more involved, since it would naturally entail a central extension and will not work out of the box with an Abelian extension.

Let us assume initially that we use a similar flattened group of maps on the $3$-ball $B^3$ and define it such that all the radial derivatives vanish:
\[
  \threeball \defeq \{ \gamma \in \map{B^3,G} :~ \partial_r^{k} \gamma (\bdr B^3) = 0 \quad \forall k\in \N \} .
\]
There is a natural automorphic action $\threeball \to \aut(\threeloop)$, but there are two issues: first, the lift to the extension will not be possible, and second, there is no group morphism $\gext{\threeloop} \to \threeball$ that would fulfill the requirements of the crossed module.

Let us redefine the $3$-loop group as follows:
\[
  \threeloop \defeq \{ \gamma \in \threeball :~ \gamma \textrm{ extends smoothly to } S^3 \textrm{ and } \gamma(\bdr B^3) = e \} .
\]
This redefinition does not change the nature of the Abelian extension above, and the corresponding Lie algebra extension is characterised by the Mickelsson-Faddeev cocycle without modifications. Note that if we now wish to identify $A \defeq \inv{f}\ext f$, we must take $f\in \threeball$.

Following Proposition~\ref{proposition:aut_lift} in Appendix~\ref{app:cohomology}, in order to lift the action of $\threeball$ to the extension $\gext{\threeloop}$ we would want the $2$-cocycle of the extension to be cohomologically invariant under such actions. It is straightforward, if a bit tedious, to check that the \emph{Lie algebra} $2$-cocycle $\theta(A;x,y)$ is cohomologically invariant under the right adjoint action of $\threeball$ on $\threeloop$. Conjugating with an arbitrary element $g\in\threeball$ one gets
\[
  \theta^g(A;x,y) = \theta(A;x,y) - \delta(\lambda)(A;x,y)
\]
where $\lambda$ is the $1$-cochain
\[
  \lambda(A;z) = c_2 \int_{S^3} \tr \left( A \omega \comm{\omega}{z} + \comm{\omega}{A} \ext z + \omega^3 z \right) ,
\]
with $\omega \defeq \ext g \inv{g}$ the right-invariant Maurer-Cartan form.

The same cohomology invariance holds (smoothly) on the group level \emph{under the assumption that the first homotopy group $\pi_1(\threeloop) \isom\pi_4(G)$ is trivial} -- see Theorem~\ref{theorem:cohom_hom} in Appendix~\ref{app:cohomology}. In any case, the invariance holds in the connected components of the $3$-loop group $\threeloop$. We can state this preliminary result as follows:
\begin{lemma}
  The Lie algebra $2$-cocycle $\theta \in \coh^2\left(\threeloopalg,\map{\threeball,i\R}\right)$ given by
  \[
    \theta(A;x,y) = c_2 \int_{S^3} \tr A \comm{\ext x}{\ext y}
  \]
  is cohomologically invariant under the right adjoint action of $\threeball$. If the first homotopy group $\pi_1(\threeloop)$ is trivial, the cohomology invariance lifts to the group $\threeloop$ with respect to the automorphic action $\threeball \to \aut(\threeloop)$.
\end{lemma}

This action does not lift to the extension, however. There is a natural action on both the base and the fibre, but the map $\threeball \to \aut(\gext{\threeloop})$ obtained by combining these is not a group homomorphism. It is therefore not enough that the defining cocycle is smoothly cohomologically invariant; compare to Proposition~\ref{proposition:aut_lift} in Appendix~\ref{app:cohomology}. Hence we need to revise the group extension in order to get a crossed module out of it.

\begin{remark}
  The homotopy group $\pi_4(G)$ vanishes for the groups $G = SU(p)$ when $p > 2$. In the case of $SU(2)$ we have a non-trivial extension of $\threeloop$ by $\Z_2$ (the Mickelsson-Faddeev extension would be trivial), and the right adjoint action of $\threeball$ lifts to this extension without complications.
\end{remark}

\subsubsection{Revised extension and the crossed module}

Let us first tackle the problem of the proposed morphism $\delta: \gext{\threeloop} \to \threeball$. We can extend the acting group $\threeball$ as follows:
\[
  \exact{\threemaps/S^1}{\gext{\threeball}}{\threeball} .
\]
This is a topologically trivial Lie group extension, and on the level of Lie algebra it is essentially characterised by the same Mickelsson-Faddeev cocycle $\theta$, with the following modification. Since now the integration domain for the $2$-cocycle is $B^3$ instead of $S^3$, we need to take the fibre modulo $S^1$. Then the coboundary $\cob \theta$ of the cocycle has a constant boundary term:
\[
  c_2 \int_{\bdr B^3} \tr \left(x\comm{\ext y}{\ext z} - y \comm{\ext z}{\ext x} + z\comm{\ext x}{\ext y} \right) ,
\]
which vanishes by this construction. Thus $\theta(A;x,y)$ fulfills the cocycle condition in this case as well.

Furthermore, there is a central extension
\[
  \exact{S^1}{\gext{\threeloop}}{\gext{\threeloop}/S^1} ,
\]
where the group $\gext{\threeball}$ acts trivially on the centre $S^1$. Indeed, $\gext{\threeloop}$ is a normal subgroup of the group $\gext{\threeball}$: the group $\threeloop$ is a normal subgroup of $\threeball$, and the fibre $\map{\threeball,S^1}$ of the acting extension is mapped onto itself under the conjugation action.

With these extended groups we have a natural morphism $\delta: \gext{\threeloop} \to \gext{\threeball}$ given by the projection
\[
  \gext{\threeloop} \to \gext{\threeloop}/S^1
\]
and the inclusion
\[
  \gext{\threeloop}/S^1 \to \gext{\threeball} .
\]

Next we need the smooth action morphism: does the action of $\gext{\threeball}$ on the Lie algebra $\gext{\threeloopalg}$ lift to a smooth action on $\gext{\threeloop}$? The answer is yes, \emph{if the group $\gext{\threeloop}$ is simply connected}. Given any simply-connected Lie group $H$, there is a group isomorphism~\cite[Cor.~9.5.11, p.~341]{HN2012}
\[
  \aut(H) \isom \aut(\lie{h}) .
\]
Furthermore, the action of $\aut(H)$ on $H$ is smooth if the automorphism group is endowed with a compatible smooth structure and the action on the Lie algebra level is smooth.
These conditions are easily fulfilled in our example when we restrict to the connected component of the identity.

\begin{proposition}\label{prop:3loop-homotopy}
  Let $\pi_5(G) = \Z$. Then the identity component of the group $\gext{\threeloop}$ is simply connected for the basic extension.
\end{proposition}
\begin{proof}
  This is a simple generalisation of the similar result for the $1$-loop group~\cite[Prop.~4.4.6, p.~50]{PS1986}. The fibration
  \[
    \exact{\threemaps}{\gext{\threeloop}}{\threeloop}
  \]
  induces the homotopy exact sequence
  \begin{align*}
    \dots &\to \pi_2(\threemaps) \to \pi_2(\gext{\threeloop}) \to \pi_2(\threeloop) \\
          &\to \pi_1(\threemaps) \to \pi_1(\gext{\threeloop}) \to 0 .
  \end{align*}
  Here $\pi_1(\threeloop) \isom \pi_4(G) = 0$ and, since (the identity component of) $\threeball$ was assumed to be simply connected earlier, $\pi_n(\threemaps) \isom \pi_n(S^1)$. Moreover, under the basic extension we can identify $\pi_2(\threeloop)$ with $\pi_1(S^1) \isom \Z$.

  Thus we have
  \[
    0 \to \pi_2(\gext{\threeloop}) \to \Z \overset{i}{\to} \Z \to \pi_1(\gext{\threeloop}) \to 0 ,
  \]
  where the map $i:\Z \to \Z$ is an isomorphism. Hence for the identity component of the extension $\pi_1(\gext{\threeloop}) = \pi_2(\gext{\threeloop}) = 0$.
\end{proof}
\begin{remark}
  The condition $\pi_5(G) = \Z$ is justified again for groups $G=SU(p)$ with $p>2$.
\end{remark}

Let us summarise the construction. Originally the aim was to build a $3$-loop crossed module on the Abelian extension
\[
	\exact{\map{\threeloop, S^1}}{\gext{\threeloop}}{\threeloop}
\]
by using the action of the group $\threeball$ on $\threeloop$. However, this action does not lift to the extension and there is no natural morphism $\gext{\threeloop} \to \threeball$ as required by the crossed module. We can then proceed by the following: first, extend the acting group by
\[
  \exact{\threemaps/S^1}{\gext{\threeball}}{\threeball} ,
\]
and impose the action of the extended group $\gext{\threeball}$ to another extension given by
\[
  \exact{S^1}{\gext{\threeloop}}{\gext{\threeloop}/S^1} .
\]
Thus we get an action groupoid $\actg{\gext{\threeloop}}{\gext{\threeball}}$ which fulfills the requirements of a smooth crossed module by the morphisms $\alpha: \gext{\threeball} \to \aut(\gext{\threeloop}) $ and $\delta: \gext{\threeloop} \to \gext{\threeball}$. This is then the $3$-loop group analogue to the $1$-loop group crossed module introduced in \cite{MRW2017}.

\begin{remark}
  There is a slightly different route to the action morphism $\alpha$
  which at the outset does not require simply-connectedness for the group $\gext{\threeloop}$. Instead we assume that the cohomology group $\coh^1(\threeloop,\threemaps)$ is trivial, which is again true for $G=SU(p)$ with $p > 2$.
  We will briefly outline this approach; see \cite{MN2019} and \cite{MW2016} for further details.

  First recall from Appendix~\ref{app:cohomology} that $A^N$ denotes the group of $N$-invariant elements of $A$, where $N$ is a normal subgroup of a Lie group $H$, and $A$ is a smooth $H$-module. Let $\coc_s^1(N,A)$ be the group of smooth $1$-cocycles on $N$ with values in $A$: under the condition of a trivial cohomology group $\coh^1(N,A)$, there is an isomorphism $\coc_s^1(N,A) \isom A/A^N$.

  We can then apply this to $A = \threemaps$, $N=\threeloop$ and $H=\threeball$, and construct the following commuting diagram:~\cite[Sec.~4]{MW2016}

  \[
  \begin{tikzcd}
  	0 \arrow[r] & \threemaps \arrow[d] \arrow[r] & \gext{\threeloop} \arrow[d] \arrow[r] & \threeloop \arrow[d] \arrow[r] & 0 \\
  	0 \arrow[r] & \threemaps/\map{S^2G,S^1} \arrow[r] & \Gamma \arrow[r] & \threeball \arrow[r] & 0
  \end{tikzcd}
  \]

  in which the group $S^2G$ consists of all smooth maps from the $2$-sphere to the group $G$, as introduced earlier at the beginning of this section. We can then identify $\threemaps^{\threeloop}$ with $\map{S^2G,S^1}$, and the group $\Gamma$ is the Lie group extension
  \[
    \exact{\threemaps/\map{S^2G,S^1}}{\Gamma}{\threeball} .
  \]
  Furthermore, there is a smooth action
  \[
    \Gamma \to \aut(\gext{\threeloop}) .
  \]
  By the smooth homomorphism
  \[
    \threemaps/S^1 \to \threemaps/\map{S^2G,S^1}
  \]
  and the trivial action of the group $\map{S^2G,S^1}$ on $\gext{\threeloop}$, we can lift the action $\Gamma \to \aut(\gext{\threeloop})$ to a smooth homomorphism
  \[
    \alpha: \gext{\threeball} \to \aut(\gext{\threeloop}) .
  \]
\end{remark}

\begin{remark}
  One of the virtues of the $1$-loop group crossed module was that it retained the possibility to a physically meaningful $S^1$-action of the group $SO(2)$. In the $3$-loop case one would naturally want the corresponding rotations of the $3$-sphere, but this is not possible in the above construction. This is due to the fact that the definition of the loops in $B^3$ requires a contracted boundary, and hence a choice of a distinguished point which cannot be made equivariant under the $S^3$-action of the group $SO(4)$.
\end{remark}

\idxloppu{loop group!$3$-loop}

\section{Bundle gerbes}\label{sec:gerbes}
\idxalku{bundle gerbe}

As principal bundles are central to the mathematical tapestry of gauge theory, it is natural to wish for higher versions thereof. Here we focus on one particular case, that of (bundle) gerbes, which can be thought of as a categorification of (Hermitian) line bundles with connection. A bundle gerbe also provides a natural geometric realisation for the third cohomology group which has a decisive role in understanding Hamiltonian anomalies -- we study this aspect more closely in Chapters~\ref{chap:ktheory}~and~\ref{chap:hanomaly}, and concentrate here on the definition and basic properties. We follow mostly the introductory works by Murray~\cite{MURRAY1996,MURRAY2010}.

Let us begin fresh with the definition. In the following we assume that the spaces we work with are smooth manifolds.
\begin{definition}\label{def:bundle_gerbe}
  Consider maps $\pi: Y \to X$ and $p: P \to Y^{[2]}$, where $Y^{[2]}$ is the fibre product of $Y$ with itself. The pair $(P,Y)$ is called a \term{bundle gerbe over $X$} if the following holds:
  \begin{enumerate}
  	\item $\pi : Y \to X$ is a surjective submersion.
  	\item $p: P \to Y^{[2]}$ is a $U(1)$-bundle.
  	\item Define smooth maps $\pi_i : Y^{[k]} \to Y^{[k-1]}$ by omitting the $i$th element from the fibre product. The multiplication
  		\[
  			\mu: \pi_3^*(P) \otimes \pi_1^*(P) \to \pi_2^*(P)
  		\]
  		defines a smooth isomorphism of $U(1)$-bundles over $Y^{[3]}$ that is associative. In other words, if we denote by $P_{(y_i,y_j)}$ the fibre of $P$ over $(y_i,y_j)$, the following diagram commutes for all $(y_1,y_2,y_3,y_4) \in Y^{[4]}$.
  		\[
  			\begin{tikzcd}
  				P_{(y_1,y_2)} \otimes P_{(y_2,y_3)} \otimes P_{(y_3,y_4)} \arrow{d} \arrow{r} &
  					P_{(y_1,y_3)} \otimes P_{(y_3,y_4)} \arrow{d} \\
  			    P_{(y_1,y_2)} \otimes P_{(y_2,y_4)} \arrow{r} &
  			    		P_{(y_1,y_4)}
  			\end{tikzcd}
  		\]
  \end{enumerate}
\end{definition}

\begin{remark}
  Originally in \cite{MURRAY1996} the projection $\pi: Y \to X$ was assumed to be a fibration. However, it is enough if the map $\pi$ has only local sections, so that for all points $x\in X$ we have an open neighbourhood $\U$ and a local section $\Gamma: \U \to Y$. Such maps are called \term{locally split}, and in the context of smooth spaces they are surjective submersions~\cite{MuS2000}.
  For the anomalous bundle gerbe introduced in Chapter~\ref{chap:hanomaly} the map $\pi$ is indeed not a fibration.
\end{remark}

A bundle gerbe is closely related to the more general notion of \term{gerbe}, which can be formally defined either as a stack of groupoids~\cite[Ch.~III Déf.~2.1.1, p.~129]{GIRAUD1971} (see for instance \cite{MOERDIJK2002} for an introduction in English) or as a Lie group -banded sheaf of groupoids~\cite[Def.~5.2.4, p.~196]{BRYLINSKI1993}, both armed with certain local properties.\footnote{Brylinski traces his definition back to Giraud's,
though it takes a few moments of pondering to see how the two are linked.} While these terms are often mixed rather liberally in the literature, the definition of a gerbe is quite a bit more general -- in this light, bundle gerbes as defined above are smooth versions of the Abelian gerbe: the band here is the group of smooth maps $\map{X,U(1)}$~\cite{MURRAY2010}.
In any case, bundle gerbes can be thought of as fibrations with groupoids as fibres. As such, they can represent obstructions for the existence of global bundles with particular properties, the details depending on the setting. In Chapter~\ref{chap:hanomaly}, we deal with such a bundle gerbe obstruction arising in quantum theory of massless fermion fields.

In order to open up the definition we need to consider the fibre products more carefully. A $k$-fold fibre product $Y^{[k]}$ is a submanifold of the Cartesian product manifold $Y^k$:
\[
  Y^{[k]} \defeq \{(y_1,\dots,y_k) :~ \pi(y_1) = \dots = \pi(y_k) \} \subset Y^k .
\]
The manifold structure is retained since $\pi$ is a submersion. One can then naturally consider differential $p$-forms $\Omega^p(Y^{[k]})$ on the fibre products; moreover, there is a boundary map
\[
  \cob: \Omega^p(Y^{[k-1]}) \to \Omega^p(Y^{[k]}) :~ \cob\omega \defeq \sum_{i=1}^{k} (-1)^{k-1} \pi_i^{*} (\omega) .
\]
From these maps one forms the \term{fundamental complex}
\[
\begin{tikzcd}
  0 \arrow[r] & \Omega^p(X) \arrow[r,"\pi^*"] & \Omega^p(Y) \arrow[r,"\cob"] & \Omega^p(Y^{[2]}) \arrow[r,"\cob"] & \dots
\end{tikzcd}
\]
which is exact for all forms of degree $p \geq 0$. Furthermore, if we have a map $g: Y^{[k-1]} \to A$ to an Abelian group $A$, we define $\cob g : Y^{[k]} \to A$ as an alternating sum
\[
  \cob g = \sum_{i=1}^k (-1)^{i-1}(g \circ \pi_i) .
\]
Finally, given an $A$-bundle $P \to Y^{[k-1]}$ there is another $A$-bundle $\cob P \to Y^{[k]}$ fulfilling
\[
  \cob(P) = \pi_1^*(P) \otimes (\pi_2^*(P))^* \otimes \pi_3^*(P) \otimes \cdots
\]
From these definitions it follows that $\cob^2 g = 1$ and that $\cob^2 P $ is trivial. Note also that the boundary map $\cob$ commutes with the de Rham derivative of forms.

The fibre product $Y^{[2]}$ in Definition~\ref{def:bundle_gerbe} has a natural groupoid structure, and the multiplication property of the bundle gerbe is constructed to be compatible with the in-built groupoid multiplication on the base $Y^{[2]}$.
By multiplication we can take a section $\sigma$ of $\cob P \to Y^{[3]}$, and furthermore a section $\cob\sigma$ of $\cob^2 P \to Y^{[4]}$; since the latter bundle is trivial by the definition above, and following the associativity of the multiplication, we settle for the natural condition $\cob \sigma = 1$.

\begin{definition}
    \idx{bundle gerbe!trivial}

  A bundle gerbe $(P,Y)$ is \termd{trivial} if there is a $U(1)$-bundle $T \to Y$ such that $(P,Y) \isom (\delta(T),Y)$. If such an isomorphism can be found, we call it a \termd{trivialisation} of $(P,Y)$.
\end{definition}

Recall that for an Abelian $G$ any principal $G$-bundle $P$ has a dual bundle denoted by $P^*$, and that the tensor product of two principal $G$-bundles is a principal $G$-bundle. These concepts extend naturally to bundle gerbes.
\begin{definition}
  Given a bundle gerbe $(P,Y)$ there is a \termd{dual bundle gerbe} $(P,Y)^* \defeq (P^*,Y)$.
\end{definition}
\begin{definition}
  Two bundle gerbes $(P,Y)$ and $(Q,Z)$ form a \termd{product bundle gerbe} $(P,Y) \otimes (Q,Z) \defeq (P\otimes Q, Y \times_X Z)$ where
  \begin{enumerate}
    \item $\pi: Y \times_X Z \to X$ is a surjective submersion.
    \item $P \otimes Q  \to (Y \times_X Z)^{[2]}$ is a $U(1)$-bundle.
  \end{enumerate}
  Here, $(Y \times_X Z)^{[2]} = Y^{[2]} \times_X Z^{[2]}$ and fibrewise
  \[
    (P\otimes Q)_{((y_1,z_1),(y_2,z_2))} = P_{(y_1,y_2)} \otimes Q_{(z_1,z_2)} .
  \]
\end{definition}

From these definitions we conveniently arrive at a suitable concept of \emph{equivalence} for bundle gerbes. Specifically and in contrast to principal bundles, not all trivial bundle gerbes are isomorphic; hence the right equivalence is the stable isomorphism, reminiscent of the stable isomorphism of vector bundles in topological K-theory (see Section~\ref{sec:ktheory}).
\begin{definition}
    \idx{bundle gerbe!equivalence of}

  Two bundle gerbes $(P,Y)$ and $(Q,Z)$ are \termd{stably equivalent} or simply \termd{equivalent} if the product $(P^*,Y) \otimes (Q,Z)$ is isomorphic to a trivial bundle gerbe.
\end{definition}

Stable equivalence is an equivalence relation of bundle gerbes~ \cite{MuS2000}. Given such an equivalence class one can define a local bundle gerbe in terms of a \emph{family} of $U(1)$-bundles over intersections $\U_i \cap \U_j$, where $\U_i, \U_j$ belong to a good cover of $X$. As such bundle gerbes can work as a categorification of transition functions of fibre bundles as follows. Recall that fibre bundles $Y\to X$ can be characterised up to an isomorphism by local transition functions; these are (complex) functions $Y^{[2]} \to \C$ which over local triple intersections satisfy the cocycle condition. If instead of complex numbers we characterise these cocycles by {$U(1)$-torsors} and replace the equations with coherent isomorphisms, we are back at the definition of the bundle gerbe.

Lastly, we note the following convenient way to illustrate a bundle gerbe $(P,Y)$ as a single diagram:
\[
  \gerbe{P}{Y}{X}
\]

\begin{remark}
  Bundle gerbes of Definition~\ref{def:bundle_gerbe} are sometimes called (principal) $U(1)$-bundle gerbes, or Abelian bundle gerbes. One can work out the theory of more general (principal) $G$-bundle gerbes as well, see \cite{ACJ2005} for the definition and applications in higher Yang-Mills theory.
\end{remark}

\subsection{Differential geometry}

A bundle gerbe is an intrinsically differential geometric object. It is therefore natural to define a connection and its curvature analogous to those of a fibre bundle -- furthermore, there are additional entities related to the bundle gerbes: the $3$-curvature and the curving.

Recall that any principal $U(1)$-bundle $P$ over $Y$ has a connection. This extends naturally to their fibre products, and thus we can formulate a connection on a bundle gerbe.
\begin{definition}
    \idx{connection!bundle gerbe}

  Let $\conn$ be a connection on $P \to Y^{[2]}$. It extends to \termd{bundle gerbe connection} if it is compatible with the bundle gerbe multiplication. In other words, if for the section $\sigma$ of $\cob P \to Y^{[3]}$ it holds that
  \[
    \sigma^*(\cob\conn) = 0 .
  \]
  The \termd{curvature form}
    \idx{curvature!$2$-form}

  $F_{A} \in \Omega^2(Y^{[2]})$ of a bundle gerbe connection is defined as $F_{A} \defeq \ext A$ for a $1$-form $A$ of the connection $\conn$. The curvature satisfies $\cob F_A = 0$.
\end{definition}
Bundle gerbe connections always exist: this follows from exactness of the fundamental complex and the fact that we can always define a connection on a principal $U(1)$-bundle.

Since $\cob F_A = 0$ and the fundamental complex is exact, it follows that there must be a $2$-form $f$ on $Y$ such that $\cob f = F_A$. Taking the de Rham derivative we get
\[
  \cob \ext f = \ext \cob f = \ext F_A = 0 ,
\]
and hence for some $3$-form $\omega$ on $X$ we can write $\ext f = \pull{\omega}$. Moreover, $\omega$ is closed,
as $\pull{\ext \omega} = \ext \pull{\omega} = \ext^2 f = 0$.
\begin{definition}
  The $2$-form $f \in \Omega^2(Y)$ defined by $\cob f = F_A$ is called the \termd{curving} of the bundle gerbe connection $\conn$, and the $3$-form $\omega \in \Omega^3(X)$ defined by $\pull{\omega} = \ext f$ is called the \termd{$3$-curvature} of the bundle gerbe connection $\conn$.
\end{definition}

The $3$-curvature can be used to represent the third cohomology of the space $X$, and hence bundle gerbes can be used as a geometric realisation of these classes.

\subsection{Geometric realisation of cohomology classes}

Giraud's motivation in \cite{GIRAUD1971} for introducing gerbes was to describe non-Abelian cohomology. Specifically, a gerbe with a Lie group band $G$ over a manifold $X$ characterises cohomology classes in $\coh^2(X,\sheaf{G})$ (in sheaf cohomology, so cochains take values in the sheaf of maps to $G$ -- see Appendix~\ref{app:cohomology}). If the band is an Abelian group one gets an isomorphism between the equivalence classes of gerbes and the second (Abelian) cohomology~\cite[Thm.~5.2.8, p.~201]{BRYLINSKI1993}. By a \emph{transgression} this gives a handle on the third integral cohomology group $\coh^3(X,\Z)$,
hence bundle gerbes offer a geometric realisation of the third cohomology.
  \idx{transgression}

An apt analogue can be given in regards to principal $U(1)$-bundles. A principal $U(1)$-bundle or a complex line bundle over a manifold $X$ can be characterised in terms of the second integral cohomology: it can be locally represented as a cocycle $c_{ij}: \U_i \cap \U_j \to U(1)$, where $\U_i, \U_j$ are elements of an open cover of $X$. The associated characteristic class of the bundle is the first Chern class in $\coh^2(X,\Z)$, and the map $c$ fulfills the familiar first order cocycle condition~\cite[Sec.~23.1]{HJJS2007}
\[
  c_{ij}c_{jk}c_{ki} = 1 \quad \textrm{ on } \U_i \cap \U_j \cap \U_k .
\]

Similarly, to a bundle gerbe we can locally attach a cocycle $c_{ijk}: \U_i \cap \U_j \cap \U_k \to U(1)$ (fulfilling the second-order cocycle condition). We then associate a class in $\coh^3(X,\Z)$ to the bundle gerbe; this is called the Dixmier-Douady class.

\begin{definition}
    \idx{Dixmier-Douady class}

  \termd{Dixmier-Douady class} as a characteristic class is a homotopy class of maps
  \[
    X \to K(\Z,3) ,
  \]
  where $X$ is a compact space. It is an element in $\coh^3(X,\Z)$ by the isomorphism
  \[
    \coh^3(X,\Z) \isom [X,K(\Z,3)] .
  \]
\end{definition}
This is a higher degree version of the first Chern class, which is a homotopy class of maps  $X \to K(\Z,2)$.~\cite[Sec.~10.1]{HJJS2007}

The Dixmier-Douady class can then be represented as a closed $3$-form with integral periods. Since any closed $3$-form appears as the $3$-curvature of a bundle gerbe, we can form an equivalence between the Dixmier-Douady classes and bundle gerbes.
\begin{proposition}
    \idx{bundle gerbe!equivalence of}

  Two bundle gerbes are equivalent (stably isomorphic) if and only if their Dixmier-Douady classes match. In particular, a bundle gerbe is trivial if and only if its Dixmier-Douady class vanishes. Hence there is an isomorphism
  \[
    \text{Stable isomorphism classes of bundle gerbes on $X$} \isom \coh^3(X,\Z) .
  \]
\end{proposition}
\begin{proof}
  Central component is the transgression
  \[
    \coh^2(X,\sheaf{U(1)}) \isom \coh^3(X,\Z) .
  \]
  For details, see \cite{MURRAY1996}.
\end{proof}

\begin{example}[Lifting bundle gerbe~{\cite[Sec.~6.1]{MURRAY2010}}]
    \idx{bundle gerbe!lifting}

  The Hamiltonian anomaly manifests as a projective bundle of Fock spaces, and this is tied to the problem of lifting the group of gauge transformations over the moduli space to its extension by an Abelian group. Anticipating this, we can as an example consider the lifting bundle gerbe related to such prolongation problems of group actions. The gist of this construction is as follows. Consider a central extension of a Lie group $G$
  \[
    \exact{U(1)}{\gext{G}}{G} ,
  \]
  and a principal $G$-bundle $Y \to X$. Does this bundle lift to a principal $\gext{G}$-bundle $\gext{Y} \to X$?
  The answer lies in topology. Transition functions of the bundle $Y \to X$
  induce a $U(1)$-valued cocycle as above, and the lift is possible if and only if the corresponding cohomology class in $\coh^2(X,\sheaf{U(1)}) \isom \coh^3(X,\Z))$ is trivial.

  Associated to the bundle $Y \to X$ we can define a map $Y^{[2]} \to G$, and the $U(1)$-bundle $\gext{G} \to G$ can be pulled back to a $U(1)$-bundle $Q \to Y^{[2]}$. One can show that there is a sound definition for the fibre multiplication, and thus a well-defined bundle gerbe $(Q,Y)$ called the \term{lifting bundle gerbe}. Its Dixmier-Douady class is by definition the obstruction to the desired lift, and thus the lifting bundle gerbe is trivial if and only if the $G$-bundle $Y\to X$ lifts to a $\gext{G}$-bundle. This result can be extended to hold for more general Lie group extensions besides the central extension -- in particular, it applies to the Abelian extension of the Hamiltonian anomaly~\cite{HMSV2013}.
\end{example}

\begin{remark}
    \idx{Kac-Moody!group}

  There is a natural connection to the affine Kac-Moody groups in terms of the canonical generator $\coh^3(G,\Z) \isom \Z$ which underlies the theory of central extensions of $\loopg{G}$ for a compact Lie group $G$~\cite[Ch.~4]{PS1986}. This can be characterised by bundle gerbes over $G$, the cohomology class of which by the transgression map gives a central extension of the loop group $\loopg{G}$. Thus a $U(1)$-bundle over the loop group can be realised as an element in $\coh^3(G,\Z)$ with the generator
  \[
    \omega = k \tr \left( \inv{g}\ext g \wedge \comm{\inv{g}\ext g}{\inv{g} \ext g} \right),
  \]
  where $k$ is the normalisation parameter and $\inv{g}\ext g$ is the Maurer-Cartan form on an adjoint representation of $G$.
  This gives the representation of the Dixmier-Douady class of the related bundle gerbe. See \cite[Ch.~6]{BRYLINSKI1993} for details and more examples regarding applications to loop spaces.
\end{remark}

\begin{remark}
  In the spirit of continuing categorification one can also introduce higher versions of the (bundle) gerbes and thus construct realisations of higher cohomology classes. Note in particular the bundle $2$-gerbe with useful applications as outlined in \cite{STEVENSON2004}.
    \footnote{See also \cite{WALDORF2007} and \cite{BUNK2017} for the $2$-category structures defined on bundle gerbes.}

  Similarly to the bundle gerbe, the $2$-gerbes provide a geometric handle to the cohomology classes of degree $4$. Whereas the cohomology classification of bundle gerbes has a direct link to the anomalous gauge algebra $2$-cocycles in quantum physics, bundle $2$-gerbes may illuminate the failing of the Jacobi identity described by gauge algebra $3$-cocycles~\cite{CMW1997,JACKIW1985}.
\end{remark}

\idxloppu{bundle gerbe}

config: arxiv


TITLE: Translating Mathematical (Universal/Existential Quantifier) Statements into English?
QUESTION [1 upvotes]: Can anyone confirm if I have these correct? Or if not, where I am going wrong?

Translate these statements into English, where $K(x)$ is '$x$ is a Kangaroo' and $H(x)$ is '$x$ hops' and the domain consists of all animals.

(a) - $\forall x(K(x)\rightarrow H(x))$ - "All animals that hop are Kangaroos."
(b) - $\forall x(K(x) \land H(x))$ - "All animals are Kangaroos and hop."
(c) - $\exists x(K(x) \rightarrow H(x))$ - "There exists a Kangaroo that hops."
(d) - $\exists x(K(x) \land H(x))$ - "There exists an animal that is a Kangaroo and Hops."

REPLY [1 votes]: If you can come up with good "test cases",
you can use them to help you decide if your translations are correct.
Consider statement (a):
$$\forall x(K(x)\rightarrow H(x)).$$
This statement is satisfied in a universe in which the only animals are
some kangaroos (every one of which hops),
some frogs (every one of which also hops), and some snakes (which never hop).
Let $x$ be any animal in this universe.
Then either $x$ is a snake, so $K(x)\rightarrow H(x)$ is true because
the antecedent is false;
or $x$ is a frog, so $K(x)\rightarrow H(x)$ is true because
the conclusion is true (also because the antecedent is false);
or $x$ is a kangaroo, so $K(x)\rightarrow H(x)$ is true because
the conclusion is true.
But it is not true in this universe that all animals that hop are kangaroos.
Indeed, there are some frogs in this universe which are animals that hop
but are not kangaroos.
An alternative way to translate $\forall x.P(x)$, where the domain of $x$
is "animals", is to start with the words
"Every animal" and continue with an English phrase meaning
"satisfies $P(x)$ where $x$ is that animal."
The expression $K(x)\rightarrow H(x)$ signifies
that if $x$ is a kangaroo, $x$ hops;
so we might translate (a) as,
"Every animal is such that if it is a kangaroo, it hops."
The preceding quoted sentence is very awkward English, however.
An alternative wording is,
"Every animal, if it is a kangaroo, hops."
But most people, I think, would have more trouble parsing this
sentence than it is really worth.
Another alternative is to recognize that a common English translation of
$K(x)\rightarrow H(x)$ is, "$x$ hops if $x$ is a kangaroo."
Therefore the fully quantified statement in (a) could be translated,
"Every animal hops if it is a kangaroo."
We can check that the universe of kangaroos, frogs, and snakes satisfies
all of the actual sentences above that start with "Every animal."
On the other hand, if we introduce a kangaroo into our universe
that cannot hop because it has no legs, the mathematical statement (a)
is falsified and so are all of the "Every animal" translations of it.
This of course does not prove that the translation is accurate,
but it is a good test case to weed out inaccurate translations.
I would use similar techniques of modeling and translation for the
other sentences, keeping in mind that sometimes a statement that starts
with $\exists x$ can be translated into English starting with the
words, "Some animal."
Also keep in mind that good models for existential quantifiers
are often minimalistic, for example, a universe with only a single frog and no other animals satisfies (c); so does a universe populated only by a single snake.
But a universe whose entire animal population consists of legless kangaroos
does not satisfy (c).


TITLE: Know if a basis is orthonormal just with a few dot products
QUESTION [0 upvotes]: I have the following dot products results

$\langle(1,0),(1,0)\rangle =4$
$\langle(1,1),(1,1)\rangle = 1$
$\langle(1,0),(3,3)\rangle =0$

I can't find a way to know if the following bases are orthonormal

i) $\{(1,0),(1,1)\}$
ii) $\{(1/2 ,0),(0,1)\}$
iii) $\{(1/\sqrt{5}\cdot(0,1) ,(-1,1)\}$
iv)
$\{1/\sqrt{5}\cdot(0,1), 1/\sqrt{20}\cdot(5,4)\}$

I just know that i) it's not, because $\langle(1,0),(1,0)\rangle=4$
How can I approach the other parts?

REPLY [0 votes]: If $(1,0)$ is orthogonal to $(3,3)$, then it is orthogonal to $(1,1)$ as well.  That's the only orthogonality you have, so you have to use it.
Now since $(1,1) - (1,0) = (0,1)$, you can find out whether $(0,1)$ is normalized or not, which will give you a partial answer to ii).
I expect they will all be solved with similar linear combinations.


TITLE: Why is the unit of hydraulic impedance/resistance kilogram per second and hypermeter?
QUESTION [3 upvotes]: I'm trying to figure out the unit of impedance in the hydraulic analogy of electronic networks. Assuming
$$Z=\frac{P}{Q},$$
with $P$ as pressure difference, $Q$ as volumetric flow rate and $Z$ as impedance, the unit would be
$$\rm \frac{Pa}{m^3/s} = \frac{N/m^2}{m^3/s} = \frac{kg/m\cdot s^2}{m^3/s} = \frac{kg}{m^4\cdot s}$$
Intuitively, this unit doesn't make sense. Did I get some part of the unit conversion wrong? If not, could someone help me understand the unit?

REPLY [0 votes]: I find it easier to understand when using mass-flow k$\rm g/s$ , after all electrical current is the flow of mass (electrons):
Then the formula would be
$$\Delta P={\rm\frac{kg}{s}}\cdot R({\rm{mass}})={\rm\frac{kg}{m\cdot s^2}}\iff R({\rm mass})=\frac{1}{{\rm m\cdot s}}$$
This means that this first formula I extracted is independent of the density.
By adding in the density we get the same as the above formula.
$$Z({\rm volume})={\rm Density}\cdot R({\rm mass})={\rm \frac{ kg}{m^3}\frac{1}{m\cdot s}=\frac{kg}{m^4\cdot s}}$$
Propane $1,7~ \rm kg/m^3,$ Water $1000~ \rm kg/m^3$
Let's examine a mass-flow of $2~\rm kg/s$ for each fluid:
For propane:
$${\rm Flow(volume)=\frac{2~\frac{kg}{s}}{1,7~\frac{kg}{m^3}}=1,176~\frac{m^3}{s}}$$
For water:
$${\rm Flow(volume)=\frac{2~\frac{kg}{s}}{1000~\frac{kg}{m^3}}=0,002~\frac{m^3}{s}}$$
So we would expect the speed of the gas to be a lot higher than of the water when using the same pipe.
So now let's assume the same pressure drop for both water and propane: 50 Pa
Which resistance could cause such a pressure drop?
$$R({\rm mass})=\frac{\Delta P}{{\rm \frac{kg}{s}}}=\frac{{\rm 50~ Pa}}{{\rm 2~\frac{kg}{s}}}=25~{\rm \frac{1}{m\cdot s}}$$
Now let's add in the densities:
Propane:
$$Z({\rm volume})=1,7~{\rm \frac{kg}{m^3}}\cdot 25~{\rm \frac{1}{m\cdot s}}=42,5~{\rm \frac{kg}{m^4\cdot s}}$$
And water:
$$Z({\rm volume})=1000~{\rm \frac{kg}{m^3}}\cdot 25~{\rm {\frac{1}{m\cdot s}}}=25000~{\rm \frac{kg}{m^4\cdot s}}$$
Now let's calculate the flow:
$${\rm Propaneflow~(volume)}=\frac{\Delta P}{Z{\rm (volume)}}={\rm \frac{50~ Pa}{42,5\frac{kg}{m^4\cdot s}}=1,176~\frac{m^3}{s}}$$
$${\rm Waterflow~(volume)}=\frac{\Delta P}{Z{\rm (volume)}}={\rm \frac{ 50 Pa}{25000~\frac{kg}{m^4\cdot s}}=0,002~\frac{m^3}{s}}$$
As we see we arrive at the right result.
From the calculations above we saw that the unit of the density-independent resistance was: ${\rm \frac{1}{m\cdot s}}$ . When it does not depend on the fluid density it can only depend on other matters. From the units we see the following:
$$\begin{align}R{\rm(mass) \cdot Area} &= {\rm Speed}\\ {\rm \frac{1}{m\cdot s}\cdot m^2} &={\rm \frac{m}{s}}\\\iff ~~~~~ {\rm \frac{1}{m\cdot s}} &={\rm \frac{\frac{m}{s}}{m^2}}\\
\iff  R{\rm (mass)} &={\rm\frac{Speed}{Area}};\end{align}$$
substituting this new formula into the old one we get:
$$Z{\rm (volume)=Density\cdot \frac{Speed}{Area}}$$
And in the unit analysis:
$${\rm \frac{kg}{m^4\cdot s}=\frac{\left(\frac{kg}{m^3}\cdot \frac{m}{s}\right)}{m^2} =\frac{kg}{m^4\cdot s}}$$
As we see this rings quite true. So, using this logic, with constant density the resistance only depends on the speed and the size of the pipe. If the size of the pipe is also a constant, we see that the pressuredrop increases exponentially to the speed. We see this from the following analysis:
$$Z{\rm (volume)=Density\cdot \frac{Speed}{Area}}$$
We multiply this with the flow in [m³/s] and get:
$$DeltaP = Density * Speed^2$$
By dividing this formule with the length in metres we arrive at an interesting result:
$$\frac{DeltaP}{Length} = \frac{Density*Speed^2}{Length}$$
And by multiplying the right side with the area in squaremetres we get:
$$\frac{DeltaP}{Length} = \frac{Density*Area*Speed^2}{Volume}$$
For a circular pipe we get the following:
$$\frac{DeltaP}{Length} = \frac{Density*(PI/4)*Diameter^2*Speed^2}{Length*(PI/4)*Diameter^2}$$
We can remove (PI/4) again. So by separating the right side a bit we get:
$$\frac{DeltaP}{Length} = \frac{Diameter^2}{Diameter^2}* \frac{Density*Speed^2}{Length} = \frac{Diameter}{Length}* \frac{Density*Speed^2}{Diameter}$$
This is actually a simplified form of the Darcy-Weisbach formula, but note that I derived it solely from using unit analysis. The only difference is that they add a friction factor. The Darcy-Weisbach formula reads:
$$\frac{DeltaP}{Length}=\frac{Frictionfactord}{2} * \frac{Density*Speed^2}{Diameter}$$
<=>
$${\rm \frac{Pa}{m}=Fd * \frac{\left(\frac{kg}{m^3}\ * (\frac{m}{s})^2\right)}{2*m} =Fd * \frac{kg}{2*m^2 * s^2\cdot}}=\frac{Fd}{2}*\frac{Pa}{m}$$
The Darcy friction factor is equal to ${Fd = \rm \frac{64}{Reynolds-number}}$ and Reynolds number is equal to:
$$Re = \frac{Density*velocity*Length}{dynamic viscosity}$$
If we have constant temperature, both the density and the viscosity are constant. When looking at a specific pipe the length too is constant. This gives us:
$$Re = Factor*velocity (no-unit)$$
Now we can calculate the Darcy friction factor:
$$Fd = \frac{64}{Factor*velocity}$$
Now we have two expressions for a constant without a unit. One is the one we calculated from the "Electric current analogy" and the other one is our Darcy friction factor at constant temperature. Now let's equate them, remembering what our factor consisted of:
$$\frac{Diameter}{Length} = \frac{\frac{64}{Factor*velocity}}{2} = \frac{64}{2*\frac{Density * Length*velocity}{dynamic-viscosity}}= \frac{32*dynamic-viscosity}{Density * Length*velocity}$$
So now we have an equation to solve when we can equate the flow of a fluid in a circular pipe to the rules of resistance in electricity:
$$\frac{Diameter*Density*velocity}{dynamic-viscosity} = 32$$
I would suggest solving it for water in excel for an array of table values and then calculating the percentage we're off from 100% right for each example.
I know your question was about units, but I think I gave some answer to that too.


TITLE: Remainder Theorem co
QUESTION [0 upvotes]: In this screenshot: https://gyazo.com/5052812379fedee11586ea1db3d55b78
It states "As the divisor is a linear expression and F(X) is a cubic polynomial then Q(X) must be a quadratic and the remainder must be a constant.
I don't understand how they determined it would be a constant by that information.

REPLY [0 votes]: For  Euclidean division in polynomial rings, the condition for the remainder $R(x)$ is ‘$R(x)=0$ or $\deg R(x)<\deg(\text{divisor})$’.
Here, for a linear divisor, it says the remainder is $0$ or has degree $0$, i.e. is a constant.


TITLE: A relatively bounded perturbation of a closed operator is a closed operator.
QUESTION [1 upvotes]: Please I need help with an example I cant figure out and which will hopefully help me to get the theory:
Let $X$ be Banach space and $A, B$ general operators. Furthermore $A$ is closed, $\operatorname{dom}(A) \subset \operatorname{dom}(B)$, there are $a,b; \quad a\ge 0, \; 0 \lt b \lt 1$
$\forall x \in \operatorname{dom} (A); \quad \lVert Bx \rVert \le a\lVert x\rVert +b\lVert Ax\rVert $
I need to:

find $\operatorname{dom}(A+B)$
show that $\forall x \in \operatorname{dom}(A), \quad \lVert Ax\rVert \le \frac{1}{1-b}\lVert (A+B)x\rVert +\frac{a}{1-b}\lVert x\rVert$
show that $A+B$ is closed

Thanks a lot, i would like to finally get it, but i probably cant do it alone.

REPLY [3 votes]: $\DeclareMathOperator{dom}{dom}$The domain of a sum $A + B$ of unbounded operators is by definition the intersection of the domains of $A$ and $B$: $\dom(A+B) = \dom A \cap \dom B$ and by hypothesis $\dom A \subseteq \dom B$, so $\dom(A + B) = \dom A$.

For all $x \in \dom{A}$ we have
$$\tag{$\ast$}
\lVert (A+B)x \rVert \leq a \lVert x \rVert + (1+b)\lVert Ax\rVert
$$
by the triangle inequality and the hypothesis that $\lVert Bx\rVert \leq a\lVert x\rVert + b\lVert Ax\rVert$. On the other hand,
$$
-a \lVert x \rVert + (1-b)\lVert Ax\rVert = \lVert Ax\rVert - (a\lVert x\rVert + b\lVert Ax\rVert) \leq \lVert Ax\rVert - \lVert -Bx\rVert \leq \lVert (A + B)x\rVert
$$
which immediately yields for all $x \in \dom A$ the inequality
$$\tag{${\ast}{\ast}$}
\lVert Ax \rVert \leq
\frac{a}{1-b}\lVert x\rVert + \frac{1}{1-b}\lVert (A+B)x\rVert
$$  of your second bullet since $1 -b \gt 0$.

Let $x_n \in \dom A = \dom(A+B)$ be a Cauchy sequence. Then $Ax_n$ is a Cauchy sequence if and only if $(A+B)x_n$ is a Cauchy sequence by $(\ast)$ and by $({\ast}{\ast})$.
The same estimates show that for a sequence $x_n \to 0$ with $x_n \in \dom A$ we have $Ax_n \to 0$ if and only if $(A+B)x_n \to 0$. In particular, $A$ is closable if and only if $A+B$ is closable.
If we now assume that $A$ is closed, then we know that $A+B$ is closable. Let us show that $A+B$ is closed.
Suppose $x_n \in \dom A$ is such that $x_n \to x$ and $(A+B)x_n \to y$. Then $(A+B)x_n$ is Cauchy, so $Ax_n$ is Cauchy, too. Therefore $Ax_n \to z$. It follows that $x \in \dom A$ and $Ax = z$ because $A$ is closed. But as $\dom A = \dom(A+B)$, we know that $x \in \dom(A+B)$ and since we already know that $(A+B)$ is closable we must have $(A+B)x = y$ (otherwise $(x_n -x)$ would be a null sequence in $\dom(A+B)$ with $(A+B)(x_n - x) \nrightarrow 0$ contradicting closability of $A+B$).

Further results in this direction and many examples can be found in Kato's Perturbation theory for linear operators, Chapter IV, §1, sections 1 and 2.


TITLE: Integer polynomials with translation of the constant term - Are these examples unique?
QUESTION [0 upvotes]: Consider the polynomial
$$(x+k_1)(x+k_2)(x+k_3)(x+k_4)+c \in \mathbb{Z}[x]$$
for given $c \in \mathbb{Z}$ and $k_1,k_2,k_3,k_4 \in \mathbb{Z}$.
In this question, a user is asked to find for which constants $k \in \mathbb{Z}$ the polynomial $(x+1)(x+2)(x+3)(x+k)+1$ is factorable as a square of a quadratic polynomial. As it turns out, the only values are $k = 0,4$. By translation, we can obviously find an infinite family of solutions with $(c,k_1,k_2,k_3,k_4)=(1,l,l+1,l+2,l+3)$ where $l \in \mathbb{Z}$.
Naturally, a generalization of this is the question above is: Are there any other examples (up to translation) aside from those already stated, $c = 1$ or not? As the discrimant of this polynomial vanishes iff a factor with multiplicity bigger than $1$ appears, I tried a couple of different spacings of the variables $k_i$ with one variable undeterminded, but that did not lead to another solution (maybe I haven't searched hard enough?).
Generalizing this even further: What about a different numbers of factors on the left side (with undeterminded variables defined accordingly) and an $n$-th power on the right side? Of course, considering the order of the polynomial on the left, $n$ has to be a divisor - other than 1 -  of the number of factors on the right side, but I haven't investigated further in this direction. Formalized this would look like:
$$?\exists n,l \in \mathbb{N}, k_1,\ldots,k_l \in \mathbb{Z}: \left(\prod\limits_{i=1}^{l}(x+k_i)\right)+c = p(x)^n$$
where $p(x) \in \mathbb{Z}[x]$.
So, my two questions on this matter are

When is $(x+k_1)(x+k_2)(x+k_3)(x+k_4)+1$ factorable as a quadratic polynomial? What about $(x+k_1)(x+k_2)(x+k_3)(x+k_4)+c$ with $c \in \mathbb{Z}$?
As a general problem: When is $\left(\prod\limits_{i=1}^{l}(x+k_i)\right)+c$ factorable as an $n$-th power of a polynomial , where $n,l \in \mathbb{N}$?

I am happy about every thought or solution on this, even though I am most comfortable with an undergraduates course worth of Abstract Algebra.
EDIT: So I have been thinking about this problem a bit more today and actually involved Mathematica to aid in the quest to more examples. For 2., I haven't actually made any advances yet; but if I'm not mistaken, I found and amazing algebraic identity which characterizes (part of) the solution set for the simplified case (in 1.).
Sparing you of any further exposition, I present the following formula, which is independent of the integral nature of my problem:
$$\forall c, n, k \in \mathbb{C}: (x+c)(x+c+n)(x+c+n+k)(x+c+2n+k) + (\frac{n^2+nk}{2})^2 = (x^2+(2n+k+2c)x+2c^2+(k+2n)c+\frac{n^2+nk}{2})$$
Making appropriate restrictions, one arrives at a far more general solution to 1. than the one I had given. This was found with the help of Nate's answer, which pointed me to the existence of solutions iff $c$ in 1. is a square. With the power of Mathematica, I observed patterns of the differences between the terms and tried to generalize as much as possible, eventually arriving at the given formula. So far, this list seems to be exhaustive, but I have no idea on how to dis/prove the uniqueness of this solution.

REPLY [1 votes]: Here's an answer to the first one:
Suppose $k_1 < k_2 < k_3 < k_4$ are integers and
$$(x+k_1)(x+k_2)(x+k_3)(x+k_4)+c = (x^2+Ax+B)^2$$
Plugging in $x = -k_1$ we see that $c$ is a perfect square, let's say $c = C^2$.  Moving this to the other side:
$$(x+k_1)(x+k_2)(x+k_3)(x+k_4) = (x^2+Ax+B)^2 - C^2 = (x^2+Ax+B-C)(x^2+Ax+B+C)$$
Note that the two factors on the right are both parabolas with symmetry at $x = -A/2$ so this means that one must be $(x+k_1)(x+k_4)$ and the other must be $(x+k_2)(x+k_3)$ and $k_1+k_4 = k_2+k_3 = A$.  By exchanging $C$ and $-C$ wlog we get $k_1k_4 = B-C$ and $k_2k_3 = B+C$.  In particular these must have the same parity.
From here we can go backwards to find solutions.  Given $k_1 < k_2 < k_3 < k_4$ satisfying $k_1+k_4 = k_2+k_3$ such that $k_1k_4 - k_2k_3$ is even. Then we can take $A = k_1+k_4$, $B = \frac{k_1k_4 + k_2k_3}{2}$, and $C = \frac{k_1k_4 - k_2k_3}{2}$.


TITLE: If $f(x + y) = f(x) + f(y)$ showing that $f(cx) = cf(x)$ holds for rational $c$
QUESTION [7 upvotes]: For $f:\mathbb{R}^n \to \mathbb{R}^m$, if $f(x + y) = f(x) + f(y)$ for then for rational $c$, how would you show that $f(cx) = cf(x)$ holds?
I tried that for $c = \frac{a}{b}$, $a,b \in \mathbb{Z}$ clearly
$$
f\left(\frac{a}{b}x\right) = f\left(\frac{x}{b}+\dots+\frac{x}{b}\right) = af\left(\frac{x}{b}\right)
$$
but I can't seem to finish it, any help?

REPLY [12 votes]: Try computing $bf(x/b){}{}{}$.

REPLY [4 votes]: Let's show that $f(x/b) = \frac{f(x)}{b}$. Denote $t= f(x/b)$. Now sum $t$ with itself exactly $b$ times:
$$b\cdot t=t+t+\dots+t=f(x/b)+f(x/b)+\dots + f(x/b)$$
$$=f(x/b+\dots+x/b)=f((bx)/b)=f(x)$$
Divide both sides by $b$ and you have $t=\frac{f(x)}{b}$.

REPLY [4 votes]: For $c \in \mathbb{Z}^*$:
$$
f(cx) = cf(x)
$$
Let's do a variable substitution: $\hat{x} = cx$:
\begin{align*}
f(\hat{x}) &= cf\left(\frac{\hat{x}}{c}\right) \\
\Rightarrow f\left(\frac{\hat{x}}{c}\right) &= \frac{1}{c} f(\hat{x})
\end{align*}
Combine this with your initial findings to get the proof for $c \in \mathbb{Q}$.


TITLE: Proving $(p∨q)∨(p∨¬q)$ is a tautology
QUESTION [0 upvotes]: I am trying to prove that $(p∨q)∨(p∨¬q)$ is a tautology.
This is a relatively easy problem because it is easy to tell that the statement will be true whenever p or q or ¬q are true. I have never written proofs before, however and I am wondering what the correct way to prove this would be.
I used the distributive law initially so that my problem could be rewritten as
p∨p ∨q∨¬q which can then simplify to $p∨q∨¬q.$
What law can I use to say from here that this is a tautology due to the fact that p or q or ¬q will always be true?

REPLY [0 votes]: An alternative approach using $\lor$ introduction and elimination rules.
\begin{array}{l}
 q\lor\neg q \\
 q\rightarrow(p\lor q)\\
 (p\lor q)\rightarrow (p\lor q)\lor(p\lor\lnot q) \\
 \lnot q\rightarrow(p\lor\neg q) \\
 (p\lor\neg q)\rightarrow(p\lor\neg q)\lor(p\lor q) \\
 (q\lor\lnot q)\rightarrow(p\lor\neg q)\lor(p\lor q) \\
 (p\lor\neg q)\lor(p\lor q)
\end{array}


TITLE: Determinants in Kerr-Schild via Sylvester theorem
QUESTION [1 upvotes]: The Kerr-Schild metric has the following form
\begin{align}
g_{\mu \nu} = \tilde{g}_{\mu \nu} + \kappa \phi l_{\mu} l_{\nu}
\end{align}
where $l^\mu l_\mu =0$. Here index are raised or lower by background metric $\tilde{g}$.
I know
\begin{align}
\det(g_{\mu \nu}) = \det(\tilde{g}_{\mu \nu})
\end{align}
And I want to prove this via Sylvester determinant theorem.
[Actually, I prove this relation via Taylor's trick of det trace properties. But I want to do using Sylvester's theorem to do more further generalization like extended Kerr Schild formalsim]
Sylvester's determinant identity :
Let $A$ and $B$ are the matrix of $m\times n$, $n\times m$ then
\begin{align}
\det(I_m + A B) = \det(I_n + BA)
\end{align}

REPLY [2 votes]: I'm not sure what you mean by "Taylor's trick", but this is just linear algebra: if $A$ is a matrix with rank $1$, then $$\det({\rm Id}+A)=1+{\rm tr}(A).$$Write $\widetilde{G} = (\widetilde{g}_{\mu\nu})$ and $G=(g_{\mu\nu})$. Then $$\det(G)=\det(\widetilde{G}+\kappa\phi\ell\ell^\top) = \det(\widetilde{G})\det({\rm Id}+\kappa\phi \widetilde{G}^{-1}\ell\ell^\top)=\det(\widetilde{G})(1+\kappa\phi \,{\rm tr}(\widetilde{G}^{-1}\ell\ell^\top)).$$But ${\rm tr}(\widetilde{G}^{-1}\ell\ell^\top)=\ell^\mu\ell_\mu=0$. Thus $\det(G)=\det(\widetilde{G})$.


TITLE: convergence and integral: a small question
QUESTION [0 upvotes]: For any $t>0$ suppose that $f_t$ is a continuous function on $\mathbb{R}$ and uniformly bounded in $t$ : $\|f_t\|_\infty \leq C$. Suppose that for any $g \in L^1(\mathbb{R})$ we have
$$
\int_\mathbb{R} f_t(x)g(x) d x \to \int_\mathbb{R} g(x) dx
$$
when $t \to 0$. How show that for any $x \in \mathbb{R}$ we have $f_t(x) \to 1$ when $t \to 0$ ?
I tried to use a function $g=K1_{B(x,r)}$ where $K$ is a constant and where $B(x,r)$ is a small ball around $x$.

REPLY [3 votes]: The Riemann-Lebesgue Lemma shows that $$f_t(x)=1+\sin(x/t)$$is a counterexample.

REPLY [0 votes]: This is a well-known theorem:
Let $1\leq p\leq \infty$.  If ${h_n}$ converges in $L^p$ norm to $h$, then
there exists a subsequence $h_{n_k}$ that converges point-wise to $h$.
Notice that you are given an $L^{1}$ convergence of the sequence of functions
$h_{t}:=f_t g$ to the limit function $h:=g$.
Warning: I made a mistake here. The statement above is accurate, but I did not read the question carefully. As @ Sangchul comments:
The question says $\|f_n\|_{L^1}\rightarrow \|f\|_{L^1}$. This is a weaker
condition than convergence in $L^{1}$ norm which is defined by $\|f_n-f\|_{L^1}\rightarrow 0$.
The counterexample given by
@David C. Ullrich. shows that the convergence of norms  $\|f_n\|_{L^1}\rightarrow \|f\|_{L^1}$ does imply point-wise convergence of the sequence $f_{n}$ even after passing to any subsequence.


TITLE: Mensuration question about a hoop resting on a staircase
QUESTION [0 upvotes]: I came across a question recently:

A hoop, as shown in the diagram, rests vertically at stair case. Note: AB = 12 cm, and BC = 8 cm. Find the radius of the hoop.
  Figure (hand-made):

This is a question that I have no clue about. I thought that AB might be the radius but well from the diagram it seems no.
A thorough Google search revealed only one site which requires a login, which I don't want to do.
Can anyone please provide a explanatory answer.
Thanks a lot.

REPLY [1 votes]: An analytic solution:
The equation of a circle, radius $R$ with centre at $(h, k)$ is:$$(x-h)^2+(y-k)^2=R^2$$If we take the point A as the origin, then the centre is at $(0, R)$ and the equation reduces to:$$(x)^2+(y-R)^2=R^2$$Note that the point $(12,8)$ lies on the circle and must satisfy the equation.


TITLE: Locally exact vs globally exact
QUESTION [0 upvotes]: Why the volume form in Sphere is locally exact but not globally exact? here the integral is integral $$\int_{S^n}w$$ with $$w = \frac{1}{r} \sum_{i=1}^{n+1} (-1)^{i-1} x_i dx_1 \cdots\hat{dx_i},\cdots dx_{n+1}. $$

REPLY [3 votes]: Well, if it were globally exact, because $\partial S^n = \emptyset$, we'd have to have $\int_{S^n}\omega = 0$ by Stokes's Theorem.


text:
\section{Proofs of Lemma~\ref{sec3-lemma-10} and Lemma~\ref{sec3-claim-1}}\label{sec-4}
\subsection{Proof of Lemma~\ref{sec3-lemma-10}}

The conditions of Theorem~\ref{sec3-th-1} imply in particular that $17K_X^2+37<p$. Also by Noether's inequality, $2\chi(\mathcal{O}_X)\leq K_X^2+6$. Therefore,
\[
12\chi(\mathcal{O}_X)+11K_X^2+1\leq 17K_X^2+37<p,
\]
and hence by~\cite[Theorem 3.2]{Tz19},
\begin{gather}\label{sec3-eq-31}
\gamma\leq 12\chi(\mathcal{O}_X)+11K_X^2\leq 17K_X^2+36
\end{gather}
and  by~\cite[Theorem 3.3]{Tz19} $D$ lifts to a vector field $D_Y$ on the minimal resolution $Y$ of $X$ and every $h$-exceptional curve is stabilized by $D_Y$.

\subsection{Proof of Lemma~\ref{sec3-claim-1}}

As has been proved in Section~\ref{sec3} there exists at most one index $j$ such that $\tilde{\Delta}_j$ is an elliptic curve. For all other $i\not=j$, $\tilde{\Delta}_i \cong \mathbb{P}^1$ and $\tilde{\Delta}_i^2\leq -3$, for all $i=1,\ldots, m$.  After a renumbering of the curves we may assume that $\tilde{\Delta}_i\cong \mathbb{P}^1$, $i=1,\ldots, m-1$, and $\tilde{\Delta}_m$ is either $\mathbb{P}^1$ or an elliptic curve. In addition all other components of $\Delta_Y$ are disjoint smooth rational curves  of self intersection $-2$. Let us assume that $\tilde{\Delta}_m$ is an elliptic curve. If this is not true and hence every component of $\Delta_Y$ is a smooth rational curve, then the proof of the claim is contained in the proof of the case when $\tilde{\Delta}_m$ is elliptic.

Then by~\cite{Art62}, there exists a birational morphism $f^{\prime} \colon Y\rightarrow Z^{\prime}$ whose exceptional set is exactly $\tilde{\Delta}_i$, $i=1,\ldots, m-1$. Let $P_i=f^{\prime}(\tilde{\Delta}_i)$, $i=1,\ldots, m-1$. Then $P_i$ are log terminal quotient singularities. In particular $K_{Z^{\prime}}$ is $\mathbb{Q}$-Cartier. I will show that $K_{Z^{\prime}}$ is nef and big. Indeed. Suppose that $\tilde{\Delta}_i^2=-d_i$. Then by the previous discussion, $d_i \geq 3$, if $i\leq m-1$, and $d_m\geq 1$. Then a straightforward calculation shows that
\begin{gather}\label{sec3-eq-3}
K_Y+\sum_{i=1}^{m-1}\frac{d_i-2}{d_i}\tilde{\Delta}_i=(f^{\prime})^{\ast}K_{Z^{\prime}},
\end{gather}
Then, since $\tilde{\Delta}_i\cdot \tilde{\Delta}_j=0$, for $i\not= j$,
\begin{gather}\label{sec3-eq-1000}
K_{Z^{\prime}}^2=K_Y^2-\sum_{i=1}^{m-1}(\frac{d_i-2}{d_i})^2\tilde{\Delta}^2_i =K_X^2+\sum_{i=1}^{m-1}\frac{(d_i-2)^2}{d_i} >0,
\end{gather}
since $K_X^2>0$. Let $C^{\prime}$ be a curve in $Z^{\prime}$ and $C$ its birational transform in $Y$. Then since $K_Y$ is ample, $K_Y \cdot C \geq 0$ and hence from the equation (\ref{sec3-eq-3}) it follows that
\[
C^{\prime} \cdot K_{Z^{\prime}}=C\cdot (f^{\prime})^{\ast}K_{Z^{\prime}}=K_Y\cdot C+\sum_{i=1}^{m-1}\frac{d_i-2}{d_i}\tilde{\Delta}_i \cdot C\geq 0,
\]
since $K_Y$ is nef and $d_i \geq 3$. Therefore, $K_{Z^{\prime}}$ is nef and big. Moreover, $K_{Z^{\prime}}\cdot C^{\prime}=0$ if and only if $C\cdot \tilde{\Delta}_i=0$, $i\leq m-1$, and $K_Y\cdot C=0$. Therefore, since $Y$ is a minimal surface of general type, $C\cong \mathbb{P}^1$ and $C^2=-2$. Moreover, since $C\cdot \tilde{\Delta}_i=0$, $i\leq m-1$, it follows that $C^{\prime}$ is in the smooth part of $Z^{\prime}$ and hence $C \cong C^{\prime} \cong \mathbb{P}^1$ and $(C^{\prime})^2=-2$.

Now since $\tilde{\Delta}_m$ is disjoint from $\tilde{\Delta}_i$, $i\not= m$,  $\Delta^{\prime}_m=f^{\prime}(\tilde{\Delta}_m)$ is contained in the smooth part of $Z^{\prime}$. Then the conditions of  Proposition~\ref{sec1-prop-7} are satisfied and hence there exists a birational map $f^{\prime\prime} \colon Z^{\prime} \rightarrow Z$ which contracts $\Delta^{\prime}_m$ to a Gorenstein simple elliptic singularity $Q\in Z$. Let $f\colon Y \rightarrow Z$ be the composition of $f^{\prime}$ and $f^{\prime\prime}$. I will show that $f$ is the map that satisfies the conditions of the lemma.

Let $F$ be an $f^{\prime\prime}$-exceptional curve different that $\Delta_m^{\prime}$. Then, since by the proof of Proposition~\ref{sec1-prop-7} $f^{\prime\prime}$ is defined by $|m(K_{Z^{\prime}}+\Delta_m^{\prime})|$, for some $m>0$, $F$ is $f^{\prime\prime}$-exceptional, if and only if $F \cdot K_{Z^{\prime}}=F\cdot \tilde{\Delta}_m=0$. Hence, from the previous discussion, $F\cong \mathbb{P}^1$, $F$ is in the smooth part of $Z^{\prime}$ and $F^2=-2$. Therefore the singularities of $Z$ are the log terminal quotient singularities coming from $Z^{\prime}$, a simple elliptic singularity obtained by contracting $\Delta_m^{\prime}$ and rational double points corresponding to the contraction of the other $f^{\prime\prime}$-exceptional curves. In particular,  by~\cite{Ha98},~\cite{MS91}, all these singularities are F-pure. Finally notice that every component of $\Delta_Y$ is contracted by $f$ since the curves $\tilde{\Delta}_i$, $i\leq m-1$, are contracted by $f^{\prime}$, $\tilde{\Delta}_m$ is contracted by $f^{\prime\prime}$ and
$\Gamma_j$ are smooth rational curves of self intersection -2 disjoint from $\tilde{\Delta}_m$ and hence from the previous description of the exceptional set of $f^{\prime\prime}$, they are contracted by $f^{\prime\prime}$. This proves part  (4) of the Lemma.

Next I will show that $K_Z$ is ample. Since $f^{\prime\prime}$ contracts $\Delta^{\prime}_m$ and smooth rational curves of self intersection $-2$ contained in the smooth part of $Z^{\prime}$ and disjoint from $\Delta_m^{\prime}$, the following adjunction formula holds
\begin{gather}\label{sec3-eq-1001}
K_{Z^{\prime}}+\Delta^{\prime}_m=(f^{\prime\prime})^{\ast}K_Z.
\end{gather}
Then, $K_Z^2=K_{Z^{\prime}}^2-(\Delta^{\prime}_m)^2>0$. Moreover, let $C \subset Z$ be an curve, Then,
\[
K_Z\cdot C=(f^{\prime\prime})^{\ast}K_Z \cdot C^{\prime}=K_{Z^{\prime}}\cdot C^{\prime} +\Delta^{\prime}_m\cdot C^{\prime},
\]
where $C^{\prime}$ is the birational transform of $C$ in $Z^{\prime}$. Since $K_{Z^{\prime}}$ is nef, $K_Z \cdot C\geq 0$.  Suppose that $K_Z \cdot C =0$. Then from the previous equation it follows that
$K_{Z^{\prime}}\cdot C^{\prime}=0$ and $C^{\prime} \cdot \Delta^{\prime}_m=0$. But it has been shown that every such curve is $f^{\prime\prime}$-exceptional, which is not the case with $C^{\prime}$. Hence $K_Z\cdot C>0$, for every $C$. Hence from Kleiman's criterion, $K_Z$ is ample.

Let $d_z$ be the index of $Z$. Since $Z^{\prime}_m$ is contained in the smooth part of $Z^{\prime}$ and $f^{\prime\prime}$ contracts $\Delta^{\prime}_m$ to a simple elliptic singularity, in particular Gorenstein, and the equation (\ref{sec3-eq-1001}) it follows that $d_z$ is the same as the index of $K_{Z^{\prime}}$. Then from the  equation (\ref{sec3-eq-3}) it follows that $d_z$ is less or equal to the least common multiple of $d_1, \ldots, d_{m-1}$. Now from the equations (\ref{sec3-eq-1}) it follows that
\begin{gather}\label{sec3-eq-4}
\sum_{i=1}^{m-2}K_Y \cdot \tilde{\Delta}_i=\sum_{i=1}^{m-2}h^{\ast}K_X \cdot \tilde{\Delta}_i=\sum_{i=1}^{m-1}K_X \cdot \Delta_i \leq \sum_{i=1}^{m}K_X \cdot \Delta_i=K_X\cdot \Delta  \leq 3K_X^2.
\end{gather}
By adjunction, $K_Y\cdot \tilde{\Delta}_i=-2-\tilde{\Delta}_i^2=-2+d_i$. Moreover, from the previous equation, since $K_X$ is ample,  it follows that $m-1 \leq 3K_X^2$. Then the equation (\ref{sec3-eq-4}) gives
\begin{gather}\label{sec3-eq-100}
\sum_{i=1}^{m-1}d_i \leq 2(m-1)+3K_X^2\leq 6K_X^2+3K_X^2=9K_X^2.
\end{gather}
Now according to Proposition~\ref{sec1-prop-10} it follows that
\[
d_z\leq e^{9K_X^2/e}=u^{9K_X^2},
\]
for $u=e^{1/e}$, as claimed. This proves part (2) of the Lemma.

I will next show part (3) of the Lemma. From the equations (\ref{sec3-eq-1001}), (\ref{sec3-eq-3}) and (\ref{sec3-eq-1}) it follows that
\begin{gather*}
K_Z^2=K_{Z^{\prime}}^2+K_{Z^{\prime}}\cdot \Delta^{\prime}_m=K_Y^2+\sum_{i=1}^{m-1}\frac{d_i-2}{d_i}K_Y\cdot \tilde{\Delta}_i+K_Y\cdot \tilde{\Delta}_m\leq K_Y^2+K_Y\cdot \Delta_Y\leq 4K_X^2.
\end{gather*}
This proves part (3) of the lemma. It remains to prove part (5). Since $f$ is birational, $D_Y$ induces a vector field $D_Z$ on $Z$. Since $D^p=D$, then $D_Z^p=D_Z$. Moreover, since $f$ contracts the divisorial part of $D_Y$, $D_Z$ has only isolated singularities. Let then $\nu \colon Z\rightarrow \hat{Z}$ be the quotient of $Z$ by the $\mu_p$-action induced by $D_Z$. Then since $D_Z$ has no divisorial part, $K_Z=\nu^{\ast}K_{\hat{Z}}$. Therefore $K_Z$ is a $D_Z$-linear rank 1 reflexive sheaf on $Z$. This concludes the proof of Lemma~\ref{sec3-claim-1}.


TITLE: Confused about limit proofs conceptually
QUESTION [2 upvotes]: In a question like this:
Prove that if $\lim_{x \to a} f(x) = l$ and $\lim_{x \to a} g(x) = m$ then $\lim_{x \to a} \max(f(x), g(x)) = \max(l, m)$
In general, when asked for proofs like this, are we supposed to find a $\delta$ or is the presupposition that the limit already exists, so that it is already true that $|h(x)−Q|<\epsilon$ for $|x−a|<\delta′$??Or do we first prove that there exists a $\delta$? But how can we do that if all we are given is a GENERAL $f(x), g(x)$?
Thanks!
I am confused because I am not sure how to prove things then.

REPLY [0 votes]: let $h=max\{f,g\}$ . let $k=max\{l,m\} $.
Choose  a common $\delta>0$ for both $ f,g$ such that $0<|x-a|<\delta$ $\implies l-\epsilon <f(x)<l+\epsilon $ and $ m-\epsilon <g(x)<m+\epsilon $
Now $f(x)<l+ \epsilon<k+\epsilon$ and  $g(x)<m+ \epsilon<k+\epsilon$
So $h(x)<k+\epsilon $
Again $l-\epsilon <f(x)$ and $m-\epsilon <g(x)$
So $h(x)>max\{l-\epsilon,m-\epsilon\}=max\{l,m\}-\epsilon =k-\epsilon$
hence proved


TITLE: How is notation $\{f_n(x)\}\nearrow x$ interpreted?
QUESTION [1 upvotes]: In the context of continuity of probabilities we define $\{A_n\}\nearrow A$ to mean that $A_1 \subseteq A_2 \subseteq A_3 \subseteq ...$ and $\cup_n A_n=A$, where $A, A_1, A_2, A_3, ... \in \mathcal{F}$ for some probability triplet $(\Omega, \mathcal{F}, P)$. I have no problems with this notation.
But how is $\{f_n(x)\}\nearrow x$ interpreted, where $f_n:[0,\infty) \rightarrow [0,\infty)$ and $x$ is a random variable? I don't see how $f_1(x) \subseteq f_2(x) \subseteq f_3(x) \subseteq ...$ considering $f_n(x)$ is not a set but a real value. A possible interpretation might be that it's range of $f_n$ that is a subset of the range of $f_{n+1}$; but then it's unclear how $\cup_n f_n(x)=x$, i.e. how the union of ranges would correspond to a real value.
(Apparently $\{f_n(x)\}\nearrow x$ is useful when deriving the expectation of general non-negative random variables.)

REPLY [4 votes]: It means that $f_n(x)\le x$ for all $n,$ and that $\lim_{n\to\infty}f(x)=x.$ Verbally, we may say that $f_n(x)$ converges to $x$ from below. Often, it also indicates that $f_n(x)$ is monotonically increasing, but this varies from source to source.