text:
\begin{document}

\title[Mackey functors and bisets]
{Mackey functors and bisets}
\author{I.~Hambleton}
\thanks{\hskip -11pt  Research partially supported by
NSERC Discovery Grant A4000 and the NSF. The authors would also like to thank the SFB 478, M\"unster, for its hospitality and support in June 2008.}
\address{Department of Mathematics \& Statistics
\newline\indent
McMaster University
\newline\indent
Hamilton, ON  L8S 4K1, Canada}
\email{ian{@}math.mcmaster.ca}
\author{L.~R.~Taylor}
\address{Department of Mathematics
\newline\indent
University of Notre Dame
\newline\indent
Notre Dame, IN 46556, USA}
\email{taylor.2@nd.edu}
\author{E.~B.~Williams}
\address{Department of Mathematics
\newline\indent
University of Notre Dame
\newline\indent
Notre Dame, IN 46556, USA}
\email{williams.4@nd.edu}

\date{Jan.~7, 2010 (revision)}
\begin{abstract}\noindent
For any finite group $G$, we define a bivariant functor from the
Dress category of finite $G$-sets to the conjugation biset category,
whose objects are subgroups of $G$, and whose morphisms are generated
by certain bifree bisets.
Any additive functor from the conjugation biset category to abelian groups
yields a Mackey functor  by composition.
We characterize the Mackey functors which arise in this way.
\end{abstract}

\maketitle
\section{Introduction}\label{one}
Let $G$ be a finite group.
A Mackey functor in the sense of Dress \cite{dress2}*{p.~301} is a bivariant functor
$$\cM \colon \DG \to \ab$$  from the category of finite left $G$-sets and
$G$-maps to abelian groups, satisfying a pull-back axiom (M1) and
an additivity axiom (M2).
These axioms express the classical Mackey properties from
representation theory, as formulated by Green \cite{green1}.  In this paper, unless otherwise mentioned, by a \emph{Mackey functor} we mean a Mackey functor in the sense of Dress.

Many of the Mackey functors encountered in applications of the theory
factor through the $G$-Burnside category $\bA(G)$, whose objects are
the subgroups $H\subset G$ and whose morphisms are generated by
$H_2$\,-$H_1$ bisets with certain properties (compare \cite{htw3}*{1.A.4}).
Let $\bAdot(G)$ denote the additive completion of this category,
defined in Section \ref{three}.

In Section \ref{six} we define a subcategory
$\subBurnsideDot(G) \subset \bAdot(G)$, where the morphisms
are restricted to be \emph{conjugation bisets}, and construct a
bivariant functor $j\colon \DG \to \subBurnsideDot(G)$
(see Definition \ref{DtoBurnside}).
A Mackey functor $\cM$ is said to be \emph{conjugation invariant} provided
that the centralizer $C_G(H)$ acts trivially on $M(G/H)$ for all $H \subset G$,
via the $G$-maps $\varphi\colon G/H \to G/H$ given by
$eH \mapsto zH$, for some $z \in C_G(H)$
(see Definition \ref{def:conjugation invariant}).

The main result (see Theorem \ref{Mackeyrecognition}) is a
recognition principle for such Mackey functors.

\begin{thma} Let $G$ be a finite group.
A Mackey functor $\cM\colon \DG \to \ab$ factors through
$j\colon \DG \to \subBurnsideDot(G)$ if and only if
$\cM$ is conjugation invariant.
\end{thma}

The applications surveyed in \cite{htw3} and \cite{htw2007},
mostly depend on the following immediate consequence of
our main result.

\begin{corb} For any additive functor $F\colon \bAdot(G) \to \ab$,
the composition $F\circ j \colon \DG \to \ab$ is a Mackey functor.
\end{corb}

This result applies to classical Mackey functors  such as the
Swan ring and the Dress ring, and  to algebraic $K$-theory
and $L$-theory functors encountered in geometric topology
and surgery theory.

\begin{remark} The topic of Mackey functors has been extensively explored over
the last 30 years, and there is a large literature (see, for example, the survey by Webb \cite{webb3}). Here are some remarks about a selection of this earlier work.

\nr{a} The idea of reformulating the classical Mackey bivariant functor  properties
as a single functor out of an intermediate category has been carried out
by several authors (see Lindner \cite{lindner1}*{Theorem 2}, Gaunce Lewis   \cite{gaunce_lewis1}, tom Dieck
\cite{tomDieck2}*{Chap.~IV.8}, Th\'evenaz-Webb \cite{thevenaz-webb2}).

\nr{b} The use of bisets as morphisms in a category  has also appeared in various settings in the literature (see, for example,  Lewis-May-McClure \cite{lewis-may-mcclure1}, Adams-Gunawardena-Miller \cite{adams-gunawardena-miller1}*{\S9, p.~454}, Hambleton-Taylor-Williams
\cite{htw3}*{1.A.4}, and Bouc \cite{bouc0}, \cite{bouc2}*{\S 2}, \cite{bouc3}).

\nr{c} In \cite{dress2}*{pp.~292, 302} Dress restricts attention to a variant of classical Mackey functors, now usually called \emph{global Mackey functors} \cite{webb2}*{p.~267}, defined on the category $\GR$  of finite groups and monomorphisms, and in particular the maps induced by conjugations
depend only on the underlying group homomorphisms.
 In \cite{dress2}*{p.~298-299}, Dress  describes the passage from global Mackey functors to his functors. The output of this process is a conjugation-invariant Mackey functor.

\nr{d}
In \cite{webb2}*{p.~271},  Webb  sketches a proof of a result analogous to Theorem A,  that a global Mackey functor is equivalent to an additive functor out of a category $\Omega_{\bZ}$ whose objects are finite groups, and whose morphisms are bifree bisets. The restriction of $\Omega_{\bZ}$ to the subgroups  of a fixed group $G$ is just the category $\bAdot(G)$, defined in our MSRI preprint (1990) \cite{htw4}*{Ex.~5.5}.

\nr{e} There is a sub-category of $\GR$ defined by restricting to the subgroups of a fixed group $G$, and to the monomorphisms generated by inclusions and conjugations. A  Mackey functor out of this sub-category  is equivalent to an additive functor out of $\subBurnsideDot(G)$. This is exactly the statement of Theorem A expressed in classical terms. In other words, Theorem A provides a direct method to check the Mackey properties, starting from  minimal input data.

\nr{f} The proof of Theorem A follows Webb's proof \cite{webb2}*{p.~271} in outline, but we supply all the technical details necessary, for example, to check that we have actually constructed a  functor $F\colon \subBurnsideDot(G) \to \Ab$, factoring a given conjugation invariant Mackey functor.
\end{remark}

\section{Mackey functors}\label{two}
We will first recall the basic definitions Dress used in his formulation of
induction theory \cite{dress2}.
Let $G$ be a finite group, and let
$\DG$ denote the category whose objects are finite,
left $G$-sets and whose morphisms are $G$-maps.
A \emph{Mackey functor} is a bivariant functor
$\cM = (\cM_\ast, \cM^\ast) \colon \cD(G) \to \ab$,
where $\ab$ denotes the category of abelian groups and groups
homomorphisms, such that $\cM_\ast(S) = \cM^\ast(S)$
for each object $S \in \DG$, and the following two  properties hold:

\medskip\noindent
\textbf{(M1)} For any pullback diagram of finite $G$-sets
$${\vcenter
{\xymatrix
{S\ar[r]^\Psi\ar[d]_\Phi&S_1\ar[d]^\varphi\\S_2\ar[r]^\psi&T}}}
$$
the induced maps give an commutative diagram
$${\vcenter
{\xymatrix
{\cM(S)\ar[r]^{\Psi_\cM}&\cM(S_1)\\
\cM(S_2)\ar[r]^{\psi_\cM}\ar[u]^{\Phi^\cM}&\cM(T)\ar[u]_{\varphi^\cM}}}}
$$
Here we denote the covariant maps by $\psi_\cM$ and the
contravariant maps by $\varphi^\cM$.

\medskip\noindent
\textbf{(M2)} The  embeddings of $S_1$ and $S_2$ into the disjoint
union $S_1\disjointunion S_2$ define an isomorphism
$\cM^\ast(S_1\disjointunion S_2) \to \cM^\ast(S_1)\oplus \cM^\ast(S_2)$.
Let $\cM (\emptyset) = 0$.

\medskip
The property (M1) is the usual double coset formula, and (M2) gives additivity.
We remark that for any bivariant functor satisfying (M1), the composition
$\cM_\ast(S_1)\oplus \cM_\ast(S_2) \to \cM_\ast(S_1\disjointunion S_2) =
\cM^\ast(S_1\disjointunion S_2) \to \cM^\ast(S_1)\oplus \cM^\ast(M_2) =
\cM_\ast(S_1)\oplus \cM_\ast(S_2)$
is just the identity matrix.
It follows that  any sub-bivariant functor of a Mackey functor is Mackey.

\begin{remark} We could replace the target category $\ab$ of abelian groups
throughout by the category $\RMod$ of $R$-modules, for any commutative ring
$R$ with unit.
\end{remark}

\section{The $G$-Burnside category $\Burnside(G)$}\label{three}
We will be interested in  Mackey functors which factor through the
$G$-Burnside category $\Burnside(G)$, whose
objects are subgroups $H\subset G$, and where
$\Hom_{\Burnside(G)}(H_1, H_2)$ is the Grothendieck construction
applied to the isomorphism classes of finite bifree $H_2$\,-$H_1$ bisets
(meaning both left and right actions are free). In contrast, the morphisms in  the $G$-Burnside category of \cite{adams-gunawardena-miller1}*{\S9, p.~454}, and our category $RG$-Morita \cite{htw3}*{1.A.4}  just have a one-sided isotropy assumption.

To make our recognition principle more precise, we will define a subcategory
$\subBurnside(G)\subset \Burnside(G)$ by restricting its  morphisms to
\emph{conjugation bisets} (see Definition \ref{conjugationbisets}).

Because of the Grothendieck construction, $\Burnside(G)$ and
$\subBurnside(G)$ are both Ab--categories: the morphism sets are
abelian groups and the compositions are bilinear  \cite{maclane2}*{I.8, p.~28}.
Let $u\colon \bA(G) \to \bAdot(G)$ and
$u\colon \subBurnside(G) \to \subBurnsideDot(G)$
denote the associated universal free additive categories, and
the universal inclusions (see  \cite{maclane2}*{VII.2, problem 6, p.~194}).

It turns out that the Mackey functors which factor through
$\subBurnsideDot(G)$ have an additional property,
called \emph{conjugation invariance}, which can be expressed
in terms of their restriction to the orbit category $\Or(G)$.
Recall that the objects of $\Or(G)$ are the subgroups $H\subset G$,
and the morphisms $\Hom_{\Or(G)}(H_1, H_2)$ are the $G$-maps
$\varphi\colon G/H_1 \to G/H_2$.
Any such $G$-map is uniquely determined by $eH_1 \mapsto gH_2$,
where $g^{-1}H_1g \subseteq H_2$.
If $H_1 = H_2 = H$, then any element $z \in C_G(H)$ in the centralizer
of $H$ gives a $G$-map $\varphi_z\colon G/H \to G/H$.

\begin{definition}\label{def:conjugation invariant}
A functor $F\colon \Or(G) \to \Ab$ is called \emph{conjugation invariant}
if  the induced maps $F(\varphi_z) = \id\colon F(H) \to F(H)$, whenever
$\varphi_z\colon G/H \to G/H$ is given by $eH \mapsto zH$, for some
$z \in C_G(H)$.
A Mackey functor is said to be conjugation invariant if its restriction to
$\Or(G)$ satisfies this condition.
\end{definition}

We now relate this condition to the conjugation homomorphisms
which will be used in the definition of $\subBurnsideDot(G)$.
If $\varphi\in \Hom_{\Or(G)}(H_1, H_2)$ is represented by $g\in G$,
we let $c_g\colon H_1 \to H_2$ denote the associated
conjugation homomorphism given by $c_g(h_1) = g^{-1}h_1g \in H_2$,
for all $h_1 \in H_1$.
Since the $G$-map $\varphi$ only depends on the coset $gH_2$,
we may vary $g \sim gh_2$, for any $h_2 \in H_2$.
The associated conjugation homomorphism
$c_{gh_2} = c_{h_2}\circ c_g$ is thus well-defined
(as a homomorphism) up to conjugation by elements of $H_2$.
Let
$$c_\varphi:= [c_g] \in \Hom(H_1, H_2)/(\text{conjugation\ in\ } H_2)$$
denote the equivalence class of the associated conjugation homomorphism
to a $G$-map $\varphi(eH_1) = gH_2$.
Two different morphisms $\varphi_1, \varphi_2 \in \Hom_{\Or(G)}(H_1, H_2)$
yield the same equivalence class $c_{\varphi_1} = c_{\varphi_2}$ if and only if
$g_1h_2g_2^{-1}$ centralizes $H_1$, for some $h_2 \in H_2$.

\begin{lemma}\label{conginv}
A functor $F\colon \Or(G) \to \Ab$ is conjugation invariant if and only if
$c_{\varphi_1} = c_{\varphi_2}$ implies $F(\varphi_1) = F(\varphi_2)$
for all morphisms
$\varphi_1, \varphi_2  \in \Hom_{\Or(G)}(H_1, H_2)$.
\end{lemma}

\begin{proof}
Suppose that $\varphi_1, \varphi_2 \in \Hom_{\Or(G)}(H_1, H_2)$ are given by
$\varphi_i(eH_1) = g_iH_2$, for $i = 1,2$.
If there exists an element $h_2 \in H_2$ such that
$z:=g_1h_2g_2^{-1}\in C_G(H_1)$,
then we see that $\varphi_1 = \varphi_2 \circ \varphi_z$.
Therefore, if $F$ is conjugation invariant and $c_{\varphi_1} = c_{\varphi_2}$
we conclude that $F(\varphi_1) = F(\varphi_2)$.
Conversely, for any subgroup $H \subset G$, and any  $z\in C_G(H)$,
the $G$-maps $\varphi_z, \id\colon G/H \to G/H$ have the property
$c_{\varphi_z} = c_{\id}$.
Therefore the given condition implies that $F(\varphi_z) = F(\id) = \id$,
and hence $F$ is conjugation invariant.
\end{proof}

The morphisms in $\Burnside(G)$ are defined by  the Grothendieck
construction with addition operation the disjoint union of bisets.
By convention, the empty biset  $\emptyset$ represents the zero element.
Composition comes from the balanced product:
$${_{H_3}}X_{H_2} \circ {_{H_2}}Y_{H_1} =
({_{H_3}}X_{H_2})\times_{H_2} ({_{H_2}}Y_{H_1}).$$
The reader should check that this is well--defined
on isomorphism classes of bisets and ``bilinear'' in that
$$({_{H_3}}X_{H_2} \disjointunion {_{H_3}}Y_{H_2})
\circ {_{H_2}}Z_{H_1}\cong ({_{H_3}}X_{H_2}
\circ{_{H_2}}Z_{H_1})\disjointunion\,
({_{H_3}}Y_{H_2}\circ {_{H_2}}Z_{H_1})$$
with a similar formula for disjoint union on the right.
The morphisms in $\bAdot(G)$ are matrices of morphisms in $\Burnside(G)$.

\begin{definition}
We define a contravariant involution
$\congtran\colon \Burnside(G) \to \Burnside(G)$,
by the identity on objects, and on morphisms it is the map induced
on the Grothendieck construction by the function
which takes the finite bifree $H_2$\,-$H_1$ biset ${_{H_2}}X_{H_1}$
to the  finite bifree $H_1$\,-$H_2$ biset ${_{H_1}}X_{H_2}$ which is
$X$ as a set and $h_1\cdot x \cdot h_2$ is defined to be $h_2^{-1}x h_1^{-1}$.
\end{definition}

The reader needs to check that isomorphic bisets are isomorphic after reversing the order,
and should also check that the transpose conjugate of a disjoint union is isomorphic to the
disjoint union of the conjugate transposes of the pieces.
This means that $\congtran$ is a functor which induces a homomorphism of Hom--sets.
It is clearly an involution, not just up to natural equivalence.
Since $\congtran$ is a homomorphism on Hom-sets, it induces an additive contravariant
involution $\congtrandot\colon \bAdot(G) \to \bAdot(G)$, called \emph{conjugate transpose},
which commutes with the functor $u\colon \bA(G) \to \bAdot(G)$.
By definition,  $\congtrandot$ acts on a matrix of morphisms by applying
$\congtran$ to each entry, and then transposing the matrix.

\section{Indecomposable $H_2$\,-$H_1$ bisets}\label{four}
We will need some more detailed information about the bifree bisets used
to define morphisms in $\Burnside(G)$. Much of this material can be found in Bouc \cite{bouc0}, but we include the details here to fix our notation and to emphasize the role of the base-points.

An $H_2$\,-$H_1$ biset is just a left $(H_2\times H^{\rm op}_1)$-set and so
any finite $H_2$\,-$H_1$-biset is a disjoint union of transitive
$(H_2\times H^{\rm op}_1)$-sets.
Since there are three groups acting
($H_1$, $H_2$ and  $H_2\times H^{\rm op}_1$), the following definition
should avoid confusion.
\begin{definition}
An $H_2$\,-$H_1$ biset  $X$ is \emph{indecomposable} if $X$ is a
transitive $(H_2\times H^{\rm op}_1)$-set.
\end{definition}

Since every $H_2$\,-$H_1$ biset is a disjoint union of indecomposable
$H_2$\,-$H_1$ bisets, it  follows that $X_1 \disjointunion Y$ is isomorphic
to $X_1 \disjointunion Y$ if and only if
$X_1$ is isomorphic to $X_2$.
One result of this remark is that the Grothendieck group of finite bifree
$H_2$\,-$H_1$ bisets is the free abelian group on the indecomposable
bifree ones.

An indecomposable $H_2$\,-$H_1$ biset $X$ is isomorphic
(via choice of base point) to a coset space
$(H_2\times H^{\rm op}_1)/S$, for some subgroup
$S \subset H_2\times H^{\rm op}_1$.
These models will be used extensively below, so we make some remarks
and introduce some notation.

\begin{enumerate}\addtolength{\itemsep}{0.3\baselineskip}
\item
To see $(H_2\times H^{\rm op}_1)/S$ as an $H_2$\,-$H_1$ biset, define the
$H_2$ action to be left multiplication in the first coordinate, but define
$(h_2,h_{1})S \cdot  g_{1}= (h_2, g^{-1}_{1} h_{1}) S$.

\item We introduce the following notation for points in $(H_2\times H^{\rm op}_1)/S$:
if $h_2\in H_2$ and $h_1\in H_1$, write
$\standardformpoint{h_2}{h_1} = (h_2,h^{-1}_1)S$.
In this notation, the $H_2$\,-$H_1$ action is the evident one:
$g_2 \standardformpoint{h_2}{h_1} g_1 = \standardformpoint{g_2 h_2}{h_1 g_1}$.

\item The left isotropy group of $(e,e)$ is just $(H_2\times\{e\}) \cap S
$, and the
right isotropy group of $(e,e)$ is just $(\{e\}\times H^{\rm op}_1) \cap S$, so
the actions are free if and only if
$(H_2\times\{e\}) \cap S = \{e\times e\}$ and
$(\{e\} \times H^{\rm op}_1)\cap S = \{e\times e\}$.

\item Equivalently, the $S$-action is bifree if the compositions
$\iota_1\colon S \subset H_2\times H^{\rm op}_1 \to H_2$ and
$S \subset H_2\times H^{\rm op}_1 \to H^{\rm op}_1$
are injective.
We will work instead with the composite
$\iota_2\colon S \to H_2\times H^{\rm op}_1 \to H^{\rm op}_1 \xrightarrow{\approx} H_1$,
where the last homomorphism takes $h_1$ to $h_1^{-1}$.

\item Let $L \subset H_2$ denote the image of the isotropy subgroup $S$
under the injection $\iota_1$, and let $K \subset H_1$ denote the image of $S$
under the injection $\iota_2$.
Define
$$\gamma\colon L  \xrightarrow{\ \iota_1^{-1}\ } S \xrightarrow{\ \iota_2\ } K$$
and notice that this is an isomorphism.

\item Conversely, given subgroups $L \subset H_2$, $K\subset H_1$,
and an isomorphism $\gamma\colon L \xrightarrow{\approx} K$,
let $\biset{H_2}{[L, \gamma, K]}{{H_1}}$
denote the indecomposable bifree biset $(H_2\times H^{\rm op}_1)/S$,
where $S \subset H_2\times H^{\rm op}_1$
is the graph of
$$\bar{\gamma} \colon  L \xrightarrow{\gamma} K \subset H_1 \xrightarrow{\iota} H_1^{\rm op},$$
and $\iota\colon H_1 \xrightarrow{\approx} H^{\rm op}_1$ is
the usual isomorphism which takes $h$ to $h^{-1}$.
\end{enumerate}

\medskip
Given an indecomposable $H_2$\,-$H_1$ biset $X$,
a choice of base-point $x\in X$ yields an isotropy
subgroup $S_x\subset H_2\times H^{\rm op}_1$
and a preferred biset isomorphism
\eqncount
\begin{equation}
\standardformtoX{x}\colon (H_2\times H^{\rm op}_1)/S_x \xrightarrow{\cong} X
\end{equation}
defined by
$\standardformtoX{x}\bigl(\standardformpoint {h_2}{ h_1}\bigr) = h_2 x h_1$.

\begin{definition}[The standard representation]\namedLabel{the standard representation}
Let $X$ be an indecomposable bifree $H_2$\,-$H_1$ biset, and $x \in X$ a base-point.
The \emph{standard representation for $X$ at $x$} is the data
\newNumber[Formula]{standard form}
$$\standardformtoX{x}\colon \biset{H_2}{[L_x, \defininghomo{X}{x}, K_x]}{{H_1}} \xrightarrow{\cong} X
\leqno(\ref{standard form})$$
given by the preferred biset isomorphism
$\standardformtoX{x}\bigl(\standardformpoint {h_2}{ h_1}\bigr) = h_2 x h_1$, where
$L_x\subset H_2$ is  the image
of the isotropy subgroup $S_x$ under $\iota_1$,  $K_x\subset H_1$
is the image of $\iota_2$, and
$$\defininghomo{X}{x}\colon L_x  \xrightarrow{\ \iota_1^{-1}\ } S_x \xrightarrow{\ \iota_2\ } K_x$$
is an isomorphism.
We noted above that $L_x$, $K_x$ and $\defininghomo{X}{x}$ determine $S_x$
as a (graph) subgroup of $ H_2\times H^{\rm op}_1$. \qed
\end{definition}
\begin{remark}
Any model coset space
$\biset{H_2}{[L, \gamma, K]}{{H_1}}$ is the standard representation for
some indecomposable bifree biset.
We let $X = (H_2\times H^{\rm op}_1)/S$, and choose $x\in X$
as the coset of the identity,
then $L_x = L$, $K_x = K$ and $\defininghomo{X}{x} = \gamma$. \qed
\end{remark}

\begin{remark}\label{define_parts}
Here is a second description of the standard representation
(compare \cite{bouc-thevenaz2}*{5.1}).
We present it so as to identify the component of any point $x$
in an $H_2$\,-$H_1$ bifree biset $X$, indecomposable or not.
Let $L_x = \{ h_2 \in H_2\ \vert\ h_2 x \in x H_1\ \}$.
Check $L_x$ is a subgroup of $H_2$.
Since the $H_1$ action is free, there exists a unique $h_1\in H_1$
such that $h_2 x = x h_1$.
Define a function $f \colon L_x \to H_1$ by $f(h_2) = h_1$.
Let $K_x$ denote the image of $f$ and let
$\defininghomo{X}{x} \colon L_x \to K_x$ denote the restriction of $f$.
Check that $f$ is a homomorphism and hence $\defininghomo{X}{x}$
is an isomorphism.
Finally check that
$\standardformtoX{x}\colon \biset{H_2}{[L_x, \defininghomo{X}{x}, K_x]}{{H_1}} \to X$
defined by \namedRef{standard form} above
is an injection of $H_2$\,-$H_1$ bisets which is
then automatically onto the component of $X$ containing $x$. \qed
\end{remark}
Hereafter we will call $\biset{H_2}{[L_x, \defininghomo{X}{x},K_x]}{H_1}$ the
\emph{standard representation of $X$ at $x$}
even if $X$ is not indecomposable.
It follows that
$$\mathop{\disjointunion}_{x\in H_2\backslash X/H_1} \
\biset{H_2}{[L_x, \defininghomo{X}{x}, K_x]}{{H_1}}
\xrightarrow{\quad\disjointunion \standardformtoX{x} \quad} X$$
is a bijection of $H_2$\,-$H_1$ bisets, where we take one $x$
in each indecomposable component of $X$.
Here are two useful properties of the standard representation for the reader to verify.

\medskip
We note the effect of the involution $\congtran\colon\bA(G) \to \bA(G)$
on the standard representations.

\begin{lemma}\namedLabel{standard representation for conjugate}
If $[L_x,\defininghomo{X}{x}, K_x]$ is
the standard representation for $X$ at $x$, then
$[K_x, \superdefininghomo{X}{x}{-1},L_x]$  is
the  standard representation for $\congtran(X)$ at $x$.
\end{lemma}

Next we consider a change of base-point.
If $y$ is another point in the same indecomposable component as $x$,
then $y = h_2 x h_1$ for some choice of
$h_2\in H_2$ and $h_1\in H_1$.
For $g\in G$ and any subgroup $K\subset G$, let $K^g = g^{-1} K g$.
If $\gamma\colon L\to K$
is a homomorphism between subgroups of $G$ and if $g_1$, $g_2\in G$, define
$\leftsub{g_2}{\gamma}^{g_1}(h) = g^{-1}_1 \gamma( g_2^{-1} h g_2) g_1
\colon L^{g_2^{-1}} \to K^{g_1}$.
\begin{lemma}\namedLabel{base point change}
With notation as above,
the standard representation for $X$ at $y = h_2 x h_1 \in X$ is
$\bigl[L^{h_2^{-1}}_x,  \leftsub{h_2}{\superdefininghomo{X}{x}{h_1}}  ,K_x^{h_1}\bigr]$.
In other words, $L_y = L^{h_2^{-1}}_x$, $K_y = K_x^{h_1}$ and
$\defininghomo{X}{y} =  \leftsub{h_2}{\superdefininghomo{X}{x}{h_1}}$.
\end{lemma}

\medskip
It will be useful later to be able to identify $X/H_1$ and $H_2\backslash X$.

\begin{lemma} \namedLabel{action orbits}
If $X$ is an indecomposable, bifree $H_2$\,-$H_1$ biset, then
the bijection of bisets
$\standardformtoX{x}\colon \biset{H_2}{[L_x, \defininghomo{X}{x}, K_x]}{{H_1}} \to X$
induces
\begin{enumerate}
\item a bijection of left $H_2$-sets $H_2/ L_x \to X/H_1$
\item a bijection of right $H_1$-sets
$K_x\backslash H_1 \to H_2\backslash X$.
\end{enumerate}
\end{lemma}
\begin{proof}
The proof is immediate.
\end{proof}

\section{Composition of bisets}\label{five}
Fix indecomposable, bifree bisets $\biset{H_2}{(X_1)}{H_1}$ and
$\biset{H_3}{(X_2)}{H_2}$ and let us analyze
$X_3 = X_2\times_{H_2} X_1$.
In general $X_3$ will not be indecomposable, and the Mackey double coset
formula enters the picture.
We begin by analyzing the standard representation at a point.
\begin{lemma}\namedLabel{balanced product components}
The standard representation for $X_3=X_2\times_{_{H_2}} X_1$ at
$\quotientpoint{x_2}{x_1}$ is given by
\begin{itemize}\namedLabel{standard representation of a product}
\item $L_{\quotientpoint{x_2}{x_1}} =
\superdefininghomo{X_2}{x_2}{-1}\bigl(K_{x_2} \cap L_{x_1}\bigr)$,
\item $K_{\quotientpoint{x_2}{x_1}}=
\defininghomo{X_3}{\quotientpoint{x_2}{x_1}}(L_{\quotientpoint{x_2}{x_1}}) =
\defininghomo{X_1}{x_1}\bigl(K_{x_2} \cap L_{x_1}\bigr)$
\item $\defininghomo{X_3}{\quotientpoint{x_2}{x_1}} =
\defininghomo{X_1}{x_1} \circ \defininghomo{X_2}{x_2}$.
\end{itemize}
\end{lemma}
\begin{proof}
Since the projection $X_2\times X_1 \to X_2\times_{H_2} X_1$ is onto,
every point in $X_2\times_{H_2} X_1$ is the image of a point in $X_2 \times X_1$.
Given $x_1\in X_1$ and $x_2\in X_2$ write $\quotientpoint{x_2}{x_1}
\in X_3 = X_2\times_{H_2} X_1$ for the image of $x_2\times x_1$.
In this notation, the $H_3$\,-$H_1$ structure is the evident one:
$h_3 \quotientpoint{x_2}{x_1} h_1 = \quotientpoint{h_3 x_2}{x_1 h_1}$.

The point $\quotientpoint{x_2}{x_1}\in X_3$ is a set of points
$\{x_2 h \times h^{-1} x_1\} \subset X_2 \times X_1$
for all $h \in H_2$.
If $h_3 (x_2 \times x_1) h_1 = x_2 h \times h^{-1} x_1 \in X_2 \times X_1$
for some $h \in H_2$ then $h_3 x_2 = x_2 h$ and $h^{-1} x_1 = x_1 h_1$.
Now if $h_3 x_2 = x_2 h$ then $h = \defininghomo{X_2}{x_2}(h_3)$
and $h_3 \in L_{x_2}$.
If $h^{-1} x_1 = x_1 h_1$ then $h_1 = \defininghomo{X_1}{x_1}(h^{-1})$
and $h^{-1} \in L_{x_1}$.
Since $L_{x_1}$ is a group, $h \in L_{x_1}$.
Since $h = \defininghomo{X_2}{x_2}(h_3)$, $h \in K_{x_2}$.
Therefore,
$h_3 \in \superdefininghomo{X_2}{x_2}{-1}\bigl(K_{x_2} \cap L_{x_1}\bigr)$.
Let $L_{\quotientpoint{x_2}{x_1}} =
\superdefininghomo{X_2}{x_2}{-1}\bigl(K_{x_2} \cap L_{x_1}\bigr)$
and define
$\defininghomo{X_3}{\quotientpoint{x_2}{x_1}} =
\defininghomo{X_1}{x_1} \circ \defininghomo{X_2}{x_2}$.
What we have seen so far is that if
$h_3\ \quotientpoint{x_2}{x_1}\ h_1 =
\quotientpoint{x_2}{x_1}$ in $X_3$, then $h_3 \in L_{\quotientpoint{x_2}{x_1}}$ and
$h_1 = \defininghomo{X_3}{\quotientpoint{x_2}{x_1}}(h_3)$.

If $h_3 \in L_{\quotientpoint{x_2}{x_1}}$, then
$h_3 (x_2 \times x_1)\defininghomo{X_3}{\quotientpoint{x_2}{x_1}}(h_3) =
x_2 \defininghomo{X_2}{x_2}(h_3) \times \defininghomo{X_2}{x_2}(h_3)^{-1} x_1$
and it further follows that for any $h_2 \in H_2$,
$h_3 (x_2 h_2 \times h^{-1}_2 x_1)\defininghomo{X_3}{\quotientpoint{x_2}{x_1}}(h_3) =
x_2 \defininghomo{X_2}{x_2}(h_3)h_2 \times h^{-1}_2\defininghomo{X_2}{x_2}(h_3)^{-1} x_1 =
x_2 \bigl(\defininghomo{X_2}{x_2}(h_3)h_2\bigr) \times
\bigl(\defininghomo{X_2}{x_2}(h_3) h_2\bigr)^{-1} x_1$.
In other words, if $L_{\quotientpoint{x_2}{x_1}}$ and
$\defininghomo{X_3}{\quotientpoint{x_2}{x_1}}$ are defined as above,
then for any $h_3 \in L_{\quotientpoint{x_2}{x_1}}$,
$h_3\ \quotientpoint{x_2}{x_1}\ \defininghomo{X_3}{\quotientpoint{x_2}{x_1}}(h_3) =
\quotientpoint{x_2}{x_1}$.
If we let $K_{\quotientpoint{x_2}{x_1}} =
\defininghomo{X_3}{\quotientpoint{x_2}{x_1}}(L_{\quotientpoint{x_2}{x_1}})$,
then the proof is complete.
\end{proof}
\begin{lemma}\namedLabel{parameter}
The double coset space $K_{x_2}\backslash H_2/L_{x_1}$ parametrizes
the indecomposable components of $X_2\times_{_{H_2}} X_1$, where $x_1\in X_1$ and $x_2\in X_2$ are base points.
\end{lemma}
\begin{proof}
To simplify the notation  let $X_2 =
\biset{H_2}{[L_3, \gamma_2, K_2]}{H_1}$, and $X_1 =
\biset{H_2}{[L_2, \gamma_1, K_1]}{H_1}$.
The set of indecomposable components is given by
$$H_3\backslash (X_2\times_{H_2} X_1)/H_1 =
(H_3\backslash X_2) \times_{H_2}(X_1/H_1).$$
By \namedRef{action orbits},
$(H_3\backslash X_2) \times_{H_2}(X_1/H_1) =
(K_2\backslash H_2) \times_{H_2} (H_2/L_2) =
K_2\backslash H_2/L_2$.
If $\{ h_2\}\in K_2\backslash H_2/L_2$ is a set of double coset representatives with $h_2 \in H_2$,
then $\quotientpoint{x_2 h_2}{x_1}$ gives a set of points, one in each
indecomposable component.
\end{proof}

Next we give an explicit formula for $X_3=X_2\times_{_{H_2}} X_1$ as a disjoint union of indecomposable bisets (compare \cite{bouc0}*{3.2}). We define a map
 $$\mackeyComponent{h_2}
 \colon \biset{H_3}{[L_{x_3},\defininghomo{X_3}{x_3},K_{x_3}]}{H_1}
 \to
\biset{H_3}{[L_{x_2}, \defininghomo{X_2}{x_2}, K_{x_2}]}{H_2} \times_{_{\scriptstyle H_2}}
\biset{H_2}{[L_{x_1}, \defininghomo{X_1}{x_1}, K_{x_1}]}{H_1}$$
by the formula
$\mackeyComponent{h_2}(\standardformpoint{h_3}{h_1})
 =
\quotientpoint{\standardformpoint{h_3}{h_2}}{\standardformpoint{e}{h_1}}$, using the notation of \namedRef{the standard representation}.

\begin{theorem}[Mackey double coset formula]\namedLabel{The Mackey double coset formula}
The  $H_3$-$H_1$ biset $X_3=X_2\times_{_{H_2}} X_1$  is  given by the disjoint union  of the left-hand vertical maps
in the diagram

$$\xymatrix{\left [\superdefininghomo{X_2}{x_2}{-1}
\bigl(K_{x_2}\cap L_{x_1}^{h_2^{-1}}\bigr),

\defininghomo{X_1}{x_1} \hskip-6pt\circ c_{h_2}\circ \defininghomo{X_2}{x_2},

\defininghomo{X_1}{x_1}\bigl(K^{h_2}_{x_2} \cap L_{x_1}\bigr)\right]

\ar@{=}[r]\ar[d]^{\mackeyComponent{h_2}}&

\biset{H_3}{[L_{x_3},
\defininghomo{X_3}{x_3},
K_{x_3}]}{H_1}
\ar[d]^{\standardformtoX{x_3}}
\\

\biset{H_3}{[L_{x_2}, \defininghomo{X_2}{x_2}, K_{x_2}]}{H_2} \times_{_{\scriptstyle H_2}}
\biset{H_2}{[L_{x_1}, \defininghomo{X_1}{x_1}, K_{x_1}]}{H_1}
\ar[r]^(0.7){\standardformtoX{x_2}\times\standardformtoX{x_1}}_(0.7){\approx}&

X_2\times_{_{H_2}} X_1
}
$$
 over the base-points $x_3 = \quotientpoint{x_2 h_2}{x_1}$, for $h_2 \in K_{x_2}\backslash H_2/L_{x_1}$.
 \end{theorem}
\begin{proof}
We first derive the explicit formula displayed in the top left corner of the diagram.
Let $[L_{x_2},\defininghomo{X_2}{x_2},K_{x_2}]$ be the standard representation for
$X_2$ at $x_2$, and let $[L_{x_1},\defininghomo{X_1}{x_1},K_{x_1}]$ be
the standard representation for $X_1$ at $x_1$.
Then the standard representation for $X_2$ at $x_2 h_2$ is given by
\namedRef{base point change}:
$[L_{x_2},\defininghomo{X_2}{x_2 h_2},K_{x_2}] =
[L_{x_2},\superdefininghomo{X_2}{x_2}{h_2},K_{x_2}^{h_2}]$.
It further follows that the standard representation for $X_3=X_2\times_{_{H_2}} X_1$
at $x_3=\quotientpoint{x_2 h_2}{x_1}$ is given by
$$[L_{x_3}, \defininghomo{X_3}{x_3}, K_{x_3}]=[L_{[x_2 h_2,x_1]}, \defininghomo{X_3}{[x_2 h_2, x_1]}, K_{[x_2 h_2,x_1]}]~.$$
By \namedType{parameter}s \ref{base point change}, \ref{balanced product components} and \ref{parameter}
\begin{itemize}
\item $L_{[x_2 h_2,x_1]} =
\superdefininghomo{X_2}{x_2 h_2}{-1}\bigl(K_{x_2 h_2} \cap L_{x_1}\bigr)
=
(\superdefininghomo{X_2}{x_2}{h_2})^{-1}\bigl(
K^{h_2}_{x_2} \cap L_{x_1}\bigr)=\superdefininghomo{X_2}{x_2}{-1}
\bigl(K_{x_2}\cap L_{x_1}^{h_2^{-1}}\bigr)$\vskip 6pt
\item $K_{[x_2 h_2,x_1]} =
\defininghomo{X_1}{x_1}\bigl(K_{x_2 h_2} \cap L_{x_1}\bigr) =
\defininghomo{X_1}{x_1}\bigl(K^{h_2}_{x_2} \cap L_{x_1}\bigr)$
\vskip 6pt
\item $\defininghomo{X_3}{[x_2 h_2, x_1]} =
\defininghomo{X_1}{x_1} \hskip-6pt\circ \defininghomo{X_2}{x_2 h_2}  =
\defininghomo{X_1}{x_1}\hskip -6pt \circ \superdefininghomo{X_2}{x_2}{h_2} =
\defininghomo{X_1}{x_1} \hskip -6pt \circ c_{h_2}\circ \defininghomo{X_2}{x_2}
$
\vskip 6pt
\end{itemize}

For $h_2\in H_2$ the function
defined by $\mackeyComponent{h_2}(\standardformpoint{h_3}{h_1}) =
\quotientpoint{\standardformpoint{h_3}{h_2}}{\standardformpoint{e}{h_1}}$
makes the following diagram commute
$$\xymatrix{
\biset{H_3}{[L_{\quotientpoint{x_2 h_2}{x_1}},
\defininghomo{X_3}{\quotientpoint{x_2 h_2}{x_1}},
K_{\quotientpoint{x_2 h_2}{x_1}}]}{H_1}
\ar[rd]_{\standardformtoX{\quotientpoint{x_2 h_2}{x_1}}}
\ar[r]^-{\mackeyComponent{h_2}}
&\biset{H_3}{[L_{x_2}, \defininghomo{X_2}{x_2}, K_{x_2}]}{H_2}
\times_{_{\scriptstyle H_2}}
\biset{H_2}{[L_{x_1}, \defininghomo{X_1}{x_1}, K_{x_1}]}{H_1}
\ar[d]^{\standardformtoX{x_2}\times\standardformtoX{x_1}}\\
&{_{_{H_3}}}\bigl(X_2\times_{_{H_2}} X_1\bigr){_{_{H_1}}}\\
}$$

Since the $\Psi$ maps are injective, it suffices to prove that the formula given for
$\mackeyComponent{h_2}$ makes the diagram commute.
Over and then down takes $\standardformpoint{h_3}{h_1}$ to
$\quotientpoint{\standardformpoint{h_3}{h_2}}{\standardformpoint{e}{h_1}}$ and then to
$\quotientpoint{h_3 x_2 h_2}{x_1 h_1}$ whereas
$\standardformtoX{\quotientpoint{x_2 h_2}{x_1}}(\standardformpoint{h_3}{h_1}) =
h_3 \quotientpoint{x_2 h_2}{x_1} h_1 = \quotientpoint{h_3 x_2 h_2}{x_1 h_1}$.

The conclusion now follows immediately from the commutativity of the displayed diagram, since the disjoint union of the right-hand vertical maps  $\standardformtoX{x_3}$ is a bijection, and the map
$\standardformtoX{x_2}\times\standardformtoX{x_1}$ is a bijection (see \namedRef{the standard representation}).

\end{proof}
\begin{remark}
The map $\defininghomo{X_3}{\quotientpoint{x_2 h_2}{x_1}}$ can be displayed as
the composition on the top row of the diagram
$$\xymatrix{
L_{\quotientpoint{x_2 h_2}{x_1}}\ar[d]
\ar[rr]^{\defininghomo{X_2}{x_2}\vrule width .5pt depth 6pt height 8pt }&
&K_{x_2}\cap L^{h_2^{-1}}_{x_1}\ar[r]^-{c_{h_2} }\ar[d]
&K^{h_2}_{x_2}\cap L_{x_1}\ar[d] \ar[rrr]^{\quad \defininghomo{X_1}{x_1}
\vrule width .5pt depth 6pt height 8pt  \quad}\ar[dr]
&&& K_{\quotientpoint{x_2 h_2}{x_1}}\ar[d]\\
L_{x_2}\ar[rr]^{\defininghomo{X_2}{x_2}} &&
K_{x_2}\ar[r]^{c_{h_2}}& K^{h_2}_{x_2}
&L_{x_1}\ar[rr]^{\quad\defininghomo{X_1}{x_1}\quad}&&K_{x_1}\\
}
$$
where $c_g(h) = g^{-1} h g$ is conjugation.
\end{remark}

\section{Conjugation bisets and $\subBurnside(G)$}\label{six}
We will now define a subcategory $\subBurnside(G) \subset \Burnside(G)$,
with the same objects (the subgroups of $G$),
but with morphisms  restricted to bifree conjugation bisets.
As before, we perform the Grothendieck construction on the isomorphism
classes of these bisets to get an $\Ab$-category.
The universal construction
$u\colon \subBurnside(G) \to \subBurnsideDot(G)$
then produces an additive subcategory (with involution) of $\bAdot(G)$.
\begin{definition}[Conjugation bisets]\label{conjugationbisets}
An indecomposable \emph{conjugation biset} is
an indecomposable bifree biset $\biset{H_2}{X}{H_1}$
so that the isomorphism in the standard representation of $X$ at $x\in X$,
$\defininghomo{X}{x} = c_g$, for some $g\in G$ such that $g^{-1}L_xg= K_x$.
A \emph{conjugation biset} is a bifree biset,
each of whose indecomposable subsets satisfies this condition.
\end{definition}

\begin{remark}\namedLabel{non-uniqueness of conjugating element}
The choice of $g$ such that $\defininghomo{X}{x} = c_g$ is not unique,
but the conjugation biset only depends on $\defininghomo{X}{x}$
as a \emph{homomorphism}.
This is the basic reason that functors out of $\subBurnside(G)$ are conjugation invariant.
By definition, $c_g(h_2) = g^{-1}h_2g\in K_x$, for all $h_2 \in L_x \subset H_2$,
where $g^{-1}L_xg=K_x\subset H_1$.
We have  $c_{g} = c_{g_1}\colon L_x \to K_x$ if and only if
$g_1g^{-1} \in C_G(L_x)$, where $C_G(L_x)$ denotes the centralizer of $L_x$ in $G$.
\end{remark}

The formula for the change of base-point in \namedRef{base point change}
shows that the definition of conjugation biset does not depend on the choice of point
$x$ used to compute
$\defininghomo{X}{x}$: the element giving the conjugation may change,
but not the fact that it is given by some conjugation.
In particular, any biset which is isomorphic to
a conjugation biset is itself a conjugation biset.

By \namedRef{The Mackey double coset formula}, the composition of two conjugation bisets
is again a conjugation biset.
In addition, by \namedRef{standard representation for conjugate},
the involution $\congtran$ restricts to give an involution
$\congtran\colon \subBurnside(G) \to \subBurnside(G)$.

\subsection{Induction and restriction}
There are two extreme cases of indecomposable bifree bisets
$\biset{H_2}{X}{H_1}$.
\begin{enumerate}
\item We say $\biset{H_2}{X}{H_1}$ is a \emph{restriction} if the $H_1$ action is
transitive.
\item We say $\biset{H_2}{X}{H_1}$ is an \emph{induction} if the $H_2$ action is
transitive.
\end{enumerate}
We say that $\biset{H_2}{X}{H_1}$ is an \emph{isomorphism} if it is both
a restriction and an induction.

\begin{proposition}
Let  $\biset{H_2}{X}{H_1}$ be an indecomposable, bifree biset.
Pick a point $x\in X$ and let $[L_x, \defininghomo{X}{x}, K_x]$ be
the standard representation.
\begin{enumerate}
\item $\biset{H_2}{X}{H_1}$ is a restriction
if and only if $K_x = H_1$.
\item $\biset{H_2}{X}{H_1}$ is an induction if and only if $L_x = H_2$.
\end{enumerate}
\end{proposition}
\begin{proof}
Immediate from \namedRef{action orbits}.
\end{proof}

\begin{proposition}\label{conjugation_isomorphism}
If $\biset{H_2}{X}{H_1}$ is an isomorphism,
$\congtran(\biset{H_2}{X}{H_1})$ is the inverse isomorphism.
Conversely, if  $\biset{H_2}{X}{H_1}$ has an inverse then it is
an isomorphism.
This justifies the terminology.
\end{proposition}
\begin{proof}
Let $\biset{H_1}{Y}{H_2}$ be the inverse for  $\biset{H_2}{X}{H_1}$.
Then  $\biset{H_2}{X}{H_1} \times_{_{H_1}}  \biset{H_1}{Y}{H_2}$ is
isomorphic to $\biset{H_2}{H_2}{H_2}$.
Since $( \biset{H_2}{H_2}{H_2})/H_2$ is a point, so is
$\bigl(\biset{H_2}{X}{H_1} \times_{_{H_1}}  \biset{H_1}{Y}{H_2}\bigr)/H_2 =
(X_1/H_1) \times (Y/H_2)$.
It follows that $Y$ is a restriction.
The other equation for the inverse shows that $Y$ is also an induction.

If $\biset{H_2}{X}{H_1}$ is both an induction and a restriction, it follows that the
standard representation is
$H_2$, $H_1$ and some isomorphism $\defininghomo{X}{x}$.
By \namedRef{standard representation for conjugate} the standard representation for
$\congtran(X)$ is $[H_1, \superdefininghomo{X}{x}{-1}, H_2]$.
By \namedRef{The Mackey double coset formula}, $\congtran(X)$ is the inverse for $X$.
\end{proof}

It further follows from \namedRef{The Mackey double coset formula} that the composition of
two restrictions is a restriction and the composition of two inductions is an induction.
More explicitly, we have the following.

\begin{proposition}
Let $[L_{x_1}, \defininghomo{X_1}{x_1}, K_{x_1}]$ be the standard representation for the
biset $\biset{H_2}{X_1}{H_1}$ at $x_1$ and let
$[L_{x_2}, \defininghomo{X_2}{x_2}, K_{x_2}]$ be the standard representation for the
biset $\biset{H_3}{X_2}{H_2}$ at $x_2$.
\begin{enumerate}
\item If $X_1$ and $X_2$ are both restrictions then so is $X_2\times_{_{H_2}} X_1$.
\item If $X_1$ and $X_2$ are both inductions then so is $X_2\times_{_{H_2}} X_1$.
\item If $X_1$ is an induction and if $X_2$ is a restriction then
$[L_{x_2},\defininghomo{X_1}{x_1} \circ \defininghomo{X_2}{x_2},K_{x_1}]$  is the
standard representation for the composition using the point $\quotientpoint{x_2}{x_1}$
\end{enumerate}
\end{proposition}
\begin{proof}
This follows from \namedRef{standard representation of a product} and some
standard set theory.
\end{proof}

\begin{definition}
We define three subcategories of $\Burnside(G)$, denoted $\Res_{\bA}(G)$,
$\Ind_{\bA}(G)$ and $\Iso_{\bA}(G)$.
The objects of any of these categories are all the objects of $\Burnside(G)$.
The morphisms in $\Res_{\bA}(G)$ are the set of all restrictions, the morphisms
in $\Ind_{\bA}(G)$ are the set of all inductions and the morphisms in
$\Iso_{\bA}(G)$ are the isomorphisms. \qed
\end{definition}

\begin{remark}
Because $\Hom_{\Burnside(G)}(H_1, H_2)$ is the free abelian group on the
indecomposable bifree $H_2$\,-$H_1$ bisets,
$\Hom_{\Res(G)}(H_1, H_2) \subset \Hom_{\Burnside(G)}(H_1, H_2)$,
and indeed it is a summand.
There is a similar statement for each of $\Ind(G)$ and $\Iso(G)$.
Moreover, $\Iso(G) = \Res(G) \cap \Ind(G)$.
\end{remark}

Let $\biset{H_2}{X}{H_1}$ be an indecomposable bifree biset.
After choosing a point $x\in X$, we can display $X$ as a composition.

\begin{definition}\label{standardmaps}
Let $[L_x, \defininghomo{X}{x}, K_x]$ be the standard representation for $X$ at $x$.
We may regard  $H_2$ is an $H_2$\,-$L_x$ biset, and  $H_1$ as a
$K_x$\,-$H_1$ biset, via group multiplication in $G$.
\begin{enumerate}
\item We define
$\standardInd{x} = \biset{H_2}{(H_2)}{L_x}$, note that it is an induction and
call it the \emph{standard induction for $X$ at $x$}.
\item We define
$\standardRes{x} = \biset{K_x}{(H_1)}{H_1}$, note that it is a restriction and
call it the \emph{standard restriction for $X$ at $x$}.
\item
There are two evident isomorphisms.
Make $K_x$ into a left $L_x$ set using $\defininghomo{X}{x}$,
and into a right $K_x$ set via group multiplication.
This makes $K_x$ into an $L_x$\,-$K_x$ biset.
Similarly make $L_x$ into an $L_x$\,-$K_x$ biset using
$\superdefininghomo{X}{x}{-1}$.
Let  $\biset{L_x}{(K_x)}{K_x}$ be denoted $\standardLIso{x}$ and let
$\biset{L_x}{(L_x)}{K_x}$ be denoted $\standardRIso{x}$.
\end{enumerate}
\end{definition}
Note that $\standardRIso{x}$ and $\standardLIso{x}$ are isomorphic
as bisets via the bijection $\defininghomo{X}{x}\colon L_x \to K_x$. The following composition formula was also observed in \cite{bouc0}*{Lemme 3} and \cite{bouc-thevenaz1}*{7.4}.

\begin{proposition}\label{factorize}
As $H_2$\,-$H_1$ bisets, an indecomposable bifree biset $X$ is isomorphic
to the composition
$\standardInd{x} \circ\,  \standardRIso{x} \circ \standardRes{x}$.
\begin{enumerate}
\item The standard restriction $\standardRes{x}$ is always a conjugation restriction.
\item The standard induction $\standardInd{x}$ is always a conjugation induction.
\item
The indecomposable bifree biset $X$ is a conjugation biset if and only if
$\standardRIso{x}$ is a conjugation isomorphism.
\item
Moreover, $X$ is a restriction if and only if $\standardInd{x}$ is the identity;
$X$ is an induction if and only if $\standardRes{x}$ is the identity.
\end{enumerate}
\end{proposition}
\begin{proof}
The standard representation for $\standardInd{x}$ is
$\biset{H_2}{[L_x, id, L_x]}{L_x}$,  so
it is clearly a conjugation biset.
The case  $\standardRes{x}$ is similar.
The standard representation for $\standardLIso{x}$ is
$\biset{L_x}{[L_x, \defininghomo{X}{x}, K_x]}{K_x}$,
and so $\standardLIso{x}$ is a conjugation biset if and only if $X$ is.

The function
$$\standardInd{x} \times_{L_x}\,  \standardRIso{x} \times_{K_x} \standardRes{x} \to X$$
which sends the image of $h_2\times \ell \times h_1$ to $h_2 \ell x h_1$ can be checked
to be a bijection of bisets.
The remarks about $X$ being a restriction or an induction are immediate.
\end{proof}

\begin{remark}\label{Vyx}
Of course the choice of $x\in X$ is not unique, so let $y = h_2 x h_1$.
Let $\standardRes{y}$, $\standardInd{y}$ and $\standardLIso{y}$ be the
corresponding bisets.
Define an $L_{y}$-${L_x}$ biset $ \changepointbiset{y}{x}$ to be $L_x$
with right multiplication by $L_x$ as the right action and use
$c_{h_2}\colon L_y \to L_x$ to define the left action.
Note that $ \changepointbiset{y}{x}$ is an isomorphism and a conjugation biset.
By \namedRef{base point change}, we have
$\standardInd{x}\cong\standardInd{y}\times_{_{L_y}}  \changepointbiset{y}{x}$.

Similarly, we define a $K_{y}$-$K_x$ biset $\changepointbisetK{y}{x}$
to be $K_x$ with right multiplication by $K_x$ and left multiplication
defined by $c_{h_1}\colon K_y\to K_x$.
We have
$\standardRes{x}\cong \changepointbisetK{y}{x}^{-1} \times_{_{K_y}} \standardRes{y}$
and
$\standardRIso{x}\cong \changepointbiset{y}{x}^{-1}  \times_{_{L_y}} \standardRIso{y}
\times_{_{K_y}}
\changepointbisetK{y}{x}$.
\end{remark}

\begin{definition}
We define three subcategories of $\subBurnsideDot(G)$, denoted
$\subInd(G)\subset \Ind_{\bA}(G)$, $\subRes(G)\subset \Res_{\bA}(G)$
and $\subIso(G)\subset\Iso_{\bA}(G)$.
These are the subcategories with the same objects as the bigger categories,
but whose morphisms
are all the morphisms which are conjugation bisets. \qed
\end{definition}

\section{Bifunctors into $\subBurnsideDot(G)$}\label{seven}
In this section we will define the bivariant functor
$\DtoAdot\colon \cD(G)\to \subBurnsideDot(G)$ used in the statement of Theorem A.
Let $\cDdot(G)$ denote the category whose objects are pairs $(X, \mathbf b)$,
consisting of a finite $G$-space $X$ and an \emph{ordered} collection
$\mathbf b = (b_1, \cdots, b_n)$ of base-points,
one for each $G$-orbit of $X$.
The morphisms are the $G$-maps (not necessarily base-point preserving).
There is a functor
$$\DdotD\colon \cDdot(G) \to \cD(G)$$
defined by forgetting the base-points.
Since every object of $\cD(G)$ is isomorphic to the image $\DdotD(X,\mathbf b)$ of an object of
$\cDdot(G)$, and $\DdotD$ induces a bijection on morphism sets, it follows that
$\DdotD$ gives an equivalence between the categories
$\cDdot(G)$ and $\cD(G)$, with inverse functor $\DtoDdot$
\cite{maclane2}*{IV.4, Theorem 1, p.~91}.
Moreover, the inverse functor $\DtoDdot$ can be chosen so that we have the additivity formula
$$\DtoDdot(X\smdisjointunion Y) = \DtoDdot(X) \smdisjointunion \DtoDdot(Y)$$
for any finite $G$-sets $X$ and $Y$.
This will be needed later to verify axiom (M2) for additive functors out of $\subBurnsideDot(G)$.

\medskip
We will now define the remaining functors in the following diagram:
\def\dup{3pt}\def\ddown{-3pt}
\eqncount
\begin{equation}
\vcenter{\xymatrix@C+4pt{
&\Or(G)\ar@<\dup>[r]^{\upi}\ar@<\ddown>[r]_{\downi}\ar[d]^{\OrtoDdot}  &\subBurnside(G)\ar[d]^{u}&\cr
\cD(G)\ar[r]^{\DtoDdot}&\cDdot(G)\ar@<\dup>[r]^{\upj}\ar@<\ddown>[r]_{\downj}&
\subBurnsideDot(G) \cr}}
\end{equation}

The functor $\OrtoDdot\colon \Or(G) \to \cDdot(G)$ sends $H$ to the $1$-tuple $(G/H, eH)$ and a
  $G$-set map $G/H \to G/K$ to the same $G$-set map.
In fact $\Or(G)$ as defined is isomorphic to a full subcategory of $\cDdot(G)$.

The functor $\downi$ is the identity on objects and sends a $G$-map $f\colon G/H \to G/K$
to the conjugation biset ${_{K}}K_{{g^{-1} H g}}$ where $f(e H) = g K$.
This is well-defined, since a different choice $g k$, for $k \in K$,
of representative yields an isomorphic biset.

Note that $1_{G/H}\colon G/H \to G/H$ goes to ${_{H}}H{_{H}}$ which is the identity.
Check that if
$f_1\colon G/H_1\to G/H_2$ and $f_2\colon G/H_2 \to G/H_3$ are $G$-maps and if
$f_1(e H_1) = g_1 H_2$ and $f_2(e H_2) = g_2 H_3$
then $f_2\circ f_1(e H_1) = (g_1 g_2) H_3$ and
$${_{H_3}}\bigl(H_3\bigr){_{g^{-1}_2 H_2 g_2}} \times_{H_2}
{_{H_2}}\bigl(H_2\bigr){_{g^{-1}_1 H_1 g_1}}$$
is isomorphic to ${_{H_3}}\bigl(H_3\bigr){_{(g_1 g_2)^{-1} H_1(g_1 g_2)}}$
by the map $(h_3, h_2) \mapsto h_3 g_2^{-1} h_2 g_2$.

The functor $\upi$ is also the identity on objects, but sends a
$G$-map
$f\colon G/H \to G/K$
to the conjugation biset ${_{g^{-1} H g}}K_{K}$ where $f(e H) = g K$.
Rather than check identity and composition, just note that
${_{g^{-1} H g}}K_{K}$ is isomorphic to
$\congtran\bigl({_{K}}K_{{g^{-1} H g}}\bigr)$ by the
function which sends $k$ to $k^{-1}$,
so $\upi = \congtran \circ \downi$ and hence $\upi$ is a contravariant functor.

We define the functor $\downj$ on objects by additivity: every object of $\cDdot(G)$ has the form
$$(X,\mathbf b)= \mathop{\disjointunion}_{i=1}^k \, (X_i,b_i),$$ where $(X_i,b_i) = G/H_i$
is an object of $\Or(G)$, and we send such an object to the ordered $n$-tuple
$(\downi(H_1), \dots, \downi(H_k))$. A morphism $\varphi\colon (X,\mathbf b) \to (Y,\mathbf c)$  in $\cDdot(G)$ is a collection $(\varphi_{i})$, $1\leqslant i \leqslant k$,  of $G$-maps of the form $\varphi_{i}\colon G/H_i \to G/K_{f(i)}$, where $f\colon \mathbf b\to \mathbf c$ is a function. Every morphism in $\subBurnsideDot(G)$ is represented a finite matrix of bifree conjugation bisets, where $\emptyset = 0$. We define $\downj(\varphi) = \alpha$ to be the morphism in $\subBurnsideDot(G)$ represented by the matrix $\alpha=  (\alpha_{ij})$, where $\alpha_{ij} = \downi(\varphi_{i})$, if $j = f(i)$, and $\alpha_{ij} =0$ otherwise.

The functor $\upj$ is defined in a similar way.
Notice that  $\downj$ factors through the subcategory $\subInd(G)$,
and $\upj$ factors through the subcategory $\subRes(G)$.

\begin{definition}\label{DtoBurnside}
We define the bivariant functor
$$\DtoAdot\colon \cD(G)\to \subBurnsideDot(G)$$
as the composition $\DtoAdot = (\upj, \downj)\circ \DtoDdot$. \qed
\end{definition}

\section{The proof of Theorem A}\label{eight}
We now state our main result, which is a more detailed version of Theorem A.

\begin{theorem}\label{Mackeyrecognition}
If $F\colon \subBurnsideDot(G) \to \Ab$ is an additive functor,
then $F\circ \DtoAdot\colon \cD(G) \to \Ab$ is
a conjugation invariant Mackey functor.
Conversely, any conjugation invariant Mackey functor factors
uniquely through an additive functor out of $\subBurnsideDot(G)$.
\end{theorem}
\begin{proof}
Suppose that $F\colon \subBurnsideDot(G)\to \Ab$ is an additive functor,
and let $\cM = F\circ \DtoAdot\colon \cD(G) \to \Ab$ denote its composition
with the bivariant functor $\DtoAdot$.
Then $\cM$ is conjugation-invariant.
The Mackey property (M2) is just additivity, so it remains to consider (M1).
Let
\eqncount
\begin{equation}\label{diagram:Mone}
{\vcenter
{\xymatrix
{S\ar[r]^\Psi\ar[d]_\Phi&S_2\ar[d]^\varphi\\S_1\ar[r]^\psi&T}}}
\end{equation}
be a pull-back diagram of finite $G$-sets.
We first remark that a
$G$-map $u \colon S \to S^\prime$ between finite $G$-sets is determined
by its restriction to the disjoint $G$-orbits in $S$.
In fact, each orbit in $S$ is mapped by $u$ into exactly one
$G$-orbit of $S^\prime$.
In particular, if $u \colon S \to S^\prime$ is an isomorphism
of finite $G$-sets, then by
Proposition \ref{conjugation_isomorphism} the induced maps
$u_\cM$ and $u^\cM$ are both isomorphisms with $ (u_\cM)^{-1} = u^\cM$.

This remark implies that property (M1) depends only on
the isomorphism class of diagram (\ref{diagram:Mone}).
More precisely, let $c\colon S_1 \xrightarrow{\approx} S_1^\prime$ and
$d\colon S_2  \xrightarrow{\approx}  S_2^\prime$ be $G$-isomorphisms,
and consider the commutative diagram of $G$-sets and $G$-maps:
$$\xymatrix{S^\prime\ar@/^1.5pc/[drrrrr]^l\ar@/_2pc/[dddr]_n\ar@/_0.5pc/@{-->}[dr]_v&&&&&\cr
&S\ar[rr]^\Psi\ar[d]^\Phi\ar@/_0.5pc/[ul]_u&&S_2\ar[rr]^d\ar[d]^\varphi&&S_2^\prime\ar[dll]^j\cr
&S_1\ar[rr]^\psi\ar[d]^c&&T&\cr
&S_1^\prime\ar[urr]_i&&}$$
where $S^\prime$ is the pullback of $i$ and $j$,
with $\psi = i \circ c$ and $\varphi = j \circ d$.
There is an induced isomorphism $u \colon S \to S^\prime$
with inverse $v$, such that  $l \circ u = d\circ \Psi$, and $c\circ \Phi = n \circ u$.
If we have the property (M1) for the pullback $(S^\prime, l, n)$,
then the property (M1) holds for the pullback $(S, \Psi, \Phi)$.
This is an easy diagram chase starting with the left-hand side of
the required formula
\eqncount
\begin{equation}\label{Mone}
\varphi^\cM\circ \psi_\cM = \Psi_\cM\circ \Phi^\cM
\end{equation}
and substituting in the expressions above for $\psi$ and $\varphi$.

By the additivity property (M2), this formula
follows from the basic cases where $S_1$, $S_2$, and $T$ are transitive
$G$-sets.
Let $S_1 = G/H_1$, $S_2=G/H_2$ and $T=G/K$,
and choose elements $g_1, g_2 \in G$ such that
$g_i^{-1}H_ig_i \subseteq K$, for $i = 1,2$.
Since $\cM$ is conjugation invariant,
we may assume that $\psi(eH_1) = g_1K$ and $\varphi(eH_2) = g_2K$
(varying the choice of $g_1$, $g_2$ will not affect the maps induced by $\cM$).

For $i=1$ and $i=2$ there are $G$-isomorphisms
$\theta_i\colon G/H_i \to G/H_i^\prime$,  given by
$eH_i \mapsto g_iH_i^\prime$ where
$H_i^\prime = g_i^{-1}H_i g_i$, with the property that  the pullback
$G/H_1^\prime \subset G/K \supset G/H_2^\prime$ is isomorphic to
$S = G/H_1 \times_{G/K} G/H_2$.
It follows (by the remarks above)
that it is enough to check formula
(\ref{Mone}) in the special case where $H_1$ and $H_2$
are actually subgroups of $K$.
Hence, we may assume that $g_1 = g_2 = e$.

We will regard
$\biset{H_2}{K}{H_1}$
as an $H_2$-$H_1$ biset via the natural actions $H_2\subset K$ and $H_1\subset K$.
The pullback $G$-set $$S = \bigl\{(x H_1, y H_2) \vv xK = y K\bigr\},$$ has the
$G$-action defined by
$g\cdot (x H_1, y H_2) = (g x H_1, g y H_2)$, for all $g\in G$.
It follows that in each $G$-orbit there is always a representative of the form
$(x H_1, e H_2)$ with $x\in K$ and
$(h_2 x h_1 H_1, e H_2) = (x H_1, e H_2)$ for all $h_2\in H_2$ and all $h_1\in H_1$.
In other words, the set of $G$-orbits in $S$ is in bijection with the quotient
$H_2\backslash K / H_1$ of the biset $\biset{H_2}{K}{H_1}$.

The isotropy subgroup
$$G_{(x H_1, e H_2)} = \bigl\{ g\in G \vv g x H_1 = x H_1,\  g H_2 = e H_2\bigr\} =
H_2 \cap H_1^{x^{-1}} $$
where $H_1^{x^{-1}} = xH_1x^{-1}$ and $x \in K$ as above.
Therefore, we have a bijection of $G$-sets
$$S = \bigsqcup_{x \in H_2\backslash K / H_1} G/ H_2 \cap H_1^{x^{-1}}\ .$$
In terms of biset morphisms, the composition
\eqncount
\begin{equation}\label{formula1}
\Psi_\cM\circ \Phi^\cM =
\sum F\bigl (\biset{H_2}{H_2}{H_2\cap H_1^{x^{-1}}}\bigr ) \circ\,
F\bigl (\biset{H_2\cap H_1^{x^{-1}}}{(H_2\cap H_1^{x^{-1}})}{H_1\cap H_2^{x}}\bigr )\circ
F\bigl (\biset{H_1\cap H_2^{x}}{H_1}{H_1}\bigr )
\end{equation}
where the middle biset is just the conjugation $G$-isomorphism
$c_{x^{-1}}\colon H_1\cap H_2^{x} \to H_2\cap H_1^{x^{-1}}$.

On the other hand, the composition
$$\varphi^\cM\circ \psi_\cM = F(\biset{H_2}{K}{H_1})\ .$$
To establish formula (\ref{Mone}),
we consider the $H_2$\,-$H_1$ biset bijection
$$\biset{H_2}{K}{H_1} = \bigsqcup_{x \in H_2\backslash K / H_1} H_2\,x\,H_1 $$
from the double coset decompostion.
The standard representation of the component
$X = H_2\,x\,H_1$ expressed in the standard form is
$$ H_2\,x\,H_1 = \biset{H_2}{[L_x, \defininghomo{X}{x}, K_x]}{H_1} =
[H_2\cap H_1^{x^{-1}},  c_{x},  H_1\cap H_2^{x}]~.$$
This follows from Remark \ref{define_parts}:
$L_x = \bigl\{h_2\in H_2\vv h_2x = xh_1, \ h_1 \in H_1\bigr\} = H_2 \cap H_1^{x^{-1}}$,
$K_x = \bigl\{h_1 \in H_1\vv xh_1 = h_2x, \ h_2 \in H_2 \bigr\} =  H_1 \cap H_2^{x}$
and
$\defininghomo{X}{x} = c_x$.

Now by Proposition \ref{factorize}  we see that
the composition
$$\varphi^\cM\circ \psi_\cM = F(\biset{H_2}{K}{H_1})=
\sum F(\standardInd{x}) \circ\,  F(\standardRIso{x})\circ F(\standardRes{x})\ .$$
But, by inspection, the right-hand side of this formula is just the expression for
$ \Psi_\cM\circ \Phi^\cM$ in formula (\ref{formula1}).
This proves the property (M1).

\smallskip
Conversely, suppose that $\cM\colon \cD(G) \to \Ab$ is
a conjugation-invariant Mackey functor.
We define the functor $F$ on objects by setting $F(H) = \cM(G/H)$,
and extending additively.
Since the bivariant functor $\DtoAdot\colon \cD(G) \to \subBurnsideDot(G)$
is surjective on objects, this formula defines $F$ uniquely on objects.

The morphisms in $\subBurnsideDot(G)$
are finite matrices of bifree conjugation $H_2$\,-$H_1$ bisets.
Any such biset is a disjoint union of indecomposable conjugation bisets $X$,
uniquely up to ordering, and after picking a base point $x\in X$
we have the standard representation
$X \cong \biset{H_2}{[L_x, \defininghomo{X}{x}, K_x]}{H_1}$
and the factorization
$$\biset{H_2}{X}{H_1}\cong\biset{H_2}{[L_x, \defininghomo{X}{x}, K_x]}{H_1} \cong \standardInd{x} \circ\,  \standardRIso{x} \circ \standardRes{x}$$
of Proposition \ref{factorize}.
By Definition \ref{standardmaps}, it follows that the morphisms in
$\subBurnsideDot(G)$ are generated by compositions of
the following three types of bifree conjugation bisets:
$$ \text{(i)\ } \standardInd{x} = \biset{H_2}{(H_2)}{L_x}, \quad
\text{(ii)\ } \standardRes{x} = \biset{K_x}{(H_1)}{H_1}, \quad \text{and}\quad
\text{(iii)\ } \standardRIso{x}=\biset{L_x}{(L_x)}{K_x}\ .$$
Recall from Remark \ref{define_parts} that
$$L_x = \{h_2\in H_2\vv h_2x = xh_1, \ h_1 \in H_1\} \subset H_2$$
and $$K_x =\{h_1 \in H_1\vv xh_1 = h_2x, \ h_2 \in H_2 \}\subset H_1\ .$$
Since $X$ is a conjugation biset, the map
$\defininghomo{X}{x}\colon L_x \to K_x$ is given by
$c_{g_x}$, where $g_{x}^{-1}L_x g_{x} = K_x$, so we have
$$\defininghomo{X}{x}(h_2) = c_{g_x}(h_2) = g_{x}^{-1}h_2g_{x} = h_1\ .$$
The element $g_x$ need not be unique
(see \namedRef{non-uniqueness of conjugating element}).
The isomorphism $\superdefininghomo{X}{x}{-1}$ gives $L_x$
the right $K_x$-action on $L_x$ used to define the third biset.

We will define $F$ on the basic morphisms of these three types,
by associating to each of these
bisets a $G$-map, and then applying the Mackey functor $\cM$:
\begin{itemize}
\item to $\standardInd{x}$ associate the $G$-map
$\GsetInd{L_x}{H_2}\colon G/L_x \to G/H_2$,
with $\GsetInd{L_x}{H_2}(e L_x) = e H_2$;
\item to $\standardRes{x}$ associate the $G$-map
$\GsetRes{H_1}{K_x}\colon G/K_x \to G/H_1$,
with $\GsetRes{H_1}{K_x}(e K_x) = e H_1$;
\item to $\standardRIso{x}$ associate the maps of $G$-sets
$\GsetRIso{K_x}{L_x}\colon G/K_x \to G/L_x$,
with $\GsetRIso{K_x}{L_x}(e K_x) = g_{x}^{-1}L_x$ for the
various choices of $g_x$.
\end{itemize}
Now define
$$\text{(i)}\ F(\standardInd{x}) = \cM_\ast(\GsetInd{L_x}{H_2})
\hskip 25pt
\text{(ii)}\ F(\standardRes{x}) = \cM^\ast(\GsetRes{H_1}{K_x})
\hskip 25pt
\text{(iii)}\ F(\standardRIso{x}) = \cM_\ast(\GsetRIso{K_x}{L_x})$$

Item (iii) is well-defined by Lemma \ref{conginv},
since $\cM$ is conjugation invariant.
\gdef\GsetRIsoA#1#2{\mathfrak L_{#1}^{#2}}
We will write $\GsetRIsoA{K_x}{L_x}$ for any of the
$\GsetRIso{K_x}{L_x}$.

\medskip

For any morphism $X$ in $\subBurnside(G)$ we define
$F(X)$ by choosing a base point $x\in X$, writing
$X = \standardInd{x}\circ \standardRIso{x} \circ \standardRes{x}$, and defining
$F(X) = F(\standardInd{x})\circ F(\standardRIso{x}) \circ F(\standardRes{x})$.
The formulas in Remark \ref{Vyx} show that the definition of $F(X)$ is
independent of the choice of base point.

The functor $F$ is defined on all the morphisms in
$\subBurnsideDot(G)$
by the additive (matrix) extension of these formulas.
Any relation in the Grothendieck group
$\Hom_{\subBurnside(G)}(H_1, H_2)$ leads to a isomorphism
$X \cong Y$ of finite bifree $H_2$\,-$H_1$ bisets.
Since both sides are canonically (up to ordering) expressed as
a disjoint union of $(H_2 \times H_1^{op})$-orbits, it is clear that
$F$ is well-defined.

Finally, we must check that $F$ is a functor.
Since $F( _HH_H) = id$,
it remains to check that compositions are preserved.
Suppose that $X_1 \in \Hom_{\subBurnside(G)}(H_1, H_2)$ and
$X_2 \in \Hom_{\subBurnside(G)}(H_2, H_3)$.
By additivity, we may assume that $X_1$ and $X_2$ are indecomposable.
We must check that
$$F(X_2\circ X_1) = F(X_2) \circ F(X_1)~.$$
Consider the left-hand side of the formula,
where
$F(X_2\circ X_1) = F(X_2\times_{H_2} X_1)$ by definition.
Pick a base point $x_1\in X_1$, and $x_2\in X_2$.
By \namedRef{parameter}, the components of $X_3 = X_2\times_{H_2} X_1$
are indexed by elements $h_2\in H_2$ representing the double cosets
$\doublecoset{K_{x_2}}{H_2}{L_{x_1}}$.
Each such component contributes a summand
$F\bigl([L_{x_3}, \defininghomo{X_3}{x_3}, K_{x_3}]\bigr)$,
where $x_3 = [x_2 h_2, x_1]$.
By \namedRef{The Mackey double coset formula}, the standard representation
at this base point is
\newNumber{basic left hand side}
$$\bigl[L_{x_3}, \defininghomo{X_3}{x_3}, K_{x_3}\bigr]
=
\left [\superdefininghomo{X_2}{x_2}{-1}
\bigl(K_{x_2}\cap L^{h_2^{-1}}_{x_1}\bigr),

\defininghomo{X_1}{x_1}\hskip -8pt \circ c_{h_2}\circ \defininghomo{X_2}{x_2},

\defininghomo{X_1}{x_1}\bigl(K^{h_2}_{x_2} \cap L_{x_1}\bigr)\right]
\leqno(\ref{basic left hand side})~.$$
The right-hand side of the formula is
$$F(X_2) \circ F(X_1) =
F(\standardInd{x_2})\circ F(\standardRIso{x_2}) \circ F(\standardRes{x_2})
\circ
F(\standardInd{x_1})\circ F(\standardRIso{x_1}) \circ F(\standardRes{x_1})~.$$

By the Mackey double coset formula for $\cM$, property (M1),
$$F(\standardRes{x_2})
\circ
F(\standardInd{x_1}) =
\sum_{h_2\in\doublecoset{K_{x_2}}{H_2}{L_{x_1}}}
F(K_{x_2})
\circ F(K_{x_2} \cap L^{h_2^{-1}}_{x_1})
\circ F(L_{x_1})$$
where
\begin{itemize}
\item $K_{x_2}$ is a $(K_{x_2})$\,-\,$\bigl(K_{x_2} \cap L^{h_2^{-1}}_{x_1}\bigr)$
biset via the two inclusions, and is a standard induction,
\vskip 2pt
\item $K_{x_2} \cap L^{h_2^{-1}}_{x_1}$ is a
$\bigl(K_{x_2} \cap L^{h_2^{-1}}_{x_1}\bigr)$\,-\,$\bigl(L_{x_1} \cap K^{h_2}_{x_2}\bigr)$
biset via the evident inclusion on the left and  conjugation by
$h_2^{-1}$ on the right, and is a standard conjugation,
\vskip 2pt
\item $L_{x_1}$ is an
$\bigl(L_{x_1} \cap K^{h_2}_{x_2}\bigr)$\,-\,$(L_{x_1})$ biset
via the two inclusions and is a standard restriction.
\end{itemize}

Hence $F(X_2) \circ F(X_1) =
\sum_{h_2\in\doublecoset{K_{x_2}}{H_2}{L_{x_1}}}
A[h_2]\circ B[h_2] \circ C[h_2]$.
where
$$\begin{matrix}
A[h_2] &=& \cmmA{\GsetInd{L_{x_2}}{H_3}}\circ
\cmmA{\GsetRIsoA{K_{x_2}}{L_{x_2}}} \circ
\cmmA{\GsetInd{\Ax}{K_{x_2}}}\hfill\\\noalign{\vskip 8pt}
B[h_2] &=& \cmmA{\GsetRIsoA{\Bx}{\Ax}}\hfill\\\noalign{\vskip 8pt}
C[h_2] &=& \cmmB{\GsetRes{L_{x_1}}{\Bx}}\circ
\cmmA{\GsetRIsoA{K_{x_1}}{L_{x_1}}} \circ
\cmmB{\GsetRes{H_1}{K_{x_1}}}\hfill\\
\end{matrix}$$

To further analyze $A[h_2]$ consider the commutative diagram
$$\begin{matrix}
\hbox to 0pt{\hss$H_3\hskip 9pt\supset\hskip 10pt$} L_{x_2}&
\xleftarrow{\superdefininghomo{X_2}{x_2}{-1}}&
K_{x_2}\\\noalign{\vskip 4pt}
\cup&&\cup\\
\superdefininghomo{X_2}{x_2}{-1}\bigl(\Ax)&
\xleftarrow{\superdefininghomo{X_2}{x_2}{-1}}{}&
\Ax\\
\end{matrix}
$$
which implies
$$A[h_2] = \cmmA{
\GsetInd{\superdefininghomo{X_2}{x_2}{-1}(\Ax)}{H_3}} \circ
\cmmA{\GsetRIsoA{\Ax}{\superdefininghomo{X_2}{x_2}{-1}(\Ax)}}
~.$$

To analyze $C[h_2]$ first note that
$\cmmA{\GsetRIsoA{K_{x_1}}{L_{x_1}}} =
\cmmB{\GsetRIsoA{L_{x_1}}{K_{x_1}}}$.
Then consider the commutative diagram
$$\begin{matrix}
L_{x_1}&\xrightarrow{\defininghomo{X_1}{x_1}}&
K_{x_1}\hbox to 0pt{\hskip 10pt$\subset\hskip 8pt H_1$\hss}\\\noalign{\vskip 4pt}
\cup&&\cup\\
\Bx&
\xrightarrow{\defininghomo{X_1}{x_1}}{}&
\defininghomo{X_1}{x_1}(\Bx)\\
\end{matrix}
$$
which implies
$$\begin{matrix}
C[h_2] &= &\cmmB{\GsetRes{L_{x_1}}{\Bx}}
\circ
\cmmB{\GsetRIsoA{L_{x_1}}{K_{x_1}}} \circ
\cmmB{\GsetRes{H_1}{K_{x_1}}}\\\noalign{\vskip 8pt}& = &
\cmmB{\GsetRIsoA{\Bx}{\defininghomo{X_1}{x_1}(\Bx)}}\circ
\cmmB{\GsetRes{H_1}{\defininghomo{X_1}{x_1}(\Bx)}}\\
\end{matrix}
$$
\medskip
To analyze the conjugations, note that after substituting
the new expressions for $A[h_2]$ and $C[h_2]$ we have three conjugations now
occurring together in $A[h_2]\circ B[h_2]\circ C[h_2]$:
$$\begin{matrix}
\cmmA{\GsetRIsoA{\Ax}{\superdefininghomo{X_2}{x_2}{-1}(\Ax)}} \circ

\cmmA{\GsetRIsoA{\Bx}{\Ax}}\circ

\cmmB{\GsetRIsoA{\Bx}{\defininghomo{X_1}{x_1}(\Bx)}} & = \\
\cmmA{\GsetRIsoA{\Ax}{\superdefininghomo{X_2}{x_2}{-1}(\Ax)}} \circ

\cmmA{\GsetRIsoA{\Bx}{\Ax}}\circ

\cmmA{\GsetRIsoA{\defininghomo{X_1}{x_1}(\Bx)}{\Bx}} & = \\
\cmmA{\GsetRIsoA{\defininghomo{X_1}{x_1}(\Bx)}{\superdefininghomo{X_2}{x_2}{-1}(\Ax)}}\hfill\\
\end{matrix}
$$

Hence
$$\begin{matrix}
A[h_2]\circ B[h_2]\circ C[h_2] =\hfill \\\noalign{\vskip 8pt}
\cmmA{
\GsetInd{\superdefininghomo{X_2}{x_2}{-1}(\Ax)}{H_3}}
\circ
\cmmA{\GsetRIsoA{\defininghomo{X_1}{x_1}(\Bx)}
{\superdefininghomo{X_2}{x_2}{-1}(\Ax)}}
\circ
\cmmB{\GsetRes{H_1}{\defininghomo{X_1}{x_1}(\Bx)}}\\
\end{matrix}
$$
It now follows from (\ref{basic left  hand side}) that
$$A[h_2]\circ B[h_2]\circ C[h_2] = F(\standardInd{x_3})\circ
F( \standardRIso{x_3}) \circ F(\standardRes{x_3})$$
for each component $x_3 = [x_2h_2,x_1]$.
Therefore $F(X_2\circ X_1) = F(X_2)\circ F(X_1)$.
\end{proof}

\providecommand{\bysame}{\leavevmode\hbox to3em{\hrulefill}\thinspace}
\providecommand{\MR}{\relax\ifhmode\unskip\space\fi MR }

\providecommand{\MRhref}[2]{
\href{http://www.ams.org/mathscinet-getitem?mr=#1}{#2}
}
\providecommand{\href}[2]{#2}
\renewcommand{\MR}[1]{}

config: arxiv


TITLE: my computation of a real integral still has an imaginary number in it,
QUESTION [3 upvotes]: I have four residues that I have found.  I multiplied each by $2\pi i$, using the Residue Theorem.  But my final answer still has an $i$ in it.  Needless to say, it is not the right answer, since the goal was to compute a real integral on $R^+$
What should I do with the four residues to try and catch my mistake?  I've expanded each one in its Euler formula, $\cos(x) + i \sin(x)$, computed a common denominator, so that I can combine all residue terms.  This led to a few cancellations, but ultimately still left me with an $i$ factor.  Am I proceeding in generally the correct way?
Thanks,
EDIT: The integral is $$\int_0^\infty \frac{x^2logx}{1+x^4}dx$$
Here's a summary of my work:  the integrand has simple poles at
$$e^{i\pi/4},e^{i3\pi/4},e^{i5\pi/4},e^{i7\pi/4}$$
and the residues are
$$-\frac{\pi}{16ie^{i\pi/4}},-\frac{3\pi}{16ie^{i3\pi/4}},-\frac{5\pi}{16ie^{i5\pi/4}},-\frac{7\pi}{16ie^{i7\pi/4}}$$
finally, multiplying each by $2\pi i$, and adding up the four terms gives me (incorrectly) $$-\frac{\pi^2i}{\sqrt{2}}$$ I added the terms by expanding out all of the Euler formulas, and then combined the terms with a common denominator.  I've checked my work about 5 times now...still can't catch my mistake...

REPLY [0 votes]: The following is an outline of the solution which is the most natural in my opinion.
Write$$A:=\int_0^\infty \frac{x^2\log x}{1+x^4}dx.$$A short calculation shows that$$\int_{-\infty}^0\frac{x^2\log x}{1+x^4}dx=A+i\pi\int_0^\infty\frac{t^2}{1+t^4}dt=A+i\pi\cdot\frac{\pi}{2\sqrt{2}}=A+\frac{i\pi^2}{2\sqrt{2}}.$$(the second left hand integral can be evaluated using either complex integration or standard real analysis methods).
Now consider paths which go along the whole real line with a small circular "jump" to avoid $0$, then draw a semicircle in the upper half plane. The contribution of the semicircle vanishes at infinite radius, and the contribution of the circular jump vanishes at $0$ radius. The integral$$\int_\gamma\frac{z^2\log z}{1+z^4}dz$$along the above described kind of path can be evaluated by considering the residues at the two upper poles of the function.


TITLE: What are some conceptualizations that work in mathematics but are not strictly true?
QUESTION [87 upvotes]: I'm having an argument with someone who thinks it's never justified to teach something that's not strictly correct. I disagree: often, the pedagogically most efficient way to make progress is to iteratively learn and unlearn along the way.
I'm looking for examples in mathematics (and possibly physics) where students are commonly taught something that's not strictly true, but works, at least in some restricted manner, and is a good way to understand a concept until one gets to a more advanced stage.

REPLY [6 votes]: Geometry, and differential geometry, are full of these kinds of shortcuts, for the simple reason that geometry is a very visual, accessible, and venerable topic, resting on a deceptively deep body of theory required to make it rigorous. You are taught concepts in high school that will require many years of topology, linear algebra, measure theory, real and functional analysis, etc. before you have the full story.
Some examples:

You are taught to calculate the volume and surface areas of solids by two kinds of slicing. Rarely is it explained why surface area requires approximating a surface with slices that are generalized cones, while generalized cylinders are good enough for volume, and I've never seen discussed why and for which kinds of surfaces these techniques don't work at all. More generally, it is often assumed that if you want to measure something about a smooth surface, you can do so by looking at a limiting sequence of discrete surfaces that converge to the smooth surface under "reasonable" refinement. Defining "reasonable" requires great care (beyond the scope of high school geometry) to avoid counterexamples like the Schwarz lantern.
Concepts like arc length, surface area, curvature, etc. are not defined in a complete or rigorous way. On the one hand you have some formulas that clearly require a certainly level of differentiability, and on the other you have no problem calculating e.g. the surface area of a polyhedron, and probably don't bat an eye at the fact that polyhedra satisfy a meaningful version of Gauss-Bonnet, despite ostensible missing two degrees of differentiability! In addition to regularity, subtleties about orientability, compactness, etc. are likely not discussed.
Many theorems in plane geometry require objects to be in "general position" (usually left vaguely defined) and the corner cases are ignored. Similarly, pathological counterexamples to the formula for the Euler characteristics of polyhedra (e.g. polyhedra with odd Euler characteristic) are usually ignored.
Finally, while not the same as teaching students incorrect facts, many geometric arguments you see in early classes implicitly rely on unproven "obvious" results like the Jordan curve theorem, invariance of domain, Sard's lemma, existence of triangulations, etc. that are far from obvious.


TITLE: Evaluate the limit integral
QUESTION [0 upvotes]: I am doing this exercises, but I am stuck.
I have to evaluate this limit
$$
\lim_{n\to \infty}\int_{\mathbb{R}} f_n(x)dx
$$
where $f_n(x)=(1+\frac{2x}{n})^ne^{-x^2}$.
I have seen that $f_n(x)\to e^{2x-x^2}$, so my idea is to use the dominated convergence theorem.
The problem is that I am not able to find a function $g(x)$ such that $|f_n(x)|\leq g(x)$ $\forall x\in \mathbb{R}$ and $g$ is integrable.
Could you give me some advices on which $g$ to use?
Thanks in advance!

REPLY [1 votes]: $|f_n(x)| \leq e^{2|x|-x^{2}}$ which is integrable. [$(1+\frac a n)^{n} \leq e^{a}$ for all $a >0$].


TITLE: A combinatorics problem involving geometry
QUESTION [0 upvotes]: Find the number of triangles that can be formed in a regular polygon of $2n+1$ sides such that the center of polygon lies inside the triangle. (The triangles are to be  formed using the vertices of the polygon).

REPLY [4 votes]: First, pick any starting point $A$. (Suppose it is at the bottom of the circle.) This can be done in $(2n+1)$ ways. Number the points starting at $A=0$. Let the second point $B$ be at position $i\leq n$. (We know that at least one of $B$ and $C$ has to be at the left side of the line $AO$.) Now, for $C$ we know that $n+1\leq C< 2n+1$, because it has to be at the right of $AO$. It also has to be above $BO$, which implies $C\leq i+n<2n+1$. Now, $n+1\leq C\leq i+n$, thus the number of possibilities for $C$ is $i+n-(n+1)+1=i$. Summing over all possible $i$ gives the total number of possibilities for $B$ and $C$:
$$
\sum_{i=1}^n i=\binom {n+1}2
$$
Now, let $A$ be a random point on the circle. This creates $2n+1$ as many solutions, but since  we count each triangle three times now (with $A$ once at each corner), we have to divide by three, resulting in:
$$
\frac{2n+1}3\binom{n+1}2
$$


TITLE: Representing one diagonal of Pascal's triangle using special sums coming from a different diagonal
QUESTION [5 upvotes]: Let $m, n$ be any fixed natural numbers. Is it true that infinitely many elements of the sequence $\binom{m+k}{m}_{k=1,2,3,...}$ ( as well as of the sequence $\left(\binom{m+k}{m}-1\right)_{k=1,2,3,...})$ are representable as the sums of different elements of the sequence $\binom{n+k}{n}_{k=1,2,3,...}$?

REPLY [4 votes]: For fixed $n$, the expression $\binom{n+k}n$ is a polynomial of degree $n$ in $k$ with no fixed prime divisor. A theorem of Kamke (referenced in the first paragraph here) says that there exists $x$ such that all sufficiently large integers are the sum of $s$ values of $\binom{n+k}n$. Of course this will include, for fixed $m$, all but finitely many values $\binom{m+j}m$.


text:
\section{Introduction}
\subsection{SageMath: An Overview}
Nowadays it is nearly impossible to avoid the use of computer algebra systems in various fields of science. They have proved to be an essential tool for the verification of scientific results and the testing of scientific hypotheses. At the same time, their use in computational aspects of mathematics has led to significant theorems within various fields of mathematics. By now, there are many computer algebra systems on the market like \texttt{Magma}, \texttt{Maple}, \texttt{Mathematica} and \texttt{MATLAB}. However, their source codes are typically under a proprietary license and can therefore not be examined, verified or altered. Consequently, mathematicians are unallowed to adapt the code for their own research, which particularly turns out to be a problem when it comes to very specialized areas.

The computer algebra system \SageMath or \Sage (\enquote{\textbf{S}ystem for \textbf{A}lgebra and \textbf{G}eometry \textbf{E}xperimentation}) aims to combine the advantages of all existing algebra systems and offers an adaptable open-source tool for research and teaching in mathematics. It makes use of nearly 100 open-source packages such as \texttt{Pynac}, \texttt{Maxima}, \texttt{GAP}, \texttt{SymPy}, \texttt{matplotlib} and \texttt{Singular}, which are all interfaced via the \Python programming language. Apart from that, \Sage contains a vast library of additional code, which provides functionalities in advanced mathematics like matroids, modular forms or manifolds. We provide a brief overview, but more information can be found on the official web page~\cite{sagemath}.

\paragraph{Some Notes on History.}
Created in 2005 by William Stein and strongly influenced by \texttt{Magma}, \Sage was designed as an open-source alternative to non-free computer algebra systems~\cite{stein:sage}. Since then, it has been adapted, improved and extended by over 271 developers from all over the world -- mostly by mathematicians~\cite{sagemath:developer}. Notably in 2013, Éric Gourgoulhon, Michał Bejger and Marco Mancini developed an additional package providing differentiable manifolds and extensive symbolic tensor algebra~\cite{gour:sagemanifolds}. The project became officially known under the name \texttt{SageManifolds} and has been predominantly used for applications in general relativity~\cite{gour:sagemanifolds:website:publications}. As of 2016, it is fully embedded into \Sage and has since been developed further. At this stage, it supports scalar fields, tensor fields, differential forms, connections and more~\cite{gour:sagemanifolds:website}.

\paragraph{Capabilities and Limitations.}
As \Sage is based on the \Python programming language, it inherits all functionalities from \Python, in particular object-oriented programming. Apart from an extensive symbolic calculus making use of a huge library of algorithms imported from \texttt{Maxima} and \texttt{Singular}, the \Sage project offers a slightly modified syntax that is closer to actual mathematics. In \Python for example, integers are represented by the type \texttt{int} and real numbers by \texttt{float}. This approach is not suitable for mathematics since there are infinitely many algebraic structures which are not feasible to implement via individual classes. Due to this reason, \Sage has been set up with a parent-element-type pattern instead. \defstyle{Parents} represent mathematical sets endowed with certain structures, while \defstyle{elements} reflect their members~\cite{sagemath:reference:primer}. Specifically, each parent belongs to a dynamically generated class, which represents the parent's mathematical category. Within this setting, \Sage performs automatic conversions prior to binary operations according to strict mathematical rules~\cite{sagemath:reference:coercions}.\footnote{In proper \Python syntax, there is typically no way to perform a common operation on objects with different types, even if both are algebraically compatible. This is due to the fact that \Python is a \emph{strongly typed} language.} Such conversions are called \defstyle{coercions} and portray the key concept of \Sage. A full record of \Sage's capabilites and limitations can be found in~\cite{sagemath:reference}.

\subsection{Aims and Scopes}
Characteristic classes play a substantial role in many parts of mathematics and physics. They are related to indices of elliptic operators and yield differential geometric as well as topological invariants. However, computations by hand are often tedious and can easily introduce errors, mostly due to the non-commutativity of the cup product. Their complexity rapidly increases with higher dimensions. Already in dimension four, the physically interesting case, such computations can be extremely difficult and time consuming. For this reason, we aim for a working computer algorithm, so that we can test our theories and verify results in a time efficient manner. One attempt has been followed by Oleksandr Iena in 2015, performing symbolic computations on the Chern roots in \texttt{Singular} by using multiplicative sequences~\cite{iena:chern}. Subsequently in 2019, Zsolt Szilágyi provided an algorithm in \texttt{Singular} for the calculation of Chern classes associated to the tensor product of vector bundles~\cite{szilgyi:chern}. These algorithms are relatively fast and provide an efficient approach for purely algebraic purposes. Yet, in order to evaluate integrals on characteristic classes, we prefer a more geometric approach yielding results in concrete coordinates. As motivated in the preceding section, the most promising candidate to realize such a project is given by \Sage; it is open-source and comes with a comprehensive symbolic tensor calculus delivered by the \SageManifolds extension.

\paragraph{Our Work.}
As part of this thesis, our goal is to implement characteristic classes into \Sage by using Chern--Weil theory. It provides a construction for characteristic classes in the de Rham cohomology of the underlying base space by inserting curvature forms into invariant polynomials. However, before we transfer this algorithm into \Sage, some preparations have to be made. On the one hand, we require vector bundles including local frames and bundle connections. On the other hand, since invariant polynomials must be applied to curvature forms, a new algebraic structure representing mixed differential forms has to be established. Meanwhile, some parts of the existing code need to be adjusted or even corrected if necessary. After laying those foundations, characteristic classes fit into the setup quite naturally. The whole development is accompanied by a critical review process maintained by other \Sage developers. A list of all modifications due to this project is given in Table~\ref{tab:intro:tickets}. The comprehensive git difference statistics can be found on \url{https://trac.sagemath.org/} and is also provided on the CD attached to this thesis.

\begin{table}[t]
	\centering
	\footnotesize
	\begin{tabular}{lccr}
		\toprule
		\textbf{Implementation Task} & \textbf{Type} & \textbf{Ticket} & \textbf{Merged in} \\
		\midrule
		Output dicts and lists copied & \textit{Bugfix} & \ticket{28563} & \texttt{Sage 9.0.beta1} \\
		Scalar fields: wedge product fix & \textit{Bugfix} & \ticket{28579} & \texttt{Sage 9.0.beta1} \\
		Better treatment of zero element & \textit{Performance} & \ticket{28562} & \texttt{Sage 9.0.beta3} \\
		Scalar fields: restrictions & \textit{Enhancement} & \ticket{28554} & \texttt{Sage 9.0.beta4} \\
		Tensor fields: \texttt{set\_restriction} fixed & \textit{Bugfix} & \ticket{28628} & \texttt{Sage 9.0.beta4} \\
		Tensor fields: consistent naming & \textit{Enhancement} & \ticket{28564} & \texttt{Sage 9.0.beta7} \\
		Automorphism fields: notation of inverse & \textit{Bugfix} & \ticket{28973} & \texttt{Sage 9.1.beta1} \\
		\arrayrulecolor{darkgray}\hline
		Vector bundles and sections & \textit{Enhancement} & \ticket{28159} & \texttt{Sage 9.0.beta3} \\
		Vector bundles: section module name fix & \textit{Bugfix} & \ticket{28690} & \texttt{Sage 9.0.beta5} \\
		\arrayrulecolor{darkgray}\hline
		Mixed differential forms & \textit{Enhancement} & \ticket{27584} & \texttt{Sage 8.8.beta3} \\
		Mixed forms: code improvements & \textit{Enhancement} & \ticket{28578} & \texttt{Sage 9.0.beta5} \\
		Mixed forms: improved coercions & \textit{Enhancement} & \ticket{28916} & \texttt{Sage 9.1.beta1} \\
		Mixed forms: better treatment of zero/one & \textit{Performance} & \ticket{28921} & \texttt{Sage 9.1.beta1} \\
		\arrayrulecolor{darkgray}\hline
		Characteristic classes & \textit{Enhancement} & \ticket{27784} & \texttt{Sage 9.0.beta8} \\
		\bottomrule
	\end{tabular}
	\caption{All implementations developed as part of this thesis. A full description and git difference statistics can be found either on \Sage's trac server \url{https://trac.sagemath.org/} or on the CD attached to this thesis.}\label{tab:intro:tickets}
\end{table}

\paragraph{Outline of this Thesis.}
In this thesis, we present our implementations delivered to \Sage. Furthermore, the thesis is partly intended as a compendium. In Chapter~\ref{ch:vec}, we start by recalling basic facts about vector bundles. We briefly explain their realization in \Sage and provide two illustrative examples. In Chapter~\ref{ch:mixed}, we introduce mixed differential forms and investigate their algebraic structure, highlighting important properties in anticipation of characteristic classes. We sketch the idea behind their algebraic realization in \Sage and give elementary examples exemplifying their usage. The subsequent Chapter~\ref{ch:char_class} addresses to characteristic classes. There, we provide a more detailed but still compressed introduction into Chern--Weil theory. Afterwards, we thoroughly discuss our algorithm in \Sage and eventually provide three simple examples. One last example is devoted to a more sophisticated computation of an $\hat{A}$-form on a Lorentzian foliation of Berger 3-spheres. This chapter is closed with a summary of our work and future prospects.

During this thesis, we assume basic knowledge about differential geometry. Introductory as well as advanced literature on that topic can be found in~\cite{baer:diffgeo, greub:con, hirsch:differential_topology, lee:smooth_manifolds, walschap:metric_structures}. To fully understand our implementation details, we require some understanding of object-oriented programming. A short course with respect to \texttt{Python 3} is dedicated to~\cite{realpython:object}. Moreover, we recommend to have a vague idea behind the \Python module structure of \Sage, especially with respect to \texttt{sage.manifolds} and \texttt{sage.tensor.modules}. The full module index is provided in~\cite{sagemath:reference:module_index}. For our \Sage examples, we expect rudimentary knowledge about \Python's syntax. Free \texttt{Python 3} introduction courses can be found in~\cite{python_everybody, python_in_10}. As aforementioned, \Sage's syntax slightly differs to that of \Python's. We therefore recommend~\cite{zimmermann:sagemath, stein:sagetut} and the reference manual~\cite{sagemath:reference}. An extensive documentation as well as a wide range of examples on the usage of manifolds in particular can be found in~\cite{gour:tensor_lecture, gour:examples, gour:sagemanifolds:website:doc, sagemath:reference:manif}.

\subsection{Conventions and Setup}

\paragraph{The Mathematical Setup.}
An $n$-dimensional manifold is a topological space $M$ obeying the following axioms:
\begin{myenum}
	\item $M$ is \defstyle{locally homeomorphic} to $\RR^n$, i.\,e.\ each point $p \in M$ has a neighborhood being homeomorphic to an open subset of $\RR^n$.
	\item $M$ is a \defstyle{Hausdorff space}, i.\,e.\ any two distinct points $p,q \in M$ have disjoint open neighborhoods.
	\item $M$ is \defstyle{second-countable}, i.\,e.\ $M$ admits a countable family of open subsets $(\mathcal{B}_i)_{i \in I}$ such that every open set on $M$ can be written as a union of members of this family.
\end{myenum}
Beware that in some literature the latter condition is replaced with paracompactness. Even though manifolds can be considered over more general fields, we want to restrict ourselves to the most common case $\RR$. If a topological manifold $M$ admits a differentiable structure, that is, $M$ is covered by charts such that all transition maps are differentiable up to a certain degree, we say $M$ is a \defstyle{differentiable manifold}. As usual, manifolds having a smooth structure are called \defstyle{smooth manifolds}. Note that any manifold endowed with a differentiable structure automatically admits a compatible \emph{smooth} structure~\cite[Thm.~2.9]{hirsch:differential_topology}. We henceforth distinguish only between topological and smooth structures.

\paragraph{The SageMath Setup.}
At the end of each chapter, we illustrate our implementations with examples in \Sage. For this, we use \texttt{Jupyter Notebook}, cf.~\cite{jupyter}, running on the \Sage kernel. The associated \texttt{ipynb}-file with reduced explanations is included on the CD. Within the notebook, inputs are framed with a gray box and designated with blue numbers in brackets:
\begin{NBin}
print('Hello World!')
\end{NBin}
The corresponding output is indicated by the same number in red:\medskip
\begin{NBout}
Hello World!
\end{NBout}
Of course, we are using the most recent version of \Sage so that all our written code is accessible:
\begin{NBin}
version()
\end{NBin}
\begin{NBout}
'SageMath version 9.1.beta1, Release Date: 2020-01-21'
\end{NBout}
Manifolds can be declared by stating the manifold's dimension and name with the following command:
\begin{NBin}
Manifold(2, 'M')
\end{NBin}
\begin{NBout}
2-dimensional differentiable manifold M
\end{NBout}
For mathematical purposes, we want to have a \LaTeX-typeset output by default. This can be achieved with the following line:
\begin{NBin}

\end{NBin}


TITLE: A (simple?) matter of notation
QUESTION [1 upvotes]: I am currently working on some power series problems, so I deal with the sequence of their (generally complex) coefficients $a_i$, $i\in\mathbb{N}$, denoted in the sequel as $\langle a_n\rangle$ following reference [1]. A sequence is simply a function
$$
\mathbb{N}\ni n\mapsto a_n\in\mathbb{C}\tag{1}\label{1}
$$
Given a finite/infinite subset $S$ of the range of $\langle a_n\rangle$, I need to analyze the indexes $n$ of each member $a_n\in S$. And now my question arises: does there exists a standard notation for the inverse image of $S$ under the function expressed by \eqref{1}?
Having read many textbooks on real and complex analysis, power series and so on, I did not find anything on such matter: however, perhaps specialists in combinatorics or other similar field are more customary with such concepts so I decided to ask this question in order to not introduce in my work useless notations.
[1] Emanuel Fisher (1983), "Intermediate Real Analysis", Springer Verlag.

REPLY [2 votes]: If we use the notation $\langle a_n \rangle$ to denote a sequence $a$, where for each $n \in \mathbb{N}$, $a_n := a(n)$, it seems fitting to write the pre-image of $S \subseteq \mathbb{C}$, $\ a^{-1}(S),$ as $\langle a_n \rangle ^{-1}(S).$
Going from what you mentioned under Fred's post, one advantage of this notation is that we are not hiding the fact that $a$ is a sequence. I am not sure whether writing the pre-image this way is standard, however.


TITLE: Maps with Hopf invariant zero are suspensions
QUESTION [11 upvotes]: Let $h:\pi_{2n-1}(S^n) \rightarrow \mathbb{Z}$ be the Hopf invariant.  I believe that in the same paper that proves his suspension theorem, Freudenthal proved that if $x \in \pi_{2n-1}(S^n)$ satisfies $h(x)=0$, then $x$ is in the image of the suspension map $\pi_{2n-2}(S^{n-1}) \rightarrow \pi_{2n-1}(S^n)$.  Observe that the usual Freudenthal suspension theorem says that the map $\pi_{2n-1}(S^n) \rightarrow \pi_{2n}(S^{n+1})$ is a surjection.
Can anyone either describe a proof of this or point me in the direction of a modern source for it?  The only one I know of is Pontryagin's book "Smooth manifolds and their applications in homotopy theory", where this is Theorem 16, but I find his proof very hard to follow.  I guess I would particularly appreciate a proof in the spirit of Pontryagin's proof, though more algebraic proofs are welcome too.

REPLY [8 votes]: The integral statement is most generally this:
For a based CW space $X$ one has a suspension map $E: X\to \Omega \Sigma X$ and
a Hopf invariant $H: \Omega \Sigma X \to \Omega \Sigma (X\wedge X)$ (construction outlined below).
The composite
$H\circ E$ is canonically null and the sequence
$$
X \overset E \to \Omega \Sigma X \overset H \to  \Omega \Sigma (X \wedge X)
$$
is a homotopy fiber sequence in a range, roughly $3r$, where $r$ is the  connectivity of $X$.
Your result will follow from this easily (take $X = S^n$).
The map $H$ is often constructed as follows: let $JX$ be the reduced free monoid on the points of $X$. Using say the Moore loops model for $\Omega \Sigma X$, one can construct a monoid homomorphism $JX \to \Omega \Sigma X$ that extends the map $E$ using the universal property of the free monoid. A homology calculation shows that this map his a weak equivalence (James did this calculation when $X$ is a sphere).
Finally, the proof the above sequence is a fibration in the range roughly thrice the connectivity of $X$ can be deduced from the following four facts:
1) $J_2 X \to JX$ is $(3r+2)$-connected, where $J_2X = X \cup (X \times X)$ is filtration two.
2) The quotient $J_2X/X$ is $X\wedge X$, so we have a cofiber sequence
$$
X \to J_2 X \overset q\to X \wedge X
$$
3) By the Blakers-Massey Theorem, the above sequence is a homotopy fibration in
the range roughly $3r$.
4) The diagram
$$
\require{AMScd}
\begin{CD}
J_2X@>q>>X\wedge X\\
@VVV @VVV\\
JX@>>H>\Omega\Sigma(X\wedge X)
\end{CD}
$$
is commutative, where the left vertical maps is the inclusion and the right one is the suspension map for $X\wedge X$.


TITLE: Speeding up the calculation of integrations of a squared(powered) term using orthogonality(orthogonalities) of trigonometric functions
QUESTION [1 upvotes]: $$  \int_{-\pi }^{\pi } \left\{ 1+ \cos^{}\left(t\right) -2 \cos^{}\left(2t\right) + 3 \cos^{}\left(3t\right) -4 \sin^{}\left(4t\right) + 5 \sin^{}\left(5t\right)  \right\} ^{2}  \,dx   $$
Of course this term can be calculated developing  $~ \left\{ 1+ \cos^{}\left(t\right) -2 \cos^{}\left(2t\right) + 3 \cos^{}\left(3t\right) -4 \sin^{}\left(4t\right) + 5 \sin^{}\left(5t\right)  \right\} ^{2} ~$  and using orthogonalities of trigonometric functions.
But isn't there some more efficient or faster way(s) to find out the value of the integrations? , like using some convenient formula(s).
At this rate, I have to handle 36 terms.

REPLY [2 votes]: Using orthogonalities of trigonometric functions is indeed the fastest way to  evaluate the integral. For this reason, all the $6\cdot 5=30$ integrals of double products are zero. Moreover, for the remaining $6$ integrals of squares, recall that $\int_{-\pi}^{\pi}1 \,dt=2\pi$ and for any integer $n\not=0$,
$$\int_{-\pi}^{\pi}\cos^{2}\left(nt\right)\,dt=\int_{-\pi}^{\pi}\sin^{2}\left(nt\right)\,dt=\pi.$$
It follows that the integral is equal to:
$$(2+1^2+(-2)^2+3^2+(-4)^2+5^2)\pi=57\pi$$


TITLE: What's the underlying particle physics of endothermic reactions?
QUESTION [6 upvotes]: I don't just mean reactions that require heat to proceed, storing surplus energy in chemical bonds. I wonder about strongly endothermic reactions that suck heat out of environment.
You take some substance A (e.g. Ammonium Nitrate), and some substance B (e.g. water), both at room temperature. You mix them together in a beaker which is room temperature, all performed in room temperature air. As the reaction begins, the two substances binding into substance C, the beaker cools down quite a bit below room temperature.
This is what we observe on macroscopic scale.
I wonder what happens on the microscopic scale - chemical particles, atoms, elementary particles, their energy. Normally entropy would suggest everything would remain in equilibrium but suddenly we have a higher energy concentration within the substance at cost of energy of the environment. It is not easily reversed. What happens that makes particles "want" to bind so much that they specifically "steal energy" from the environment just so that they can react?

REPLY [2 votes]: I don't just mean reactions that require heat to proceed, storing surplus energy in chemical bonds. I wonder about strongly endothermic reactions that suck heat out of environment

A reaction that requires heat to proceed, a reaction that sucks heat out of the environment, and an endothermic reaction are all the same.  These are all just descriptions of reactions that occur because entropy (S) increases, despite that fact that enthalpy (H) also increases.
$G = H -TS$
A reaction is favorable if Gibbs free energy decreases.  The reaction can still be favorable, despite enthalpy (H) increasing, if entropy (S) increases enough.

You take some substance A (e.g. Ammonium Nitrate), and some substance B (e.g. water)... I wonder what happens on the microscopic scale

you have to separate molecules/ions/atoms of A from others of A. This could involve breaking apart ionic interactions in a crystal, or intermolecular interactions for example.  This takes energy.
you have to seperate molecules/ions/atoms of B from others of B.  For a solvent, this would involve breaking apart intermolecular interactions such as dipole-dipole interactions. This takes energy.
new interactions between A and B form.  This releases energy.

If the energy released in step 3 is less than the energy required for steps 1 and 2, the process is endothermic.


TITLE: How to calculate pressure exerted by a rocket's exhaust on a surface from a certain altitude?
QUESTION [5 upvotes]: How do you calculate the pressure that is exerted on a surface (say from a rocket engine) at an arbitrary altitude? For example, pretend I know the exit velocity and pressure of a rocket engine at an altitude of 100 m, how would I calculate the pressure exerted on the surface by the rocket exhaust? I assume you would need to know the ambient pressure as that affects the pressure exerted on a surface. I know how to calculate pressure exerted on a surface which is calculated with $P = {{F}\over{A}}$. However that only works if the object is touching the surface and thus, does not apply for rocket engines and its exhaust.

REPLY [0 votes]: Intro
If we take the aggressive assumptions that the exhaust stream maintains its shape and faces zero drag going back to the surface, then we can just use $F=ma$ to decelerate the jet, or even conservation of energy. Or the similarly large assumption that we know it’s speed, $v_f$, when reaching the ground.
This just calculates what level of force streams put on the ground when hitting. In a vacuum, or very low altitude, or with particle exhaust, the results would give estimates worth considering.
F= ma
Assume we know the speed when it hThe force to stop a column of exhaust (from $v=v_f$ to $v=0$) over some time $\Delta t$ on a mass $M$. At $t=0$ the amount of column being accelerated is $m=0$, and by $t=\Delta t$, the whole column is finishing up being accelerated.
$$F= (m) (\frac{dv}{dt})=(m_{average})(\frac{\Delta v}{\Delta t}) =(\frac{M}{2})(\frac{\Delta v}{\Delta t}) $$ $$= \tfrac{1}{2}(\frac{M}{\Delta t})(v_f-0) = \tfrac{1}{2} \dot{m}v_f$$ $$P=\frac{F}{A}=\frac{\dot{m}v_f}{2A} $$
Of course $M$ and $\Delta t$ were arbitrary and dropped out.
Conservation of energy
Alternatively, we can do work on the exhaust column of mass $M$ and length $L$ in a frame where the column is still, and use that work to impart kinetic energy (decelerate it to $v=0$ in a still frame, a frame where the jet is doing work on the ground in impact and damage and hence jet is losing energy).
$$\tfrac{1}{2}M(v_f^2-0) = (F)(L)= (F)(v_f~dt)$$
$$\implies F= \tfrac{1}{2}\dot{m}v_f$$
$$P= \frac{\dot{m} \sqrt{v_i^2+ 2gh}}{2A} $$
The two methods agree.
Notes:
The good thing about the energy derivation is that it implies we could adjust if the velocity profile is not a constant velocity across the exhaust jet.
Here is prof Skakya discussing a jet hitting a wall. He does only the first method ($F=ma$ to decelerate the stream) and gets a similar answer but does not break down the column as I did to verify, nor do conservation of energy: https://youtu.be/kNL5w73owzU
If we have particles:
Assuming they fall to actually gain speed, let $v_i$ be the exhaust’s velocity relative to the earth when it leaves the rocket (That might be the normal exhaust stream speed, minus the current velocity of the rocket.. if those are what we know.) $v$ is the exhaust speed when it hits the earth. Then the speed at earth is:
$$\tfrac{1}{2} m~(v_f^2-v_i^2) = mgh \implies v_f= \sqrt{v_i^2+ 2gh}$$
If we know the area of impact, it’s not necessary to assume that it maintains its shape, but there is still the strong assumptions that it stays evenly dispersed, all of it reaches the impact area, and most of all that it didn’t lose speed due to drag (actually gained speed from falling). If the speeds are high, we can estimate loss, if we know particle sizes, from $$F_D = \tfrac{1}{2} \rho v_f^2 C_D a$$
Where $\rho$ is for air.


TITLE: $p$-value, significance level, and the size
QUESTION [0 upvotes]: Basically, I am having a hard time understanding the difference between the $p$-value, the significance level and the size of a test.
In doing questions about hypothesis testing, I am somehow able to get the right answers when asked to derive these quantities. But at the end of the day, it seems to me that all three of them represent the probability of rejecting the null hypothesis?
Any enlightenment would be much appreciated.

REPLY [1 votes]: A hypothesis test's significance level $\alpha$ should be set before undertaking the test, and should be the desired probability of the test rejecting the null hypothesis, given that it were true.
This has sometimes also been called the size of the test, but that usage is now less common since size is also used for other terms such as sample size and effect size; alternatively, sometimes it has been used to measure the actual probability of being in the critical rejection region (e.g. a count which is assumed to have a binomial distribution may not exactly match the significance level, and so the size could be slightly less than the significance level) and more generally, size has been used when the actual critical rejection region is chosen before the significance level
The $p$-value is the probability under the null hypothesis of seeing a result as extreme or more extreme than the observed result, with the alternative hypothesis suggesting what extreme means; the $p$-value is calculated after seeing the observation.  If the $p$-value is less or equal to $\alpha$ then the test is statistically significant in terms of rejecting the null hypothesis with a significance level of $\alpha$ (not of $p$)


TITLE: highest power of Prime
QUESTION [1 upvotes]: How to find the highest power of prime $p$ in $N!$, when
.$p^r-1<N<p^r+p$
$p^r-p<N<p^r$
I know  that the highest power of prime contained in $N!$ is given by:
$$s_p(N!) = \left \lfloor \frac{N}{p} \right \rfloor + \left \lfloor \frac{N}{p^2} \right \rfloor + \left \lfloor \frac{N}{p^3} \right \rfloor + \cdots$$
But how to apply the restrictions?

REPLY [0 votes]: $p$ is a prime number, therefore it cannot be decomposed, and it appears as a whole in each parenthesis:
$$n\cdot(n-1)\cdot(n-2)...(kp+1)\cdot kp\cdot(kp-1)...\cdot\;3\;\cdot\;2\;\cdot\;1$$
We are calculating the exponents:
If $\Bigg\lfloor\frac{n}{p}\Bigg\rfloor=k\ne 0,$ it means there are $k$ $x$-es $\rightarrow$integer multiples of $p$, such that $x\leq n,\;k,x\in \mathbb N$
If $\Bigg\lfloor\frac{n}{p^2}\Bigg\rfloor=l\ne 0$, it means there are $l$ $y$-s$\rightarrow$ integer multiples of $p^2,$ such that $y\leq n, \;l,y\in \mathbb N$
I hope it's correct.
Some of $y$-s might have been included among $x$-es, but the coefficients in front of $p$ in those terms (when we decompose the composite numbers) are equal to $p$.
The algorithm continues until:$$\Bigg\lfloor\frac{n}{p^i}\Bigg\rfloor=0.$$
In general:$$\forall u,v \in [ap^m, (a+1)p^m-1\rangle: \Bigg\lfloor\frac{u}{p^m}\Bigg\rfloor=\Bigg\lfloor\frac{v}{p^m}\Bigg\rfloor$$


TITLE: If a seimigroup's left identity is unique, can it be two-side identity?
QUESTION [1 upvotes]: If a seimigroup's left identity is unique, can it be two-side identity?

The answer is true if we talk it in a ring. Like construct $(be-b+e)a=a$. But in a semigroup I can't image how to construct a equation so we can use the condition uniqueness. So I am wondeting if the statement is false. But I also couldn't get a counterexample.

REPLY [0 votes]: Consider a semigroup $S$ of words on letters $e, x$ of the form $x^n$ for $n>0 $ and $x^ne$ for $n\ge0$, defining
$$x^ne\cdot x^m:=x^{n+m}\quad\quad(n\ge0, m>0)\\
x^ne\cdot x^me:=x^{n+m}e\quad\qquad(n,m\ge0)$$
Here $e$ is the unique left identity, but it's not right identity as $xe\ne x$.
Basically, $S$ is the quotient of the free semigroup of words on letters $e,x$, by the relations $ex\sim x,\ e^2\sim e$.